# Negative ${\mathcal{R}}$-parity scissors modes in the Two-Rotors Model

Fabrizio Palumbo  INFN Laboratori Nazionali di Frascati, 00044 Frascati, Italy

###### Abstract

In a previous work I constructed ${\mathcal{R}}$-negative scissors states in
the Two-Rotors Model in which the rotors are two-sided classical bodies. Such
states have $|K|=1,0$ components and negative space parity. Here I extend this
work including reflections in a plane through the rotors symmetry axes
obtaining also ${\mathcal{R}}$-negative states of positive parity. I then
associate reflections and ${\mathcal{R}}$-operations with corresponding
operations on internal noncollective variables of the rotors. I evaluate the
em transition strengths of these states and compare them with the
corresponding strengths of ${\mathcal{R}}$-positive scissors modes.

###### pacs:

24.30.Cz,24.30.Gd,21.10.Re, 03.65.Ud

Introduction. The Two-Rotors Model (TRM) describes the dynamics of two rigid
bodies rotating with respect to each other under an attractive force around
their centers of mass. It was devised as a model for deformed atomic nuclei,
in which case the rigid bodies represent the proton and neutron systems [LoIu,
]. The collective $1^{+}$ state, the scissors mode, predicted by this model
was firstly observed [Bohle, ] in 154Gd and then in all deformed atomic nuclei
[LoIu1, ]. Recently also the prediction[LoIu, ] of the $J=2$ member of its
rotational band has been confirmed [Beck, ].

By analogy similar collective excitations were predicted in several other
systems [Guer, ] and clearly observed in Bose-Einstein condensates [Mara, ]
and very recently in dipolar quantum droplets of Dysprosium atoms [Ferr, ] and
in supersolids [Rocc, ; Tanz, ]. Moreover an application of the TRM to the
evaluation of the magnetic susceptibility of single domain magnetic
nanoparticles stuck in rigid matrices has given results compatible with a vast
body of experimental data with an agreement in some cases surprisingly good
[Hata, ]. Later it was found that, according to the Brink hypothesis [Brin, ],
scissors modes live also on excited nuclear levels [Krti, ; Adek, ].

There is by now an extensive literature on the subject [LoIu1, ]: in all these
papers the rotors were assumed to have axial symmetry and to be invariant
under inversion of their axes. Their wf should therefore be even or odd under
such an inversion, but only even states were considered. Inversion is defined
as a rotation-reflection [Hame, ]: Rotation by $\pi$ about an axis
perpendicular to the axis of the rotor assumed to have axial symmetry, called
${\mathcal{R}}$, and reflection, which I call $\sigma$, in the plane
perpendicular to the axis of this rotation. Inversion on a rotor coincides
with a parity operation; parity on the two-rotors system is the product of
inversions on the rotors.

Recently I investigated [Palu, ] the effect of relaxing the restriction to
positive ${\mathcal{R}}$ parity and I constructed ${\mathcal{R}}$-negative
states of the TRM. Because I was actually interested in nuclear states of
negative parity, I assumed the rotors to be even under the reflection $\sigma$
that was therefore altogether omitted in the notation. The negative parity
states so constructed turned out to be superpositions of $K=\pm 1,0$
components.

Evidence of $K$ mixing in the rare earths was reported already long ago [Zilg,
], but my interest in the subject was renewed by a very recent work [Beck1, ]
in which a $1^{-}$ state in 164Dy at the energy of the scissors mode has been
observed. In addition it was found that the states of a $1^{+}$ doublet, also
at the energy of the scissors mode, have $|K|=0,1$ components of comparable
amplitude. It was not reported, however, whether the $1^{-}$ state has or not
a $K=0$ component. So even though these findings may suggest a connection with
my results just outlined, they do not bear at present a strict connection with
them. Nevertheless they open perhaps a more realistic perspective for my work
on negative parity scissors modes, of which they suggest an extension. In my
work indeed negative parity states have vanishing electric dipole transitions,
because roto-oscillations of classical objects cannot generate an electric
dipole. But these classical objects are two-sided rotors with a quantal
${\mathcal{R}}$ number, which might require a role of noncollective internal
variables. The wave functions in collective models depend in fact on
collective as well as noncollective internal variables. The latter include the
particle position variables subject to constraints which compensate for the
redundancy. Classical formulations are the method of canonical transformation
[Vill, ] and the method of redundant variables [Sche, ]. In several cases it
is acceptable to keep the single particle variables without any constraint.
For instance the noncollective internal variables appear in transition matrix
elements and they were systematically set equal to 1 in all works on the TRM
and in particular in my paper [Palu, ] on negative parity states, but this may
be not appropriate in the presence of negative ${\mathcal{R}}$ symmetry. The
inversion of the axis of one rotor, in fact, can be obtained acting either on
the collective coordinates or performing a reflection on the internal
noncollective coordinates in the plane perpendicular to the symmetry axis of
the rotor through the center, and one must require that the two operations
give the same result. For negative ${\mathcal{R}}$ symmetry the rotor is a
two-sided object which may need a quantal description.

In the present work I evaluate the em transition strengths of scissors modes
with negative ${\mathcal{R}}$-symmetry and both negative and positive parity,
allowing the rotors to be odd under the reflection $\sigma$ defined above. I
thus find that the negative parity scissors mode I previously constructed can
decay to the ground state by an electric dipole transition. This is a purely
quantal effect, because roto-oscillations of classical rotors cannot produce
any electric moment. In addition I evaluate magnetic dipole strengths of
scissors modes of positive parity and negative ${\mathcal{R}}$ symmetry.

I will distinguish the rotors by the labels $1,2$, that will be later
identified with the proton and neutron fluids respectively.

The Two-Rotors Model. The classical Hamiltonian of the TRM is

$H=\frac{1}{2{\mathcal{J}}_{1}}{\vec{L}}_{1}^{2}+\frac{1}{2{\mathcal{J}}_{2}}{\vec{L}}_{2}^{2}+V$
(1)

where ${\vec{L}}_{1},{\vec{L}}_{2}$, ${\mathcal{J}}_{1},{\mathcal{J}}_{2}$ are
the angular momenta and moments of inertia of the rotors and $V$ the potential
interaction, which depends only on the angle $2\theta$ between the rotor axes.
If one denotes by ${\hat{\zeta}}_{1},{\hat{\zeta}}_{2}$ the unit vectors in
the direction of the rotors axes,
$\cos(2\theta)={\hat{\zeta}}_{1}\cdot{\hat{\zeta}}_{2}$. It is convenient to
replace them by the Euler angles $\alpha,\beta,\gamma$ that describe the
orientation of the system of the rotors as a whole plus the variable $\theta$.
Euler angles are associated with the direction cosines of the axes of the
intrinsic frame

$\displaystyle{\hat{\xi}}$ $\displaystyle=$
$\displaystyle\frac{{\hat{\zeta}}_{2}\times{\hat{\zeta}}_{1}}{2\sin\theta},\,{\hat{\eta}}=\frac{{\hat{\zeta}}_{2}-{\hat{\zeta}}_{1}}{2\sin\theta},\,{\hat{\zeta}}=\frac{{\hat{\zeta}}_{2}+{\hat{\zeta}}_{1}}{2\cos\theta}\,.$
(2)

One must then express all the operators in terms of the new variables. Setting

${\vec{I}}={\vec{L}}_{1}+{\vec{L}}_{2},\,\,\,{\vec{S}}={\vec{L}}_{1}-{\vec{L}}_{2}$
(3)

where ${\vec{I}}$ is the total orbital angular momentum acting on the Euler
angles, while ${\vec{S}}$ is not an angular momentum, one has the
representation [LoIu, ]

$S_{\xi}=i\frac{\partial}{\partial\theta},\,\,\,S_{\eta}=-\cot\theta
I_{\zeta},\,\,\,S_{\zeta}=-\tan\theta I_{\eta}\,.$ (4)

In terms of these variables inversion of the rotor is defined as a rotation-
reflection[Hame, ]: rotation through $\pi$ about the $\xi$-axis followed by a
reflection $\sigma$ in the $\eta$-, $\zeta$-plane

${\mathcal{I}}_{1}=\sigma_{1}\mathcal{R}_{1}\,,\,\,\,{\mathcal{I}}_{2}=\sigma_{2}\mathcal{R}_{2}$
(5)

where

$\mathcal{R}_{1}=R_{\zeta}(\pi)R_{\xi}(\frac{\pi}{2})R_{\theta}\,,\,\,\,\,\mathcal{R}_{2}=R_{\eta}(\pi)R_{\xi}(\frac{\pi}{2})R_{\theta}$
(6)

and

$R_{\theta}f(\theta)=f(\pi/2-\theta)=\stackrel{{\scriptstyle\circ}}{{f}}(\theta)\,.$
(7)

Physical states must be eigenstates of the inversion operators
${\mathcal{I}}_{1},{\mathcal{I}}_{2}$

${\mathcal{I}}_{1}\Psi=r_{1}\Psi\,,\,\,\,{\mathcal{I}}_{2}\Psi=r_{2}\Psi\,,\,\,\,r_{i}=\pm
1\,.$ (8)

For the whole system parity is the product of the inversions on the rotors

${\mathcal{P}}={\mathcal{I}}_{1}{\mathcal{I}}_{2}\,.$ (9)

The intrinsic Hamiltonian of the Two-Rotors Model. For the potential it was
assumed a harmonic approximation, with restoring force constant $C$. The range
of $\theta$ was divided into 2 regions:
$0<\theta<\frac{\pi}{4},\,\frac{\pi}{4}<\theta<\frac{\pi}{2}$ and 2 functions
were defined, $s_{I},s_{II}$, which are respectively equal to 1 inside the 2
regions and zero outside. Introducing the moment of inertia of the two-rotors
system as a whole

${\mathcal{J}}={\mathcal{J}}_{1}{\mathcal{J}}_{2}/({\mathcal{J}}_{1}+{\mathcal{J}}_{2}).$
(10)

and the parameters

$\theta_{0}^{2}=\frac{1}{\sqrt{{\mathcal{J}}C}},\,\,\,\,\omega=\sqrt{\frac{C}{{\mathcal{J}}}}$
(11)

the potential was written

$V\approx{1\over 2}\omega\Big{[}x^{2}s_{I}+y^{2}s_{II}\Big{]}$ (12)

where $x=\theta/\theta_{0},\,y=(\pi/2-\theta)/\theta_{0}$. The harmonic
approximation can be justified for $\theta_{0}<<1$. In the rare earth region
$\theta_{0}<10^{-1}$.

The quantized Hamiltonian, setting $\hbar=c=1$, can then be put in the
approximate form [DeFr, ]

$H=\frac{{\vec{I}}^{2}}{2{\mathcal{J}}}+{\mathcal{H}}_{0}+{\mathcal{H}}_{1}$
(13)

where

$\displaystyle{\mathcal{H}}_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\omega\,s_{I}\left[-\frac{\partial^{2}}{\partial
x^{2}}+\frac{1}{x^{2}}\Big{(}I_{\zeta}^{2}-\frac{1}{4}\Big{)}+x^{2}\right]$
(14) $\displaystyle+$
$\displaystyle\frac{1}{2}\omega\,s_{II}\left[-\frac{\partial^{2}}{\partial
y^{2}}+\frac{1}{y^{2}}\Big{(}I_{\eta}^{2}-\frac{1}{4}\Big{)}+y^{2}\right]$
$\displaystyle{\mathcal{H}}_{1}$ $\displaystyle=$ $\displaystyle
C_{0}\,s_{I}\left[-\frac{1}{2x}\Big{(}I_{\zeta}I_{\eta}+I_{\eta}(I_{\zeta}\Big{)}+iI_{\xi}\frac{\partial}{\partial
x}\right]$ (15) $\displaystyle+$ $\displaystyle
C_{0}\,s_{II}\left[-\frac{1}{2y}\Big{(}I_{\zeta}I_{\eta}+I_{\eta}(I_{\zeta}\Big{)}-iI_{\xi}\frac{\partial}{\partial
y}\right]$

and
$C_{0}=({\mathcal{J}}_{2}-{\mathcal{J}}_{1})/{{(\mathcal{J}}_{1}+{\mathcal{J}}_{2})}\,\omega\theta_{0}$.
The difference in sign in front of the derivatives in ${\mathcal{H}}_{1}$ is
due to the fact that
$\frac{\partial}{\partial\theta}=(\theta_{0})^{-1}(s_{I}\frac{\partial}{\partial
x}-s_{II}\frac{\partial}{\partial y})$. ${\mathcal{H}}_{1}$ is a small term
that until now has been neglected, but its effect should be considered for
${\mathcal{R}}$-negative states because they are opposite in sign depending on
which rotor is ${\mathcal{R}}$-negative. The total Hamiltonian is the sum of
the rotational Hamiltonian of the two-rotors system as a whole plus an
intrinsic Hamiltonian. The above expression clearly shows that the roto-
oscillations occur about both the $\zeta$\- and $\eta$-axes.

The lowest eigenfunctions of ${\mathcal{H}}_{0}$, normalized to $1/2$, are

$\displaystyle\varphi_{K}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{(K+1)\theta_{0}}}\,x^{K+\frac{1}{2}}\,e^{-{1\over
2}x^{2}}$ $\displaystyle\varphi_{K}(y)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{(K+1)\theta_{0}}}\,y^{K+\frac{1}{2}}\,e^{-{1\over
2}y^{2}}$ (16)

in region I and II with energy eigenvalues $\omega(1+|K|)$ respectively. I
will omit the functions $s_{I},s_{II}$ when I will explicitly write the
arguments of the functions. The wf of the ground and the $2^{+}$ states in
region I are

$\mathcal{F}^{0}_{00}\phi_{0}(x)\,,\,\,\,\,\mathcal{F}^{2}_{M0}\phi_{0}(x)$
(17)

where

$\mathcal{F}^{I}_{MK}=\sqrt{{2I+1}\over
16(1+\delta_{K0})\pi^{2}}\left({\mathcal{D}}^{I}_{MK}+(-)^{I}{\mathcal{D}}^{I}_{M-K}\right)\,.$
(18)

The corresponding wf in region II involve the eigenstates of $I_{\eta}^{2}$
with zero eigenvalue which are

$\mathcal{F}^{0}_{00}\,,\,\,\,\,\,\mathcal{G}^{2}_{M0}=\frac{1}{2}(\mathcal{F}^{2}_{M0}+\sqrt{3}\,\mathcal{F}^{2}_{M2})\,.$
(19)

Neglecting ${\mathcal{H}}_{1}$, the condition of positive $\mathcal{R}_{1}$
and $\mathcal{R}_{2}$ symmetry gives the entanglement between the 2 regions

$\displaystyle\Psi_{0^{+}}$ $\displaystyle=$
$\displaystyle\mathcal{F}^{0}_{00}(\phi_{0}(x)+\phi_{0}(y))$
$\displaystyle\Psi_{2^{+}M}$ $\displaystyle=$
$\displaystyle\mathcal{F}^{2}_{M0}\phi_{0}(x)-\mathcal{G}^{2}_{M0}\phi_{0}(y).$
(20)

The mixing of terms with different K-values is due to the breaking of axial
symmetry of the nucleus because of the relative motion of the rotors axes.

In the present work I study the em decay of ${\mathcal{R}}$-negative scissors
modes of positive and negative parity to the above states. I will compare with
the corresponding strengths of the ${\mathcal{R}}$-positive scissors modes
whose wf, neglecting ${\mathcal{H}}_{1}$ are [LoIu, ]

$\Psi_{1^{+}M}=\mathcal{F}^{1}_{M1}(\phi_{1}(x)-\phi_{1}(y))\,.$ (21)

${\mathcal{R}}$-negative scissors modes. The structure of
${\mathcal{R}}$-negative scissors modes, always neglecting
${\mathcal{H}}_{1}$, is [Palu, ]

$\Lambda^{\pi}_{MK}=F^{1}_{MK}(\alpha,\beta,\gamma)\chi_{K}(\theta)f^{\pi}$
(22)

where

$F^{1}_{M1}=\sqrt{{3}\over
16\pi^{2}}\left({\mathcal{D}}^{1}_{M1}+{\mathcal{D}}^{1}_{M-1}\right)\,,\,\,\,\,\,F^{1}_{M0}=\mathcal{F}^{1}_{M0}\,,$
(23) $\chi_{1}=s_{I}\,\varphi_{1}\,,\,\,\,\,\,\chi_{0}=s_{II}\varphi_{1}\,.$
(24)

The factor $f^{\pi}$ is a function of internal noncollective variables which I
introduce here according to the previous discussion. Notice the difference in
sign wr to the italic $\mathcal{F}$’s of Eq.(18). There are 2 degenerate
states distinguished by the $K$-quantum number: The states $\Lambda_{M1}$ and
$\Lambda_{M0}$ describe a precession about the $\eta,\zeta$-axis and live in
region $I,II$ respectively. They have the same energy as the
${\mathcal{R}}$-positive scissors modes, apart from the unknown energy
necessary to excite one rotor to its ${\mathcal{R}}$-negative and $\sigma=\pm
1$ state, which will be neglected.

The action of a separate ${\mathcal{R}}$-operation on the above wave functions
is

$\displaystyle{\mathcal{R}}_{1}\Lambda_{1M1}=i\,\Lambda_{1M0},\,\,\,\,\,{\mathcal{R}}_{1}\Lambda_{1M0}=-\,i\Lambda_{1M1}$
$\displaystyle{\mathcal{R}}_{2}\Lambda_{1M1}=-i\Lambda_{1M0},\,\,\,{\mathcal{R}}_{2}\Lambda_{1M0}=i\,\Lambda_{1M1}.$
(25)

One should notice that the operators ${\mathcal{R}}_{1},{\mathcal{R}}_{2}$ do
not merely invert the axes $\zeta_{1},\zeta_{2}$, but they also interchange
the $\Lambda$’s with each other. So the $\Lambda$’s are not eigenstates of
${\mathcal{R}}_{1},{\mathcal{R}}_{2}$ separately, but by repeated application
of the above equations one finds that their total ${\mathcal{R}}$ symmetry is
negative

${\mathcal{R}}\,\Lambda_{1M1}={\mathcal{R}}\,\Lambda_{1M0}=-1$ (26)

where ${\mathcal{R}}={\mathcal{R}}_{1}{\mathcal{R}}_{2}$. If each rotor must
be in a definite ${\mathcal{R}}$ state, however, so must be the total wf. One
can construct such a wf, normalized to 1, in terms of the $\Lambda$’s

$\Psi^{\pi}_{1Mr_{1}}=\left(\Lambda_{M1}+ir_{1}\Lambda_{M0}\right)f^{\pi}$
(27)

where

${\mathcal{R}}_{1}\Psi_{1M\,r_{1}}=r_{1}\,\Psi_{1M\,r_{1}}\,,\,\,\,{\mathcal{R}}_{2}\Psi_{1M\,r_{1}}=-r_{1}\,\Psi_{1M\,r_{1}}$
(28)

that show that in these states the rotors must have opposite ${\mathcal{R}}$
symmetry. In the above notation it is understood that all the scissors modes
in this paper are $\mathcal{R}$-negative, with the exception of Eq.(47).
Following the notation of Ref.[Beck1, ] I will sometimes set

$\Psi^{+}_{1r_{1}=-1}=1^{+}_{sc,1}\,,\,\,\,\,\Psi^{+}_{1r_{1}=+1}=1^{+}_{sc,2}\,.$
(29)

Instead of the quadruplets [Palu, ] with fully classical rotors there is a
doublet, a state in which rotor 1 is ${\mathcal{R}}$-even and rotor 2 is
${\mathcal{R}}$-odd and vice versa. For each such state there is a doublet
whose members have opposite reflection symmetry $\sigma$ in the $\zeta$,
$\eta$ plane and then opposite parity. (The actual existence of such states
obviously depends on the internal physics of the rotors, whether only one or
both or none of them can be excited to a $\mathcal{R}$ and/or
$\sigma$-negative state).

In the $1^{+}_{sc,1}$ it is rotor 1 which is ${\mathcal{R}}$-negative. For
consistency as said at the beginning $f^{\pi}$ must then satisfy the condition
$f^{\pi}(x_{1},y_{1},z_{1})=-f^{\pi}(x_{1},-y_{1},-z_{1})$, where
$x_{1},y_{1}z_{1}$ are the components of the position vector inside rotor 1.
The parity of the total wf is then equal to the inversion symmetry of rotor 1,
which is negative/positive if this rotor is even/odd under the reflection
$\sigma_{1}$. Therefore I set

$f^{\pi}(\vec{r}_{1})=s(z_{1})g^{\pi}(x_{1}).$ (30)

where $g^{\pi}(x_{1})=1$ for $\pi=-1$ and $g^{\pi}(x_{1})=s(x_{1})$ for
$\pi=1$. The wf of noncollective variables of the ground and the $2^{+}$
states of Eqs.(20) are set equal to 1. The modifications for $1^{+}_{sc,2}$
are obvious.

The magnetic moments of rotor i in the intrinsic frame are

$\displaystyle({\mathcal{M}}_{i}^{\pi})^{\prime}(M\lambda,\nu)=\int
d{\vec{r}}\rho_{M,i}(R_{\xi}^{-1}(\theta){\vec{r}})\,r^{\lambda}Y^{\lambda}_{\nu}({\hat{r}})f^{\pi}({\vec{r}}_{i})$
$\displaystyle\,\,\,\,\,\,\,\,\,\,\,\,=\int
d{\vec{r}}\rho_{M,i}({\vec{r}})\,r^{\lambda}Y^{\lambda}_{\nu}(R_{\xi}(\theta){\hat{r}})f^{\pi}({R_{\xi}(\theta)\vec{r}}_{i})$
(31)

where

$\rho_{M,i}=-eg_{i}\frac{1}{2}div(\rho\,r\vec{v})$ (32)

is the magnetic density, $g_{i}$ being the gyromagnetic ratio of rotor i,
$\rho$ its mass density and $\vec{v}$ its particle velocity. The electric
moments are obtained by changing the magnetic density with the electric
density. Only the ${\mathcal{R}}$-negative rotor contributes to em transition
matrix elements, because the integral over internal coordinates vanishes when
one evaluates the contribution of the ${\mathcal{R}}$-positive rotor.

Decay of the ${\mathcal{R}}-\mbox{negative},1^{-}_{sc}$ state. This state can
decay only through electric transitions, which requires that the
$\mathcal{R}$-negative rotor be charged. I therefore assume that rotor 1 has a
negative $\mathcal{R}$-symmetry and it is made of protons. Moreover it must
have negative parity, so that it must be $\sigma$-even. The changes to be made
if the neutrons have nonzero effective charges are obvious. To first order in
$\theta$, using $f^{-}(\vec{r}_{1})$ of Eq.(30) in the moments definition,
Eq.(31)

$({\mathcal{M}}_{p}^{-})^{\prime}(E1,\nu)\approx i\theta
Q_{p1}<10|I_{\xi}|1\nu>$ (33)

where $I_{\xi}$ is the $\xi$-component of the angular momentum in the
intrinsic frame and

$Q_{p1}=\frac{3}{8}\sqrt{\frac{3}{4\pi}}\,e_{p}N_{p}R\,.$ (34)

In the above equation $N_{p}$ the number of protons and $R$ the nuclear
radius. This result has been obtained by assuming that all the protons
contribute. It is known, however, that only protons on the surface do
contribute, which is accounted for in the Two-Rotors model by assuming that
because of superfluidity only the part of the rotor external to the maximum
sphere inscribed in the rotor participates in the motion [LoIu3, ]. In this
way one obtains

$Q_{p1}\rightarrow
Q_{p\delta}=\frac{1}{8}\sqrt{\frac{3}{4\pi}}\,e_{p}N_{p}R\,\delta$ (35)

where $\delta$ the deformation parameter. There is a reduction by a factor
$\frac{2}{3}\delta$ which means that the effective number of protons is
$\frac{2}{3}N_{p}\delta$. Because $<10|I_{\xi}|10>=0$, the contribution to the
electric decay comes entirely from the $|K|=1$ components of the wf so that

$<\Lambda_{1^{-},M,1}|\mathcal{M}(E1,\mu)|\Psi^{0}_{00}>=i\frac{1}{6}Q_{p\delta}\theta_{0}\delta_{M\mu}$
(36)

which yields the strength

$B(E1;1_{sc}^{-}\rightarrow 0^{+})=\frac{1}{12}(Q_{p\delta}\theta_{0})^{2}\,.$
(37)

In a similar way one can evaluate the strength of decay to the $2^{+}$ member
of the ground state rotational band

$B(E1;1_{sc}^{-}\rightarrow 2^{+})=\frac{1}{24}(Q_{p\delta}\theta_{0})^{2}$
(38)

which also comes entirely from the $|K|=1$ components of the scissors mode.

Decay of the ${\mathcal{R}-}\mbox{negative}1^{+}_{sc,i}$ states. There can be
two scissors states of positive parity, depending on which rotor is
${\mathcal{R}}$-negative. In order to have positive parity the
${\mathcal{R}}$-negative rotor must be negative also under $\sigma$ inversion.
Using $f^{+}(\vec{r}_{1})$ of Eq.(30) in the moments definition, Eq.(31), one
finds the components of the intrinsic magnetic dipole moment for
$\mathcal{R}$-negative transitions

$\displaystyle({{\mathcal{M}}_{i}^{+}})^{\prime}(M1,\pm)$ $\displaystyle=$
$\displaystyle\mp\frac{2}{\pi}{\mathcal{C}}_{i}S_{3}$
$\displaystyle({{\mathcal{M}}_{i}^{+}})^{\prime}(M1,0)$ $\displaystyle=$
$\displaystyle\frac{2\sqrt{2}}{\pi}{\mathcal{C}}_{i}S_{1}$ (39)

where

${\mathcal{C}}_{i}=\frac{1}{2}\sqrt{\frac{3}{8\pi}}\,\frac{eg_{i}}{2m}$ (40)

$g_{i}$ being the gyromagnetic ratio of rotor i. Notice that the components
$S_{1},S_{3}$ are cartesian, while the components
$({{\mathcal{M}}_{i}^{+}})^{\prime}(M1,\pm),({{\mathcal{M}}_{i}^{+}})^{\prime}(M1,0)$
are spherical. In their evaluation I used the approximations
$Y^{\lambda}_{\nu}(R_{\xi}(\theta){\hat{r}})\approx
Y^{\lambda}_{\nu}({\hat{r}}),f^{\pi}(R_{\xi}(\theta){\vec{r}})\approx
f^{\pi}({\vec{r}})$.

Let me first neglect the term ${\mathcal{H}}_{1}$ of the intrinsic
Hamiltonian. Consider the decay to the gs. The operator $S_{3}$ annihilates a
state of zero angular momentum. It is then easy to see that only the $K=0$
component of the intrinsic magnetic moment contributes with strength

$\displaystyle B(M1;1^{+}_{sc,i}\rightarrow
0^{+})=\frac{1}{16\,\pi^{3}}\frac{1}{\theta_{0}^{2}}\left(\frac{eg_{i}}{2m}\right)^{2}\,.$
(41)

In a similar way one fins that also the decay to the $2^{+}$ gets contribution
only from the $K=0$ component of the scissors mode with branching ratio

$\displaystyle\frac{B({\mathcal{R}}=-1,1^{+}_{sc,i}\rightarrow
2^{+})}{B({\mathcal{R}}=-1,1^{+}_{sc,i}\rightarrow
0^{+})}=\Big{(}\frac{1}{2}\frac{\sqrt{5}\,C^{10}_{2010}}{C^{10}_{0010}}\Big{)}^{2}=\frac{1}{2}\,.$
(42)

It differs from that of the Alaga [Alag, ] rule by the factor $1/2$ in the
second member of the equality, and this factor comes from the normalization of
${\mathcal{F}^{2}_{M0}}$ in Eq.(19), namely from $K$ mixing in the wf of the
$2^{+}$(see the Supplemental Material). In the same way, but after inclusion
of the term ${\mathcal{H}}_{1}$ one gets to order $\alpha$ the strength of the
transition to the $2^{+}$(in which there is also a $|K|=1$ contribution)

$B(M1;1^{+}_{sc,i}\rightarrow
2^{+})\approx\frac{1}{2}(1-r_{i}\alpha)B(M1;1^{+}_{sc,i}\rightarrow 0^{+})$
(43)

where

$\alpha=\frac{5}{2}\sqrt{\pi}\,\frac{{\mathcal{J}}_{2}-{\mathcal{J}}_{1}}{{\mathcal{J}}_{1}+{\mathcal{J}}_{2}}\,\theta_{0}\,.$
(44)

I evaluated (see the Supplemental material) also the electric quadrupole decay
strength

$B(E2;1^{+}_{sc,i}\rightarrow 2^{+})=\frac{1}{4}\theta_{0}^{2}Q_{i2}^{2}$ (45)

where $Q_{i2}=-\frac{1}{\pi}\sqrt{\frac{3}{10\pi}}e_{i}N_{i}R^{2}$, $e_{i}$
being the effective electric charges and $N_{i}$ the number of particles of
rotor i. It also comes entirely from the $K=0$ component. It can be compared
with the corresponding strength $B(E2;1^{+}_{sc,i},\mathcal{R}=1\rightarrow
2^{+})$ of the decay of the $\mathcal{R}$-positive scissors mode[LoIu, ].

Finally I report by comparison the magnetic decay strengths of the
$\mathcal{R}$-positive scissors mode. The intrinsic magnetic dipole moment
components in this case are

$\displaystyle\mathcal{M}^{\prime}(M1,\mathcal{R}=+1,\pm)$ $\displaystyle=$
$\displaystyle\mp({\mathcal{C}}_{1}-{\mathcal{C}}_{2})\,(S_{1}\pm S_{2})$
$\displaystyle\mathcal{M}^{\prime}(M1,\mathcal{R}=+1,0)$ $\displaystyle=$
$\displaystyle\sqrt{2}\,({\mathcal{C}}_{1}-{\mathcal{C}}_{2})\,S_{3}$ (46)

because both rotors contribute. The branching ratio of the transition
strengths to the $2^{+}$ and the ground state, neglecting ${\mathcal{H}}_{1}$
is (see the Supplemental Material) is

$\displaystyle B(M1;1^{+},\mathcal{R}=+1\rightarrow 2^{+})$ $\displaystyle=$
$\displaystyle\frac{1}{2}B(M1;1^{+},\mathcal{R}=+1\rightarrow 0^{+})$ (47)
$\displaystyle=$
$\displaystyle\frac{1}{16\,\pi}\frac{1}{\theta_{0}^{2}}\left(\frac{e(g_{1}-g_{2})}{2m}\right)^{2}.$

The first strength has been evaluated in the present work. A large amount of
experimental data is available [Wess, ,Pitz, ,Beck2, ], which are in
reasonable agreement with the above result. It comes about in a quite
nontrivial way. Both regions $I$ and $II$ defined before Eq.(10) give the same
contribution, due to the $K=\pm 1$ components of the intrinsic moment: The
terms coming from the operators $S_{2},S_{3}$ in the intrinsic moments, Eqs.
(46), are negligible ($O(\theta_{0}))$ or cancel out. Moreover both rotors
contribute, each one with with its gyromagnetic ratio, while in the decay of
the $\mathcal{R}$-negative scissors mode only the $\mathcal{R}$-negative rotor
contributes.

Conclusion. In a previous work [Palu, ] I determined negative $\mathcal{R}$
and space parity scissors states of the TRM. Such states were obtained by
assuming that each rotor might exist in a positive as well as negative
$\mathcal{R}$ parity state. In spite of their two-sided structure, the rotors
were assumed to be fully classical objects. In the present work I treat the
rotors as quantal objects as regard of their internal noncollective position
variables, allowing also negative parity wr to reflections in a plane through
their symmetry axes. In this way one can construct states of negative
$\mathcal{R}$ symmetry with both negative and positive space parity.

Now a few comments concerning the findings of Ref.[Beck, ]. Assuming
$\mathcal{R}$ invariance for each rotor

i) The $1_{sc}^{-}$ state I constructed might be related to the $1^{-}$ state
reported in Ref.[Beck, ] only if the latter has $|K|=0,1$ components of
comparable amplitude

ii) The $1_{sc,i}^{+}$ of the doublet I constructed (the proton or neutron
rotor being $\mathcal{R}$-negative) have $|K|=0,1$ components of equal
amplitude, in qualitative agreement with the doublet of $1^{+}$ states of
Ref.[Beck, ].These latter states have energies E= 3.159 and 3.173 kev and
corresponding branching ratios $B_{1\rightarrow 2}/B_{1\rightarrow
0}=(1+0.36)/2,(1-0.24)/2$: The corrections are of opposite sign in the 2
states in agreement with Eq.(43). This is not trivial: It is a consequence of
the structure of the $1^{+}_{sc,i}$ states, Eq.(27). In order to have a
quantitative agreement with the observed values of the B(M1) strengths,
however, the parameter $\alpha$ of Eq.(44) should be about 0.3, while with
current values of $\theta_{0}$, $\alpha<0.1$.

iii) If instead one requires $\mathcal{R}$ invariance only for the whole two
rotors system, any superposition of the states (22) is possible in the TRM,
because according to Eq.(26) each of them is odd under the total
${\mathcal{R}}$ operation.

As already noticed there are several $J=1$ states with energies in the range
of the scissors mode and evidence of $K$ mixing [Zilg, ; Beck1, ]. Perhaps to
fully understand the ${\mathcal{R}}$ nature of such states it might be
necessary to measure the em form factors, which should be sensitive to the
phase factor $f^{\pi}$ of Eq. (30).

## Acknowledgements

I thank T. Beck and N. Pietralla for a most fruitful correspondence and for
communicating some of their results before the publication.

## References

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# The probability flow in the Stock market and Spontaneous symmetry breaking
in Quantum Finance

Ivan Arraut
Lee Shau Kee School of Business and Administration
The Open University of Hong Kong,
30 Good Shepherd Street, Homantin, Kowloon
<EMAIL_ADDRESS>
Jo$\tilde{\bf a}$o Alexandre Lobo Marques
FBL, University of Saint Joseph
Estrada Marginal da Ilha Verde, 14-17, Macao, China
Sergio Gomes
FBL, University of Saint Joseph
Estrada Marginal da Ilha Verde, 14-17, Macao, China

<EMAIL_ADDRESS>

###### Abstract

The Spontaneous Symmetry breaking in Quantum Finance considers the martingale
condition in the stock market as a vacuum state if we express the financial
equations in the Hamiltonian form. The original analysis for this phenomena
ignores completely the kinetic terms in the neighborhood of the minimal of the
potential terms. This is correct in most of the cases. However, when we deal
with the Martingale condition, it comes out that the kinetic terms can also
behave as potential terms and then reproduce a shift on the effective location
of the vacuum (Martingale). In this paper we analyze the effective symmetry
breaking patterns and the connected vacuum degeneracy for these special
circumstances. Within the same scenario, we analyze the connection between the
flow of information and the multiplicity of martingale states, providing in
this way powerful tools for analyzing the dynamic of the stock market.

_K_ eywords Martingale condition $\cdot$ Vacuum condition $\cdot$ spontaneous
symmetry breaking $\cdot$ degenerate vacuum

##  1 Introduction

When we express the Financial equations of the Stock market in the Hamiltonian
form, the flow of information through the market can be quantified by
analyzing the flow of probability. The financial Hamiltonians are in general
non-Hermitian and then they do not preserve the information. However, under
some special combination of the free-parameters of the market, the financial
Hamiltonian can become Hermitian, preserving then the information
(probability). The flow of information in the market, is connected with the
definition of the martingale or equilibrium state, which is the equilibrium
state where there is no flow of information. In probability theory, a
martingale process is the one where the future expectation value of a random
variable is just the present value [1]. In Quantum Finance, the martingale
condition corresponds to a risk-neutral evolution, consistent with the most
basic concepts taken from probability theory [2]. A neutral evolution, is free
from any possibility of arbitrage [2, 3]. Arbitrage in finance gives the
possibility of investors to operate in different markets [3]. For example, in
general it is possible for a broker to buy sharings in New York and then sell
them in Hong Kong, getting in this way, some income from the differences in
the prices in both markets over the same product. This can be done in a market
outside the equilibrium condition (outside martingale). In addition, only big
corporations can get benefits with some income from arbitrage due to the high
volume in their inversions. Individuals cannot receive enough earnings by
using arbitrage due to the fee charges in transactions which would cancel any
possibility of income [3]. The arbitrage process helps the market to arrive to
the equilibrium state. Once the market is in this equilibrium condition, any
possibility or arbitrage is lost. It is at this point when we have a
martingale process with a risk-neutral evolution. In fact, the existence of a
martingale condition is known as the fundamental theorem of finance [3, 4]. In
a previous paper, some of the authors formulated the Spontaneous Symmetry
breaking in Quantum Finance [5]. Within this formalism, the Martingale
condition (state) appears as a vacuum condition which becomes degenerate under
some circumstances [3, 5]. In the most general sense, this vacuum condition is
non-unique for the Black-Scholes (BS) and the Merton-Garman (MG) cases when we
explore the symmetries under change of prices and the symmetry under changes
in volatility for the MG case. These symmetries come out to be spontaneously
broken [6] (their generators do not annihilate the vacuum state), except for
some combination of parameters, which guarantee the vacuum (martingale state)
to be unique ( non-degenerate). For the regime analyzed in [5], the same
mentioned combination of parameters guarantee the no-flow of information
through the boundaries of the system. This is an interesting connection
between spontaneous symmetry breaking and flow of information, already
perceived in a different context in [7]. In [5] it was also formulated an
extended martingale condition for the case of the MG equation, which considers
the martingale state not only as a function of the prices of the stock, but
also on the stochastic volatility. For this case again the vacuum, taken as
the martingale state, is degenerate and then the corresponding symmetries are
spontaneously broken. For this extended concept of Martingale, under the
regimes explored in [5], the conditions for the vacuum to be single come out
to be not only the necessity of no-flow of information through the boundaries
of the system, but also the absence of any white noise coming from the
stochastic volatility. These two conditions guarantee the martingale state
(vacuum) to be single, restoring then the symmetries for the vacuum state.
Another interesting situation was analyzed in [5]. It corresponds to an ideal
case where for the MG and the BS case, additional non-derivative terms are
included such that the martingale condition is still satisfied. These
potential terms are different to those analyzed before by some other authors
[3, 8]. The potential terms added in [5] are also different to the standard
case studied in [9], where a double slit constraint was explored. Indeed, the
potential terms analyzed in [5] correspond to collective decisions or
collective behavior. Although the results obtained in [5] are correct, they
were always focused on the analysis of situations ignoring the kinetic
contributions in the neighborhood of the Martingale state. This corresponds to
the strong-field regime to be defined in this paper. This regime, although
valid, does not represent the whole scenario. For this reason, in this paper
we also explore other regimes by extending the results obtained in [5]. We
then explore situations where the kinetic terms behave as additional potential
terms (weak-field regime and intermediate regimes). This certainly happens in
the reality when we consider the Martingale condition in the BS equation as
well as for the MG equation. Indeed, the kinetic terms cannot be ignored in
general around the neighborhood of the martingale state when we consider its
standard definition. Still we can say that there are regimes where the results
obtained in [5] are valid and regimes where the additional results obtained in
this paper represent a more accurate picture of the reality. Finally, in this
paper we fully connect the notions of spontaneous symmetry breaking and flow
of information (probability) in the stock market. The paper is organized as
follows: In Sec. (2), we describe the BS equation and we express it in its
Hamiltonian form. In Sec. (3), we explain the MG equation and its
corresponding Hamiltonian. In Sec, (4), we explain the meaning of a martingale
condition from the classical perspective in Finance. In Sec. (5), we explain
why the martingale condition, for the evolution of an Option, is equivalent to
a vacuum condition from the perspective of Quantum Mechanics. In Sec. (6), we
introduce some potential terms in the financial Hamiltonians and we analyze
under which conditions they preserve the martingale condition. In Sec. (7), we
extend the results obtained in [5] where it was shown that the symmetries
under a changes in the prices are spontaneously broken. This situation occurs,
except for some particular combinations of the parameters of the Hamiltonian.
The same situation appears for the symmetries under changes of volatility for
the MG equation if we extend the notion of martingale states in order to
include the volatility as one of the variables. In the cases studied in this
paper, we consider the kinetic term contributions which were ignored in [5].
This means that in this paper we explore both, weak and strong regimes for the
Quantum field representing the series expansion of the martingale state. The
martingale state then corresponds to the vacuum condition for this quantum
field denoted by $\phi_{vac}$. Its explicit result depends on the regime under
analysis as well as the order of the series expansion when we express the
Hamiltonian as a function of Quantum fields. In Sec. (8), we analyze the
details about the extended martingale condition which depends not only on the
prices of the Options but also on the stochastic volatility. This is the
section where the explicit results for the MG case are analyzed. In Sec….
Probability flow

Finally, in Sec. (9), we conclude.

##  2 The Black-Scholes equation

The stock price $S(t)$ is normally taken as a random stochastic variable
evolving in agreement to a stochastic differential equation given by

$\frac{dS(t)}{dt}=\phi S(t)+\sigma SR(t).$ (1)

Here $\phi$ is the expected return of the security, $R(t)$ is the Gaussian
white noise with zero mean and $\sigma$ is the volatility [3]. Note that this
simple equation contains one derivative term on the left-hand side and non-
derivative terms on the right-hand side. The fundamental analysis of Black and
Scholes, exclude the volatility such that we can guarantee the evolution of
the price of the stock with certainty [10]. In this way, by imposing
$\sigma=0$, we obtain a simple solution for the equation (1) as

$S(t)=e^{\phi t}S(0).$ (2)

The possibility of arbitrage is excluded if we can make a perfect hedged
portfolio. In this sense, any possibility of uncertainty is excluded and we
can analyze the evolution of the price free of any white noise [3]. We can
consider the following portfolio

$\Pi=\psi-\frac{\partial\psi}{\partial S}S.$ (3)

This is a portfolio where an investor holds the option and then short sells
the amount $\frac{\partial\psi}{\partial S}$ for the security $S$. By using
the Ito calculus (stochastic calculus) [3], it is possible to demonstrate that

$\frac{d\Pi}{dt}=\frac{\partial\psi}{\partial
t}+\frac{1}{2}\sigma^{2}S^{2}\frac{\partial^{2}\psi}{\partial S^{2}}.$ (4)

Here the change in the value of $\Pi$ does not have any uncertainty associated
to it [3]. The random term has disappeared due to the choice of portfolio.
Since here we have a risk-free rate of return for this case (no arbitrage)
[11, 12], then the following equation is satisfied

$\frac{d\Pi}{dt}=r\Pi.$ (5)

If we use the results (3) and (4), together with the previous equation, then
we get

$\frac{\partial\psi}{\partial t}+rS\frac{\partial\psi}{\partial
S}+\frac{1}{2}\sigma^{2}S^{2}\frac{\partial^{2}\psi}{\partial S^{2}}=r\psi.$
(6)

This is the Black-Scholes equation [13, 14, 15], which is independent of the
expectations of the investors, defined by the parameter $\phi$, which appears
in eq. (1). In other words, in the Black-Scholes equation, the security
(derivative) price is based on a risk-free process. The Basic Assumptions of
the Black-Scholes equation are:
1). The spot interest rate $r$ is constant.
2). In order to create the hedged portfolio $\Pi$, the stock is infinitely
divisible, and in addition it is possible to short sell the stock.
3). The portfolio satisfies the no-arbitrage condition.
4). The portfolio $\Pi$ can be re-balanced continuously.
5). There is no fee for transaction.
6). The stock price has a continuous evolution.

###  2.1 Black-Scholes Hamiltonian formulation

We will explain how the eq. (6), can be expressed as an eigenvalue problem
after a change of variable. The resulting equation will be the Schrödinger
equation with a non-Hermitian Hamiltonian. For eq. (6), consider the change of
variable $S=e^{x}$, where $-\infty<x<\infty$. In this way, the BS equation
becomes

$\frac{\partial\psi}{\partial t}=\hat{H}_{BS}\psi,$ (7)

where we have defined the operator

$\hat{H}_{BS}=-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}+\left(\frac{1}{2}\sigma^{2}-r\right)\frac{\partial}{\partial x}+r.$ (8)

as the BS Hamiltonian. Note that the resulting Hamiltonian is non-Hermitian
since $\hat{H}\neq\hat{H}^{+}$ [8]. In addition, note that since the spot
interest rate $r$ is constant, then the potential term is just a constant
term. This means that the vacuum condition is trivial for this case. Under the
BS Hamiltonian, the evolution in time of the Option is non-unitary in general
(in addition, the Hamiltonian non-necessarily obeys the $PT$ symmetry). This
means that the probability is not necessarily preserved in time, although it
is certainly well-defined and its total value is equal to one. In general,
there are some cases in ordinary Quantum Mechanics, as well as in Quantum
Field Theory, where it is interesting to explore non-Hermitian Hamiltonians
(Lagrangians) [16]. Based on the previous explanations, we cannot expect the
financial market to obey unitarity. The reason for this is simply because the
market is not a closed system and there are many external factors influencing
its behavior as for example it is the amount of people and organizations
trading at some instant of time. Then the assumption of unitarity makes no
sense at all and the Hamiltonian must be non-Hermitian. In this paper however,
when we add some potential terms to the BS and MG equations, we will impose
the Hermiticity condition on them. When the symmetry is spontaneously broken
and we are working at the vacuum level, all the terms in the original BS
equation become irrelevant, giving then importance to (only) the potential
terms. In this way we can follow the standard formalism suggested in [16]. The
same conclusions apply to any other equation only containing kinetic terms as
it is the case of the Merton-Garman (MG) equation to be analyzed shortly.
Inside the MG case however, the conclusions about what is happening during the
breaking of the symmetry of the market (spontaneously) are more relevant than
in the BS case. We will return back to this point later in this paper.

##  3 The Merton-Garman equation: Preliminaries and derivation

We can consider a more general case where the security and the volatility are
both stochastic. In such a case, the market is incomplete [3]. Although
several stochastic processes have been considered for modeling the case with
stochastic volatility [17], here we consider the generic case, defined by the
set of equations [3]

$\displaystyle\frac{dS}{dt}=\phi Sdt+S\sqrt{V}R_{1}$
$\displaystyle\frac{dV}{dt}=\lambda+\mu V+\zeta V^{\alpha}R_{2}.$ (9)

Here the volatility is defined through the variable $V=\sigma^{2}$ and $\phi$,
$\lambda$, $\mu$ and $\zeta$ are constants [18]. The Gaussian noises $R_{1}$
and $R_{2}$, corresponding to each of the variables under analysis, are
correlated in the following form

$<R_{1}(t^{\prime})R_{1}(t)>=<R_{2}(t^{\prime})R_{2}(t)>=\delta(t-t^{\prime})=\frac{1}{\rho}<R_{1}(t)R_{2}(t^{\prime})>.$
(10)

Here $-1\leq\rho\leq 1$, and the bra-kets $<AB>$ correspond to the correlation
between $A$ and $B$. If we consider a function $f$, depending on the Stock
price, the time, as well as on the white noises; with the help of the Ito
calculus, it is possible to derive the total derivative in time of this
function as

$\displaystyle\frac{df}{dt}=\frac{\partial f}{\partial t}+\phi S\frac{\partial
f}{\partial S}+(\lambda+\mu V)\frac{\partial f}{\partial
V}+\frac{\sigma^{2}S^{2}}{2}\frac{\partial^{2}f}{\partial S^{2}}+\rho
V^{1/2+\alpha}\zeta\frac{\partial^{2}f}{\partial S\partial
V}+\frac{\zeta^{2}V^{2\alpha}}{2}\frac{\partial^{2}f}{\partial V^{2}}+\sigma
S\frac{\partial f}{\partial S}R_{1}+\zeta V^{\alpha}\frac{\partial f}{\partial
V}R_{2}.$ (11)

This equation can be expressed in a more compact form which separates the
stochastic terms from the non-stochastic ones as follows

$\frac{df}{dt}=\Theta+\Xi R_{1}+\psi R_{2}.$ (12)

Here we have defined

$\displaystyle\Xi=\sigma S\frac{\partial f}{\partial
S},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\psi=\zeta V^{\alpha}\frac{\partial
f}{\partial V},$ $\displaystyle\Theta=\frac{\partial f}{\partial t}+\phi
S\frac{\partial f}{\partial S}+(\lambda+\mu V)\frac{\partial f}{\partial
V}+\frac{\sigma^{2}S^{2}}{2}\frac{\partial^{2}f}{\partial S^{2}}+\rho
V^{1/2+\alpha}\zeta\frac{\partial^{2}f}{\partial S\partial
V}+\frac{\zeta^{2}V^{2\alpha}}{2}\frac{\partial^{2}f}{\partial V^{2}},$ (13)

keeping in this way the notation used in [3] for convenience.

###  3.1 Derivation of the Merton-Garman equation

If we consider two different options defined as $C_{1}$ and $C_{2}$ on the
same underlying security with strike prices and maturities given by $K_{1}$,
$K_{2}$, $T_{1}$ and $T_{2}$ respectively. It is possible to create a
portfolio

$\Pi=C_{1}+\Gamma_{1}C_{2}+\Gamma_{2}S.$ (14)

If we consider the result (12), then we can define the total derivative with
respect to time as for the folio as

$\frac{d\Pi}{dt}=\Theta_{1}+\Gamma_{1}\Theta_{2}+\Gamma_{2}\phi
S+\left(\Xi_{1}+\Gamma_{1}\Xi_{2}+\Gamma_{2}\sigma
S\right)R_{1}+\left(\psi_{1}+\Gamma_{1}\psi_{2}\right)R_{2}.$ (15)

Note that this result is obtained after recognizing $f(t)=C_{1}$ or
$f(t)=C_{2}$ in eq. (12) when it corresponds. It has been demonstrated that
even in this case of stochastic volatility, it is still possible to create a
hedged folio and then at the end we arrive again to the condition (5), after
finding special constraints for $\Gamma_{1}$ and $\Gamma_{2}$ such that the
white noises are removed. The solution for $\Pi$ is a non-trivial one for this
case, and then it requires the definition of the parameter

$\displaystyle\beta(S,V,t,r)=\frac{1}{\partial C_{1}/\partial
V}\left(\frac{\partial C_{1}}{\partial t}+(\lambda+\mu V)\frac{\partial
C_{1}}{\partial S}+\frac{VS^{2}}{2}\frac{\partial^{2}C_{1}}{\partial
S^{2}}+\rho V^{1/2+\alpha}\zeta\frac{\partial^{2}C_{1}}{\partial S\partial
V}+\frac{\zeta^{2}V^{2\alpha}}{2}\frac{\partial^{2}C_{1}}{\partial
V^{2}}-rC_{1}\right)$ $\displaystyle=\frac{1}{\partial C_{2}/\partial
V}\left(\frac{\partial C_{2}}{\partial t}+(\lambda+\mu V)\frac{\partial
C_{2}}{\partial S}+\frac{VS^{2}}{2}\frac{\partial^{2}C_{2}}{\partial
S^{2}}+\rho V^{1/2+\alpha}\zeta\frac{\partial^{2}C_{2}}{\partial S\partial
V}+\frac{\zeta^{2}V^{2\alpha}}{2}\frac{\partial^{2}C_{2}}{\partial
V^{2}}-rC_{2}\right)$ (16)

This parameter does not appear for the case of the BS equation. Indeed $\beta$
in the MG equation is defined as the market price volatility risk because the
higher its value is, the lower is the intention of the investors to risk. Take
into account that in the MG equation the volatility is a stochastic variable.
Since the volatility is not traded in the market, then it is not possible to
make a direct hedging process over this quantity [3]. In this way, when we
have stochastic volatility, it is necessary to consider the expectations of
the investors. This effect appears through the parameter $\beta$. It has been
demonstrated in [19] that the value of $\beta$ in agreement with eq. (3.1) is
a non-vanishing result. In general, it is always assumed that the risk of the
market (in price) has been included inside the MG equation. The MG equation is
then obtained by rewriting the equation (3.1) in the form

$\frac{\partial C}{\partial t}+rS\frac{\partial C}{\partial S}+(\lambda+\mu
V)\frac{\partial C}{\partial V}+\frac{1}{2}VS^{2}\frac{\partial^{2}C}{\partial
S^{2}}+\rho\zeta V^{1/2+\alpha}S\frac{\partial^{2}C}{\partial S\partial
V}+\zeta^{2}V^{2\alpha}\frac{\partial^{2}C}{\partial V^{2}}=rC,$ (17)

where the effects of $\beta$ now appear contained inside the modified
parameter $\lambda$ in this equation. In other words, we have shifted the
parameter $\lambda\to\lambda-\beta$ in eq. (17). Later in this paper, we will
express this equation in the Hamiltonian form, which is the ideal one for
understanding the concept of spontaneous symmetry breaking in Quantum Finance.

###  3.2 Hamiltonian form of the Merton-Garman equation

The previously analyzed MG equation can be formulated as a Hamiltonian
(eigenvalue) equation. We can define a change of variable defined as

$\displaystyle S=e^{x},\;\;\;\;\;-\infty<x<\infty,$
$\displaystyle\sigma^{2}=V=e^{y},\;\;\;\;\;-\infty<y<\infty,$ (18)

and then the MG equation (17) becomes [3, 20, 21]

$\frac{\partial C}{\partial t}+\left(r-\frac{e^{y}}{2}\right)\frac{\partial
C}{\partial x}+\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right)\frac{\partial
C}{\partial y}+\frac{e^{y}}{2}\frac{\partial^{2}C}{\partial x^{2}}+\rho\zeta
e^{y(\alpha-1/2)}\frac{\partial^{2}C}{\partial x\partial
y}+\zeta^{2}e^{2y(\alpha-1)}\frac{\partial^{2}C}{\partial y^{2}}=rC.$ (19)

If we express this equation as an eigenvalue problem in the same form as in
eq. (7) for the BS case, then we have

$\frac{\partial C}{\partial t}=\hat{H}_{MG}C,$ (20)

with the MG Hamiltonian defined as

$\hat{H}_{MG}=-\frac{e^{y}}{2}\frac{\partial^{2}}{\partial
x^{2}}-\left(r-\frac{e^{y}}{2}\right)\frac{\partial}{\partial x}-\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right)\frac{\partial}{\partial
y}-\rho\zeta e^{y(\alpha-1/2)}\frac{\partial^{2}}{\partial x\partial
y}-\zeta^{2}e^{2y(\alpha-1)}\frac{\partial^{2}}{\partial y^{2}}+r.$ (21)

Exact solutions for the MG equation have been found for the case $\alpha=1$ in
[20] by using path-integral techniques. The same equation has been solved in
[17] for the case $\alpha=1/2$ by using standard techniques of differential
equations. Note that the equation has two degrees of freedom. Later we will
see that when we have spontaneous symmetry breaking, it becomes irrelevant to
know the exact solution of this equation.

##  4 The martingale condition in finance

The martingale condition is required for having a risk-neutral evolution for
the price of an Option. This means that the price of a financial instrument is
free of any possibility of arbitrage. In probability theory, the risk-free
evolution is modeled inside a stochastic process. Assume for example $N+1$
random variables $X_{i}$, with a joint probability distribution defined as
$p(x_{1},x_{2},...,x_{N+1})$. Then the martingale process is simply defined as
the condition under which

$E[X_{n+1}|x_{1},x_{2},...,x_{n}]=x_{n},$ (22)

is satisfied [3]. Note that $E[X_{i}]$ is the expectation value of the random
variable. Eq. (22) suggests that the expected value of a subsequent
observation of a random variable is simply the present value. For the purpose
of this paper, the random variables correspond to the future prices of the
stock given by $S_{1}$, $S_{2}$, …, $S_{N+1}$, which are defined at different
times $t_{1}$, $t_{2}$, …, $t_{N+1}$. We can then apply the same martingale
condition to the stocks if we make the corresponding discounts in order to
compare prices defined at different moments [2, 3]. We can assume that the
future value of an equity is defined as $S(t)$. If there is a free-risk
evolution of the discounted price defined as

$e^{-\int_{0}^{t}r(t^{\prime})dt^{\prime}}S(t).$ (23)

Then the value follows the martingale process [22]. In this way the
conditional probability for the present price is the actual value given by
$S(0)$. The martingale condition can then be expressed as [3]

$S(0)=E\left[e^{-\int_{0}^{t}r(t^{\prime})dt^{\prime}}S(t)|S(0)\right],$ (24)

and this result is general. Equivalent expressions have been used for the
analysis of the evolution of forward rates [3]. The importance of martingales
is analyzed in [23]. The interpretation of eq. (24) is clear. The left-hand
side is just the present price of the security. The right-hand side is the
expected value of the discounted price of the security at the time $t$.
Discounted means that the quantity evaluated at the time $t$ has to be
extrapolated to the present value. Both quantities must be equivalent under
the martingale condition.

##  5 The martingale condition as a vacuum condition for a Hamiltonian

The previous section dealt with the martingale condition. This section deals
with the equivalent formulation of the same principle but from the perspective
of The Hamiltonian formulation. Consider as before an option on a security
$S=e^{x}$ that matures at time $T$ with the corresponding pay-off function
$g(x)$. In this way we can describe the risk-free evolution of the option as

$C(t,x)=\int_{-\infty}^{\infty}dx^{\prime}<x|e^{-(T-t)\hat{H}}|x^{\prime}>g(x^{\prime}).$
(25)

By using the previous definition of martingales, for this case we have

$S(t)=E\left[e^{-(t_{*}-t)r}S(t_{*})|S(t)\right].$ (26)

If we introduce $S(x)$ (the price of the security) in eq. (25), then under the
martingale condition, we have

$S(t,x)=\int_{-\infty}^{\infty}dx^{\prime}<x|e^{-(t_{*}-t)\hat{H}}|x^{\prime}>S(x^{\prime}).$
(27)

This equation can be re-expressed in Dirac notation as

$<x^{\prime}|S>=\int_{-\infty}^{\infty}dx^{\prime}<x|e^{-(t_{*}-t)\hat{H}}|x^{\prime}><x^{\prime}|S>.$
(28)

If we take the base $|x^{\prime}>$ as a complete set of states, then the
condition $\hat{I}=\int dx^{\prime}|x^{\prime}><x^{\prime}|$ ($\hat{I}$ is the
identity matrix) is satisfied and then the previous expression is simplified
as

$|S>=e^{-(t_{*}-t)\hat{H}}|S>.$ (29)

Then there is no Hamiltonian (time) evolution for the state $|S>$ under the
previous conditions. It also comes out that the Hamiltonian annihilates the
same state as follows

$\hat{H}|S>=0.$ (30)

Interestingly, the BS Hamiltonian given in eq. (8) as well as the MG
Hamiltonian giving in eq. (21) satisfy the martingale condition in the form
defined in eq. (30).

##  6 Non-derivative terms introduced in the financial Hamiltonians

It is possible to introduce potential terms to the BS equation as well as to
the MG one as it is explained in [3]. It has been demonstrated that the
martingale condition can still be maintained if the potential satisfies some
special conditions. In general, a potential term will appear as

$\hat{H}_{BS,MG}^{eff}=\hat{H}_{BS,MG}+\hat{V}(x),$ (31)

with the potential term $\hat{V}$ containing non-derivative terms depending on
the security $S$. Since we usually have a change of variables in the
Hamiltonian formulation, this functional dependence is indirect. In the
previous equation, $\hat{H}_{BH,MG}^{eff}$ is the effective Hamiltonian
including the potential contribution. Some barrier options as well as some
path-dependent options admit the inclusion of potential terms for their deep
understanding [24]. On the other hand, for the case of the Black-Scholes
Hamiltonian, the martingale condition is maintained if the potential appears
in the Hamiltonian in the following form [3]

$\hat{H}_{BS}^{eff}=-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}+\left(\frac{1}{2}\sigma^{2}-V(x)\right)\frac{\partial}{\partial
x}+V(x).$ (32)

Then an effective Hamiltonian expressed in this way can be used for pricing
the option. The discount in these general cases depends on the price of the
option itself. Then the security discount defined in eq. (23) is modified as
[3]

$e^{-\int_{0}^{t}r(t^{\prime})dt^{\prime}}S(t)\to
e^{-\int_{0}^{t}V(x(t^{\prime}))dt^{\prime}}S(t).$ (33)

In [3] it is argued that the usual discounting of a security using the spot
interest rate $r$ is determined by the argument of no arbitrage involving
fixed deposits in the money market account. Studies about viable potentials
matching with the reality of the market are under analysis. It is important to
notice that the Hamiltonian (32) can be converted to a Hermitian operator by
using a similarity transformation as has been reported in [3]. Then we can
define

$\hat{H}_{BS}^{eff}=e^{s}\hat{H}_{Herm}e^{-s}.$ (34)

Here the Hermitian Hamiltonian is defined as

$\hat{H}_{Herm}=-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}+\frac{1}{2}V^{\prime}(x)+\frac{1}{2\sigma^{2}}\left(V+\frac{1}{2}\sigma^{2}\right)^{2},$
(35)

and $s=x/2-(1/\sigma^{2})\int^{x}_{0}dyV(y)$. This result can be obtained by
replacing (32) and (35) in eq. (34). From the Hermitian Hamiltonians, it is
possible to construct a complete basis and then we can find real eigenvalues
associated to this Hamiltonian. Note in particular that in the Black-scholes
case $V(x)=r$ is constant. It is a simple task to demonstrate that the
Hermitian Hamiltonian obtained by similarity transformation can be also
expressed as

$\hat{H}_{Herm}=e^{\alpha
x}\left(-\frac{\sigma^{2}}{2}\frac{\partial^{2}}{\partial
x^{2}}+\gamma\right)e^{-\alpha x},$ (36)

with

$\gamma=\frac{1}{2\sigma^{2}}\left(r+\frac{1}{2}\sigma^{2}\right)^{2}\;\;\;\;\;\alpha=\frac{1}{\sigma^{2}}\left(\frac{1}{2}\sigma^{2}-r\right).$
(37)

Among the trivial examples of potentials already analyzed in the literature,
we find the Down-and-Out barrier option, where the stock price has to be over
some minimal value, below which it becomes worthless. This behavior can be
guarantee with an infinite potential barrier boundary condition imposed for
the value of the corresponding price. This case can be worked out directly
from the non-Hermitian Hamiltonian defined in eq. (31). Another example of
potential corresponds to the Double-Knock-Out barrier option, where it is
easier to work with the Hermitian Hamiltonian part as it is defined in eq.
(34) but including the potential part as follows

$\hat{H}_{DB}=\hat{H}_{BS}+\hat{V}(x)=e^{s}\left(\hat{H}_{Herm}+\hat{V}(x)\right)e^{-s}.$
(38)

This definition is used for analyzing cases where the stock has to be
maintained between a maximal and a minimal value. Note that the definition
(38) can be also used for the analysis of the Down-and-Out barrier if we focus
on the non-Hermitian part (the term in the middle of the equation) with the
corresponding potential. Note that these examples of potentials representing
real situations are trivial cases. More details about these examples can be
found in [3].

##  7 Deeper analysis for the Black-Scholes and the Merton-Garman equation

If we analyze the martingale condition, we can notice that its interpretation
as a vacuum condition is not perfect. The reason is that although the
Hamiltonian annihilates the martingale state $|S>$, as can be seen from eq.
(30), the momentum operator corresponding to the prices of the options
($\hat{p}_{x}$) does not annihilate the vacuum perfectly. This can be seen
from their definitions as follows

$\hat{p}_{x}|S>=e^{x}|S>.$ (39)

This means that the symmetry under translations of the prices, carried out
from the security $S$ is spontaneously broken. An exception is the case where
$S\to 0$ and then $x\to-\infty$ as can be seen from eq. (3.2). This however,
would give us a trivial value for the security as $S=0$. For general values,
the symmetry under translations of prices is spontaneously broken. This means
that the different values of $S$ represent different possible vacuums or
ground states. Different vacuums however, would have different amount of
information because under the ideal conditions, all the information of the
market is stored inside the prices. The action of $\hat{p}_{x}$ over $|S>$,
then maps one vacuum toward another one defined through the selected value
$x$. Such operation could be seen as a rotation in a complex plane if we make
the transformation of variables $x\to in\theta$. In such a case, we have

$\hat{p}_{x}|S>=e^{in\theta}|S>=|S^{\prime}>\neq|S>.$ (40)

Here $n$ is just a number and $\theta$ is a dimensionless phase. After this
change of variable, the action of $\hat{p}_{x}$ is to map one vacuum into
another one through a rotation defined by the phase $\theta$. This condition
shows the vacuum degeneracy. The same condition has different meanings
depending on whether we consider the BS or the MG equation. This important
detail about the Martingale condition deserves more attention.

###  7.1 Reinterpretation of the Martingale condition

We can re-define the martingale state $S(x,t)$ as a Quantum Field by doing the
following change of variable
$S(x,t)=e^{x}=\sum_{n=0}^{\infty}x^{n}/n!=\sum_{n=0}^{\infty}\phi^{n}(x,t)$,
as it was proposed in [5]. Then we can express the BS equation as a function
of the Quantum Field $\phi(x,t)$ by replacing this definition inside eq. (8),
obtaining in this way

$\hat{H}_{BS}S(x,t)=-\frac{\sigma^{2}}{2}n(n-1)\phi^{n-2}+\left(\frac{1}{2}\sigma^{2}-r\right)n\phi^{n-1}+r\phi^{n}(x,t)=V(\phi)=0,$
(41)

which is valid in the neighborhood of the martingale state (vacuum). Here we
omit the sum symbol because the comparison has to be done term by term in the
series. This means that we have one equation for each term in the series
expansion for $\phi$. In eq. (41), we can see that all the terms in the
Hamiltonian can now be considered to be potential terms. Note that the result
(41) is valid if $d\phi/dx=\sum_{n}\phi/n$. In some regimes (especially inside
the weak-field approximation), the kinetic terms in eq. (41) play a
fundamental role in the identification of the vacuum state. The vanishing
condition in eq. (41) is a natural consequence of the Martingale state
definition, which being a vacuum condition, is annihilated by the Hamiltonian.
This is the case because the symmetries under time-translation are not
spontaneously broken. We can solve the martingale condition in eq. (41),
obtaining then

$\phi_{vac}=\frac{n}{2}\left(1-\frac{\sigma^{2}}{2r}\right)\left(1\pm\sqrt{1+\frac{8r\sigma^{2}}{(\sigma^{2}-2r)^{2}}\frac{(n-1)}{n}}\right),\;\;\;n\neq
0.$ (42)

This is the definition of the martingale state in the sense of the Quantum
Field $\phi$, as a function of the free-parameters of the theory. Note that
for $n=0$, the result (42) is not well-defined, and then we impose the
condition $n\neq 0$. Another special case is $n=1$, for which, eq. (42)
provides the following solutions

$\phi_{vac}=\frac{1}{2}\left(1-\frac{\sigma^{2}}{2r}\right)\left(1\pm
1\right).$ (43)

The result for this previous equation gives a trivial non-degenerate vacuum
$\phi_{vac}^{n=1}=0$, but it can also give the following non-trivial result

$\phi_{vac}^{n=1}=1-\frac{\sigma^{2}}{2r}.$ (44)

Note that this result is the same obtained in [5] when the BS eq. was
analyzed. This is the case because in [5], the kinetic terms were ignored
(Strong-field regime), which is precisely what happens in eq. (41) when $n=1$.
However, note that the result (44) appeared for $n=2$ instead of $n=1$ in [5].
The reason for this is that in [5], the derivative of the potential was
considered instead of the potential itself as it is the case here. In any
case, this only demonstrates that the regimes analyzed in this paper are more
general than the simple case studied in [5]. All the other results, with
$n\neq 0,1$ are non-trivial. Eq. (41) is by itself a potential function
$V(\phi)$ and we can demonstrate that any derivative of this potential will
not give any information different to the one obtained from eq. (41). Then for
example, if we find find the extreme condition for eq. (41) as $\partial
V(\phi)/\partial\phi=0$, we get

$\phi^{2}+\left(\frac{1}{2}\frac{\sigma^{2}}{r}-1\right)(n-1)\phi-\frac{\sigma^{2}}{2r}(n-1)(n-2)=0,$
(45)

which is just essentially the same result in eq. (41) but with the coefficient
terms shifted. Then for this case, when $n=1$, the previous equation gives a
trivial result $\phi=0$. In this way, since any higher-order derivative
provides the same information as the one obtained from eq. (41), for the
purpose of analysis, we will just focus on eq. (41) for understanding the
different regimes. Finally, note that although $\hat{H}_{BS}\phi^{n}=0$, still
$\hat{p}\phi^{n}=n\phi^{n-1}\neq 0$, considering $n\neq 0$. Then the vacuum is
only annihilated by the momentum if the field $\phi$ is itself trivial.
Something which happens for some combination of parameters. Then for example,
eq. (44) gives a trivial result when $\sigma^{2}=2r$, which is the same
condition for the Hamiltonian to be Hermitian. Then, at least for this
particular case, there is a direct connection between flow of information and
spontaneous symmetry breaking. When the martingale state is unique, no random
fluctuations can change the prices in the market. However, when the martingale
(vacuum) state is degenerate, then a multiplicity of possible ground states
emerge and then any fluctuation generates changes in the information of the
system and as a consequence, changes on the prices of the Option under
analysis.

####  7.1.1 Vacuum conditions for $n\neq 0,1$: Weak and strong field regimes

1). Weak-field regime: $\phi^{n}<<\phi$

For this regime, from eq. (41), we obtain

$-\frac{\sigma^{2}}{2}(n-1)+\left(\frac{1}{2}\sigma^{2}-r\right)\phi\approx
0,$ (46)

which can be easily solved to obtain

$\phi_{vac}\approx\frac{\sigma^{2}}{\sigma^{2}-2r}(n-1).$ (47)

Note that this result is trivial for $n=1$ for any combination of parameters.
This means that in these situations, again random fluctuations cannot change
the prices in the market. The case (47) is also trivial when $\sigma^{2}=0$
(zero volatility). Zero volatility means zero fluctuations and then zero
possibility of changing the prices in the market.

2). Strong Field regime:$\phi^{n}>>\phi$

In this regime, eq. (41) gives the following condition

$r\phi^{2}+\left(\frac{1}{2}\sigma^{2}-r\right)n\phi\approx 0.$ (48)

We have two solutions; one of them is the trivial $\phi_{vac}\approx 0$ and
the second solution is

$\phi_{vac}=\left(1-\frac{1}{2}\frac{\sigma^{2}}{r}\right)n.$ (49)

Here again, this result is trivial if $\sigma^{2}=2r$, which is again the
condition for the Hamiltonian to be Hermitian and then preserve information.
Then it seems that this is the combination of parameters which avoids changes
of information due to random fluctuations. Before going to the next section,
we have to remark that the results obtained for the BS case will be valid for
the MG case because the standard definition of martingale is independent of
the stochastic volatility. The only change to be done for the solutions inside
the MG equation, when we consider $\phi_{vac}$, is in the definition of the
volatility $\sigma^{2}=e^{y}$, with $y$ representing the variable connected to
the stochastic volatility.

##  8 A more general condition for the symmetry breaking in the Merton Garman
equation

When we analyze the MG equation, the martingale condition is normally taken
such that it is independent on the stochastic volatility. The volatility is a
function of the variable $y$, as it is defined in eq. (3.2). If the martingale
condition is taken as independent of $y$, then any term with derivative with
respect to this variable, will annihilate the state $S(x,t)$ defined
previously. From this perspective, $y$ can be taken as fixed when we are
determining the vacuum conditions. In this portion of the paper, we would like
to define a more general martingale condition, such that the possible changes
in $y$ can be considered. We will take the martingale state as

$\hat{H}_{MG}e^{x+y}=\hat{H}_{MG}S(x,y,t)=0.$ (50)

Here we extend the arguments showed in eq. (3.2), considering the extensions
of the original martingale state $S(x,t)=e^{x}$. The condition (50) will be
considered here as the extended martingale condition with $S(x,y,t)=e^{x+y}$.
By using the Hamiltonian (21) and the result (50), we can observe that the
condition for the Hamiltonian to annihilate the martingale state is

$\lambda+e^{y}\left(\mu+\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}+\rho\zeta
e^{y(\alpha-1/2)}\right)=0,$ (51)

as far as $e^{x+y}\neq 0$, avoiding then the trivial solution. This previous
condition is necessary for the state $e^{x+y}$ to be considered as the
martingale state and it represents a constraint among the free-parameters
involved in the MG equation.

###  8.1 The extended martingale condition and the flow of information

Previously, when we studied the ordinary martingale condition, we could
demonstrate that it can be also considered as a vacuum state. The vacuum in
the BS case came out to be single if there is no flow of information through
the boundaries of the system and it was degenerate if the information flows
through the boundaries of the system. When the vacuum is single, the momentum,
defined as the generator of the changes in prices, is a perfect symmetry. On
the other hand, when the vacuum is degenerate, then the same symmetry is
spontaneously broken because although the Hamiltonian annihilates the ground
state (martingale condition), the momentum does not do it. The interesting
point here is the connection between spontaneous symmetry breaking and flow of
information, which is connected with the changes on the prices in the market
[3]. Something which has been suggested before in [7] in a different context.
By using the previously proposed extended Martingale state, we can express it
as a function of Quantum fields in the form
$S(x,y,t)=e^{x+y}=\left(\sum_{n=0}^{\infty}\phi_{x}^{n}\right)\left(\sum_{m=0}^{\infty}\phi_{y}^{m}\right)$.
By replacing this expression in eq. (21), we get

$\displaystyle-\frac{e^{y}}{2}n(n-1)\phi_{x}^{n-2}\phi_{y}^{m}-\left(r-\frac{e^{y}}{2}\right)n\phi_{x}^{n-1}\phi_{y}^{m}-\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right)m\phi_{x}^{n}\phi_{y}^{m-1}$
$\displaystyle-\rho\zeta
e^{y(\alpha-1/2)}nm\phi_{x}^{n-1}\phi_{y}^{m-1}-\zeta^{2}e^{2y(\alpha-1)}m(m-1)\phi_{x}^{n}\phi_{y}^{m-2}+r\phi_{x}^{n}\phi_{y}^{m}=0.$
(52)

The vanishing condition comes out from the definition of martingale. The
previous result is valid if $d\phi_{x}/dx=\sum_{n}\phi_{x}/n$ and
$d\phi_{y}/dy=\sum_{n}\phi_{y}/n$. The equation (8.1) is a quadratic equation
for both fields, namely, $\phi_{x}$ and $\phi_{y}$. Then we can decide to
solve it with respect to either field if convenient. In this paper we will not
do it; we will rather analyze a few interesting cases and regimes in order to
visualize the structure of the extended martingale condition for the MG
equation. With the new definition of martingale, it is known that the
symmetries under changes of price as well as the symmetries under changes of
volatility are both spontaneously broken. This means that the information of
the system changes due to fluctuations in the prices as well as fluctuations
on the volatility. These situations are represented Mathematically as [5]

$\hat{p}_{x}|S>\neq 0,\;\;\;\;\;\hat{p}_{y}|S>\neq 0.$ (53)

with $|S>$ representing the ket related to the martingale state $S(x,y,t)$.
Additionally, $<x,y|\hat{p}_{x}|S>=\partial_{x}S(x,y,t)$ and
$<x,y|\hat{p}_{y}|S>=\partial_{y}S(x,y,t)$; representing in this way the
action of the operators $\hat{p}_{x,y}$ as the generators of translations in
the prices and volatility related to the options. We can now analyze some
important cases related to the condition (8.1). Note for example that for
$n=m=0$, we do not get anything interesting from this equation. Indeed, for
this case, $r=0$ (zero interest rate). On the other hand, we can analyze the
following interesting cases

####  8.1.1 Extended martingale with n=0 and m=1

For this case, eq. (8.1) gives us the following result

$\phi_{y\;vac}=\frac{1}{r}\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right),$ (54)

with an arbitrary value for $\phi_{x\;vac}$. This means that for this
particular case, the symmetry under changes in the volatility is spontaneously
broken, except when the condition

$\lambda e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}=0,$ (55)

is satisfied. This condition complemented with eq. (51), gives the result

$e^{y(\alpha-3/2)}=-\frac{\rho}{\zeta},$ (56)

which is a constraint related to important parameters connected to the
stochastic volatility [3]. Note that in this case, the field $\phi_{x}$ is
arbitrary. This means that the symmetry under changes of the prices is also
spontaneously broken. Note in addition that the condition (55) corresponds to
the condition for recovering the Hermiticity with respect to the variable $y$.
As a consequence of this, the unique vacuum condition with respect to $y$,
here represented by $\phi_{y\;vac}=0$, is connected to the no-flow of
information with respect to volatility variable $y$. This means that when (55)
is valid, then the random fluctuations coming from the stochastic volatility
do not affect do not produce changes on the information of the system. Still
for this case however, the information is not completely preserved due to the
random fluctuations on the prices, which disappear when $r=e^{y}/2$,
eliminating in this way all the non-Hermiticities in the MG Hamiltonian in eq.
(21).

####  8.1.2 Extended martingale with n=1 and m=0

For this case, eq. (8.1) gives us the solution

$\phi_{x\;vac}=1-\frac{e^{y}}{2r},$ (57)

with $\phi_{y}$ arbitrary. Compare eq. (57) with the case $n=1$ of the BS
result in eq. (44). The result is basically the same but with the re-defined
volatility as a stochastic variable $\sigma^{2}=e^{y}$. Here again the single
vacuum condition $\phi_{x\;vac}=0$ is related to the no-flow of information
but this time with respect to the variable $x$ which is a function of the
price of the stock. For this case, $\phi_{y\;vac}$ is arbitrary and then still
the symmetries with respect to the changes in volatility are spontaneously
broken. In this way, only the random fluctuations of the volatility affect the
information content of the market.

####  8.1.3 Extended martingale with n=1 and m=1

For this case, eq. (8.1) gives us the solution

$-\left(r-\frac{e^{y}}{2}\right)\phi_{y}-\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right)\phi_{x}-\rho\zeta
e^{y(\alpha-1/2)}+r\phi_{x}\phi_{y}=0.$ (58)

We can solve this expression for either Quantum field, $\phi_{y}$ or
$\phi_{x}$. One interesting situation emerges when the Hermiticity condition
for the Hamiltonian (21) is satisfied. This occurs when eq. (55) is satisfied
and when $e^{y}=2r$. In such a case we have from eq. (58)

$\phi_{x\;vac}\phi_{y\;vac}=\frac{\rho\zeta e^{y(\alpha-1/2)}}{r}.$ (59)

If in addition $\rho=0$ or $\zeta=0$, then we would have $\phi_{x}\phi_{y}=0$,
and then either, the symmetry is spontaneously broken with respect to
translation in prices but not with respect the variations in volatility or on
the contrary, the symmetry is spontaneously broken with respect to changes in
the volatility but we still have a perfect symmetry with respect to changes in
the prices of the options. The same result also suggests that the simultaneous
changes in the price with a subsequent changes in volatility (and viceversa),
represent a perfect symmetry for the system as far as $\rho=0$ or $\zeta=0$.
This means that for this combined symmetry, simultaneous random fluctuations
of the volatility and prices keep the information of the system unchanged.
Note that $\rho$ is a variable representing the correlation between the
Gaussian noises $R_{1}$ and $R_{2}$ as it was defined in eq. (10). Then
$\rho=0$ means zero correlation between the Gaussian noises. On the other
hand, if $\zeta=0$, then the Gaussian noise $R_{2}$ decouples and it comes out
that it does not affect the changes in the stochastic volatility. This last
issue can be observed from eq. (3).

####  8.1.4 Strong Field condition $\phi_{x}^{n}>>\phi_{x}$ and
$\phi_{y}^{m}>>\phi_{y}$; for any value of $n$, $m$

For this case, from eq. (8.1), we obtain

$\phi_{x\;vac}\phi_{y\;vac}\approx\left(1-\frac{e^{y}}{2r}\right)n\phi_{y\;vac}+\left(\frac{\lambda}{r}e^{-y}+\frac{\mu}{r}-\frac{\zeta^{2}}{2r}e^{2y(\alpha-1)}\right)m\phi_{x\;vac},$
(60)

which can be approximated to zero under the weak-field approximation if we
ignore cubic order terms in eq. (8.1) in comparison to the fourth-order terms.
Even if we do not ignore such terms, eq. (60) is exactly zero under the
Hermiticity conditions (no-flow of information) defined by eq. (55) and the
constraint $e^{y}=2r$. In any case, this result shows that for strong field
approximation for both fields, connected to the symmetries related to changes
in prices and volatility, it is not possible to have a single vacuum condition
related to both symmetries. However, it is possible to have a perfect vacuum
condition after the simultaneous application of both symmetries, namely, the
symmetries under changes in the price plus the symmetries under changes in
volatility. This result comes out to be similar to the one analyzed in eq.
(59), if we have $\rho=0$ or $\zeta=0$ as it was explained previously.

####  8.1.5 Weak-field condition $\phi_{x}^{n}<<\phi_{x}$ and
$\phi_{y}^{m}<<\phi_{y}$; for any value of $n$, $m$

For this regime, eq. (8.1) gives the approximation

$-\frac{e^{y}}{2}n(n-1)\phi_{y}^{2}-\rho\zeta
e^{y(\alpha-1/2)}nm\phi_{x}\phi_{y}-\zeta^{2}e^{2y(\alpha-1)}m(m-1)\phi_{x}^{2}\approx
0,$ (61)

which for $n=m=1$ would give again $\phi_{x\;vac}\phi_{y\;vac}\approx 0$ as
before. We can certainly solve this previous equation as a quadratic equation
in either, $\phi_{x}$ or $\phi_{y}$. If we decide to solve for $\phi_{y}$, we
get

$\phi_{y\;vac}\approx\frac{\rho\zeta
e^{y(\alpha-3/2)}m\phi_{x\;vac}}{(1-n)}\left(1\pm\sqrt{1-\frac{2(m-1)(n-1)}{\rho^{2}nm}}\right).$
(62)

Note the coupling between $\phi_{y\;vac}$ and $\phi_{x\;vac}$, such that the
triviality of one field implies the triviality of the other. However, it is
still possible to have a trivial value for $\phi_{y\;vac}$ and still a non-
trivial one for $\phi_{x\;vac}$ if $\zeta=0$ or if $m=1$. The coupling between
fields means that the broken symmetry with respect to the changes in the
prices is connected to the broken symmetry with respect to changes in
volatility. This also means that the random fluctuations of the prices are
tied to the random fluctuations of the volatility. This naturally is connected
to the flow of information inside the system.

####  8.1.6 Weak-Strong-field approximation: $\phi_{x}^{n}>>\phi_{x}$ and
$\phi_{y}^{m}<<\phi_{y}$

For this case, eq. (8.1) becomes

$\left(\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}\right)\phi_{x}^{2}\phi_{y}-\zeta^{2}e^{2y(\alpha-1)}(m-1)\phi_{x}^{n}\phi_{y}^{m-2}\approx
0,$ (63)

which gives a trivial solution for $\phi_{x\;vac}\approx 0$ and the following
result for $\phi_{y\;vac}$

$\phi_{y\;vac}\approx\frac{\zeta^{2}e^{2y(\alpha-1)}(1-m)}{\lambda
e^{-y}+\mu-\frac{\zeta^{2}}{2}e^{2y(\alpha-1)}}.$ (64)

Then for this case, the symmetry under changes in the prices is not broken but
the one related to the changes in volatility is spontaneously broken, except
when $m=1$ or $\zeta=0$. Then in the most general situations, the random
fluctuations cannot affect the prices of the market under this approximation.
Still, the random fluctuations generate changes in the information content of
the system because the location of the equilibrium condition changes due to
the random fluctuations on the volatility. Once again, a value of $\zeta=0$
decouples the Gaussian noise from the evolution in time of the volatility.

####  8.1.7 Weak-Strong-field approximation: $\phi_{x}^{n}<<\phi_{x}$ and
$\phi_{y}^{m}>>\phi_{y}$

For this case, eq. (8.1) is approximated to

$\frac{e^{y}}{2}(n-1)\phi_{y}^{2}+\left(r-\frac{e^{y}}{2}\right)\phi_{x}\phi_{y}^{2}\approx
0.$ (65)

This equation gives a trivial result for $\phi_{y\;vac}\approx 0$ and a non-
trivial one for $\phi_{x\;vac}$ as

$\phi_{x\;vac}\approx\frac{(1-n)\frac{e^{y}}{2}}{r-\frac{e^{y}}{2}},$ (66)

which vanishes when $n=1$ or when $y\to-\infty$ ($\sigma^{2}\to 0$). Then for
this case, in general, the symmetry under changes of the volatility is a
perfect symmetry, meanwhile, the symmetry under changes of the prices is
spontaneously broken. Then the random fluctuations only affect the the
information contained on the prices in the market in these special
circumstances. Interestingly, when $y\to\infty$, the previous expression
becomes

$\phi_{x\;vac}\approx n-1,$ (67)

which is independent from any parameter and only depends on the order of the
term in the series expansion under analysis.

##  9 Conclusions

In this paper we have analyzed the degeneracy of the martingales state on
regimes which were not considered before in previous papers. We have also
connected the concept of spontaneous symmetry breaking in Quantum Finance with
the flow of information through the market. The analysis has been done for
both, the BS equation as well as for the case where the volatility is
stochastic, namely, for the MG equation. For the BS equation as well as for
the MG one, the symmetries under changes of the prices of the Options cannot
annihilate the vacuum (taken as the martingale state). These symmetries are
then spontaneously broken in both cases. This means that random fluctuations
in the market generate changes in the information of the system, forcing it to
move toward another equilibrium condition. We found the conditions where the
vacuum, in these two situations (MG and BS equations), becomes unique. These
come out to be the same conditions for which there is no flow of information
through the boundaries of the system (Market), or equivalently, the same
conditions correspond to the cases where the Hamiltonian becomes Hermitian.
Hermitian Hamiltonians in general preserve the information of the system.
Under the ordinary definition of martingale, the connection between flow of
information and spontaneous symmetry breaking, is exact. We could then extend
the martingale condition for including the symmetries under changes of the
stochastic volatility in a similar fashion to what is done in [5]. We defined
then the momentum for the volatility as the generator of these
transformations. We could also find the conditions under which the martingale
state (vacuum) becomes unique for different regimes and for the different
series expansion terms. A unique vacuum or martingale state, means that the
random fluctuations cannot change the information of the system under such
special situations. In [5] the authors only focused on the strong field
approximation. Here however, we could complete the scenario by analyzing the
weak-field approximation for the BS as well as for the MG case. In the weak-
field approximation, the kinetic terms are the most relevant. Since the
kinetic terms, expressed as a function of a Quantum Field, behave as potential
terms; then their existence affects the vacuum location and behavior. For the
case of the MG equation, we have to define two different Quantum Fields, one
corresponding to the prices in the market and the other related to the
volatility. In this paper we analyzed the regimes where both fields in the MG
equation are either, strong or weak. Finally, we analyzed the situations where
one field is strong and the other is weak. In these cases, it came out that
one of the symmetries involved were perfect but the other one was
spontaneously broken in general, except for some specific combination of
parameters. Perfect symmetries mean conservation of the information under
random fluctuations. This is a very important statement for the stock market
analysis.

Acknowledgement I. A. would like to thank to Prof. Yu Chen from the University
of Tokyo for providing useful references and bibliography, as well as very
useful discussions.

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