no less than $\rho$ and that no point in $L(\tau)$ is a proper distance
greater than $\rho$ from all $x_{i}$. In addition, for small $\delta$, the set
of point $L(\tau,\rho)$ will consist of the points of $L(\tau-\delta,\rho)$,
each rescalled by a factor of $1+\frac{\delta}{\tau}$ with any resulting hole
in $L(\tau)$ more than a proper distance of $\rho$ from the rescaled points of
$L(\tau-\delta,\rho)$ filled by an additional point chosen randomly according
to the invariant measure on $L(\tau)$.

##  XX  Residual Entanglement Again

Let $|\psi(\tau)\rangle\in\mathcal{H}$ be a branch not immediately subject to
further branching. Let $Z$ be some large spatial region and $W$ the
intersection of $Z$ with $L(\tau,\rho)$. Let $X$ be the intersection of the
complement of $Z$ with $L(\tau,\rho)$. Let $\mathcal{Q}$ be the subspace of
$\mathcal{H}$ spanned by the set of field operators with support in $W$ acting
on the vacuum and $\mathcal{R}$ the subspace of $\mathcal{H}$ spanned by the
set of field operators with support in $X$ acting on the vacuuum. Although for
reasons already briefly mentioned in Section XIV, $\mathcal{H}$ will not
simply be isomorphic to a tensor product of $\mathcal{Q}$ and $\mathcal{R}$,
the non-vacuum sectors of these spaces remain linearly independent. Let
$\\{|\psi_{\mathcal{Q}i}\rangle\\}$ and $\\{|\psi_{\mathcal{R}i}\rangle\\}$ be
orthonormal bases for $\mathcal{Q}$ and $\mathcal{R}$, respectively. Let
$S_{i}$ and $T_{i}$ be operators with support in $W$ and in $X$, respectively,
such that

$\displaystyle|\psi_{\mathcal{Q}i}\rangle$ $\displaystyle=$ $\displaystyle
S_{i}|\Omega\rangle,$ (184a) $\displaystyle|\psi_{\mathcal{R}i}\rangle$
$\displaystyle=$ $\displaystyle T_{i}|\Omega\rangle.$ (184b)

Then any branch $|\psi(\tau)\rangle$ has a unique expansion of the form

$|\psi(\tau)\rangle=\sum_{ij}\alpha_{ij}S_{i}T_{j}|\Omega\rangle,$ (185)

A polar decomposition of the matrix $\alpha_{ij}$ then yields a Schmidt
decomposition for $|\psi(\tau)\rangle$

$\displaystyle|\psi(\tau)\rangle$ $\displaystyle=$
$\displaystyle\sum_{i}\lambda_{i}U_{i}V_{i}|\Omega\rangle,$ (186a)
$\displaystyle\langle\Omega|{U^{\dagger}}_{i}U_{j}|\Omega\rangle$
$\displaystyle=$ $\displaystyle\delta_{ij},$ (186b)
$\displaystyle\langle\Omega|{V^{\dagger}}_{i}V_{j}|\Omega\rangle$
$\displaystyle=$ $\displaystyle\delta_{ij},$ (186c)

where $U_{i}$ and $V_{i}$ are operators with support in $W$ and $X$,
respectively.

According to a version of the hypothesis in Section X copied over to
relativistic complexity and branching, the state $|\psi(\tau)\rangle$ will be
entangled only over bounded regions in $L(\tau,\rho)$, so that for
sufficiently large $Z$, the sum in Eq. (186a) nearly reduces to a single term

$|\psi(\tau)\rangle\approx\lambda(\tau)U(\tau)V(\tau)|\Omega\rangle.$ (187)

Now define $\mathcal{Q}^{\perp}$ to be the projection of $\mathcal{Q}$
orthogonal to $\mathcal{R}$. Since the union of $W$ and $X$ is all of
$L(\tau,\rho)$, $\mathcal{Q}^{\perp}$ will hold the degrees of freedom present
in $\mathcal{H}$ but missing from $\mathcal{R}$. We therefore expect

$\mathcal{H}=\mathcal{Q}^{\perp}\otimes\mathcal{R}.$ (188)

For sufficiently large $Z$, an alternative version of Eq. (191) becomes

$|\psi(\tau)\rangle\approx\lambda(\tau)|\psi_{\mathcal{Q}^{\perp}}\rangle\otimes|\psi_{\mathcal{R}}\rangle,$
(189)

where

$\displaystyle|\psi_{\mathcal{Q}^{\perp}}\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{Q}^{\perp},$ (190a)
$\displaystyle|\psi_{\mathcal{R}}\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{R}.$ (190b)

The difference between $|\psi_{\mathcal{Q}^{\perp}}\rangle$ and
$U(\tau)|\Omega\rangle$ and between $|\psi_{\mathcal{R}}\rangle$ and
$V(\tau)|\Omega\rangle$, however, should be only near the boundary of $Z$. For
sufficiently large $Z$, Eq. (189) then becomes

$|\psi(\tau)\rangle\approx\lambda(\tau)[U(\tau)|\Omega\rangle]\otimes[V(\tau)|\Omega\rangle].$
(191)

##  XXI $\rho$ Small, $\tau\rightarrow\infty$

Now assume that $\rho$ has been made much smaller than any of the length
scales occuring in the following. For notational simplicity we will drop
$\rho$ as an agument of the various functions in which it appears. The example
of Section VIII shows that as $\tau\rightarrow\infty$, for any system not
confined to a bounded region, branch splitting will continue without stop. To
frame a plausible hypothesis for the behavior of branching as
$\tau\rightarrow\infty$ which takes continued splitting into account, we
define a branch labelling scheme which then facilitates the definition of a
set of summed branches.

For a system beginning in some initial state $|\psi\rangle$ with complexity
close to 0 at proper time $\tau_{0}$, consider the set of branch states which
result from minimizing $Q(\tau,\\{|\psi_{i}\rangle\\})$ for
$\tau\geq\tau_{0}$. Let $E$ be the set of all branching events. As discussed
in Section VI on non-relativistic branching, we will assume $E$ consists
almost entirely of splits of branches into pairs of sub-branches. In addition,
however, $E$ may also include rare events for isolated subsystems of the
universe for which associated branches split and then later recombine, leaving
no record in the rest of the universe. It will turn out to be technically
convenient to adopt a rule according to which recombining branches remain
distinct so that all splits effectively become permanent. A corresponding rule
will then be needed to handle subsequent splits of the remaining distinct
branches.

The hypothesis that all splitting events in $E$ consist of some branch
splitting permanently into a pair of sub-branches yields a labeling scheme for
branches. Each branch state $|\psi(s,\tau)\rangle$ can be labelled with a set
of pairs

$s=\\{(e_{0},\ell_{0}),\ldots(e_{n-1},\ell_{n-1})\\},n>0,$ (192)

giving a corresponding history of splitting events $e_{i}\in E$ and branch
indices $\ell_{i}\in\\{0,1\\}$. For a splitting event $e\in E$ at time $\tau$
of a state $|\psi(w,\tau)\rangle$ with history

$w=\\{w_{0},\ldots w_{n-1}\\},$ (193)

the resulting branch states $|\psi(u,\tau)\rangle,|\psi(v,\tau)\rangle$ have

$\displaystyle u$ $\displaystyle=$ $\displaystyle\\{w_{0},\ldots
w_{n-1},(e,0)\\},$ (194a) $\displaystyle v$ $\displaystyle=$
$\displaystyle\\{w_{0},\ldots w_{n-1},(e,1)\\}.$ (194b)

As a consequence of allowing branch states to pass through recombination
events unaffected, it follows that each branch in the optimal set
$\\{|\psi_{i}\rangle\\}$ at $\tau$ will be a sum of some corresponding set
$S_{i}$ of orthogonal branch states

$|\psi_{i}\rangle=\sum_{w\in S_{i}}|\psi(w,\tau)\rangle.$ (195)

Suppose $|\psi_{i}\rangle$ is a branch state at $\tau$ with $w\in S_{i}$ and
suppose that the branching event $e$ at $\tau$ splits $|\psi_{i}\rangle$ into
branches $|\phi_{0}\rangle$ and $|\phi_{1}\rangle$. The rule needed to
compensate for having ignored events in which branches can rejoin is then

$\displaystyle|\psi(u,\tau)\rangle$ $\displaystyle=$
$\displaystyle\langle\psi(w,\tau)|\phi_{0}\rangle|\phi_{0}\rangle,$ (196a)
$\displaystyle|\psi(v,\tau)\rangle$ $\displaystyle=$
$\displaystyle\langle\psi(w,\tau)|\phi_{1}\rangle|\phi_{1}\rangle,$ (196b)

The initial state we assign branch index 0 of an initial null branching event
$\emptyset\in E$ at $\tau_{0}$. Thus $|\psi\rangle$ at $\tau_{0}$ becomes
$|\psi[\\{(\emptyset,0)\\},\tau_{0}]\rangle$. Each $s$ can also be viewed as a
map from some subset of $E$ into $\\{0,1\\}$. For $s$ of Eq. (192), define
$|s|$ to be $n$.

For $\tau_{0}$ sufficiently large, $Q(\tau,\\{|\psi_{i}\rangle\\})$ will be
nearly translationally invariant over the spatial region contributing to any
branching event. Since $Q(\tau,\\{|\psi_{i}\rangle\\})$ is also Lorentz
invariant, it seems reasonable to assume at the least that for each branching
event and resulting branch in any Poincaré frame there will be a corresponding
branching event and branch in any other frame. The relation between
corresponding branches will be considered in more detail in Section XXII.

For any $\tau\geq\tau_{0}$, let $S(\tau)$ be the set of $s$ corresponding to
the set of branches which minimize $Q(\tau,\\{|\psi_{i}\rangle\\})$. Each
$S(\tau)$ can be viewed as a set of maps, each map in the set taking a subset
of $E$ into $\\{0,1\\}$. Define $S$ to be the set of all such maps, each
taking some subset of $E$ into $\\{0,1\\}$. Appending, for the moment, a
reference frame label $f$ to $S_{f}(\tau)$, the set $S$ then contains at least

$S\supseteq\cup_{f}\cup_{\tau}S_{f}(\tau).$ (197)

For any $s\in S$, and any $\tau$, define $|\chi(s,\tau)\rangle$ to be the sum
of all the $\tau$ branches with histories containing $s$

$|\chi(s,\tau)\rangle=\sum_{s^{\prime}\in S(\tau),s^{\prime}\supseteq
s}|\psi(s^{\prime},\tau)\rangle.$ (198)

For any $\tau$, there will be a corresponding $n_{\tau}$ such that

$|\chi(s,\tau)\rangle=0,|s|>n_{\tau}.$ (199)

On the other hand, for every $s\in S$ there is a $\tau_{s}$ such that

$|\chi(s,\tau)\rangle\neq 0,\tau>{\tau}_{s}.$ (200)

For any $\tau_{0}\leq\tau_{1}\leq\tau$, selecting the $s^{\prime}\in
S(\tau_{1})$ which are descendents of some $s\in S(\tau_{0})$ yields

$|\chi(s,\tau)\rangle=\sum_{s^{\prime}\in S(\tau_{1}),s^{\prime}\supseteq
s}|\chi(s^{\prime},\tau)\rangle.$ (201)

For any pair of distinct $s,s^{\prime}\in S(\tau)$, for any
$\tau^{\prime}\geq\tau$, the states $|\chi(s,\tau^{\prime})\rangle$ and
$|\chi(s^{\prime},\tau^{\prime})\rangle$ are orthogonal. For any $s\in
S(\tau)$, the only $s^{\prime}\in S(\tau)$ which satisfies
$s^{\prime}\supseteq s$ is $s^{\prime}=s$ itself, in which case

$|\chi(s,\tau)\rangle=|\psi(s,\tau)\rangle.$ (202)

Let $U(\tau)$ be the unitary operator on the full relativistic Hilbert space
$\mathcal{H}^{R}$ which takes the Schroedinger-like representation of a state
on the hyperboloid $L(\tau)$ to the representation of that state on the
hyperplane with $x^{0}=0$. Define $|\hat{\chi}(s,\tau)\rangle$ to be the
$x^{0}=0$ representation of $|\chi(s,\tau)\rangle$

$|\hat{\chi}(s,\tau)\rangle=U(\tau)|\chi(s,\tau)\rangle.$ (203)

For any $\tau_{0}\leq\tau_{1}\leq\tau$, Eq. (201) implies

$|\hat{\chi}(s,\tau)\rangle=\sum_{s^{\prime}\in
S(\tau_{1}),s^{\prime}\supseteq s}|\hat{\chi}(s^{\prime},\tau)\rangle.$ (204)

For any $\tau\geq\tau_{0}$

$|\hat{\chi}[\\{(\emptyset,0)\\},\tau]\rangle=U(\tau_{0})|\psi\rangle.$ (205)

The example of Section VIII shows that as $\tau\rightarrow\infty$, for any
system not confined to a bounded region, branch splitting will continue
without stop and the values of $|s|$ for $s\in S(\tau)$ will grow without
bound. Thus there is no fixed $s\in S$, for which $|\psi(s,\tau)\rangle$
remains defined as $\tau\rightarrow\infty$. For every $s\in S$, however, the
summed branch $|\hat{\chi}(s,\tau)\rangle$ remains defined and potentially has
a limit as $\tau\rightarrow\infty$.

The discussion in Section V.3 of the evolution with increasing $t$ of the
optimal branch decomposition arising from the non-relativistic
$Q(\\{|\psi_{i}\rangle\\})$, now applied to $Q(\tau,\\{|\psi_{i}\rangle\\})$,
implies the evolution with increasing $\tau$ of the optimal relativistic
branch decomposition will be piecewise continuous. The discontinuous piece,
according to Sections VI and XIX, will consist almost entirely of permanent
splitting of some branch into a pair of sub-branches. The continuous piece,
according to Section V.3, for sufficiently large $b$ will consist almost
entirely of continuous unitary evolution with $\tau$ of the branches which do
not split. If branch splitting is entirely permanent splitting into pairs of
sub-branches and branches which don’t split change purely by unitary evolution
in $\tau$, then each $|\hat{\chi}(s,\tau)\rangle$ of Eq. (203), for any $s\in
S(\tau^{\prime})$ for any $\tau\geq\tau^{\prime}$, will be constant in $\tau$.
Thus the existence, for any $s\in S$, of the limit

$\lim_{\tau\rightarrow\infty}|\hat{\chi}(s,\tau)\rangle=|\hat{\chi}(s)\rangle$
(206)

appears to be a plausible hypothesis.

##  XXII  Translational Covariance

Let $|\psi\rangle\in\mathcal{H}^{R}$ be the representation of a state on the
$x^{0}=0$ hyperplane and $\\{|\hat{\chi}(s)\rangle\\}$ the corresponding set
of $\tau\rightarrow\infty$ branches. Let $P^{\mu},0\leq\mu<4,$ be the momentum
operator on $\mathcal{H}^{R}$. Let $|\psi_{z}\rangle$ be a copy of
$|\psi\rangle$ translated by $z_{\mu}$

$|\psi_{z}\rangle=\exp(-iz_{\mu}P^{\mu})|\psi\rangle.$ (207)

Let $\\{|\hat{\chi}_{z}(s)\rangle\\}$ be the set of $\tau\rightarrow\infty$
branches arising from $|\psi_{z}\rangle$. We now give an argument in support
of the hypothesis that for every $s\in S$

$|\hat{\chi}_{z}(s)\rangle=\exp(-iz_{\mu}P^{\mu})|\hat{\chi}(s)\rangle.$ (208)

If Eq. (208) holds for $z_{\mu}$ in a neighborhood of 0, it holds for all
$z_{\mu}$. We will assume $z_{\mu}$ small in the following.

Let $|\psi(\tau)\rangle$ be $|\psi\rangle$ represented on $L(\tau)$

$|\psi(\tau)\rangle=U^{\dagger}(\tau)|\psi\rangle,$ (209)

and let $|\psi(s,\tau)\rangle$ be the corresponding branch decomposition

$|\psi(\tau)\rangle=\sum_{s}|\psi(s,\tau)\rangle.$ (210)

Let $\\{Y_{j}\\}$ be a partition of space into disjoint regions, $W_{j}$ the
intersection of $Y_{j}$ with $L(\tau)$.

Let $\mathcal{Q}_{j}$ be the subspace of $\mathcal{H}^{R}$ spanned by the
operators with support in $W_{j}$ acting on the vacuum. Assume the
$\\{W_{j}\\}$ are all sufficiently large that Eq. (191) yields for each
$|\psi(s,\tau)\rangle$

$\displaystyle|\psi(s,\tau)\rangle$ $\displaystyle\approx$
$\displaystyle\lambda(s,\tau)\bigotimes_{j}|\psi_{j}(s,\tau)\rangle,$ (211a)
$\displaystyle|\psi_{j}(s,\tau)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{Q}_{j},$ (211b)

so that

$|\psi(\tau)\rangle\approx\sum_{s}\lambda(s,\tau)\bigotimes_{j}|\psi_{j}(s,\tau)\rangle.$
(212)

Let $E_{j}(\tau)$ be the set of events in the causal past of $W_{j}$. Any
$s\in S(\tau)$ is a union of the overlapping sets

$\displaystyle s$ $\displaystyle=$ $\displaystyle\cup_{j}s_{j}$ (213a)
$\displaystyle s_{j}$ $\displaystyle=$ $\displaystyle
s\cap\Bigl{(}E_{j}(\tau)\times\\{0,1\\}\Bigr{)},$ (213b)

Let $S_{j}(\tau)$ be the set of $s_{j}$ arising from $s\in S(\tau)$. For the
$W_{j}$ sufficiently large in comparison to the size of the entangled region
driving any branching event, the set of events with ambiguous classification
according to these definitions is a small fraction of the size of each set. We
will assume the vector $|\psi_{j}(s,\tau)\rangle$ actually has the form
$|\psi_{j}(s_{j},\tau)\rangle$ since it is unchanged by parts of $s$ outside
$s_{j}$. If each of the $W_{j}$ is sufficiently large,
$|\psi_{j}(s_{j},\tau)\rangle$ and
$|\psi_{j^{\prime}}(s_{j^{\prime}},\tau)\rangle,j^{\prime}\neq j,$ will be
orthogonal. The orthogonality of distinct branches implies in addition

$\langle\psi_{j}(s_{j},\tau)|\psi_{j}(s^{\prime}_{j},\tau)\rangle=\delta_{s_{j}s^{\prime}_{j}}.$
(214)

Let $|\psi_{z}(\tau)\rangle$ be

$\displaystyle|\psi_{z}(\tau)\rangle$ $\displaystyle=$ $\displaystyle
U^{\dagger}(\tau)\exp(-iz_{\mu}P^{\mu})|\psi\rangle,$ (215a) $\displaystyle=$
$\displaystyle
U^{\dagger}(\tau)\exp(-iz_{\mu}P^{\mu})U(\tau)|\psi(\tau)\rangle.$ (215b)

If $\tau$ is sufficiently large that each of the regions $W_{j}$ is nearly
flat, we will now argue that Eqs. (212) and (215b) should give

$\displaystyle|\psi_{z}(\tau)\rangle$ $\displaystyle\approx$
$\displaystyle\sum_{s}\lambda(s,\tau)\bigotimes_{j}|\psi_{zj}(s_{j},\tau)\rangle,$
(216a) $\displaystyle|\psi_{zj}(s_{j},\tau)\rangle$ $\displaystyle=$
$\displaystyle\exp(-iz_{\mu j}P^{\mu})|\psi_{j}(s_{j},\tau)\rangle,$ (216b)
$\displaystyle z_{\mu j}$ $\displaystyle=$ $\displaystyle a_{\mu
j}^{\nu}z_{\nu},$ (216c)

where $a_{\mu j}^{\nu}$ is the Lorentz boost which takes points in the
hyperplane with $x^{0}=x^{0}_{j}$ to points in the hyperplane holding $W_{j}$,
where $x^{0}_{j}$ is the time component of the center point of $W_{j}$.

An argument in support of Eqs. (216a) - (216c) is as follows. Suppose first
that only a single $|\psi_{j}(s_{j},\tau)\rangle$ differs from the vacuum and
that the corresponding $W_{j}$ is entirely flat. Then the effect of $U(\tau)$
on $|\psi_{j}(s_{j},\tau)\rangle$ should consist of a boost which takes states
represented in the the hyperplane holding $W_{j}$ to states represented in the
hyperplane with $x^{0}=x^{0}_{j}$ followed by a time development operator
taking states represented in the $x^{0}=x^{0}_{j}$ hyperplane to states in the
$x^{0}=0$ hyperplane. But since time development itself is assumed
translationally covariant, only the boost components of $U(\tau)$ and
$U^{\dagger}(\tau)$ in Eq. (215b) will have an effect on
$\exp(iz_{\mu}P^{\mu})$. The time development parts of $U(\tau)$ and
$U^{\dagger}(\tau)$ will commute through. The effect of the boost components
of $U(\tau)$ and $U^{\dagger}(\tau)$ on the translation operator is then given
Eqs. (216a) - (216c). If each of the $W_{j}$ is sufficiently large, since
$|\psi_{j}(s_{j},\tau)\rangle$ and
$|\psi_{j^{\prime}}(_{j^{\prime}}s,\tau)\rangle,j^{\prime}\neq j,$ will be
orthogonal, $U(\tau)$ should act nearly independently on each
$|\psi_{j}(s_{j},\tau)\rangle$. The result is Eqs. (216b) - (216c) for the
full expansion of $|\psi_{z}(\tau)\rangle$ in Eq. (216a).

Consider the $|\chi_{z}(s,\tau)\rangle$ found from the branches of
$|\psi_{z}(\tau)\rangle$. Suppose again to begin that in Eq. (216a) for
$|\psi_{z}(\tau)\rangle$ only a single $|\psi_{zj}(s_{j},\tau)\rangle$ differs
from the vaccum, suppose that $W_{j}$ is entirely flat, and ignore for the
moment the $z_{0j}$ component of $z_{\mu j}$. For $W_{j}$ entirely flat, the
net complexity function $Q(\tau,\\{|\psi_{i}\rangle\\})$ is translationally
covariant. It follows that the $W_{j}$ factor of any branch arising from
$|\psi_{z}(\tau)\rangle$ will be the translation the $W_{j}$ factor of a
corresponding branch of $|\psi(\tau)\rangle$, and similarly for the summed
branches

$|\chi_{zj}(s_{j},\tau)\rangle=\exp(-i\sum_{\mu>0}z_{\mu
j}P^{\mu})|\chi_{j}(s_{j},\tau)\rangle.$ (217)

Now ignore the $z_{\mu j},\mu>0,$ and assume $z_{0j}>0$. Then
$|\psi_{zj}(s_{j},\tau)\rangle$ will be $|\psi_{j}(s_{j},\tau)\rangle$
developed forward to $\tau+z_{0}$ and thus potentially subject to additional
branching. According to Section V.2, each factor of a tensor product of states
branches independently, and therefore even if more than one
$|\psi_{j}(s_{j},\tau)\rangle$ differs from the vacuum, the branches of
$|\psi_{j}(s_{j},\tau+z_{0})\rangle$ will be of the form
$\mu(s^{\prime}_{j})|\psi_{j}(s^{\prime}_{j},\tau+z_{0})\rangle$ for
$s^{\prime}_{j}\supseteq s_{j}$ and some set of nonnegative real
$\mu(s^{\prime}_{j})$. We therefore have

$\displaystyle|\chi_{zj}(s_{j},\tau)\rangle$ $\displaystyle=$
$\displaystyle\sum_{s_{j}^{\prime}\supseteq
s_{j}}\mu(s_{j}^{\prime})|\psi_{j}(s_{j}^{\prime},\tau+z_{0})\rangle,$ (218a)
$\displaystyle=$
$\displaystyle\exp(-iz_{0j}P^{\mu})|\psi_{j}(s_{j},\tau)\rangle,$ (218b)

so that by Eq. (202) for $|\chi_{j}(s_{j},\tau)\rangle$

$|\chi_{zj}(s_{j},\tau)\rangle=\exp(-iz_{0j}P^{\mu})|\chi_{j}(s_{j},\tau)\rangle.$
(219)

Combing Eqs. (217) and (219) gives

$|\chi_{zj}(s_{j},\tau)\rangle=\exp(-iz_{\mu
j}P^{\mu})|\chi_{j}(s_{j},\tau)\rangle,$ (220)

for $z_{0j}>0,s_{j}\in S_{j}(\tau)$.

On the other hand, for $z_{0j}<0$, the derivation of Eq. (219) can be repeated
but with the roles of $|\psi_{z}(\tau)\rangle$ and $|\psi(\tau)\rangle$
reversed, so that

$|\chi_{j}(s_{zj},\tau)\rangle=\exp(iz_{\mu
j}P^{\mu})|\chi_{zj}(s_{zj},\tau)\rangle,$ (221)

and therefore

$|\chi_{zj}(s_{zj},\tau)\rangle=\exp(-iz_{\mu
j}P^{\mu})|\chi_{j}(s_{zj},\tau)\rangle,$ (222)

for $s_{zj}\in S_{zj}(\tau)$.

The argument for Eq. (218a) implies for both positive and negative $z_{0j}$

$S_{zj}(\tau)=S_{j}(\tau+z_{0j}).$ (223)

Thus the sets of possible $s_{zj}$ for Eq. (220) and for Eq. (222) differ.
This difference can be removed by choosing some third
$\tau^{\prime}<\tau-|z_{0j}|$ and then using Eq. (201) to obtain Eqs. (220)
and (222) both for $s_{zj}\in S_{zj}(\tau^{\prime})$.

If the $W_{j}$ are all sufficiently large, which is possible for $\tau$
sufficiently large, if all of the $|\psi_{zj}(s_{j},\tau)\rangle$ and
$|\psi_{j}(s_{j},\tau)\rangle$ differ from the vaccum, Eqs. (220) and (222)
should still apply to each independently. We then have for an arbitrary
$|\psi\rangle$ and displaced $\exp(-iz_{\mu}P^{\mu})|\psi\rangle$ the summed
branches

$\displaystyle|\chi(s,\tau)\rangle$ $\displaystyle\approx$
$\displaystyle\lambda(s)\bigotimes_{j}|\chi_{j}(s_{j},\tau\rangle,$ (224a)
$\displaystyle|\chi_{z}(s,\tau)\rangle$ $\displaystyle\approx$
$\displaystyle\lambda(s)\bigotimes_{j}|\chi_{zj}(s_{j},\tau\rangle,$ (224b)

remain related by Eqs. (220) and (222). The derivation of Eqs. (215a) - (216c)
then implies

$U(\tau)|\chi_{z}(s,\tau)\rangle\approx\exp(-iz_{\mu}P^{\mu})U(\tau)|\chi(s,\tau)\rangle,$
(225)

and therefore

$|\hat{\chi}_{z}(s,\tau)\rangle\approx\exp(-iz_{\mu}P^{\mu})|\hat{\chi}(s,\tau)\rangle.$
(226)

If the limit in Eq. (206) exists as assumed, the $\tau\rightarrow\infty$ limit
of Eq. (226) is then Eq. (208).

##  XXIII  Born Rule As an Invariant Measure on Branching Histories

To begin, assume a particular Poincaré frame, $f$. Consider an infinite
sequence $s_{i}$ such that

$\displaystyle s_{i}$ $\displaystyle\in$
$\displaystyle\cup_{\tau}S_{f}(\tau),$ (227a) $\displaystyle|s_{i}|$
$\displaystyle=$ $\displaystyle i,$ (227b) $\displaystyle s_{i}$
$\displaystyle\subset$ $\displaystyle s_{i+1}.$ (227c)

A version of the Born rule based on asymptotic late time branches says the
probability a state with history which begins as $s_{i}$ at the next branching
event lands in $s_{i+1}$ is

$P(s_{i+1}|s_{i})=\frac{\langle\hat{\chi}(s_{i+1})|\hat{\chi}(s_{i+1})\rangle}{\langle\hat{\chi}(s_{i})|\hat{\chi}(s_{i})\rangle}.$
(228)

The Born rule we now formulate as a measure on the set of branching histories,
each extending over all time, beginning from some initial state
$|\psi\rangle$. An all-time branching history $\hat{s}$ is an infinite set of
pairs which assigns each event $e\in E$ to a corresponding branch index
$i\in\\{0,1\\}$.

$\hat{s}=\\{(e_{0},i_{0}),(e_{1},i_{1}),...\\}.$ (229)

Let $\hat{S}$ be the set of all such all-time histories $\hat{s}$. For every
$s\in S$, let $v(s)\subset\hat{S}$ be the collection of $\hat{s}\in\hat{S}$
which are supersets of $s$,

$v(s)=\\{\hat{s}\in\hat{S}|\hat{s}\supset s\\}.$ (230)

For every such $v(s)$ define the function $\mu[v(s)]$ to be

$\mu[v(s)]=\langle\hat{\chi}(s)|\hat{\chi}(s)\rangle.$ (231)

Let $\Sigma$ be the $\sigma$-algebra of sets in $\hat{S}$ generated by all
$v(s)$ for $s\in S$. The complement of any $v(s)$ is given by the finite union

$v(s)^{c}=\cup_{s^{\prime}\in c(s)}v(s^{\prime}),$ (232)

where $c(s)$ is the set of $s^{\prime}$ each consisting of exactly one of the
events in $s$ but with branch index reversed

$c[\\{(e_{0},i_{0}),...(e_{n-1},i_{n-1})\\}]=\\\ \Bigl{\\{}\\{(e_{0},\neg
i_{0})\\},...\\{(e_{n-1},\neg i_{n-1})\\}\Bigr{\\}}.$ (233)

In addition, for any $s,s^{\prime}\in S$,

$v(s)\cap v(s^{\prime})=v(s\cup s^{\prime}).$ (234)

It follows that every element of $\Sigma$ is given by a union of a countable
collection of pairwise disjoint $v(s)$. For every countable collection of
pairwise disjoint sets $\\{v(s_{i})\\}$, define

$\mu[\cup_{i}v(s_{i})]=\sum_{i}\mu[v(s_{i})].$ (235)

Eq. (235) turns $\mu$ into a probability measure on $\Sigma$.

Eq. (228) follows from Eq. (231). Since the $|\hat{\chi}(s)\rangle$ are
Poincaré covariant and the algebra $\Sigma$ is frame independant, the measure
$\mu$ is Poincaré invariant. The Born rule can then be formulated as the
hypothesis that world’s history of branching events is an $\hat{s}\in\hat{S}$
chosen randomly according to the measure $\mu$.

##  XXIV  Time Dependent View of Branching History

The Poincaré covariant set of $\tau\rightarrow\infty$ branches
$|\hat{\chi}(s)\rangle$ and corresponding branching history $\hat{s}$ chosen
according to the Born measure of Section XXIII we take to be the physical
objects underlying macroscopic reality. From these, a view of branching
history unfolding in time in any particular Poincaré frame can be constructed.

In any particular frame, for any all-time history of events $\hat{s}$, there
is a corresponding sequence of partial branch histories $s_{n}\in S,n\geq 1,$
with

$\displaystyle|s_{n}|$ $\displaystyle=$ $\displaystyle n,$ (236a)
$\displaystyle s_{n}$ $\displaystyle\subset$ $\displaystyle s_{n+1},$ (236b)
$\displaystyle\cup_{n}s_{n}$ $\displaystyle=$ $\displaystyle\hat{s},$ (236c)

ordered in such a way that for every $n$ the last event in $s_{n}$ occurs
after the last event in $s_{n-1}$. Let $|\hat{\chi}(s_{n})\rangle$ be the
corresponding sequence of states represented on the $x^{0}=0$ hyperplane. From
these define $|\psi_{n}(\tau)\rangle$ to be

$|\psi_{n}(\tau)\rangle=U^{\dagger}(\tau)|\hat{\chi}(s_{n})\rangle,$ (237)

where $U^{\dagger}(\tau)$ is the unitary operator taking states represented on
the $x^{0}=0$ hyperplane to their representation on the $L(\tau)$ hyperbeloid.
The system begins at $\tau_{0}$ evolution from the initial state
$|\psi\rangle$

$|\psi_{1}(\tau_{0})\rangle=|\psi\rangle,$ (238)

then at a sequence of proper times $\tau_{n},n\geq 1$, successively branches
from $|\psi_{n}(\tau_{n})\rangle$ to $|\psi_{n+1}(\tau_{n})\rangle$.

The $\tau_{n}$ can be found as follows. Define $|\psi_{n}(\tau)\rangle$ and
$\rho_{n}$ to be

$\displaystyle|\phi_{n}(\tau)\rangle$ $\displaystyle=$
$\displaystyle|\psi_{n}(\tau)\rangle-|\psi_{n+1}(\tau)\rangle$ (239a)
$\displaystyle\rho_{n}$ $\displaystyle=$
$\displaystyle\frac{\langle\psi_{n+1}(\tau)|\psi_{n+1}(\tau)\rangle}{\langle\psi_{n}(\tau)|\psi_{n}(\tau)\rangle}.$
(239b)

From these define

$\Delta_{n}(\tau)=[C(|\psi_{n}(\tau)\rangle)]^{2}-\\\
\rho_{n}[C(|\psi_{n+1}(\tau)\rangle)]^{2}-(1-\rho_{n})[C(|\phi_{n}\rangle)]^{2}\\\
+b\rho_{n}\ln(\rho_{n})+b(1-\rho_{n})\ln(1-\rho_{n}).$ (240)

Each $\tau_{n}$ will then be the smallest $\tau$ such that

$\Delta_{n}(\tau)\geq 0.$ (241)

By choice of the $s_{n}$, the sequence of $\tau_{n}$ is guaranteed to be
increasing.

##  XXV Conclusion

In Section II we argued that the branching which follows from environmentally
induced decoherence by itself looks like it’s missing something. The present
article consists of a series of conjectures which propose to fill in what’s
missing.

What are the odds these various guesses might be right? With the exception of
the experiments proposed in Section XI, the various conjectures can all, at
least in principle, be tested by numerical experiments. Among the hypotheses
which could be checked numerically are the proposal in Section VI that
branching is almost always a permanent split of a single branch into a pair of
sub-branches, the proposal in Section X that branches on the large scale are
nearly tensor products each factor of which is entangled only over limited
distance, and the conjecture in Section XXI that the infinite proper time
limit exists for the $|\hat{\chi}(s,\tau)\rangle$. On the other hand, the
complexities needed for the experiments in Section XI might also, at least in
principle, be filled in numerically leading to a realizable attempt to
estimate $b$.

The non-relativistic version of the proposal here is something like the Von
Neumann-Wigner interpretation Vonneumann ; Wigner ; London  turned on its
head. Instead of conscious observation causing branching events, branching
events occur with or without an observer but those which include a sufficient
set of an observer’s degrees of freedom register as a transition in thought. A
somewhat related possibility is that distinct mental states might in all cases
be associated with distinct branches because the complexity arising from the
superposition of distinct mental states is itself sufficient to cause
branching. The sequence of mental transitions associated with a trajectory of
thought might then be a sequence of successive branching events.

###### Acknowledgements.

Thanks to Jess Riedel for an extended debate over an earlier version of this
work and to an anonymous reviewer for comments leading to many improvements
incorporated in the present version.

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##  Appendix A  Truncated Hermitian Operator Hilbert Space

Let $\mathcal{H}_{x}^{n}$ be the subspace of $\mathcal{H}_{x}$ with less than
$n$ bosons. The dimension $d_{n}$ of each $\mathcal{H}_{x}^{n}$ is finite. Let
$\mathcal{H}^{n}$ be the product over $x$ of all $\mathcal{H}^{n}_{x}$

$\mathcal{H}^{n}=\bigotimes_{x}\mathcal{H}_{x}^{n}.$ (242)

For any site $x$, let $\mathcal{F}^{n}_{x}$ consist of all Hermitian $f_{x}$
on $\mathcal{H}^{n}_{x}$ with finite

$\parallel f_{x}\parallel^{2}=\mathrm{Tr}_{x}(f_{x})^{2},$ (243)

and vanishing trace

$\mathrm{Tr}_{x}f_{x}=0.$ (244)

For any pair of nearest neighbor sites $\\{x,y\\}$, let $\mathcal{F}^{n}_{xy}$
consist of all Hermitian $f_{xy}$ on
$\mathcal{H}^{n}_{x}\otimes\mathcal{H}^{n}_{y}$ with finite

$\parallel f_{xy}\parallel^{2}=\mathrm{Tr}_{xy}(f_{xy})^{2},$ (245)

and vanishing traces

$\displaystyle\mathrm{Tr}_{x}f_{xy}$ $\displaystyle=$ $\displaystyle 0,$
(246a) $\displaystyle\mathrm{Tr}_{y}f_{xy}$ $\displaystyle=$ $\displaystyle
0.$ (246b)

Inner products on $\mathcal{F}^{n}_{x}$ and $\mathcal{F}^{n}_{xy}$ are

$\displaystyle\langle f_{x},f^{\prime}_{x}\rangle$ $\displaystyle=$
$\displaystyle\mathrm{Tr}_{x}(f_{x}f^{\prime}_{x}),$ (247a)
$\displaystyle\langle f_{xy},f^{\prime}_{xy}\rangle$ $\displaystyle=$
$\displaystyle\mathrm{Tr}_{xy}(f_{xy}f^{\prime}_{xy}).$ (247b)

Operators $f_{x}\in\mathcal{F}^{n}_{x}$ and $f_{xy}\in\mathcal{F}^{n}_{xy}$
can be made into operators on $\mathcal{H}^{n}$ by

$\displaystyle\hat{f}_{x}$ $\displaystyle=$ $\displaystyle
f_{x}\bigotimes_{q\neq x}I_{q},$ (248a) $\displaystyle\hat{f}_{xy}$
$\displaystyle=$ $\displaystyle f_{xy}\bigotimes_{q\neq x,y}I_{q},$ (248b)

where $I_{q}$ is the identity operator on $\mathcal{H}^{n}_{q}$. As usual, we
now drop the hat and use the same symbol for operators on
$\mathcal{H}^{n}_{x}$, $\mathcal{H}^{n}_{x}\otimes\mathcal{H}^{n}_{y}$, and
the corresponding operators on $\mathcal{H}^{n}$.

Let $K^{n}$ be the vector space over the reals of linear operators $k$ on
$\mathcal{H}^{n}$ given by sums of the form

$k=\sum_{xy}f_{xy}+\frac{1}{\sqrt{d_{n}}}\sum_{x}f_{x}$ (249)

for any collection of $f_{xy}\in\mathcal{F}^{n}_{xy}$ for a set of nearest
neighbor pairs $\\{x,y\\}$ and any collection of $f_{x}\in\mathcal{F}^{n}_{x}$
in a set of sites $x$. The inner product on $K$ is

$\langle k,k^{\prime}\rangle=\sum_{xy}\langle
f_{xy},f^{\prime}_{xy}\rangle+\sum_{x}\langle f_{x},f^{\prime}_{x}\rangle.$
(250)

An equivalent inner product on $K^{n}$, which is a version of the inner
product on operator Hilbert space in [10], is

$\langle
k,k^{\prime}\rangle=\frac{\mathrm{Tr}(kk^{\prime})}{d_{n}^{n_{L}-2}},$ (251)

where $\mathrm{Tr}$ is the trace on all of $\mathcal{H}^{n}$ and $n_{L}$ is
the number of sites in the lattice $L$. As a result of the factor of
$\frac{1}{\sqrt{d_{n}}}$ in Eq. (249), if $d_{n}$ is made large, matrix
elements of $k$ given by Eq. (249) will approach those of $k$ given by Eq.
(15) and $K^{n}$ will become equivalent to the operator space $K$ of Section
III.2.

##  Appendix B  Lower Bound on the Complexity of Entangled States

The proof of Eq. (37) proceeds as follows. The trajectories $k(\nu)\in K$ and
$U_{k}(\nu)$ which determine any $C(|\psi\rangle,|\omega\rangle)$, according
to Eqs. (17a) - (20), we characterize by a corresponding set of trajectories
of Schmidt spectrum vectors. We then find the rotation matrices which govern
the motion of these vectors as $\nu$ changes. A bound on the time integral of
the angles which occur in these matrices by a time integral of $\parallel
k(\nu)\parallel$ yields Eq. (37).

###  B.1  Schmidt Spectra

Consider some entangled $n$-fermion $|\psi\rangle$ of form Eq. (36). For a
trajectory $k(\nu)\in K$, let $U_{k}(\nu)$ be the solution to Eqs. (17a) and
(17b). Define $|\omega(\nu)\rangle$ to be

$|\omega(\nu)\rangle=U_{k}(\nu)|\omega\rangle,$ (252)

for some product state $|\omega\rangle$ and assume that $k(\nu)$ has been
chosen to give

$|\omega(1)\rangle=\xi|\psi\rangle,$ (253)

for a phase factor $\xi$. Since all $k(\nu)$ conserve fermion number,
$|\omega\rangle$ according to Eq. (9) must have the form

$|\omega\rangle=d_{f}^{\dagger}(p_{n-1})...d_{f}^{\dagger}(p_{0})\times\\\
d_{b}^{\dagger}(q_{m-1})...d_{b}^{\dagger}(q_{0})|\Omega\rangle,$ (254)

for some number of bosons $m$.

We now divide the lattice $L$ into a collection of disjoint regions and define
a corresponding collection of Schmidt decompositions of the trajectory of
states which determine any $C(|\psi\rangle,|\omega\rangle)$. Divide $L$ into
subsets $L^{e},L^{o}$, with, respectively, even or odd values of the sums of
components $\hat{x}_{i}$. The sites in each subset have nearest neighbors only
in the other. Let $D^{e}_{ij},D^{o}_{ij},D^{e},D^{o}$ be

$\displaystyle D^{e}_{ij}$ $\displaystyle=$ $\displaystyle L^{e}\cap D_{ij}$
(255a) $\displaystyle D^{o}_{ij}$ $\displaystyle=$ $\displaystyle L^{o}\cap
D_{ij}$ (255b) $\displaystyle D^{e}$ $\displaystyle=$
$\displaystyle\cup_{ij}D^{e}_{ij}.$ (255c) $\displaystyle D^{o}$
$\displaystyle=$ $\displaystyle\cup_{ij}D^{o}_{ij}.$ (255d)

Between $D^{e}$ and $D^{o}$ choose the larger, or either if they are equal.
Assume the set chosen is $D^{e}$. Among the $nm$ spins $s_{ij}$, at least
$\frac{nm}{2}$ will have the same value and therefore correspond to $D_{ij}$
which do not intersect. The corresponding collection of $D^{e}_{ij}$ will then
include at least $\frac{nmV}{4}$ points.

From this set of $D^{e}_{ij}$ construct a set of subsets $E_{\ell}$ each
consisting of $2n$ distinct points chosen from $2n$ distinct $D^{e}_{ij}$. The
total number of $E_{\ell}$ will then be at least $\frac{mV}{8}$. We will
consider only the first $\frac{mV}{8}$ of these.

The Hilbert space $\mathcal{H}$ is given by a tensor product

$\mathcal{H}=\mathcal{H}^{f}\otimes\mathcal{H}^{b},$ (256)

of a fermion space $\mathcal{H}^{f}$ and a boson space $\mathcal{H}^{b}$.
Similarly the space $\mathcal{H}_{x}$ at each $x$ is given by a tensor product

$\mathcal{H}_{x}=\mathcal{H}_{x}^{f}\otimes\mathcal{H}_{x}^{b},$ (257)

of a fermion space $\mathcal{H}_{x}^{f}$ and a boson space
$\mathcal{H}_{x}^{b}$. The dimensions of $\mathcal{H}_{x}^{f}$ and
$\mathcal{H}_{x}^{b}$ are, respectively, 4 and $\infty$.

For each set $E_{\ell}$ form the tensor product spaces

$\displaystyle\mathcal{Q}_{\ell}$ $\displaystyle=$
$\displaystyle\bigotimes_{x\in E_{\ell}}\mathcal{H}_{x}^{f},$ (258a)
$\displaystyle\mathcal{R}_{\ell}$ $\displaystyle=$
$\displaystyle\mathcal{H}^{b}\bigotimes_{q\neq E_{\ell}}\mathcal{H}_{q}^{f}.$
(258b)

It follows that $\mathcal{Q}_{\ell}$ has dimension $4^{2n}$ and

$\mathcal{H}=\mathcal{Q}_{\ell}\otimes\mathcal{R}_{\ell}.$ (259)

A Schmidt decomposition of $|\omega(\nu)\rangle$ according to Eq. (259) then
becomes

$|\omega(\nu)\rangle=\sum_{j}\lambda_{j\ell}(\nu)|\phi_{j\ell}(\nu)\rangle|\chi_{j\ell}(\nu)\rangle,$
(260)

where

$\displaystyle|\phi_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{Q}_{\ell}$ (261a)
$\displaystyle|\chi_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{R}_{\ell},$ (261b)

for $0\leq j<4^{2n}$ and real non-negative $\lambda_{j\ell}(\nu)$ which
fulfill the normalization condition

$\sum_{j}[\lambda_{j\ell}(\nu)]^{2}=1.$ (262)

Each $|\phi_{j\ell}(\nu)\rangle$ is a pure fermion state while the
$|\chi_{j\ell}(\nu)\rangle$ can include both fermions and bosons.

The fermion number operators $N[\mathcal{Q}_{\ell}]$ and
$N[\mathcal{R}_{\ell}]$ commute and $|\omega(\nu)\rangle$ is an eigenvector of
the sum with eigenvalue $n$. It follows that the decomposition of Eq. (260)
can be done with $|\phi_{j\ell}(\nu)\rangle$ and $|\chi_{j\ell}(\nu)\rangle$
eigenvectors of $N[\mathcal{Q}_{\ell}]$ and $N[\mathcal{R}_{\ell}]$,
respectively, with eigenvalues summing to $n$. Let $|\phi_{0\ell}\rangle$ be
$|\Omega_{\ell}\rangle$, the vacuum state of $\mathcal{Q}_{\ell}$, and let
$|\phi_{i\ell}(\nu)\rangle,1\leq i\leq 4n$, span the $4n$-dimensional subspace
of $\mathcal{Q}_{\ell}$ with $N[\mathcal{Q}_{\ell}]$ of 1. We assume the
corresponding $\lambda_{i\ell}(\nu),1\leq i\leq 4n$, are in nonincreasing
order. Consider Eq. (260) for $\nu=1$. By Eq. (253), for any choice of $\ell$
there will be a set of $2n$ nonzero orthogonal
$|\phi_{1\ell}(1)\rangle,...|\phi_{2n\ell}(1)\rangle$ with

$\lambda_{j\ell}(1)=\sqrt{\frac{1}{mV}},$ (263)

for $1\leq j\leq 2n$.

On the other hand, for $\nu=0$, Eq. (260) becomes a decomposition of the
product state $|\omega\rangle$. The boson part of $|\omega(0)\rangle$ will
occur as the same overall tensor factor in each
$|\chi_{1\ell}(0)\rangle,...|\chi_{n\ell}(0)\rangle$. The fermion part of
$|\omega(0)\rangle$ is a product of $n$ independent single fermion states, the
space spanned by the projection of these into some $\mathcal{Q}_{\ell}$ is at
most $n$ dimensional, and as a result at most $n$ orthogonal
$|\phi_{1\ell}(0)\rangle,...|\phi_{n\ell}(0)\rangle$ can occur. Therefore at
$\nu=0$, there will be at most $n$ nonzero
$\lambda_{1\ell}(0),...\lambda_{n\ell}(0)$. For $n<j\leq 2n$, we have

$\lambda_{j\ell}(0)=0.$ (264)

But according to Eq. (262), for each fixed value of $\ell$ the set of
components $\\{\lambda_{j\ell}(\nu)\\}$ indexed by $j$ is a unit vector. Eqs.
(264) and (263) then imply that as $\nu$ goes from $0$ to $1$,
$\\{\lambda_{j\ell}(\nu)\\}$ must rotate through a total angle of at least
$\arcsin(\sqrt{\frac{n}{mV}})$.

For the small interval from $\nu$ to $\nu+\delta\nu$ let $\mu_{j\ell}(\nu)$
and $\theta_{\ell}(\nu)$ be

$\displaystyle\lambda_{j\ell}(\nu+\delta\nu)$ $\displaystyle=$
$\displaystyle\lambda_{j\ell}(\nu)+\delta\nu\mu_{j\ell}(\nu),$ (265a)
$\displaystyle\theta_{\ell}(\nu)^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}[\mu_{j\ell}(\nu)]^{2}.$ (265b)

We then have

$\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu\geq\arcsin\left(\sqrt{\frac{n}{mV}}\right).$
(266)

Summed over the $\frac{mV}{8}$ values of $\ell$, Eq. (266) becomes

$\sum_{\ell}\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu\geq\frac{mV}{8}\arcsin\left(\sqrt{\frac{n}{mV}}\right),$
(267)

and therefore

$\sum_{\ell}\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu\geq\frac{1}{4\pi}\sqrt{mnV}.$
(268)

###  B.2  More Schmidt Spectra

Replacing the subsets $E_{\ell}$ defined in Appendix B.1, with subsets of $L$
obtained from the $S_{\ell}$ of Section IV leads to an additional bound
similar to Eq. (268).

For each $0\leq\ell<q$, of the two subsets of $L$ defined by $S_{\ell}$, let
$T_{\ell}$ be the subset which, for each $0\leq i<m$, holds $n_{0}$ of the
sets $D_{ij},0\leq j<n$. Redefine $\mathcal{Q}_{\ell},\mathcal{R}_{\ell}$ of
Eqs. (258a) and (258b), to be

$\displaystyle\mathcal{Q}^{T}_{\ell}$ $\displaystyle=$
$\displaystyle\bigotimes_{x\in T_{\ell}}\mathcal{H}_{x}^{f},$ (269a)
$\displaystyle\mathcal{R}^{T}_{\ell}$ $\displaystyle=$
$\displaystyle\mathcal{H}^{b}\bigotimes_{q\neq T_{\ell}}\mathcal{H}_{q}^{f}.$
(269b)

For each $0\leq\ell<q$ there is again a corresponding Schmidt decomposition of
$|\omega(\nu)\rangle$ of Eqs. (252) and (253)

$|\omega(\nu)\rangle=\sum_{j}\lambda^{T}_{j\ell}(\nu)|\phi^{T}_{j\ell}(\nu)\rangle|\chi^{T}_{j\ell}(\nu)\rangle,$
(270)

where

$\displaystyle|\phi^{T}_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{Q}^{T}_{\ell},$ (271a)
$\displaystyle|\chi^{T}_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{R}^{T}_{\ell}.$ (271b)

Each $|\phi^{T}_{j\ell}(\nu)\rangle$ is a pure fermion state while the
$|\chi^{T}_{j\ell}(\nu)\rangle$ can include both fermions and bosons. For
$\nu=1$, for every $0\leq\ell<q$, the sum over $j$ in Eq. (270) has $m$
nonzero entries each with

$\lambda^{T}_{j\ell}(1)=\frac{1}{\sqrt{m}},$ (272)

with $|\phi^{T}_{j\ell}(1)\rangle$ carrying fermion number $n_{0}$ and
$|\chi^{T}_{j\ell}(1)\rangle$ carrying fermion number $n_{1}$.

Duplicating the discussion of Appendix B.1, a trajectory of angles
$\theta^{T}_{\ell}(\nu)$ can be defined which rotates the unit vector
$[\lambda^{T}_{j\ell}(0)]$ arising from the product state $|\omega(0)\rangle$
into the unit vector $[\lambda^{T}_{j\ell}(1)]$ of Eq. (272). For each
$0\leq\ell<q$, a version of the lower bound of Eq. (266) can be obtained by
finding the product state $|\omega(0)\rangle$ which gives
$[\lambda^{T}_{j\ell}(0)]$ closest to $[\lambda^{T}_{j\ell}(1)]$ for the set
of $0\leq j<m$ corresponding to $|\phi^{T}_{j\ell}(0)\rangle$ and
$|\chi^{T}_{j\ell}(0)\rangle$ with fermion numbers $n_{0}$ and $n_{1}$,
respectively.

According to Eq. (9), the product state $|\omega(0)\rangle$ includes $n$
fermion creation operators $d^{\dagger}_{f}(p_{i})$ given by Eq. (8a). Since
$|\omega(1)\rangle$ and therefore $|\omega(0)\rangle$ are normalized to 1, we
can require the $p_{i}(x,s)$ to be othronomal. The simplest way to insure
$n_{0}$ and $n_{1}$, respectively, for $|\phi_{0\ell}(0)\rangle$ and
$|\chi_{0\ell}(0)\rangle$ is for the support of $p_{i}(x,s)$ to be entirely
within $T_{\ell}$ for $0\leq i<n_{0}$ and entirely outside $T_{\ell}$ for
$n_{0}\leq i<n$. The Schmidt decomposition of Eq. (270) then yields a vector
$[\lambda^{T}_{j\ell}(0)]$ with only a single nonzero entry and therefore

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)=\frac{1}{\sqrt{m}}.$
(273)

A larger value of the sum in Eq. (273) is possible only if an even number of
$p_{i}(x,s)$ have support both within $T_{\ell}$ and outside $T_{\ell}$. For
some $r\leq n_{0},n_{1},$, define $z$ to be the set

$z=\\{i|0\leq i<r\\}\cup\\{i|n_{0}\leq i<n_{0}+r\\}.$ (274)

Then for $i\in z$, suppose

$p_{i}(x,s)=p^{0}_{i}(x,s)+p^{1}_{i}(x,s),$ (275)

where the $p^{0}_{i}(x,s)$ have support entirely within $T_{\ell}$ and the
$p^{1}_{i}(x,s)$ have support entirely outside $T_{\ell}$. Since $p_{i}(x,s)$
is normalized and the support of $p^{0}_{i}(x,s)$ is disjoint from the support
of $p^{1}_{i}(x,s)$ we have

$\parallel|p^{0}_{i}\rangle\parallel^{2}+\parallel|p^{1}_{i}\rangle\parallel^{2}=1.$
(276)

The piece $|\hat{\omega}(0)\rangle$ of $|\omega(0)\rangle$ with fermion number
$n_{0}$ on $T_{\ell}$ and $n_{1}$ outside $T_{\ell}$ is given by

$|\hat{\omega}(0)\rangle=\\\ \sum_{u}[\bigotimes_{i\in
u}|p^{0}_{i}\rangle\bigotimes_{j\in z-u}|p^{1}_{j}\rangle]\bigotimes_{i\notin
z}|p_{i}\rangle,$ (277)

where the sum is over all $r$ element subsets $u\subset z$.

The vector $[\lambda^{T}_{j\ell}(0)]$ corresponding to
$|\hat{\omega}(0)\rangle$ will have at most $\frac{(2r)!}{(r!)^{2}}$ nonzero
entries, one for each of the sets $u$ in the sum in Eq. (277). The Cauchy-
Schwartz inequality then yields

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)\leq\\\
\sqrt{\frac{(2r)!}{m(r!)^{2}}}\parallel|\hat{\omega}(0)\rangle\parallel$ (278)

Let $|\hat{p}^{0}_{i}\rangle$ be the projection of $|p^{0}_{i}\rangle$
orthogonal to all $|\hat{p}^{0}_{j}\rangle,j<i$, and let
$|\hat{p}^{1}_{i}\rangle$ be the projection of $|p^{1}_{i}\rangle$ orthogonal
to all $|\hat{p}^{1}_{j}\rangle,j<i$. Substituting
$\\{|\hat{p}^{0}_{i}\rangle\\}$ and $\\{|\hat{p}^{1}_{i}\rangle\\}$ for
$\\{|p^{0}_{i}\rangle\\}$ and $\\{|p^{1}_{i}\rangle\\}$, respectively, in Eq.
(277) leaves $|\hat{\omega}(0)\rangle$ unchanged. The value of
$\parallel|\hat{\omega}(0)\rangle\parallel$ will then be maximized if the
resulting $\parallel|\hat{p}^{0}_{i}\rangle\parallel$ and
$\parallel|\hat{p}^{1}_{i}\rangle\parallel$ are increased as needed to satisfy
Eq. (276).

Suppose now that $\parallel|\hat{\omega}(0)\rangle\parallel$ has been
maximized with respect to $\parallel|\hat{p}^{0}_{i}\rangle\parallel$ and
$\parallel|\hat{p}^{1}_{i}\rangle\parallel$ for all $0\leq i<r$, expect some
pair of values $j,k$. The remaining dependence on
$\parallel|\hat{p}^{0}_{i}\rangle\parallel$ and
$\parallel|\hat{p}^{1}_{i}\parallel$ for $i=j,k,$ is maximized at

$\displaystyle\parallel|\hat{p}^{0}_{j}\rangle\parallel$ $\displaystyle=$
$\displaystyle\parallel|\hat{p}^{0}_{k}\rangle\parallel,$ (279a)
$\displaystyle\parallel|\hat{p}^{1}_{j}\rangle\parallel$ $\displaystyle=$
$\displaystyle\parallel|\hat{p}^{1}_{k}\rangle\parallel.$ (279b)

If $\parallel|\hat{\omega}(0)\rangle\parallel$ is then maximized with respect
to the remaining $i$ independent $\parallel|\hat{p}^{0}_{i}\rangle\parallel$
and $\parallel|\hat{p}^{1}_{i}\rangle\parallel$, Eq. (278) becomes

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)\leq\sqrt{\frac{(2r)!}{m2^{r}(r!)^{2}}}$
(280)

Suppose $m$ has the form $\frac{(2r)!}{(r!)^{2}}$. For any $r^{\prime}\leq r$
Eq. (280) becomes

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)\leq\frac{(2r^{\prime})!}{\sqrt{m}2^{r^{\prime}}(r^{\prime}!)^{2}}.$
(281)

An induction argument then shows that Eq. (281) is an increasing function of
$r^{\prime}$. For $r^{\prime}>r$ on the other hand, the sum in Eq. (281)
becomes

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)\leq\frac{\sqrt{m}}{2^{r^{\prime}}},$
(282)

which is a decreasing function of $r^{\prime}$. The maximum of Eq. (281) will
therefore be at $r^{\prime}=r$.

Now suppose $m$ lies between $\frac{(2r)!}{(r!)^{2}}$ and
$\frac{(2r+2)!}{[(r+1)!]^{2}}$. For $r^{\prime}\leq r$ the maximum Eq. (281)
will still be at $r^{\prime}=r$ and given by

$\displaystyle\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)$
$\displaystyle\leq$ $\displaystyle\frac{(2r)!}{\sqrt{m}2^{r}(r!)^{2}},$ (283a)
$\displaystyle<$ $\displaystyle\sqrt{\frac{(2r)!}{2^{2r}(r!)^{2}}}.$ (283b)

For $r^{\prime}=r+1$, Eq. (281) becomes

$\displaystyle\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)$
$\displaystyle\leq$ $\displaystyle\frac{\sqrt{m}}{2^{r+1}},$ (284a)
$\displaystyle<$ $\displaystyle\sqrt{\frac{(2r+2)!}{2^{2r+2}[(r+1)!]^{2}}}.$
(284b)

A further induction argument shows that Eqs. (283b) and (284b) are decreasing
functions of $r$. Thus for $m\geq 2$, we have

$\sum_{j}\lambda^{T}_{j\ell}(0)\lambda^{T}_{j\ell}(1)\leq\frac{1}{\sqrt{2}}.$
(285)

A duplicate of the argument leading to Eq. (268) then yields

$\sum_{\ell}\int_{0}^{1}|\theta^{T}_{\ell}(\nu)|d\nu\geq\frac{\pi q}{4}.$
(286)

###  B.3  Schmidt Rotation Matrix

A lower bound on $C(|\psi\rangle)$ follows from Eqs. (268) and (286).
Appendices B.3 and B.4 derive the consequence of Eq. (268). A derivation of
the additional terms in the bound on $C(|\psi\rangle)$ which follow from Eq.
(286) is briefly summarized in Appendix B.5.

The rotation of $\lambda_{j\ell}(\nu)$ during the interval from $\nu$ to
$\nu+\delta\nu$ will be determined by $k(\nu)$. For each $f_{xy}$ in Eq. (15)
for $k(\nu)$ which can contribute to a nonzero value of $\theta_{\ell}(\nu)$,
the nearest neighbor pair $\\{x,y\\}$ has one point, say $x$ in $E_{\ell}$.
Since $E_{\ell}\subset D^{e}$ and the nearest neighbors of all points in
$D^{e}$ are in $D^{o}$, $y$ can not be in $E_{\ell}$. Let $g_{\ell}(\nu)$ be
the sum of all such $f_{xy}$. The effect of all other terms in Eq. (15) on the
Schmidt decomposition of Eq. (260) will be a unitary transformation on
$\mathcal{R}_{\ell}$ and identity on $\mathcal{Q}_{\ell}$. All other terms
will therefore leave $\lambda_{j\ell}(\nu)$ unchanged.

The effect of $g_{\ell}(\nu)$ on $\lambda_{j\ell}(\nu)$ over the interval from
$\nu$ to $\nu+\delta\nu$ can be determined from the simplification

$|\omega(\nu+\delta\nu)\rangle=\exp[i\delta\nu
g_{\ell}(\nu)]|\omega(\nu)\rangle.$ (287)

From $|\omega(\nu+\delta\nu)\rangle\langle\omega(\nu+\delta\nu)|$ of Eq.
(287), construct the density matrix $\rho(\nu+\delta\nu)$ by a partial trace
over $\mathcal{R}_{\ell}$, using the basis for $\mathcal{R}_{\ell}$ from the
Schmidt decomposition in Eq. (260) of $|\omega(\nu)\rangle$ at $t$

$\rho(\nu+\delta\nu)=\sum_{j}[\langle\chi_{j\ell}(\nu)|\omega(\nu+\delta\nu)\rangle\times\\\
\langle\omega(\nu+\delta\nu)|\chi_{j\ell}(\nu)\rangle].$ (288)

An eigenvector decomposition of $\rho(\nu+\delta\nu)$ exposes the
$\lambda_{j\ell}(\nu+\delta\nu)$

$\rho(\nu+\delta\nu)=\sum_{j}[\lambda_{j\ell}(\nu+\delta\nu)^{2}\times\\\
|\phi_{j\ell}(\nu+\delta\nu)\rangle\langle\phi_{j\ell}(\nu+\delta\nu)|].$
(289)

A power series expansion through first order in $\delta\nu$ applied to Eqs.
(287), (288) and (289) then gives for $\mu_{j\ell}(\nu)$ of Eq. (265a)

$\mu_{j\ell}(\nu)=\sum_{k}r_{jk\ell}(\nu)\lambda_{k\ell}(\nu),$ (290)

for the rotation matrix $r_{jk\ell}(\nu)$

$r_{jk\ell}(\nu)=-\operatorname{Im}[\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|\\\
g_{\ell}(\nu)|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle].$ (291)

###  B.4  Rotation Angle Bounds

Since the $f_{xy}$ contributing to $g_{\ell}(\nu)$ conserve total fermion
number $N$, $g_{\ell}(\nu)$ can be expanded as

$\displaystyle g_{\ell}(\nu)$ $\displaystyle=$
$\displaystyle\sum_{xy}g_{\ell}(x,y,\nu),$ (292a) $\displaystyle
g_{\ell}(x,y,\nu)$ $\displaystyle=$
$\displaystyle\sum_{i=0,1}a^{i}(x,y,\nu)z^{i}(x,y,\nu)$ (292b)

where $z^{0}(x,y,\nu)$ acts only on states with
$N(\mathcal{H}_{x}\otimes\mathcal{H}_{y})$ of 0, $z^{1}(x,y,\nu)$ acts only on
states with $N(\mathcal{H}_{x}\otimes\mathcal{H}_{y})$ strictly greater than
0, and the $z^{i}(x,y,\nu)$ are normalized by

$\parallel z^{i}(x,y,\nu)\parallel=1.$ (293)

The operator $z^{0}(x,y,\nu)$ will be

$\displaystyle z^{0}(x,y,\nu)$ $\displaystyle=$ $\displaystyle
z^{0f}(x,y)\otimes g^{b}(x,y,\nu),$ (294a) $\displaystyle z^{0f}(x,y,\nu)$
$\displaystyle=$ $\displaystyle P^{f}(x,y)\bigotimes_{q\neq x,y}I_{q},$ (294b)

where $P^{f}(x,y)$ projects onto the vacuum state of
$\mathcal{H}^{f}_{x}\otimes\mathcal{H}^{f}_{y}$ and $g^{b}(x,y,\nu)$ is a
normalized Hermitian operator acting on
$\mathcal{H}^{b}_{x}\otimes\mathcal{H}^{b}_{y}$

Combining Eqs. (265b),(290) - (292b) gives

$\displaystyle|\theta_{\ell}(\nu)|$ $\displaystyle\leq$
$\displaystyle\sum_{xyi}|\theta^{i}_{\ell}(x,y,\nu)|$ (295a)
$\displaystyle[\theta^{i}_{\ell}(x,y,\nu)]^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}[\mu^{i}_{j\ell}(x,y,\nu)]^{2},$ (295b)

with the definitions

$\mu^{i}_{j\ell}(x,y,\nu)=-a^{i}(x,y,\nu)\sum_{k}\operatorname{Im}\\{\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|\\\
z^{i}(x,y,\nu)|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle\lambda_{k\ell}(\nu)\\}.$
(296)

Since the $|\phi_{j\ell}(\nu)\rangle$ are orthonormal, $g^{b}(x,y,\nu)$ is
Hermitian and the $\lambda_{k\ell}(\nu)$ are real, we have

$\operatorname{Im}\\{\langle\phi_{j\ell}(\nu)|\phi_{k\ell}(\nu)\rangle\\\
\langle\chi_{j\ell}(\nu)|g^{b}(x,y,\nu)|\chi_{k\ell}(\nu)\rangle\lambda_{k\ell}(\nu)\\}=0.$
(297)

Eq. (296) for $i=0$ can then be turned into

$\mu^{0}_{j\ell}(x,y,\nu)=\\\
a^{0}(x,y,\nu)\sum_{k}\operatorname{Im}\\{\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|[I-z^{0f}(x,y)]\\\
g^{b}(x,y,\nu)|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle\lambda_{k\ell}(\nu)\\}.$
(298)

But in addition

$|\omega(\nu)\rangle=\sum_{k}|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle\lambda_{k\ell}(\nu).$
(299)

Also $I-z^{0f}(x,y)$ is a projection operator so that

$[I-z^{0f}(x,y)]^{2}=I-z^{0f}(x,y).$ (300)

The normalization condition on $z^{0}(x,y,\nu)$ implies $[g^{b}(x,y,\nu)]^{2}$
has trace 1 as an operator on $\mathcal{H}^{b}_{x}\otimes\mathcal{H}^{b}_{y}$
and therefore all eigenvalues bounded by 1. Eqs. (294a), (295b), (298), (299),
and (300) then give

$[\theta^{0}_{\ell}(x,y,\nu)]^{2}\leq\\\
[a^{0}(x,y,\nu)]^{2}\langle\omega(\nu)|[I-z^{0f}(x,y)]|\omega(\nu)\rangle.$
(301)

For $\mu^{1}_{j\ell}(x,y,\nu)$, since $z^{1}(x,y,\nu)$ is nonzero only on the
subspace with $N(\mathcal{H}_{x}\otimes\mathcal{H}_{y})$ nonzero, we have

$\mu^{1}_{j\ell}(x,y,\nu)=-a^{1}(x,y,\nu)\operatorname{Im}\\{\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|\\\
z^{1}(x,y,\nu)[I-z^{0f}(x,y)]|\omega(\nu)\rangle\\}.$ (302)

Eqs. (295b) and (302) give

$[\theta^{1}_{\ell}(x,y,\nu)]^{2}\leq[a^{1}(x,y,\nu)]^{2}\langle\omega(\nu)|[I-z^{0f}(x,y)]\\\
[z^{1}(x,y,\nu)]^{2}[I-z^{0f}(x,y)]|\omega(\nu)\rangle.$ (303)

But by Eq. (293), $[z^{1}(x,y,\nu)]^{2}$ as an operator on
$\mathcal{H}_{x}\otimes\mathcal{H}_{y}$, has trace 1 and therefore all
eigenvalues bounded by 1. Thus Eq. (303) implies

$[\theta^{1}_{\ell}(x,y,\nu)]^{2}\leq\\\
[a^{1}(x,y,\nu)]^{2}\langle\omega(\nu)|[I-z^{0f}(x,y)]|\omega(\nu)\rangle.$
(304)

By construction of $D^{e}$, each nearest neighbor pair $\\{x,y\\}$ with $x\in
D^{e}$ must have $y\in D^{o}$. Also any $x\in D^{e}$ is contained in at most a
single $E_{\ell}$. As a result Eqs. (295a), (301) and (304) imply

$\sum_{\ell}|\theta_{\ell}(\nu)|\leq\sum_{x\in D^{e},y\in
D^{o}}\\{[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]\times\\\
\sqrt{\langle\omega(\nu)|[I-z^{0f}(x,y)]|\omega(\nu)\rangle}\\}.$ (305)

The Cauchy-Schwartz inequality then gives

$[\sum_{\ell}|\theta_{\ell}(\nu)|]^{2}\leq\sum_{x\in D^{e},y\in
D^{o}}[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]^{2}\times\\\ \sum_{x\in D^{e},y\in
D^{o}}\langle\omega(\nu)|[I-z^{0f}(x,y)]|\omega(\nu)\rangle.$ (306)

The state $|\omega(\nu)\rangle$ can be expanded as a linear combination of
orthogonal states each with $n$ fermions each at a single position. A state
with fermions at $n$ positions will survive the projection $I-z^{0f}(x,y)$
only if at least one of the fermions is either at $x$ or $y$. Each $x\in
D^{e}$ can be the member of only a single such pair of nearest neighbor
$\\{x,y\\}$. A $y\in D^{o}$ can be in 6 $x,y$ pairs for an $x\in D^{e}$. Thus
a term with $n$ fermion positions in the expansion of $|\omega(\nu)\rangle$
will pass $I-z^{0f}(x,y)$ for at most $6n$ pairs of $x$ and $y$. Therefore

$\sum_{x\in D^{e},y\in
D^{o}}\langle\omega(\nu)|[I-z^{0f}(x,y)]|\omega(\nu)\rangle\leq 6n.$ (307)

By Eq. (16)

$\parallel k(\nu)\parallel^{2}\geq\sum_{\ell,x\in D^{e},y\in D^{o}}\parallel
g_{\ell}(x,y,\nu)\parallel^{2}$ (308)

In addition, $z^{0}(x,y,\nu)$ is orthogonal to $z^{1}(x,y,\nu)$. It follows
that

$\parallel k(\nu)\parallel^{2}\geq\sum_{x\in D^{e},y\in
D^{o}}[|a^{0}(x,y,\nu)|^{2}+|a^{1}(x,y,\nu)|^{2}].$ (309)

Assembling Eqs. (306), (307) and (309) gives

$\parallel k(\nu)\parallel^{2}\geq\frac{1}{2}\sum_{x\in D^{e},y\in
D^{o}}[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]^{2}\\\
\geq\frac{1}{12n}[\sum_{\ell}|\theta_{\ell}(\nu)|]^{2}$ (310)

Eq. (268) then implies

$\int_{0}^{1}\parallel k(\nu)\parallel\geq\frac{1}{\pi}\sqrt{\frac{mV}{192}},$
(311)

and therefore

$C(|\psi\rangle,|\omega\rangle)\geq\frac{1}{\pi}\sqrt{\frac{mV}{192}}.$ (312)

Since Eq. (312) holds for all product $|\omega\rangle$ we obtain

$C(|\psi\rangle)\geq\frac{1}{\pi}\sqrt{\frac{mV}{192}}.$ (313)

###  B.5  Additional Terms

The nearest neighbor $\\{x,y\\}$ which contribute to each
$\theta^{T}_{\ell}(\nu)$ in Eq. (286) are all distinct from the pairs which
contribute to $\theta_{\ell}(\nu)$ in Eq. (268). A repeat of the steps leading
to Eq. (310) yields

$\parallel
k(\nu)\parallel^{2}\geq\frac{1}{12n}[\sum_{\ell}|\theta_{\ell}(\nu)|+\sum_{\ell}|\theta^{T}_{\ell}(\nu)|]^{2}.$
(314)

Eq. (313) becomes

$C(|\psi\rangle)\geq\frac{1}{\pi}\sqrt{\frac{mV}{192}}+\frac{\pi
q}{\sqrt{192}}.$ (315)

##  Appendix C  Upper Bound on the Complexity of Entangled States

An upper bound on $C(|\psi\rangle)$ of the $n$-particle entangled state of Eq.
(36) is given by $C(|\psi\rangle,|\omega\rangle)$ for any $n$-particle product
state $|\omega\rangle$, for which in turn an upper bound is given by

$C(|\psi\rangle,|\omega\rangle)\leq\int_{0}^{1}dt\parallel k(\nu)\parallel,$
(316)

for any trajectory $k(\nu)\in K$ fulfilling Eqs. (252) and (253). Beginning
with an $|\omega\rangle$ consisting of $n$ particles each at one of a
corresponding set of $n$ single points, we construct a sufficient $k(\nu)$ in
three stages. First,$|\omega\rangle$ is split into a sum of $m$ orthogonal
product states, each again consisting of $n$ particles one at each of a
corresponding set of $n$ single points. Then the position of each of the
particles in the product states is moved to the center of the wave function of
one of the single particle states of Eq. (35). Finally, by approximately
$\ln(V)/\ln(8)$ iterations of a fan-out tree, the $mn$ wave functions
concentrated at points are spread over the $mn$ cubes $D_{ij}$.

###  C.1 Product State to Entangled State

Define the set of positions $x_{ij}$ to be

$\displaystyle(x_{ij})^{1}$ $\displaystyle=$ $\displaystyle i+(x_{00})^{1},$
(317a) $\displaystyle(x_{ij})^{2}$ $\displaystyle=$ $\displaystyle
j+(x_{00})^{2},$ (317b) $\displaystyle(x_{ij})^{3}$ $\displaystyle=$
$\displaystyle(x_{00})^{3},$ (317c)

for $0\leq i<m,0\leq j<n$ and arbitrary base point $x_{00}$. Let the set of
$n$-particle product states $|\omega_{i}\rangle$ be

$|\omega_{i}\rangle=\prod_{0\leq j<n}\Psi^{\dagger}(x_{ij},1)|\Omega\rangle.$
(318)

The entangle $n$-particle state $|\chi\rangle$

$|\chi\rangle=\sqrt{\frac{1}{m}}\sum_{i}|\omega_{i}\rangle$ (319)

we generate from a sequence of unitary transforms acting on
$|\omega\rangle=|\omega_{0}\rangle$.

Let $k_{0}$ acting on $\mathcal{H}_{x_{00}}\otimes\mathcal{H}_{x_{01}}$ have
matrix elements

$\langle\Omega|\Psi(x_{00},-1)\Psi(x_{01},-1)k_{0}\\\
\Psi^{\dagger}(x_{00},1)\Psi^{\dagger}(x_{01},1)|\Omega\rangle=-i,$ (320)
$\langle\Omega|\Psi(x_{00},1)\Psi(x_{01},1)k_{0}\\\
\Psi^{\dagger}(x_{00},-1)\Psi^{\dagger}(x_{01},-1)|\Omega\rangle=i,$ (321)

and extend $k_{0}$ to $\mathcal{H}$ by Eq. (14). We then have

$\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\sqrt{\frac{m-1}{m}}\prod_{0\leq
j<n}\Psi^{\dagger}(x_{0j},s_{1j})|\Omega\rangle,$ (322)

where

$\theta_{0}=\arcsin(\sqrt{\frac{m-1}{m}}),$ (323)

and the set of spin indices $s_{ij},0\leq i,j<n$ is

$\displaystyle s_{ij}$ $\displaystyle=$ $\displaystyle-1,j\leq i,$ (324a)
$\displaystyle s_{ij}$ $\displaystyle=$ $\displaystyle 1,j>i.$ (324b)

Now let $k_{1}$ acting on $\mathcal{H}_{x_{01}}\otimes\mathcal{H}_{x_{02}}$
have matrix elements

$\langle\Omega|\Psi(x_{01},-1)\Psi(x_{02},-1)k_{1}\\\
\Psi^{\dagger}(x_{01},-1)\Psi^{\dagger}(x_{02},1)|\Omega\rangle=-i,$ (325)
$\langle\Omega|\Psi(x_{01},-1)\Psi(x_{02},1)k_{1}\\\
\Psi^{\dagger}(x_{01},-1)\Psi^{\dagger}(x_{02},-1)|\Omega\rangle=i,$ (326)

and extend $k_{1}$ to $\mathcal{H}$ by Eq. (14). We then have

$\exp(i\theta_{1}k_{1})\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\sqrt{\frac{m-1}{m}}\prod_{0\leq
j<n}\Psi^{\dagger}(x_{0j},s_{2j})|\Omega\rangle,$ (327)

for $\theta_{1}$ given by $\frac{\pi}{2}$.

Continuing in analogy to Eqs. (320) - (327), for a sequence of operators
$k_{j}$, $0\leq j<n-1$, acting on
$\mathcal{H}_{x_{0j}}\otimes\mathcal{H}_{x_{0j+1}}$, and corresponding
$\theta_{j}$ we obtain

$\exp(i\theta_{n-2}k_{n-2})...\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\sqrt{\frac{m-1}{m}}\prod_{0\leq
j<n}\Psi^{\dagger}(x_{0j},-1)|\Omega\rangle,$ (328)

Let $k_{n-1}$ acting on $\mathcal{H}_{x_{00}}\otimes\mathcal{H}_{x_{10}}$ have
matrix elements

$\displaystyle\langle\Omega|\Psi(x_{10},1)k_{n-1}\Psi^{\dagger}(x_{00},-1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle-i,$ (329a)
$\displaystyle\langle\Omega|\Psi(x_{00},-1)k_{n-1}\Psi^{\dagger}(x_{10},1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle i,$ (329b)

extend $k_{n-1}$ to $\mathcal{H}$ by Eq. (14), and let $\theta_{n-1}$ be
$\frac{\pi}{2}$. Applying $\exp(i\theta_{n-1}k_{n-1})$ to Eq. (328), followed
by a similar sequence of $\exp(i\theta_{j}k_{j}),n\leq j<2n-1$ acting on
$\mathcal{H}_{x_{0(j-n+1)}}\otimes\mathcal{H}_{x_{1(j-n+1)}}$ gives

$\exp(i\theta_{2n-2}k_{2n-2})...\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\sqrt{\frac{m-1}{m}}|\omega_{1}\rangle$
(330)

Multiplying Eq. (330) by $\exp(i\theta_{j}k_{j}),2n-1\leq j<3n-2$ on
$\mathcal{H}_{x_{1(j-2n+1)}}\otimes\mathcal{H}_{x_{1(j-2n+2)}}$, and then
$\exp(i\theta_{j}k_{j}),3n-2\leq j<4n-2$ on
$\mathcal{H}_{x_{1(j-3n+2)}}\otimes\mathcal{H}_{x_{2(j-3n+2)}}$ gives

$\exp(i\theta_{4n-3}k_{4n-3})...\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\sqrt{\frac{1}{m}}|\omega_{1}\rangle+\sqrt{\frac{m-2}{m}}|\omega_{2}\rangle.$
(331)

The end result of a sequence of $2mn-m$ such steps is $|\chi\rangle$ of Eq.
(319)

$\exp(i\theta_{2mn-m-1}k_{2mn-m-1})...\exp(i\theta_{0}k_{0})|\omega_{0}\rangle=\\\
\sqrt{\frac{1}{m}}\sum_{i}|\omega_{i}\rangle.$ (332)

The $k_{i}$ and $\theta_{i}$ of Eq. (332) have

$\displaystyle\parallel k_{i}\parallel$ $\displaystyle=$
$\displaystyle\sqrt{2},$ (333a) $\displaystyle|\theta_{i}|$
$\displaystyle\leq$ $\displaystyle\frac{\pi}{2}.$ (333b)

Thus Eq. (332) implies

$C(|\chi\rangle,|\omega\rangle)\leq\sqrt{2}\pi m(n-\frac{1}{2}).$ (334)

###  C.2 Entangled State Repositioned

Let $y_{ij}$ be the center of cube $D_{ij}$ of Eq. (35), $s_{ij}$ the spins of
Eq. (35) and $\zeta_{i}$ the phases of Eq. (36). Define the entangled
$n$-particle state $|\phi\rangle$ be

$|\phi\rangle=\sum_{i}\zeta_{i}\prod_{j}\Psi^{\dagger}(y_{ij},s_{ij})|\Omega\rangle.$
(335)

For each $0\leq i<m,0\leq j<n$, let $z^{0}_{ij},z^{1}_{ij}...z^{r_{ij}}_{ij}$
be the shortest sequence of nearest neighbor sites such that

$\displaystyle z^{0}_{ij}$ $\displaystyle=$ $\displaystyle x_{ij},$ (336a)
$\displaystyle z^{r_{ij}}_{ij}$ $\displaystyle=$ $\displaystyle y_{ij},$
(336b)

for the $x_{ij}$ in Eqs. (317a) - (319) and such that all $z^{\ell}_{ij}$ for
distinct $\ell,i,j,$ are themselves distinct. For each $0\leq\ell<r_{ij}-1$,
for nearest neighbor pair $z^{\ell}_{ij},z^{\ell+1}_{ij}$, let $k^{\ell}_{ij}$
acting on $\mathcal{H}_{z^{\ell}_{ij}}\otimes\mathcal{H}_{z^{\ell+1}_{ij}}$
have matrix elements

$\displaystyle\langle\Omega|\Psi(z^{\ell+1}_{ij},1)k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell}_{ij},1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle-i,$ (337a)
$\displaystyle\langle\Omega|\Psi(z^{\ell}_{ij},1)k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell+1}_{ij},1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle i,$ (337b)

and extend $k^{\ell}_{ij}$ to $\mathcal{H}$ by Eq. (14). For each $i,j$ pair
with $j<n-1$, for the final nearest neighbor step
$\exp(ik^{\ell}_{ij}),\ell=r_{ij}-1,$ Eqs. (337a) and (337b) are modified to
produce spin orientation $s_{ij}$ at $y_{ij}$

$\displaystyle\langle\Omega|\Psi(z^{\ell+1}_{ij},s_{ij})k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell}_{ij},1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle-i,$ (338a)
$\displaystyle\langle\Omega|\Psi(z^{\ell}_{ij},1)k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell+1}_{ij},s_{ij})|\Omega\rangle$
$\displaystyle=$ $\displaystyle i,$ (338b)

and for $j=n-1$ for the final $\exp(ik^{\ell}_{in-1}),\ell=r_{in-1}-1,$ Eqs.
(337a) and (337b) are modified in addition to generate the phase $\zeta_{i}$

$\displaystyle\langle\Omega|\Psi(z^{\ell+1}_{ij},s_{ij})k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell}_{ij},1)|\Omega\rangle$
$\displaystyle=$ $\displaystyle-i\zeta_{i},$ (339a)
$\displaystyle\langle\Omega|\Psi(z^{\ell}_{ij},1)k^{\ell}_{ij}\Psi^{\dagger}(z^{\ell+1}_{ij},s_{ij})|\Omega\rangle$
$\displaystyle=$ $\displaystyle i\zeta^{*}_{i}.$ (339b)

Define $r$ to be

$r=\max_{ij}r_{ij},$ (340)

and for each $i,j$ pair define

$k^{\ell}_{ij}=0,r_{ij}\leq\ell<r.$ (341)

Let $k^{\ell}$ be

$k^{\ell}=\sum_{ij}k^{\ell}_{ij}.$ (342)

Then we have

$\prod_{ij}[\exp(i\frac{\pi}{2}k^{s-1})...\exp(i\frac{\pi}{2}k^{0})]|\chi\rangle=|\phi\rangle,$
(343)

for $|\chi\rangle$ of Eq. (319).

The $k^{\ell}$ of Eqs. (342), (337a) - (339b) have

$\parallel k^{\ell}_{ij}\parallel\leq\sqrt{2mn}.$ (344)

Thus Eq. (343) implies

$C(|\phi\rangle,|\chi\rangle)\leq\frac{\pi\sqrt{mn}r}{\sqrt{2}}.$ (345)

We now minimize $r$ over the base point $x_{00}$

$\hat{r}=\min_{x_{00}}r,$ (346)

with the result

$C(|\phi\rangle,|\chi\rangle)\leq\frac{\pi\sqrt{mn}r}{\sqrt{2}},$ (347)

where we have dropped the hat on $r$.

###  C.3 Fan-Out

The state $|\phi\rangle$ with particles at the centers of the cubes $D_{ij}$
we now fan-out to the state $|\psi\rangle$ with particle wave functions spread
uniformly over the cubes $D_{ij}$. For sufficiently small lattice spacing $a$
nearly all of the complexity of the bound on $C(|\psi\rangle)$ is generated in
this step.

Let $d$ be the length of the edge of the $D_{ij}$. Each edge of $D_{ij}$ then
consists of $d+1$ sites. The volume $V$ is then $d^{3}$. We begin with case

$d=2^{p},$ (348)

for some integer $p$. For simpilicity we present the fan-out applied to a
prototype single particle state $|\upsilon_{0}\rangle$ on prototype cube $G$
with edge length $d$, and center at some point $y$

$|\upsilon_{0}\rangle=\Psi^{\dagger}(y,1)|\Omega\rangle.$ (349)

The first stage of the fan-out process consists of splitting
$|\upsilon_{0}\rangle$ into a pair of components displaced from each other in
lattice direction 1. For integer $-2^{p-2}\leq i\leq 2^{p-2}$ define $y(i)$ to
be $y$ incremented by $i$ nearest neighbor steps in lattice direction 1. For
$1\leq j\leq 2^{p-2}$ define $k_{j}$ on
$\mathcal{H}_{y(j-1)}\otimes\mathcal{H}_{y(j)}$ to have matrix elements

$\displaystyle\langle\Omega|\Psi[y(j),1]k_{j}\Psi^{\dagger}[y(j-1),1]|\Omega\rangle$
$\displaystyle=$ $\displaystyle-i,$ (350a)
$\displaystyle\langle\Omega|\Psi[y(j-1),1]k_{j}\Psi^{\dagger}[y(j),1]|\Omega\rangle$
$\displaystyle=$ $\displaystyle i.$ (350b)

For $-2^{p-2}\leq j\leq-1$ define $k_{j}$ by Eqs. (350a) and (350b) but with
$j+1$ in place of $j-1$. Then define $\bar{k}_{j}$ by

$\displaystyle\bar{k}_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(k_{1}+k_{-1}),$ (351a)
$\displaystyle\bar{k}_{j}$ $\displaystyle=$ $\displaystyle k_{j}+k_{-j},2\leq
j\leq 2^{p-2}.$ (351b)

With these definitions it then follows that

$|\upsilon_{1}\rangle=\exp(i\frac{\pi}{2}\bar{k}_{m})...\exp(i\frac{\pi}{2}\bar{k}_{1})|\upsilon_{0}\rangle,$
(352)

for $m=2^{p-2}$, is given by

$|\upsilon_{1}\rangle=\frac{1}{\sqrt{2}}\sum_{i=-2^{p-2},2^{p-2}}\Psi^{\dagger}[y(i),1]|\Omega\rangle.$
(353)

Eqs. (351a) and (351b) imply

$\displaystyle\parallel\bar{k}_{1}\parallel$ $\displaystyle=$
$\displaystyle\sqrt{2},$ (354a) $\displaystyle\parallel\bar{k}_{j}\parallel$
$\displaystyle=$ $\displaystyle 2,2\leq j\leq 2^{p-2}.$ (354b)

It then follows that

$C(|\upsilon_{1}\rangle,|\upsilon_{0}\rangle)<2^{p-2}\pi,$ (355)

where for simplicity we have used an overestimate for
$\parallel\bar{k}_{1}\parallel$.

The next stage of the fan-out consists of splitting each of the 2 components
of $|\upsilon_{1}\rangle$ but now in lattice direction 2. For
$\bar{k}_{j},2^{p-2}<j\leq 2^{p-1}$, defined by adapting of Eqs. (350a) -
(351b), we have

$|\upsilon_{2}\rangle=\exp(i\frac{\pi}{2}\bar{k}_{m})...\exp(i\frac{\pi}{2}\bar{k}_{1})|\upsilon_{0}\rangle,$
(356)

with $m=2^{p-1}$, given by

$|\upsilon_{2}\rangle=\\\
\frac{1}{2}\sum_{i=-2^{p-2},2^{p-1}}\sum_{j=-2^{p-2},2^{p-2}}\Psi^{\dagger}[y(i,j),1]|\Omega\rangle,$
(357)

for $y(i,j)$ defined to be $y(i)$ displaced $j$ steps in lattice direction 2.
Eqs. (351a) and (351b) adapted to the fan-out in direction 2 give
$\bar{k}_{j},2^{p-2}<j\leq 2^{p-1}$ each acting on twice as many sites as was
the case for the direction 1 fan-out and therefore

$\displaystyle\parallel\bar{k}_{2^{p-2}+1}\parallel$ $\displaystyle=$
$\displaystyle 2,$ (358a) $\displaystyle\parallel\bar{k}_{j}\parallel$
$\displaystyle=$ $\displaystyle 2\sqrt{2},2^{p-2}+2\leq j\leq 2^{p-1}.$ (358b)

It then follows that

$C(|\upsilon_{2}\rangle,|\upsilon_{1}\rangle)<2^{p-2}\sqrt{2}\pi.$ (359)

Splitting yet again, now in lattice direction 3, yields

$|\upsilon_{3}\rangle=\\\
\exp(i\frac{\pi}{2}\bar{k}_{m})...\exp(i\frac{\pi}{2}\bar{k}_{1})|\upsilon_{0}\rangle,$
(360)

for $m=2^{p-1}+2^{p-2}$, given by

$|\upsilon_{3}\rangle=\frac{1}{\sqrt{8}}\sum_{i=-2^{p-2},2^{p-1}}\\\
\sum_{j=-2^{p-2},2^{p-2}}\sum_{\ell=-2^{p-2},2^{p-2}}\Psi^{\dagger}[y(i,j,\ell),1]|\Omega\rangle,$
(361)

for $y(i,j,\ell)$ defined to be $y(i,j)$ displaced $\ell$ steps in lattice
direction 3.

Eqs. (351a) and (351b) adapted to the fan-out in direction 3 give
$\bar{k}_{j},2^{p-1}<j\leq 2^{p-1}+2^{p-2}$, each acting on twice as many
sites as was the case for the direction 2 fan-out and therefore

$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\parallel\bar{k}_{2^{p-1}+1}\parallel$
$\displaystyle=$ $\displaystyle 2\sqrt{2},$ (362a)
$\displaystyle\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!\parallel\bar{k}_{j}\parallel$
$\displaystyle=$ $\displaystyle 4,2^{p-1}+2\leq j\leq 2^{p-1}+2^{p-2}.$ (362b)

It then follows that

$C(|\upsilon_{3}\rangle,|\upsilon_{2}\rangle)<2^{p-1}\pi.$ (363)

The weight originally concentrated in $|\upsilon_{0}\rangle$ at the center
point $y$ of $G$, with edge length $d$, in $|\upsilon_{3}\rangle$ is
distributed equally over the center points of 8 sub-cubes of $G$ each with
edge length $\frac{d}{2}$. Combining Eqs. (355), (359) and (363) gives

$C(|\upsilon_{3}\rangle,|\upsilon_{0}\rangle)<(3+\sqrt{2})2^{p-2}\pi.$ (364)

The fan-out process of Eqs. (352) - (364) we now repeat a total of $p-1$
iterations arriving at a state $|\upsilon_{3p-3}\rangle$ with weight equally
distributed over the center points of $2^{3p-3}$ cubes each with edge length
$2$. Eqs. (364) rescaled for iteration $\ell$ give

$C(|\upsilon_{3\ell}\rangle,|\upsilon_{3\ell-3}\rangle)<(3+\sqrt{2})2^{p-\ell-1}2^{\frac{3\ell-3}{2}}\pi.$
(365)

The term $2^{p-\ell-1}$ counts the decreasing number of lattice steps between
cube centers as the fan-out process is iterated, while the term
$2^{\frac{3\ell-3}{2}}$ counts the growing number of cubes and therefore of
sites which each subsequent operator $\bar{k}(i)$ acts on simultaneously.

To complete the fan-out process, the weight at the center of each of the cubes
with edge length 2 needs to be distributed to the 26 points forming its
boundary. This process can be carried out in 3 additional steps thereby
defining $|\upsilon_{3p-2}\rangle,|\upsilon_{3p-1}\rangle$ and
$|\upsilon_{3p}\rangle$.

To obtain $|\upsilon_{3p-2}\rangle$ from $|\upsilon_{3p-3}\rangle$, the weight
at the center of each edge length 2 cube is distributed simultaneously and
equally to the points at the centers of the 6 edge length 2 squares forming
the cube’s boundary. This process itself is done simultaneously across all
$2^{3p-3}$ cubes. The result is

$C(|\upsilon_{3p-2}\rangle,|\upsilon_{3p-3}\rangle)\leq\frac{\pi}{2}2^{\frac{3\ell-3}{2}}.$
(366)

To obtain $|\upsilon_{3p-1}\rangle$ from $|\upsilon_{3p-2}\rangle$, the weight
at the center of each edge length 2 square is distributed simultaneously and
equally to the center point of the 4 length 2 lines forming the boundary of
that square. This process itself is done simultaneously across all faces of
all $2^{3p-3}$ cubes. The result is

$C(|\upsilon_{3p-1}\rangle,|\upsilon_{3p-2}\rangle)\leq\frac{\sqrt{3}\pi}{2\sqrt{2}}2^{\frac{3\ell-3}{2}}.$
(367)

To obtain $|\upsilon_{3p}\rangle$ from $|\upsilon_{3p-1}\rangle$, the weight
at the center of each length 2 line is distributed simultaneously and equally
to that line’s pair of end points. This process itself is done simultaneously
across all lines forming the boundaries of the faces of all $2^{3p-3}$ cubes.
The result is

$C(|\upsilon_{3p}\rangle,|\upsilon_{3p-1}\rangle)\leq\frac{\pi}{2}2^{\frac{3\ell-3}{2}}.$
(368)

The bound on $C(|\upsilon_{3p}\rangle,|\upsilon_{3p-3}\rangle)$ obtained by
summing Eqs. (366) - (368) turns out to be less than the bound in Eq. (365)
for $\ell=p$. We therefore sum Eq. (365) from $\ell$ of 1 to $p$ and obtain

$C(|\upsilon_{3p}\rangle,|\upsilon_{0}\rangle)<\frac{(3+\sqrt{2})(2+\sqrt{2})}{4\sqrt{2}}\pi
2^{\frac{3p}{2}}.$ (369)

Substituting $V$ for $2^{3p}$, we then have

$C(|\upsilon_{3p}\rangle,|\upsilon_{0}\rangle)<\frac{(3+\sqrt{2})(2+\sqrt{2})}{4\sqrt{2}}\pi\sqrt{V}.$
(370)

The bound of Eq. (369) is derived assuming Eq. (348) giving the edge $d$ of
cube $G$ as an even power of 2. Consider now the case

$2^{p-1}<d<2^{p}.$ (371)

Assume again that at each iteration $\ell$ of the fan-out process, each edge
length of each parent cube is split as evenly as possible into halves to
produce 8 child cubes with all edges nearly equal. Suppose $d$ is $2^{p}-1$.
After iteration $\ell$ has been completed, the total number of cubes will
still be $2^{3\ell}$. Orthogonal to each direction, the cubes can be grouped
into $2^{\ell}$ planes, each holding $2^{2\ell}$ cubes. But for each direction
one of these orthogonal planes will have an edge in that direction which is
one lattice unit shorter than the corresponding edge of the other $2^{\ell}$
planes. It follows that the update process in each direction can proceed with
$2^{p-\ell-1}-1$ steps occuring simultanously across all cubes, and one final
update skipped for the cubes with a single edge in that direction one lattice
unit shorter. The bound of Eq. (365) will hold without modification. For $d$
given by $2^{p}-2$, after iteration $\ell$, for each direction, there will be
two planes of $2^{2\ell}$ cubes each with the edge in that direction one
lattice unit shorter. The bound of Eq. (365) will continue to hold. Similarly
for $d$ given by $2^{p}-q$ for any $q<2^{p-1}$.

For $d$ of Eq. (371), when $\ell$ reaches $p-1$ the resulting cubes (no longer
exactly cubes) will have a mix of edges of length $2$ and of length $1$. The
argument leading to Eqs. (366) - (368) can be adapted to show they continue to
hold for the final pass with $\ell$ of $p$. The bound of Eq. (369) remains in
place for the net result of the entire fan-out process. By assumption,
according to Eq. (371) we have

$2d>2^{p}.$ (372)

Then since $V$ is $d^{3}$, Eq. (369) gives

$C(|\upsilon_{3p-1}\rangle,|\upsilon_{0}\rangle)<\frac{(3+\sqrt{2})(2+\sqrt{2})}{2}\pi\sqrt{V},$
(373)

which is weaker than Eq. (370) and therefore holds whether or not $d$ is an
even power of 2.

The bound of Eq. (370) applies to a fan-out process on a single prototype
state on cube $G$. Assume the process repeated in parallel on the $mn$ cubes
$D_{ij}$, thereby generating $|\psi\rangle$ of Eq. (36). For $|\phi\rangle$ of
Eq. (335) we then have

$C(|\psi\rangle,|\phi\rangle)\leq\frac{(3+\sqrt{2})(2+\sqrt{2})}{2}\pi\sqrt{mnV}.$
(374)

From Eqs. (334) and (347), it follows that for a product state
$|\omega\rangle$ we have

$C(|\psi\rangle,|\omega\rangle)\leq c_{1}\sqrt{mnV}+c_{2}mn+c_{3}\sqrt{mn}r,$
(375)

where

$\displaystyle c_{1}$ $\displaystyle=$
$\displaystyle\frac{(3+\sqrt{2})(2+\sqrt{2})}{2}\pi,$ (376a) $\displaystyle
c_{2}$ $\displaystyle=$ $\displaystyle\sqrt{2}\pi,$ (376b) $\displaystyle
c_{3}$ $\displaystyle=$ $\displaystyle\frac{\pi}{\sqrt{2}},$ (376c)

for $r$ of Eq. (347). Eq. (38) then follows.

##  Appendix D  Complexity Group

We now show that the topological closure of the group $G$ of all $U_{k}(1)$
realizable as solutions to Eqs. (17a) and (17b) has as a subgroup the direct
product

$\hat{G}=\times_{n}SU(d_{n}),$ (377)

where $SU(d_{n})$ acts on the subspace of $\mathcal{H}$ with eigenvalue $n$ of
the fermion number operator $N$, $d_{n}$ is the dimension of this subspace,
and the product is over the range $0\leq n\leq 16B^{3}$.

###  D.1  Lie Algebras

The $8B^{3}$ sites of the lattice $L$ we reorder as a 1-dimensional array of
distinct sites, successive pairs of which are nearest neighbors with respect
to the original lattice $L$. The new array of sites we label with an integer
valued index $z$ ranging from 0 to $8B^{3}-1$.

For any pair of nearest neighbor $\\{z,z^{\prime}\\}$, let
$\mathcal{F}_{zz^{\prime}}$ be the set of operators of the form

$f_{zz^{\prime}}=g_{zz^{\prime}}\bigotimes_{q\neq z,z^{\prime}}I_{q},$ (378)

where $I_{q}$ is the identity operator on $\mathcal{H}_{q}$ and
$g_{zz^{\prime}}$ is a traceless Hermitian operator acting on
$\mathcal{H}_{z}\otimes\mathcal{H}_{z^{\prime}}$ which commutes with
$N_{zz^{\prime}}$, the fermion number operator on
$\mathcal{H}_{z}\otimes\mathcal{H}_{z^{\prime}}$. Let $K_{p}$ be the vector
space over the reals of operators of the form

$k=\sum_{zz^{\prime}}f_{zz^{\prime}},$ (379)

for any collection of $f_{zz^{\prime}}\in\mathcal{F}_{zz^{\prime}}$ for
$z,z^{\prime}\leq p$.

Let $G_{p}$ be the group on $\mathcal{H}$ of all $U_{k}(1)$ realizable as
solutions to Eq. (17a) for $k(\nu)\in K_{p}$. The topological closure of the
group $G_{p}$ consists of all operators of the form $\exp(ih)$ for $h\in
L_{p}$, where $L_{p}$ is the Lie algebra generated by $K_{p}$ [21]. Said
differently, $L_{p}$ is the smallest set of operators such that
$K_{p}\subseteq L_{p}$ and, in addition, for any $h_{0},h_{1}\in L_{p}$, and
any real $r_{0},r_{1}$, there are $h_{2},h_{3}\in L_{p}$ given by

$\displaystyle h_{2}$ $\displaystyle=$ $\displaystyle r_{0}h_{0}+r_{1}h_{1},$
(380a) $\displaystyle h_{3}$ $\displaystyle=$ $\displaystyle i[h_{0},h_{1}].$
(380b)

The requirement that $L_{p}$ be closed under sums in Eq. (380a) follows from
the Trotter product formula applied to the large $t$ limit

$\exp(ir_{0}h_{0}+ir_{1}h_{1})=\\\
\lim_{t\rightarrow\infty}[\exp(it^{-1}r_{0}h_{0})\exp(it^{-1}r_{1}h_{1})]^{t}.$
(381)

The requirement that $L_{p}$ be closed under commutation in Eq. (380b) follows
from the Baker-Campbell-Hausdorff formula applied to the large $t$ limit

$\exp([h_{0},h_{1}])=\\\
\lim_{t\rightarrow\infty}[\exp(it^{-1/2}h_{0})\exp(-it^{-1/2}h_{1})\times\\\
\exp(-it^{-1/2}h_{0})\exp(it^{-1/2}h_{1})]^{t}.$ (382)

The requirement of taking a topological closure of the group generated by
$U_{k}(1)$ in order to generate $L_{p}$ is a consequence of the appearance of
limits in Eqs. (381) and (382).

###  D.2  Induction

For any integer $0<p\leq 8B^{3}$ \- 1, divide $\mathcal{H}$ into the product

$\displaystyle\mathcal{Q}_{p}$ $\displaystyle=$
$\displaystyle\bigotimes_{q\leq p}\mathcal{H}_{q},$ (383a)
$\displaystyle\mathcal{R}_{p}$ $\displaystyle=$
$\displaystyle\bigotimes_{q>p}\mathcal{H}_{q},$ (383b)
$\displaystyle\mathcal{H}$ $\displaystyle=$
$\displaystyle\mathcal{Q}_{p}\otimes\mathcal{R}_{p}.$ (383c)

By induction on $p$, we will show that the closure of $G_{p}$ includes the
subgroup $\hat{G}_{p}$

$\displaystyle\hat{G}_{p}$ $\displaystyle=$
$\displaystyle\times_{n}\hat{G}_{pn},$ (384a) $\displaystyle\hat{G}_{pn}$
$\displaystyle=$ $\displaystyle SU(d_{pn})\bigotimes_{z>p}I_{z},$ (384b)

where $SU(d_{pn})$ acts on the subspace $\mathcal{Q}_{pn}$ of
$\mathcal{Q}_{p}$ with eigenvalue $n$ of the total number operator $N$, and
$d_{pn}$ is the dimension of $\mathcal{Q}_{pn}$. The product in Eq. (384a) is
over $0\leq n\leq 2p+2$. Eqs. (384a) (384b) for the case $p=8B^{3}-1$ become
Eq. (377).

The set of $g_{zz^{\prime}}$ in Eq. (378) is a subset of the set of $f_{xy}$
in Eq. (14) of Section III. Thus $\hat{G}_{p}$ for $p=8B^{3}-1$ is a subgroup
of the group $G$ of Section III. Proof of Eq. (384a) therefore implies Eq.
(18) of Section III.

For $p=1$, Eqs. (384a) and Eqs. (384b) follow immediately from the definition
of $K_{p}$. Assuming Eqs. (384a) and Eqs. (384b) for some $p-1$, we will prove
them for $p$.

Let $S_{pn}$ be an orthonormal basis for $\mathcal{Q}_{pn}$ consisting of all
$n$-fermion, $m$-boson, $m\leq b_{max}(p+1)$, vectors of the form

$\displaystyle|\psi\rangle$ $\displaystyle=$ $\displaystyle\prod_{i\leq
n}\Psi^{\dagger}(z^{f}_{i},s_{i})\prod_{j\leq
m}\Phi^{\dagger}(z^{b}_{j})|\Omega\rangle$ (385a) $\displaystyle s_{i}$
$\displaystyle\in$ $\displaystyle\\{-1,1\\},$ (385b) $\displaystyle
z^{f}_{i},z^{b}_{j}$ $\displaystyle\leq$ $\displaystyle p,$ (385c)

for any list of $n$ distinct pairs of $(z^{f}_{i},s_{i})$ and any list of $m$
integers $z^{b}_{j}$ such that each $z^{b}_{j}$ coincides with at most
$b_{max}-1$ other $z^{b}_{j^{\prime}}$. For any pair of distinct
$|\psi_{0}\rangle,|\psi_{1}\rangle\in S_{pn}$, and 2 x 2 traceless Hermitian
$h$, define

$H(|\psi_{0}\rangle,|\psi_{1}\rangle,h)=\sum_{ij}|\psi_{i}\rangle\langle\psi_{j}|h_{ij},$
(386a) $H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)=\\\
H(|\psi_{0}\rangle,|\psi_{1}\rangle,h)\bigotimes_{z>p}I_{z}.$ (386b)

The set of all such $H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)$ is a linear
basis for the Lie algebra $L_{pn}$ of the group $\hat{G}_{pn}$ of Eq. (384b).

Thus to prove Eqs. (384b) and (384a) for $p$ it is sufficient to show that any
$H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)$ for some
$|\psi_{0}\rangle,|\psi_{1}\rangle\in S_{pn}$ and 2 x 2 traceless Hermitian
$h$, given the induction hypothesis, is contained in the Lie algebra generated
by $L_{p-1n^{\prime}}$ for some $n^{\prime}$ and $\mathcal{F}_{p-1p}$.

###  D.3  Without Bosons

We consider first $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ both with $m$ of 0
in Eqs. (385a) - (385c). We will work backwards starting from some
$H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)$ for
$|\psi_{0}\rangle,|\psi_{1}\rangle\in S_{pn}$. Since $|\psi_{0}\rangle$ and
$|\psi_{1}\rangle$ have the same value of total $N$ on the region $z\leq p$,
it follows that a $U_{0}$ can be found in $\hat{G}_{p-1}$ such that

$\displaystyle|\psi_{2}\rangle$ $\displaystyle=$ $\displaystyle
U_{0}|\psi_{0}\rangle,$ (387a) $\displaystyle|\psi_{3}\rangle$
$\displaystyle=$ $\displaystyle U_{0}|\psi_{1}\rangle$ (387b)

are orthogonal vectors in $S_{pn}$, their restrictions to the region $p-1\leq
z\leq p$ are also orthogonal but have equal total particle counts on $p-1\leq
z\leq p$. The particle count difference between $|\psi_{0}\rangle$ and
$|\psi_{1}\rangle$ at point $p$ is at most 2, and equal and opposite to the
difference between the corresponding totals on the region $z\leq p-1$. This
compensating difference can be moved by $U_{0}$ to the point $p-1$.

A $k$ in $\mathcal{F}_{p-1p}$ can then be found such that

$\displaystyle|\psi_{4}\rangle$ $\displaystyle=$
$\displaystyle\exp(ik)|\psi_{2}\rangle,$ (388a)
$\displaystyle|\psi_{5}\rangle$ $\displaystyle=$
$\displaystyle\exp(ik)|\psi_{3}\rangle,$ (388b)
$\displaystyle|\psi_{4}\rangle$ $\displaystyle=$
$\displaystyle|\psi_{6}\rangle\otimes|\upsilon\rangle$ (388c)
$\displaystyle|\psi_{5}\rangle$ $\displaystyle=$
$\displaystyle|\psi_{7}\rangle\otimes|\upsilon\rangle,$ (388d)

for some $|\upsilon\rangle\in\mathcal{H}_{p}$, with particle number
$n_{\upsilon}$ and $|\psi_{6}\rangle$ and $|\psi_{7}\rangle$ orthogonal
vectors in $S_{(p-1)m}$ with $m=n-n_{\upsilon}$.

It is then possible to find a $U_{2}$ in $\hat{G}_{p-1}$ such that

$\displaystyle|\psi_{8}\rangle$ $\displaystyle=$ $\displaystyle
U_{2}|\psi_{4}\rangle,$ (389a) $\displaystyle|\psi_{9}\rangle$
$\displaystyle=$ $\displaystyle U_{2}|\psi_{5}\rangle,$ (389b)
$\displaystyle|\psi_{8}\rangle$ $\displaystyle=$
$\displaystyle|\chi\rangle\otimes|\phi_{0}\rangle\otimes|\upsilon\rangle,$
(389c) $\displaystyle|\psi_{9}\rangle$ $\displaystyle=$
$\displaystyle|\chi\rangle\otimes|\phi_{1}\rangle\otimes|\upsilon\rangle,$
(389d) $\displaystyle|\phi_{0}\rangle$ $\displaystyle=$
$\displaystyle\Psi^{\dagger}(p-1,-1)|\Omega\rangle,$ (389e)
$\displaystyle|\phi_{1}\rangle$ $\displaystyle=$
$\displaystyle\Psi^{\dagger}(p-1,1)|\Omega\rangle,$ (389f)

for a some $|\chi\rangle$ in $S_{(p-2)(m-1)}$.

Combining Eqs. (387a) - (389f), the induction hypothesis implies the existence
of $U_{0},U_{2}\in\hat{G}_{p-1}$ and $k\in\mathcal{F}_{(p-1)p}$ such that

$U_{2}\exp(ik)U_{0}H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)U_{0}^{\dagger}\exp(-ik)U_{2}^{\dagger}=\\\
|\chi\rangle\langle\chi|\otimes\sum_{ij}|\phi_{i}\rangle\langle\phi_{j}|h_{ij}\otimes|\upsilon\rangle\langle\upsilon|.$
(390)

The expression on the right-hand side of Eq. (390) can then be obtained from a
commutator between an operator $k\in\mathcal{F}_{(p-1)p}$ and an operator
$g\in L_{(p-1)m}$ for $m=n-n_{\upsilon}$. For 2 x 2 traceless Hermitian
$k_{ij}$, define

$k=\sum_{ij}|\phi_{i}\rangle\langle\phi_{j}|k_{ij}\otimes|\upsilon\rangle\langle\upsilon|\bigotimes_{q\neq
p-1,p}I_{q},$ (391)

and for a 2 x 2 traceless Hermitian $g_{ij}$, define

$g=|\chi\rangle\langle\chi|\otimes\sum_{ij}|\phi_{i}\rangle\langle\phi_{j}|g_{ij}\bigotimes_{q>p-1}I_{q}.$
(392)

For any traceless, Hermitian 2 x 2 $h_{ij}$, there are $k_{ij}$ and $g_{ij}$
such that

$h=i[k,g].$ (393)

Combining Eqs. (390), (391), (392) and (393) then gives

$H_{p}(|\psi_{0}\rangle,|\psi_{1}\rangle,h)=\\\
U_{0}^{\dagger}\exp(-ik)U_{2}^{\dagger}i[k,g]U_{2}\exp(ik)U_{0},$ (394)

which completes the induction step and for $|\psi_{0}\rangle$ and
$|\psi_{1}\rangle$ with $m$ of 0 in Eqs. (385a) - (385c).

###  D.4  With Bosons

We consider next $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ both with nonzero
$n$ and $m$ in Eqs. (385a) - (385c).

Suppose $0<n<2p+2$.

If the boson factors in $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ are
identical, then by a combination of a rotation by a $U_{0}$ in $\hat{G}_{p-1}$
and by a $U_{1}$ in the group generated by $k\in\mathcal{F}_{(p-1)p}$ the
boson factors can both be turned into the case $m$ of 0, already covered in
Appendix D.3.

Suppose the boson factors in $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ are not
identical but the fermion factors in $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$
are identical. Then again, but a combination of a rotation by a $U_{0}$ in
$\hat{G}_{p-1}$ and by a $U_{1}$ in the group generated by
$k\in\mathcal{F}_{(p-1)p}$ the boson factors can both be turned into the case
$m$ of 0 but with orthogonal fermion factors in $|\psi_{0}\rangle$ and
$|\psi_{1}\rangle$. Thus back to the case covered in Appendix D.3.

Suppose both the fermion factors and the boson factors in $|\psi_{0}\rangle$
and $|\psi_{1}\rangle$ are not identical. The induction step of Appendix D.3
shows that the action of $\hat{G}_{p}$ is available at least on the fermion
factors in $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$. A $U_{0}$ in
$\hat{G}_{p}$ can therefore be found which makes the fermion factors in
$|\psi_{0}\rangle$ and $|\psi_{1}\rangle$ distinct both on the region $p-1\leq
z\leq p$ and on the region $0\leq z\leq p-1$. It follows that a $U_{1}$ in
$\hat{G}_{p-1}$ and a $U_{2}$ in the group generated by
$k\in\mathcal{F}_{(p-1)p}$ can then be found which take $|\psi_{0}\rangle$ and
$|\psi_{1}\rangle$ back to $m$ of 0.

Suppose finally either $n$ is 0 and $|\psi_{0}\rangle$ and $|\psi_{1}\rangle$
have only fermions or $n$ is $2p+2$ and all sites are filled with fermions. In
either case, $\hat{G}_{p-1}$ and $\hat{G}_{p}$ act purely on boson states. The
induction step to show that the Lie algebra of $\hat{G}_{p}$ is generated by
the Lie algebra of $L_{p-1n^{\prime}}$, either for $n^{\prime}$ of 0 or
$n^{\prime}$ of $2p$, and $\mathcal{F}_{p-1p}$ becomes nearly a direct
translation of the induction step in Appendix D.3 from fermion states to boson
states. We omit the details.

##  Appendix E  Auxiliary Field Algebra

We will construct a Hilbert space $\mathcal{H}^{B}$ generated by the algebra
$B$ of polynomials in the $\Sigma_{i}(x,s)$ and $\Upsilon_{i}(x)$ acting
purely as creation operators on $|\Omega^{B}\rangle$ and satisfying Eqs.
(153a) - (153d).

Let $B^{\Sigma}$ be the algebra generated by the set of all $\Sigma_{i}(x,s)$,
for any $x,s$ and $i$, and let $B^{\Upsilon}$ be the algebra generated by the
set of all $\Upsilon_{i}(x)$, for any $x$ and $i$. Since every $a\in
B^{\Sigma}$ commutes with every $b\in B^{\Upsilon}$, the algebra $B$ is the
tensor product

$B=B^{\Sigma}\otimes B^{\Upsilon}.$ (395)

For every $x$, let $B^{\Sigma}_{x}$ be the algebra generated by the set of
$\Sigma_{i}(x,s)$, for any $s$ and $i$, and let $B^{\Upsilon}_{x}$ be the
algebra generated by the set of $\Upsilon_{i}(x)$ for any $i$. Then for every
$x\neq y$, every $a_{x}\in B^{\Sigma}_{x}$ commutes or anticommutes with every
$a_{y}\in B^{\Sigma}_{y}$, and every $a_{x}\in B^{\Upsilon}_{x}$ commutes with
every $a_{y}\in B^{\Upsilon}_{y}$. Therefore the algebras $B^{\Sigma}$ and
$B^{\Upsilon}$ are the products

$\displaystyle B^{\Sigma}$ $\displaystyle=$
$\displaystyle\bigotimes_{x}B^{\Sigma}_{x},$ (396a) $\displaystyle
B^{\Upsilon}$ $\displaystyle=$ $\displaystyle\bigotimes_{x}B^{\Upsilon}_{x}.$
(396b)

Now let $\eta_{x}$ be a boost that takes the point $x$ to the point
$(\tau,0,0,0)$. For Eqs. (155a) and (155b) to be covariant, $\Sigma_{0}(x,s)$
has to transform under boosts like $\Psi(x,s)$ and $\Sigma_{1}(x,s)$ has to
transform under boosts like $\Psi^{\dagger}(x,s)$. Let $S^{x}_{ss^{\prime}}$
and $\bar{S}^{x}_{ss^{\prime}}$ be the spin transformation matrices
corresponding to $\eta_{x}$ and define $\hat{\Sigma}_{0}(x,s)$ and
$\hat{\Sigma}_{1}(x,s)$ to be

$\displaystyle\hat{\Sigma}_{0}(x,s)=\sum_{s^{\prime}}S^{x}_{ss^{\prime}}\Sigma_{0}(x,s^{\prime}),$
(397a)
$\displaystyle\hat{\Sigma}_{1}(x,s)=\sum_{s^{\prime}}\bar{S}^{x}_{ss^{\prime}}\Sigma_{1}(x,s^{\prime}).$
(397b)

For each $x$ and $s$, let $B^{\Sigma}_{xs}$ be the algebra generated by
$\hat{\Sigma}_{0}(x,s)$ and $\hat{\Sigma}_{1}(x,s)$. Then for $s\neq
s^{\prime}$, every $a_{xs}\in B^{\Sigma}_{xs}$ either commutes or anticommutes
with every $a_{xs^{\prime}}\in B^{\Sigma}_{xs^{\prime}}$. Therefore the
algebra $B^{\Sigma}_{x}$ is the product

$B^{\Sigma}_{x}=\bigotimes_{s}B^{\Sigma}_{xs}.$ (398)

Eq. (395) implies $\mathcal{H}^{B}$ is a tensor product

$\mathcal{H}^{B}=\mathcal{H}^{\Sigma}\otimes\mathcal{H}^{\Upsilon},$ (399)

of a space generated by $B^{\Sigma}$ acting on $|\Omega^{B}\rangle$ and a
space generated by $B^{\Upsilon}$ acting on $|\Omega^{B}\rangle$ and Eqs.
(396a) and (396b) imply $\mathcal{H}^{\Sigma}$ and $\mathcal{H}^{\Upsilon}$
are themselves products of spaces $\mathcal{H}^{\Sigma}_{x}$ and
$\mathcal{H}^{\Upsilon}_{y}$ generated, respectively, by $B^{\Sigma}_{x}$ and
$B^{\Upsilon}_{x}$ acting on $|\Omega^{B}\rangle$

$\displaystyle\mathcal{H}^{\Sigma}$ $\displaystyle=$
$\displaystyle\bigotimes_{x}\mathcal{H}^{\Sigma}_{x},$ (400a)
$\displaystyle\mathcal{H}^{\Upsilon}$ $\displaystyle=$
$\displaystyle\bigotimes_{x}\mathcal{H}^{\Upsilon}_{x}.$ (400b)

Similarly, Eq. (398) implies $\mathcal{H}^{\Sigma}_{x}$ is a product of
$\mathcal{H}^{\Sigma}_{xs}$ generated by $B^{\Sigma}_{xs}$ acting on
$|\Omega^{B}\rangle$

$\mathcal{H}^{\Sigma}_{x}=\bigotimes_{s}\mathcal{H}^{\Sigma}_{xs}.$ (401)

For the pair of operators $\hat{\Sigma}_{0}(x,s)$ and $\hat{\Sigma}_{1}(x,s)$
which generate $B^{\Sigma}_{xs}$, Eqs. (153a) and (153c) become

$\displaystyle[\hat{\Sigma}_{0}(x,s)]^{2}$ $\displaystyle=$ $\displaystyle 0,$
(402a) $\displaystyle[\hat{\Sigma}_{1}(x,s)]^{2}$ $\displaystyle=$
$\displaystyle 0,$ (402b)
$\displaystyle\\{\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{0}(x,s)\\}$
$\displaystyle=$ $\displaystyle\gamma^{0}_{ss}.$ (402c)

Eqs. (402a) - (402c) combined with approximate Lorentz and charge conjugation
invariance of the complexity of states in $\mathcal{H}$ imply that for the
field polynomials $P_{i}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]$

$\displaystyle P_{0}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]$
$\displaystyle=$ $\displaystyle 1,$ (403a) $\displaystyle
P_{1}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]$ $\displaystyle=$
$\displaystyle u\hat{\Sigma}_{0}(x,s),$ (403b) $\displaystyle
P_{2}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]$ $\displaystyle=$
$\displaystyle u\hat{\Sigma}_{1}(x,s),$ (403c) $\displaystyle
P_{3}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]$ $\displaystyle=$
$\displaystyle v[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)],$ (403d)

where $u$ and $v$ are normalization constants independant of $x$ and $s$, an
orthonormal basis for $\mathcal{H}^{\Sigma}_{xs}$ must have the form

$|x,s,i\rangle=P_{i}[\hat{\Sigma}_{0}(x,s),\hat{\Sigma}_{1}(x,s)]|\Omega^{B}\rangle,$
(404)

up to an overall unitary rotation of the basis. Eqs. (402a) - (402c) imply the
result of any other polynomial in $\hat{\Sigma}_{0}(x,s)$ and
$\hat{\Sigma}_{1}(x,s)$ acting on $|\Omega^{B}\rangle$ is equal to some
corresponding linear combination of the $|x,s,i\rangle$ of Eq. (404). The
complexity of a state in $\mathcal{H}^{B}$ is independent of overall
normalization, however, so $u$ can be arbitrarily set to 1. The remaining
constant $v$ determines the contribution to complexity arising from sites
occupied by more than a single fermion. In the continuum limit of complexity,
if a continuum limit exists, the weight of multiply occupied sites in any
state will go to 0. The continuum limit should therefore be independent of
$v$.

For the pair of operators $\Upsilon_{0}(x)$ and $\Upsilon_{1}(x)$ which
generate $B^{\Upsilon}_{x}$, Eq. (153d) becomes

$[\Upsilon_{0}(x),\Upsilon_{1}(x)]=i.$ (405)

Eq. (405) combined with approximate Lorentz and charge conjugation invariance
of the complexity of states in $\mathcal{H}$ imply that, up to an overall
unitary rotation of the basis, an orthonormal basis for
$\mathcal{H}^{\Upsilon}_{x}$ will consist of a family of states
$\\{|x,n_{0},n_{1}\rangle^{\Upsilon}\\}$ labeled by a pair of nonnegative
integers $n_{0},n_{1}$. For each $n_{0},n_{1}$ pair
$P_{n_{0}n_{1}}[\Upsilon_{0}(x),\Upsilon_{1}(x)]$ is an ordered product,
independent of $x$, of $n_{0}$ copies of $\Upsilon_{0}(x)$ and $n_{1}$ copies
of $\Upsilon_{1}(x)$ subject to the requirement

$P_{n_{0}n_{1}}[\Upsilon_{0}(x),\Upsilon_{1}(x)]=P_{n_{1}n_{0}}[\Upsilon_{1}(x),\Upsilon_{0}(x)].$
(406)

The $\\{|x,n_{0},n_{1}\rangle^{\Upsilon}\\}$ are given by

$|x,n_{0},n_{1}\rangle^{\Upsilon}=u_{n_{0}n_{1}}P_{n_{0}n_{1}}[\Upsilon_{0}(x),\Upsilon_{1}(x)]|\Omega\rangle^{B},$
(407)

where the $u_{n_{0}n_{1}}$ are normalization constants independant of $x$ and
symmetric in the indices $n_{0},n_{1}$. Eq. (405) implies the result of any
other polynomial in $\Upsilon_{0}(x)$ and $\Upsilon_{1}(x)$ acting on
$|\Omega^{B}\rangle$ is equal to some corresponding linear combination of the
$|x,n_{0},n_{1}\rangle$ of Eq. (407). To be consistent with the normalization
choice for fermions, $u_{00}$, $u_{01}$ and $u_{10}$ will be set to 1. The
remaining $u_{n_{0}n_{1}}$ determine the contribution to complexity arising
from sites occupied by more than a single boson and should have no effect on
the continuum limit of complexity, if a continuum limit exists.

Eq. (154) implies the $P_{n_{0}n_{1}}[\Upsilon_{0}(x),\Upsilon_{1}(x)]$
identically vanish for $n_{0}\geq n$ or $n_{1}\geq n$.

The end result of Eqs. (399) - (401) is an $\mathcal{H}^{B}$ generated by the
algebra $B$ acting on $|\Omega^{B}\rangle$ which is an ordered tensor product

$\mathcal{H}^{B}=\bigotimes_{x}\mathcal{H}^{B}_{x},$ (408)

on which, according to Eqs. (402a) - (407), the
$\Sigma_{i}(x,s),\Upsilon_{i}(x),$ satisfy Eqs. (153a) - (153d).

It is convenient to define at this point an orthonormal basis $P$ for $B$. In
particular, no linear combination of elements of $P$ is 0 as a result of the
anticommutation and commutation relations of Eqs. (153a) - (153d). Each $p\in
P$ consists of a product of a $p^{\Sigma}\in P^{\Sigma}$ and a
$p^{\Upsilon}\in P^{\Upsilon}$, where $P^{\Sigma}$ and $P^{\Upsilon}$ are
orthonormal bases for the fermion field algebra $B^{\Sigma}$ and the boson
field algebra $B^{\Upsilon}$, respectively. Each $p^{\Sigma}$ is defined to be
a product over all distinct $x$ and $s$ of one of the fermion field
combinations in Eqs. (403a) - (403d). Each $p^{\Upsilon}$ is defined to be a
product over all distinct $x$ of one of the normalized boson field
combinations $u_{n_{0}n_{1}}P_{n_{0}n_{1}}[\Upsilon_{0}(x),\Upsilon_{1}(x)]$.

##  Appendix F  Lower Bound on the Complexity of Entangled Relativistic States

The proof of Eq. (180) bounding from below the complexity of the entangled
relativistic state $|\psi^{B}\rangle$ of Eq. (178b) is a version of the proof
in Appendix B of a lower bound on the complexity of the entangled non-
relativistic state of Eq. (36), but with the regular lattice of Section III.1
replaced by the random lattice of Section XIII and with the inclusion in
$\mathcal{H}^{B}$ of anti-fermion states. The proof in Appendix B can be
adapted to the presence of anti-fermion states in $\mathcal{H}^{B}$ by
treating fermion-anti-fermion pairs in $\mathcal{H}^{B}$ following the
treatment of bosons in Appendix B. To do this we convert the complexity
calculation in $\mathcal{H}^{B}$ into an equivalent complexity calculation in
yet another auxiliary Hilbert space.

###  F.1  More Auxiliary Hilbert Spaces

For a trajectory $k^{B}(\nu)\in K^{B}$, let $U_{k^{B}}(\nu)$ be the solution
to

$\displaystyle\frac{dU_{k^{B}}(\nu)}{d\nu}$ $\displaystyle=$ $\displaystyle-
ik^{B}(\nu)U_{k^{B}}(\nu),$ (409a) $\displaystyle U_{k^{B}}(0)$
$\displaystyle=$ $\displaystyle I.$ (409b)

Define $|\omega(\nu)^{B}\rangle$ to be

$|\omega^{B}(\nu)\rangle=U_{k^{B}}(\nu)|\omega^{B}\rangle.$ (410)

for a product state $|\omega^{B}(0)\rangle\in\mathcal{H}^{B}$

$|\omega^{B}(0)\rangle=d_{f}(p_{j-1})...d_{f}(p_{0})d_{\bar{f}}(q_{k-1})...d_{\bar{f}}(q_{0})\times\\\
d_{b}(r_{\ell-1})...d_{b}(r_{0})|\Omega^{B}\rangle,$ (411)

with $j$ fermions, $k$ anti-fermions, and $\ell$ bosons, Assume that
$|\omega^{B}(0)\rangle$ and $k^{B}(\nu)$ have been chosen to give

$|\omega(1)^{B}\rangle=\xi|\psi^{B}\rangle,$ (412)

for a phase factor $\xi$. Fermion number conservation by $k^{B}(\nu)$ implies
$j-k$ must equal the fermion number $n$ of $|\psi^{B}\rangle$.

To deal with the presence of anti-fermions in $\mathcal{H}^{B}$, we will make
use of yet one more auxiliary Hilbert space, $\mathcal{H}^{C}$, which consists
purely of fermion states generated by all polynomials in an auxiliary field
$\Sigma^{C}_{1}(x,s)$ acting on an auxiliary vacuum $|\Omega^{C}\rangle$. The
tensor product $\mathcal{H}^{C}\otimes\mathcal{H}^{B}$ we name
$\mathcal{H}^{D}$.

There is a natural map $M$ from $\mathcal{H}^{D}$ to $\mathcal{H}^{B}$ defined
by

$M[P(\Sigma_{1}^{C})|\Omega^{\mathcal{C}}\rangle\otimes|\psi^{B}\rangle]=P(\Sigma_{1})|\psi^{B}\rangle,$
(413)

where $P(\Sigma_{1}^{C})$ is a polynomial in the field $\Sigma_{1}^{C}(x,s)$,
$P(\Sigma_{1})$ is the corresponding polynomial but in the field
$\Sigma_{1}(x,s)$ and $|\psi^{B}\rangle$ is any state in $\mathcal{H}^{B}$.
The map $M$ takes a subspace of $\mathcal{H}^{D}$ to the null vector in
$\mathcal{H}^{B}$ and thus does not have an inverse.

Corresponding to the decomposition of $\mathcal{H}^{D}$ and $\mathcal{H}^{B}$
as tensor products over all sites

$\displaystyle\mathcal{H}^{D}$ $\displaystyle=$
$\displaystyle\bigotimes_{x}\mathcal{H}^{D}_{x},$ (414a)
$\displaystyle\mathcal{H}^{B}$ $\displaystyle=$
$\displaystyle\bigotimes_{x}\mathcal{H}^{B}_{x},$ (414b)

the map $M$ is given by the product

$M=\prod_{x}M_{x},$ (415)

where each $M_{x}$ maps $\mathcal{H}^{D}_{x}$ to $\mathcal{H}^{B}_{x}$. The
maps $M_{x}$ and $M_{y}$ for distinct $x$ and $y$ commute.

Let $K^{D}$ be the Hilbert space of Hermitian operators of Section XV for
$\mathcal{H}^{D}$ in place of $\mathcal{H}^{B}$ and with the additional
requirement that $k^{D}\in K^{D}$ separately conserve both the fermion number
$N^{B}$ of $\mathcal{H}^{B}$ and the fermion number $N^{C}$ of
$\mathcal{H}^{C}$.

We now convert $k^{B}(\nu)\in K^{B},|\omega^{B}(\nu)\rangle\in\mathcal{H}^{B}$
connecting

$|\omega^{B}(1)\rangle=\xi|\psi^{B}\rangle,$ (416)

for a phase factor $\xi$, to the product state $|\omega^{B}(0)\rangle$ into
corresponding $k^{D}(\nu)\in K^{D},|\omega^{D}(\nu)\rangle\in\mathcal{H}^{D}$
connecting some $|\omega^{D}(1)\rangle$ to a product state
$|\omega^{D}(0)\rangle$ along a path such that for $0\leq\nu\leq 1$

$\displaystyle M|\omega^{D}(\nu)\rangle$ $\displaystyle=$
$\displaystyle|\omega^{B}(\nu)\rangle,$ (417a) $\displaystyle\parallel
k^{D}(\nu)\parallel$ $\displaystyle\leq$ $\displaystyle 9\parallel
k^{B}(\nu)\parallel.$ (417b)

In addition, while $|\omega^{B}(\nu)\rangle$ is an eigenvector of $N^{B}$ with
eigenvalue $n$, $|\omega^{D}(\nu)\rangle$ is an eigenvector of $N^{B}$ with
eigenvalue 0 and of $N^{C}$ with eigenvalue $n$. Eq. (417b) implies

$C^{D}[|\omega^{D}(1)\rangle,|\omega^{D}(0)\rangle]\leq\\\
9C^{B}[|\omega^{B}(1)\rangle,|\omega^{B}(0)\rangle].$ (418)

Thus a lower bound on $C^{D}[|\omega^{D}(1)\rangle,|\omega^{D}(0)\rangle]$
will give a lower bound on
$C^{B}[|\omega^{B}(1)\rangle,|\omega^{B}(0)\rangle]$.

Let the product state $|\omega^{D}(0)\rangle$ be
$|\omega^{C}\rangle\otimes|\omega^{B}\rangle$ where

$|\omega^{C}\rangle=d^{C}_{f}(p_{n+m-1})...d^{C}_{f}(p_{m})|\Omega^{C}\rangle$
(419a)
$|\omega^{B}\rangle=d_{f}(p_{m-1})...d_{f}(p_{0})d_{\bar{f}}(q_{m-1})...d_{\bar{f}}(q_{0})\times\\\
d_{b}(r_{\ell-1})...d_{b}(r_{0})|\Omega^{B}\rangle.$ (419b)

for $d_{f}(p_{i})$, $d_{\bar{f}}(q_{i})$ and $d_{b}(r_{i})$ from Eq. (411),
and $d_{f}^{C}(p_{i})$ constructed from $d_{f}(p_{i})$ of Eq. (411) by
substituting $\Sigma_{1}^{C}(x,s)$ for $\Sigma_{1}(x,s)$.

Eqs. (416) and (417a) imply the state $|\omega^{D}(1)\rangle$ will satisfy

$M|\omega^{D}(1)\rangle=\xi|\psi^{B}\rangle.$ (420)

In addition, since the trajectory $k^{D}(\nu)$ conserves $N^{B}$ and $N^{C}$
and $|\omega^{D}(0)\rangle$, by Eqs. (419a) and (419b), has $N^{B}$ of 0 and
$N^{C}$ of $n$, $|\omega^{D}(1)\rangle$ must have these same eigenvalues.
Also, since $M$ acts only on the $\Sigma_{1}^{C}(x,s)$ fermion content of
$|\omega^{D}(1)\rangle$ and $|\psi\rangle^{B}$, by Eqs. (178a) and (178b), has
no boson content and no $\Sigma_{0}^{B}(x,s)$ anti-fermion content,
$|\omega^{D}(1)\rangle$ must have no boson content, no $\Sigma_{0}^{B}(x,s)$
and $\Sigma_{1}^{B}(x,s)$ content and be given instead by

$|\omega^{D}(1)\rangle=\xi|\psi^{C}\rangle\otimes|\Omega^{B}\rangle,$ (421)

where $|\psi^{C}\rangle$ is

$\displaystyle q^{C}$ $\displaystyle=$ $\displaystyle
z^{-1}m^{-\frac{1}{2}}\sum_{0\leq i<m}\zeta_{i}p^{C}_{i},$ (422a)
$\displaystyle|\psi^{C}\rangle$ $\displaystyle=$ $\displaystyle
q^{C}|\Omega^{C}\rangle,$ (422b)

for $p^{C}_{i}$ given by

$p^{C}_{i}=V^{-\frac{n}{2}}\prod_{0\leq j<n}\left[\sum_{x\in
D_{ij},k}u^{k}(x)\Sigma^{C}_{1}(x,k)\right],$ (423)

for the same $\zeta_{i},u^{k}(x)$ and $D_{ij}$ in Eqs. (175) - (178b) for
$|\psi^{B}\rangle$.

For both the nonrelativistic version of complexity in Section III.3 and the
relativistic version in Section XVI, $C(|\psi\rangle)$ is actually independent
of the normalization of $|\psi\rangle$. We can therefore safely set $z$ to 1
in Eq. (422a). The result is that $|\omega^{D}(1)\rangle$ in Eq. (421) is
normalized to 1, which for consistency we now assume also for
$|\omega^{D}(0)\rangle$.

Now approximate Eqs. (17a), (17b), (252) and (253) for
$|\omega^{B}(\nu)\rangle$ by a series of discrete steps

$|\omega^{B}(\nu+\delta)\rangle=[1-i\delta
k^{B}(\nu)]|\omega^{B}(\nu)\rangle.$ (424)

We will prove Eqs. (417a) and (417b) by induction in $\nu$. Eqs. (413), (411),
(419a) and (419b) give Eq. (417a) and (417b) for $\nu=0$

Now assume $k^{D}(\nu)$ satisfying Eq. (417b) has been found for
$\nu<\nu^{\prime}$ such that $|\omega^{D}(\nu)\rangle$ given by

$|\omega^{D}(\nu+\delta)\rangle=[1-i\delta
k^{D}(\nu)]|\omega^{D}(\nu)\rangle,$ (425)

satisfies Eq. (417a) for $\nu\leq\nu^{\prime}$. We will show that a
$k^{D}(\nu^{\prime})$ exists satisfying Eq. (417b) and extending Eq. (417a) to
$\nu^{\prime}+\delta$.

According to Eq. (249), $k^{B}$ in Eq. (424) consists of a sum of operators of
the form

$\hat{f}^{B}_{xy}=f^{B}_{xy}\bigotimes_{q\neq x,y}I_{q},$ (426)

where $f^{B}_{xy}$ is a Hermitian operator on
$\mathcal{H}^{B}_{x}\otimes\mathcal{H}^{B}_{y}$ for a pair of nearest neighbor
sites $\\{x,y\\}$ which conserves $N^{B}$ and has vanishing partial traces for
both $\mathcal{H}^{B}_{x}$ and $\mathcal{H}^{B}_{y}$. We assume the dimension
$d_{\mathcal{H}}$ of $\mathcal{H}^{D}_{x}$, and the corresponding slightly
smaller dimension of $\mathcal{H}^{B}_{x}$, are large enough that the
contribution to $k^{B}$ of single site operators of the form given in Eq.
(248a) can be neglected.

Then the required $k^{D}(\nu^{\prime})$ can be found if for every allowed
$\hat{f}^{B}_{xy}$ there is a $\hat{f}^{D}_{xy}$ of the form

$\hat{f}^{D}_{xy}=f^{D}_{xy}\bigotimes_{q\neq x,y}I^{D}_{q},$ (427)

where $f^{D}_{xy}$ is a Hermitian operator on
$\mathcal{H}^{D}_{x}\otimes\mathcal{H}^{D}_{y}$ which has vanishing partial
traces for both $\mathcal{H}^{D}_{x}$ and $\mathcal{H}^{D}_{y}$, conserves
$N^{B}$ and $N^{C}$ and for which

$\displaystyle M\hat{f}^{D}_{xy}|\omega^{D}(\nu^{\prime})\rangle$
$\displaystyle=$
$\displaystyle\hat{f}^{B}_{xy}|\omega^{B}(\nu^{\prime})\rangle,$ (428a)
$\displaystyle\parallel f^{D}_{xy}\parallel$ $\displaystyle\leq$
$\displaystyle 9\parallel f^{B}_{xy}\parallel.$ (428b)

To find the required $f^{D}_{xy}$, decompose
$\mathcal{H}^{D}_{x}\otimes\mathcal{H}^{D}_{y}$ into a direct sum of subspaces

$\mathcal{H}^{D}_{x}\otimes\mathcal{H}^{D}_{y}=\oplus_{mn}\mathcal{H}^{D}_{mn},$
(429)

with eigenvalues $m$ and $n$ of $N^{B}$ and $N^{C}$, respectively. Similarly,
decompose $\mathcal{H}^{B}_{x}\otimes\mathcal{H}^{B}_{y}$ into a direct sum of
subspaces

$\mathcal{H}^{B}_{x}\otimes\mathcal{H}^{B}_{y}=\oplus_{m}\mathcal{H}^{B}_{m},$
(430)

with eigenvalue $m$ of $N^{B}$.

Let $P^{D}_{mn}$ be the projection operator onto $\mathcal{H}^{D}_{mn}$.
Define $M_{mn}$ to be

$M_{mn}=M_{x}M_{y}P^{D}_{mn},$ (431)

for $M_{x}$ and $M_{y}$ from Eq. (415). Then $M_{mn}$ maps
$\mathcal{H}^{D}_{mn}$ onto $\mathcal{H}^{B}_{m+n}$. Let
$\mathcal{H}^{D\perp}_{mn}$ be the orthogonal complement of the subspace of
$\mathcal{H}^{D}_{mn}$ which is mapped to 0 by $M_{mn}$. For each
$|\psi^{B}\rangle\in\mathcal{H}^{B}_{m+n}$ there is a unique
$|\psi^{D}\rangle\in\mathcal{H}^{D\perp}_{mn}$ such that

$M_{x}M_{y}|\psi^{D}\rangle=|\psi^{B}\rangle.$ (432)

For each such $|\psi^{B}\rangle$, define $M_{mn}^{-1}$

$M_{mn}^{-1}|\psi^{B}\rangle=|\psi^{D}\rangle,$ (433)

and for any $|\psi^{B}\rangle\in\mathcal{H}^{B}_{\ell}$ with $\ell$ other than
$m+n$

$M_{mn}^{-1}|\psi^{B}\rangle=0.$ (434)

Eqs. (432) - (434) imply

$M_{x}M_{y}M_{mn}^{-1}=P^{B}_{m+n},$ (435)

where $P^{B}_{m+n}$ is the projection operator onto $\mathcal{H}^{B}_{m+n}$.
Define $g^{D}_{xy}$ to be

$g^{D}_{xy}=\sum_{mn}M_{mn}^{-1}f^{B}_{xy}M_{mn}.$ (436)

By Eq. (431), $g^{D}_{xy}$ maps each $\mathcal{H}^{D}_{mn}$ into itself and
therefore conserves both $N^{C}$ and $N^{B}$.

We then have

$M_{x}M_{y}g^{D}_{xy}=\sum_{mn}P^{B}_{m+n}f^{B}_{xy}M_{mn},\\\
=f^{B}_{xy}\sum_{mn}M_{mn},\\\ =f^{B}_{xy}M_{x}M_{y},$ (437)

where the first line follows from by Eq. (435) and the second follows because
$M_{mn}$ maps onto $\mathcal{H}^{B}_{m+n}$ and $f^{B}_{xy}$ conserves $N_{B}$.
Eqs. (415), (437) and the induction hypothesis, Eq. (417a) for $\nu^{\prime}$,
give

$M\hat{g}^{D}_{xy}|\omega^{D}(\nu^{\prime})\rangle=\hat{f}^{B}_{xy}M|\omega^{D}(\nu^{\prime})\rangle\\\
=\hat{f}^{B}_{xy}|\omega^{B}(\nu^{\prime})\rangle,$ (438)

which is Eq. (428a).

In addition, since $M_{mn}^{-1}$ maps into $\mathcal{H}^{D}_{mn}$, Eqs. (431)
and (435) imply

$M_{m^{\prime}n^{\prime}}M_{mn}^{-1}=\delta_{m^{\prime}m}\delta_{n^{\prime}n}P^{B}_{m+n}.$
(439)

We then have

$\mathrm{Tr}^{D}_{xy}(g^{D}_{xy})^{2}=\sum_{mn}\mathrm{Tr}^{B}_{xy}[P^{B}_{m+n}f^{B}_{xy}P^{B}_{m+n}f^{B}_{xy}]\\\
=\sum_{mn}\mathrm{Tr}^{B}_{xy}[P^{B}_{m+n}(f^{B}_{xy})^{2}]$ (440)

where the first line follows from Eqs. (436) and (439) and the second holds
because $f^{B}_{xy}$ conserves $N^{B}$. Since the index $s$ of
$\Sigma_{1}(x,s)$ is in the range $0\leq s<4$, the maximum possible value of
$N^{B}$ for $x$ and $y$ together is 8. As a result there are at most 9
different combinations of $m$ and $n$ giving any value of $m+n$. Eq. (440)
then implies

$\mathrm{Tr}^{D}_{xy}(g^{D}_{xy})^{2}\leq
9\mathrm{Tr}^{B}_{xy}(f^{B}_{xy})^{2},$ (441)

which is Eq. (428b).

Finally, $g^{D}_{xy}$ can be split into

$g^{D}_{xy}=f^{D}_{xy}+\frac{1}{\sqrt{d_{\mathcal{H}}}}I_{x}\otimes
f^{D}_{y}+\\\ \frac{1}{\sqrt{d_{\mathcal{H}}}}f^{D}_{x}\otimes
I_{y}+\frac{1}{d_{\mathcal{H}}}f^{D}I_{x}\otimes I_{y},$ (442)

where

$\displaystyle\mathrm{Tr}^{D}_{x}f^{D}_{xy}$ $\displaystyle=$ $\displaystyle
0,$ (443a) $\displaystyle\mathrm{Tr}^{D}_{y}f^{D}_{xy}$ $\displaystyle=$
$\displaystyle 0,$ (443b) $\displaystyle\mathrm{Tr}^{D}_{y}f^{D}_{y}$
$\displaystyle=$ $\displaystyle 0,$ (443c)
$\displaystyle\mathrm{Tr}^{D}_{x}f^{D}_{x}$ $\displaystyle=$ $\displaystyle
0,$ (443d)

and $d_{\mathcal{H}}$ is the dimension of each $\mathcal{H}^{D}_{x}$. Eqs.
(442) - (443d) imply

$\mathrm{Tr}^{D}_{xy}(g^{D}_{xy})^{2}=\mathrm{Tr}^{D}_{xy}(f^{D}_{xy})^{2}+\mathrm{Tr}^{D}_{y}(f^{D}_{y})^{2}+\\\
\mathrm{Tr}^{D}_{x}(f^{D}_{x})^{2}+(f^{D})^{2}.$ (444)

Eqs. (444) and (441) imply $f^{D}_{xy}$ satisfies Eq. (428b). For
$d_{\mathcal{H}}$ large enough, Eqs. (442) and (438) imply $\hat{f}^{D}_{xy}$
satisfies Eq. (428a).

Which completes the induction step of the proof of Eqs. (417a) and (417b). Eq.
( 418) follows. To obtain a lower bound on
$C^{B}[|\omega^{B}(1)\rangle,|\omega^{B}(0)\rangle]$ we now derive a lower
bound on $C^{D}[|\omega^{D}(1)\rangle,|\omega^{D}(0)\rangle]$.

###  F.2  Schmidt Spectra Again

From each region $D_{ij}$, extract a subset $\hat{D}_{ij}$, consisting of the
center points $x$ of all cells $c(x)$ reached by starting at $y_{ij}$ and
traveling along a geodesic in $L(\tau,\sigma)$ in the positive or negative
$x_{1}$ direction a number $\leq\frac{d}{2\rho}$ of discrete steps each of
proper length $2\rho$, then traveling along a geodesic in the positive or
negative $x_{2}$ direction a number $\leq\frac{d}{2\rho}$ of discrete steps
each of proper length $2\rho$, then traveling along a geodesic in the positive
or negative $x_{3}$ direction a number $\leq\frac{d}{2\rho}$ of discrete steps
each of proper length $2\rho$. Since each $c(x)$ is contained within a sphere
of radius $\rho$ around its center point, none of the points in $\hat{D}_{ij}$
will be nearest neighbors and for large $d$, the total number of points in
each $\hat{D}_{ij}$ will be nearly $\frac{d^{3}}{\rho^{3}}$. Since $V$ is
between $\frac{48d^{3}}{\pi\rho^{3}}$ and $\frac{6d^{3}}{\pi\rho^{3}}$, the
number of points in each $\hat{D}_{ij}$ is $zV$ for $z$ between
$\frac{\pi}{6}$ and $\frac{\pi}{48}$. We will assume $V$ is large enough that
we can ignore the statistical uncertainty in the number of points in each
$\hat{D}_{ij}$.

From this set of $\hat{D}_{ij}$ construct a set of subsets $E_{\ell}$ each
consisting of $2n$ distinct points chosen from $2n$ distinct $\hat{D}_{ij}$.
Since there are $mn$ sets $\hat{D}_{ij}$, there will be $\frac{zmV}{2}$ sets
$E_{\ell}$.

For each $E_{\ell}$ form the tensor product spaces

$\displaystyle\mathcal{Q}_{\ell}$ $\displaystyle=$
$\displaystyle\bigotimes_{x\in E_{\ell}}\mathcal{H}_{x}^{C},$ (445a)
$\displaystyle\mathcal{R}_{\ell}$ $\displaystyle=$
$\displaystyle[\bigotimes_{x\neq
E_{\ell}}\mathcal{H}_{x}^{C}]\otimes\mathcal{H}^{B}.$ (445b)

The space $\mathcal{Q}_{\ell}$ has dimension $16^{2n}$ and $\mathcal{H}^{D}$
becomes

$\mathcal{H}^{D}=\mathcal{Q}_{\ell}\otimes\mathcal{R}_{\ell}.$ (446)

A Schmidt decomposition of $|\omega^{D}(\nu)\rangle$ according to Eq. (446)
then becomes

$|\omega^{D}(\nu)\rangle=\sum_{j}\lambda_{j\ell}(\nu)|\phi_{j\ell}(\nu)\rangle|\chi_{j\ell}(\nu)\rangle,$
(447)

where

$\displaystyle|\phi_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{Q}_{\ell},$ (448a)
$\displaystyle|\chi_{j\ell}(\nu)\rangle$ $\displaystyle\in$
$\displaystyle\mathcal{R}_{\ell},$ (448b)

for $0\leq j<16^{2n}$ and real non-negative $\lambda_{j\ell}(\nu)$ which
fulfill the normalization condition

$\sum_{j}[\lambda_{j\ell}(\nu)]^{2}=1.$ (449)

Each $|\phi_{j\ell}(\nu)\rangle$ is a pure fermion state while the
$|\chi_{j\ell}(\nu)\rangle$ can include fermions, antifermions and bosons.

The fermion number operators $N^{C}[\mathcal{Q}_{\ell}]$ and
$N^{C}[\mathcal{R}_{\ell}]$ commute and $|\omega^{D}(\nu)\rangle$ is an
eigenvector of the sum with eigenvalue $n$. It follows that the decomposition
of Eq. (447) can be done with $|\phi_{j\ell}(\nu)\rangle$ and
$|\chi_{j\ell}(\nu)\rangle$ eigenvectors of $N^{C}[\mathcal{Q}_{\ell}]$ and
$N^{C}[\mathcal{R}_{\ell}]$, respectively, with eigenvalues summing to $n$.
Let $|\phi_{0\ell}\rangle$ be $|\Omega_{\ell}\rangle$, the vacuum state of
$\mathcal{Q}_{\ell}$, and let $|\phi_{i\ell}(\nu)\rangle,1\leq i\leq 8n$, span
the $8n$-dimensional subspace of $\mathcal{Q}_{\ell}$ with
$N^{C}[\mathcal{Q}_{\ell}]$ of 1. We assume the corresponding
$\lambda_{i\ell}(\nu),1\leq i\leq 8n$, are in nonincreasing order. Consider
Eqs. (421) - (423) for $|\omega^{D}(1)\rangle$. For any choice of $\ell$ there
will be a set of $2n$ nonzero orthogonal
$|\phi_{1\ell}(1)\rangle,...|\phi_{2n\ell}(1)\rangle$ with

$\lambda_{j\ell}(1)=\sqrt{\frac{1}{mV}}.$ (450)

On the other hand, for the product state $|\omega^{D}(0)\rangle$ in Eqs.
(419a) and (419b), the $|\phi_{j\ell}(\nu)\rangle$ come entirely from
$|\omega^{C}\rangle$, which is a product of $n$ independent single fermion
states. The space spanned by the projection of these into some
$\mathcal{Q}_{\ell}$ is at most $n$ dimensional, and as a result at most $n$
orthogonal $|\phi_{1\ell}(0)\rangle,...|\phi_{n\ell}(0)\rangle$ can occur.
Therefore at $\nu=0$, there will be at most $n$ nonzero
$\lambda_{1\ell}(0),...\lambda_{n\ell}(0)$. For $n<j\leq 8n$, we have

$\lambda_{j\ell}(0)=0.$ (451)

Since $\\{\lambda_{j\ell}(\nu)\\}$ is a unit vector, Eqs. (450) and (451)
imply that as $\nu$ goes from $0$ to $1$, $\\{\lambda_{j\ell}(\nu)\\}$ must
rotate through ¡a total angle of at least $\arcsin(\sqrt{\frac{n}{mV}})$.

For the small interval from $\nu$ to $\nu+\delta\nu$ let $\mu_{j\ell}(\nu)$
and $\theta_{\ell}(\nu)$ be

$\displaystyle\lambda_{j\ell}(\nu+\delta\nu)$ $\displaystyle=$
$\displaystyle\lambda_{j\ell}(\nu)+\delta\nu\mu_{j\ell}(\nu),$ (452a)
$\displaystyle\theta_{\ell}(\nu)^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}[\mu_{j\ell}(\nu)]^{2}.$ (452b)

We then have

$\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu\geq\arcsin(\sqrt{\frac{n}{mV}}).$ (453)

Summed over the $\frac{zmV}{2}$ values of $\ell$, Eq. (453) becomes

$\sum_{\ell}\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu\geq\frac{zmV}{2}\arcsin(\sqrt{\frac{n}{mV}}),$
(454)

and therefore

$\displaystyle\sum_{\ell}\int_{0}^{1}|\theta_{\ell}(\nu)|d\nu$
$\displaystyle\geq$ $\displaystyle\frac{r}{\pi}\sqrt{mnV},$ (455a)
$\displaystyle\geq$ $\displaystyle\frac{1}{48}\sqrt{mnV},$ (455b)

since $z$ is greater than $\frac{\pi}{48}$.

###  F.3  Rotation Matrix and Rotation Angle Bounds

The rest of the proof of the lower bound on relativistic complexity, Eq.
(180), is a copy of Sections B.3 and B.4 of the proof in Appendix B of the
non-relativistic lower bound, Eq. (37), but with the non-relativistic fermion
charge $N$ and Hilbert spaces $\mathcal{H}^{f}$ and $\mathcal{H}^{b}$
replaced, respectively, by $N^{C}$, $\mathcal{H}^{C}$ and $\mathcal{H}^{B}$.

As in Appendix B.3, the rotation of $\lambda_{j\ell}(\nu)$ during the interval
from $\nu$ to $\nu+\delta\nu$ will be determined by the sum
$g^{D}_{\ell}(\nu)$ of all contributions to $k^{D}(\nu)$ of Eq. (425) arising
from $f^{D}_{xy}$ for nearest neighbor pairs $\\{x,y\\}$ with one point, say
$x$, in $E_{\ell}$. By construction of the $E_{\ell}$, if $x$ is in
$E_{\ell}$, $y$ can not be in $E_{\ell}$ or any distinct $E_{\ell^{\prime}}$.
A repeat of the derivation of Eqs. (287) - (289) then leads to

$\mu_{j\ell}(\nu)=\sum_{k}r_{jk\ell}(\nu)\lambda_{k\ell}(\nu),$ (456)

for the rotation matrix $r_{jk\ell}(\nu)$

$r_{jk\ell}(\nu)=-\operatorname{Im}[\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|\\\
g^{D}_{\ell}(\nu)|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle],$ (457)

for $|\phi_{k\ell}(\nu)\rangle$ and $|\chi_{k\ell}(\nu)\rangle$ of Eq. (447)
and $\mu_{j\ell}(\nu)$ of Eq. (452a).

Since the $f^{D}_{xy}$ contributing to $g^{D}_{\ell}(\nu)$ conserve total
fermion number $N^{C}$, $g^{D}_{\ell}(\nu)$ can be expanded as

$\displaystyle g^{D}_{\ell}(\nu)$ $\displaystyle=$ $\displaystyle\sum_{x\in
E_{\ell},y\notin E_{\ell}}g^{D}_{\ell}(x,y,\nu),$ (458a) $\displaystyle
g^{D}_{\ell}(x,y,\nu)$ $\displaystyle=$
$\displaystyle\sum_{i=0,1}a^{i}(x,y,\nu)z^{i}(x,y,\nu)$ (458b)

where $z^{0}(x,y,\nu)$ acts only on states with
$N^{C}(\mathcal{H}^{D}_{x}\otimes\mathcal{H}^{D}_{y})$ of 0, $z^{1}(x,y,\nu)$
acts only on states with
$N^{C}(\mathcal{H}^{D}_{x}\otimes\mathcal{H}^{D}_{y})$ strictly greater than
0, and the $z^{i}(x,y,\nu)$ are normalized by

$\parallel z^{i}(x,y,\nu)\parallel=1.$ (459)

The operator $z^{0}(x,y,\nu)$ will be

$\displaystyle z^{0}(x,y,\nu)$ $\displaystyle=$ $\displaystyle
z^{0C}(x,y)\otimes g^{B}(x,y,\nu),$ (460a) $\displaystyle z^{0C}(x,y,\nu)$
$\displaystyle=$ $\displaystyle P^{C}(x,y)\bigotimes_{q\neq x,y}I_{q},$ (460b)

where $P^{C}(x,y)$ projects onto the vacuum state of
$\mathcal{H}^{C}_{x}\otimes\mathcal{H}^{C}_{y}$ and $g^{B}(x,y,\nu)$ is a
normalized Hermitian operator acting on
$\mathcal{H}^{B}_{x}\otimes\mathcal{H}^{B}_{y}$

Combining Eqs. (452b),(456) - (458b) gives

$\displaystyle|\theta_{\ell}(\nu)|$ $\displaystyle\leq$
$\displaystyle\sum_{x\in E,y\notin E,i}|\theta^{i}_{\ell}(x,y,\nu)|$ (461a)
$\displaystyle[\theta^{i}_{\ell}(x,y,\nu)]^{2}$ $\displaystyle=$
$\displaystyle\sum_{j}[\mu^{i}_{j\ell}(x,y,\nu)]^{2},$ (461b)

with the definitions

$\mu^{i}_{j\ell}(x,y,\nu)=-a^{i}(x,y,\nu)\sum_{k}\operatorname{Im}\\{\langle\phi_{j\ell}(\nu)|\langle\chi_{j\ell}(\nu)|\\\
z^{i}(x,y,\nu)|\phi_{k\ell}(\nu)\rangle|\chi_{k\ell}(\nu)\rangle\lambda_{k\ell}(\nu)\\}.$
(462)

A duplicate of the proof of Eqs. (297) - (303) then yields

$[\theta^{0}_{\ell}(x,y,\nu)]^{2}\leq\\\
[a^{0}(x,y,\nu)]^{2}\langle\omega(\nu)|[I-z^{0C}(x,y)]|\omega(\nu)\rangle,$
(463a) $[\theta^{1}_{\ell}(x,y,\nu)]^{2}\leq\\\
[a^{1}(x,y,\nu)]^{2}\langle\omega(\nu)|[I-z^{0C}(x,y)]|\omega(\nu)\rangle,$
(463b)

which combined with Eq. (461a) imply

$\sum_{\ell}|\theta_{\ell}(\nu)|\leq\sum_{x\in E,y\notin
E}\\{[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]\times\\\
\sqrt{\langle\omega(\nu)|[I-z^{0C}(x,y)]|\omega(\nu)\rangle}\\}.$ (464)

where

$E=\cup_{\ell}E_{\ell}$ (465)

The Cauchy-Schwartz inequality then gives

$[\sum_{\ell}|\theta_{\ell}(\nu)|]^{2}\leq\sum_{x\in E,y\notin
E}[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]^{2}\times\\\ \sum_{x\in E,y\notin
E}\langle\omega(\nu)|[I-z^{0C}(x,y)]|\omega(\nu)\rangle.$ (466)

A repeat of the argument leading to Eq. (307) implies

$\sum_{x\in E,y\notin
E}\langle\omega(\nu)|[I-z^{0C}(x,y)]|\omega(\nu)\rangle\leq Mn,$ (467)

where $M$ is the maximum number of nearest neighbors of any lattice point $x$.
An upper bound on $M$ can be found as follows. Recall any $c(x)$ is contained
in a sphere with center $x$ and radius $\rho$ and contains a sphere with
center $x$ and radius $\frac{\rho}{2}$. It follows that $M$ is less than or
equal to the number of disjoint spheres of radius $\frac{\rho}{2}$ that can
placed with centers on a sphere with center $x$ and radius $2\rho$. For each
of the $\frac{\rho}{2}$ spheres, a slice through its center orthogonal to the
line from its center to $x$ will be contained in a sphere with center $x$ and
radius $\frac{\sqrt{17}}{2}$. The area of each of these slices is
$\frac{\pi\rho^{2}}{4}$, the area of the radius $\frac{\sqrt{17}}{2}$ sphere
is $\frac{68\pi\rho^{2}}{4}$, and therefore

$M\leq 68.$ (468)

By Eq. (16)

$\parallel k^{D}(\nu)\parallel^{2}\geq\sum_{\ell,x\in E,y\notin E}\parallel
g^{D}_{\ell}(x,y,\nu)\parallel^{2}$ (469)

In addition, $z^{0}(x,y,\nu)$ is orthogonal to $z^{1}(x,y,\nu)$. It follows
that

$\parallel k^{D}(\nu)\parallel^{2}\geq\sum_{x\in E,y\notin
E}[|a^{0}(x,y,\nu)|^{2}+|a^{1}(x,y,\nu)|^{2}].$ (470)

Assembling Eqs. (466), (467), (468) and (470) gives

$\parallel k^{D}(\nu)\parallel^{2}\geq\frac{1}{2}\sum_{x\in E,y\notin
E}[|a^{0}(x,y,\nu)|+|a^{1}(x,y,\nu)|]^{2}\\\
\geq\frac{1}{136n}[\sum_{\ell}|\theta_{\ell}(\nu)|]^{2}$ (471)

Eq. (455b) then implies

$\int_{0}^{1}\parallel k(\nu)\parallel\geq\frac{1}{2348}\sqrt{mV},$ (472)

and therefore

$C^{D}(|\omega^{D}(1)\rangle,|\omega^{D}(0)\rangle)\geq\frac{1}{2348}\sqrt{mV},$
(473)

which by Eqs. (418) and (416) yields

$C^{B}(|\psi^{B}\rangle,|\omega^{B}(0)\rangle)\geq\frac{1}{21132}\sqrt{mV}.$
(474)

Since Eq. (474) holds for all product $|\omega^{B}(0)\rangle$ we finally
obtain

$C^{B}(|\psi^{B}\rangle)\geq\frac{1}{21132}\sqrt{mV}.$ (475)

##  Appendix G  Upper Bound on the Complexity of Entangled Relativistic States

The proof of Eq. (181) bounding from above the complexity of the entangled
relativistic state $|\psi^{B}\rangle$ of Eq. (178b) follows the proof in
Appendix C of an upper bound on the complexity of the entangled non-
relativistic state of Eq. (36), but with the regular lattice of Section III.1
replaced by the random lattice of Section XIII and $\mathcal{H}$ replaced by
$\mathcal{H}^{B}$.

An upper bound on $C^{B}(|\psi^{B}\rangle)$ is given by
$C^{B}(|\psi^{B}\rangle,|\omega^{B}\rangle)$ for any product state
$|\omega^{B}\rangle$, for which in turn an upper bound is given by

$C^{B}(|\psi^{B}\rangle,|\omega^{B}\rangle)\leq\int_{0}^{1}dt\parallel
k^{B}(\nu)\parallel,$ (476)

for any trajectory $k^{B}(\nu)\in K^{B}$ fulfilling

$\displaystyle\frac{d\omega^{B}(\nu)}{d\nu}$ $\displaystyle=$ $\displaystyle-
ik^{B}(\nu)\omega^{B}(\nu),$ (477a) $\displaystyle\omega^{B}(0)$
$\displaystyle=$ $\displaystyle|\omega^{B}\rangle,$ (477b)
$\displaystyle\omega^{B}(1)$ $\displaystyle=$
$\displaystyle\xi|\psi^{B}\rangle,$ (477c)

for a phase factor $\xi$.

As in Appendix C, construction of a sufficient $k^{B}(\nu)$ begins with an
$|\omega^{B}\rangle$ consisting of $n$ fermions each at one of a corresponding
set of $n$ single points. Then $|\omega^{B}\rangle$ is split into a sum of $m$
orthogonal product states, each again consisting of $n$ fermions one at each
of a corresponding set of $n$ single points. Then the position of each of the
fermions in the product states is moved to the center of of one of the
monomials of Eq. (177). Finally, by approximately $\ln(V)/\ln(8)$ iterations
of a fan-out tree, the $mn$ wave functions concentrated at points are spread
over the $mn$ cubes $D_{ij}$.

###  G.1  Cell Count Bound

The bound on $C^{B}(|\psi^{B}\rangle)$ relies on a bound we will derive first
on the number of distinct cells $c(x),x\in L(\tau,\sigma,\rho)$, which
intersect a geodesic $g(\lambda)\in
L(\tau,\sigma),0\leq\lambda\leq\lambda_{max},$ of length $\lambda_{max}$.

Let $\bar{g}$ be the set of all points within a proper distance $2\rho$ of any
point on $g(\lambda)$. Since every $c(x)$ is contained in a sphere with center
$x$ and radius $\rho$, it follows that $\bar{g}$ contains all $c(x)$ which
intersect $g(\lambda)$. On the other hand, each $c(x)$ within $\bar{g}$
contains a sphere with center $x$ and radius $\frac{\rho}{2}$ which is
disjoint from all other $c(x^{\prime})$ contained in $\bar{g}$. The total
volume occupied by the collection of disjoint radius $\frac{\rho}{2}$ spheres
has to be less than the total volume of $\bar{g}$. The number
$p(\lambda_{max})$ of $c(x)$ which intersect $g(\lambda)$ is therefore bounded
by

$p(\lambda_{max})\leq 24\frac{\lambda_{max}}{\rho}+64.$ (478)

###  G.2  Product State to Entangled State

For each value of $0\leq i<m$, let $x_{i0}$ be the center point of the cell
found by traveling from an abritrarily chosen starting point, $x_{00}$, along
a geodesic in the $x^{1}$ direction a proper distance of $4i\rho$. Then from
each $x_{i0}$ travel along a geodesic in the $x^{2}$ direction. For each
$0<j<n$, let $x_{ij}$ be the center point of the cell the geodesic beginning
at $x_{i0}$ enters after leaving the cell with center point $x_{ij-1}$. All
points on the geodesics beginning at $x_{i0}$ and at $x_{i^{\prime}0}$ for
$i\neq i^{\prime}$ will be at least a distance of $4\rho$ apart. As a result
each $x_{ij}$ will be both distinct from and not a nearest neighbor of each
$x_{i^{\prime}j^{\prime}}$ with $i\neq i^{\prime}$. The gap between $x_{ij}$
and $x_{i^{\prime}j^{\prime}}$ accomplishes the goal of making it possible,
despite the random placement of cells, to insure that $x^{B}_{ij}$ and
$x^{B}_{ij+1}$ are nearest neighbors as will turn out to be required.

Let the set of $n$-particle product states $|\omega^{B}_{i}\rangle$ be

$|\omega^{B}_{i}\rangle=\prod_{0\leq
j<n}[\sum_{k}u^{k}(x_{ij})\Sigma_{1}(x_{ij},k)]|\Omega^{B}\rangle.$ (479)

The entangle $n$-particle state $|\chi^{B}\rangle$

$|\chi^{B}\rangle=\sqrt{\frac{1}{m}}\sum_{i}|\omega^{B}_{i}\rangle$ (480)

we generate from a sequence of unitary transforms acting on
$|\omega^{B}\rangle=|\omega^{B}_{0}\rangle$.

The sequence of $k^{B}$ which convert the product state $|\omega^{B}\rangle$
into the entangled state $|\chi^{B}\rangle$ follows the sequence of $k$
mapping the product state $|\omega\rangle$ to the entangled state
$|\chi\rangle$ in Appendix C.1, with the non-relativistic fermion operator
$\Psi^{\dagger}(x,s)$ replaced by the relativistic $\hat{\Sigma}_{1}(x,s)$.

From $k^{B}_{0},...k^{B}_{n-2}$ in place of $k_{0},...k_{n-2}$ of Eqs. (320) -
(328) we obtain

$\exp(i\theta^{B}_{n-2}k^{B}_{n-2})...\exp(i\theta_{0}^{B}k^{B}_{0})|\omega^{B}_{0}\rangle=\sqrt{\frac{1}{m}}|\omega_{0}\rangle+\\\
\sqrt{\frac{m-1}{m}}\prod_{0\leq
j<n}[\sum_{k}v^{k}(x_{0j})\Sigma_{1}(x_{0j},k)]|\Omega^{B}\rangle,$ (481)

with

$\displaystyle\parallel k_{i}^{B}\parallel$ $\displaystyle=$
$\displaystyle\sqrt{2},$ (482a) $\displaystyle|\theta_{i}^{B}|$
$\displaystyle\leq$ $\displaystyle\frac{\pi}{2},$ (482b)

as in Eqs. (333a) and (333b) and therefore total cost

$\sum_{0\leq j\leq n-2}|\theta^{B}_{j}|\parallel
k^{B}_{j}\parallel\leq\frac{\pi(n-1)}{\sqrt{2}}.$ (483)

The spinor $v^{k}(x)$ in Eq. (481) , as defined in Section XVII, is orthogonal
to $u^{k}(x)$ of Eq. (479) and obtained, as is $u^{k}(x)$, by boosting from
the origin of $L(\tau,\sigma)$ to $x$ a spin state of a free fermion at rest
at the origin of $L(\tau,\sigma)$.

Then from $k^{B}_{n-1},...k^{B}_{n-1+p},3n-2\geq p<48n^{2}+159n$, in place of
$k_{n-1},...k_{2n-2}$ of Eqs. (329a) - (330) we obtain