# Transferring quantum entangled states between multiple single-photon-state
qubits and coherent-state qubits in circuit QED

Qi-Ping Su1 Hanyu Zhang1 Chui-Ping Yang1,2<EMAIL_ADDRESS>1Department of
Physics, Hangzhou Normal University, Hangzhou, Zhejiang 311121, China
2Quantum Information Research Center, Shangrao Normal University, Shangrao,
Jiangxi 334001, China

###### Abstract

We present a way to transfer maximally- or partially-entangled states of $n$
single-photon-state (SPS) qubits onto $n$ coherent-state (CS) qubits, by
employing $2n$ microwave cavities coupled to a superconducting flux qutrit.
The two logic states of a SPS qubit here are represented by the vacuum state
and the single-photon state of a cavity, while the two logic states of a CS
qubit are encoded with two coherent states of a cavity. Because of using only
one superconducting qutrit as the coupler, the circuit architecture is
significantly simplified. The operation time for the state transfer does not
increase with the increasing of the number of qubits. When the dissipation of
the system is negligible, the quantum state can be transferred in a
deterministic way since no measurement is required. Furthermore, the higher-
energy intermediate level of the coupler qutrit is not excited during the
entire operation and thus decoherence from the qutrit is greatly suppressed.
As a specific example, we numerically demonstrate that the high-fidelity
transfer of a Bell state of two SPS qubits onto two CS qubits is achievable
within the present-day circuit QED technology. Finally, it is worthy to note
that when the dissipation is negligible, entangled states of $n$ CS qubits can
be transferred back onto $n$ SPS qubits by performing reverse operations. This
proposal is quite general and can be extended to accomplish the same task, by
employing a natural or artificial atom to couple $2n$ microwave or optical
cavities.
Keywords entangled state, single-photon-state qubit, coherent-state qubit,
circuit QED

###### pacs:

03.67.Bg, 42.50.Dv, 85.25.Cp

##  I Introduction

Discrete- and continuous-variable methods to optical quantum communication and
quantum information processing (QIP) rely on different encoding qubits for
their realization [1-3]. For the discrete-variable approaches, single photons
are involved [4,5], the photonic qubits live in a two-dimensional Hilbert
space formed, for instance, by orthogonal polarizations or the absence and
presence of a single photon. For convenience, we define such photonic qubits
as single-photon-state (SPS) qubits. In contrast, for the continuous-variable
approaches, encoding is implemented in the quadrature components of a light
field with an inherently infinite-dimensional space. Quantum information
carried by a qubit can be indicated as an arbitrary superposition of coherent
states or an arbitrary superposition of cat states of light waves [6,7]. In
recent years, besides progress achieved with SPS qubits, considerable
theoretical and experimental activity have also been shifted to QIP with
coherent-state (CS) or cat-state (C-S) qubits. There exists a number of works
for QIP using qubits encoded with coherent states [6,8-10] or encoded with
cat-state qubits [7,11-16].

Both encodings of SPS qubits and CS/C-S qubits have advantages and drawbacks
[17]. CS/C-S qubits benefit from unconditional operations, unambiguous state
discrimination, high detection efficiencies, and more practical interfacing
with conventional information technology. However, it is commonly recognized
that quantum states of multiphoton wave fields are difficult to manipulate and
suffer from intrinsically limited fidelities. On the other hand, quantum
states of SPS qubits are relatively easy to operate, but usually suffer from
decoherence due to the loss of a single photon. A combination of qubits with
different encodings, that is, in a hybrid architecture [18], may offer great
advantages [19]. Some operations may be more efficient using SPS qubits while
others may better take advantage of CS/C-S qubits. In this endeavour, it is a
key requirement to transfer information or entanglement between qubits with
different encodings.

The goal of this work is aimed at transferring the following type of entangled
state

$c\left|i_{1}\right\rangle\left|i_{2}\right\rangle...\left|i_{n}\right\rangle+d\left|\overline{i}_{1}\right\rangle\left|\overline{i}_{2}\right\rangle...\left|\overline{i}_{n}\right\rangle$
(1)

between $n$ SPS qubits and $n$ CS qubits. For $c\neq d,$ the state (1) is a
partially-entangled state. For $c=d=1/\sqrt{2}$, the state (1) is a maximally-
entangled state, which is called Bell state when $n=2$ or
Greenberger–Horne–Zeilinger (GHZ) state when $n\geq 3.$ As is well known, Bell
states and GHZ states are crucial resources and play an important role in
quantum communication and QIP. On the other hand, transferring entangled
states between qubits of different encodings is of fundamental interest in
quantum mechanics, and may be necessary and important in the hybrid quantum
communication and QIP. For instance, in a hybrid QIP based on SPS qubits and
CS qubits, entanglement transfer may be necessary between SPS-qubit-based
quantum processors and CS-qubit-based quantum processors.

The new and rapidly growing field of circuit QED, consisting of microwave
radiation fields and fixed artificial atoms, offers extremely exciting
prospects for solid-state QIP [20-28]. In the following, we will present an
approach to transfer the entangled state (1) between $n$ SPS qubits and $n$ CS
qubits, in a circuit QED system that consists of $2n$ microwave cavities
coupled to a superconducting flux qutrit. In our proposal, the two logic
states $\left|i\right\rangle$ and $\left|\overline{i}\right\rangle$ of a SPS
qubit are denoted as the vacuum state $\left|0\right\rangle$ and the single-
photon state $\left|1\right\rangle$ of a cavity; while the two logic states
$\left|i\right\rangle$ and $\left|\overline{i}\right\rangle$ of a CS qubit are
indicated by the two coherent states $\left|\pm\alpha\right\rangle$ of a
cavity, which can be quasi-orthogonal for a sufficiently large $\alpha$ (for
instance,
$\left|\left\langle\alpha\right|\left.-\alpha\right\rangle\right|=e^{-2\left|\alpha\right|^{2}}\sim
10^{-3}$ for $\alpha=1.86$). As is well known, coherent states are eigenstates
of the photon annihilation operator and thus tolerant to single-photon loss
[11].

In this work, we will explicitly show how to transfer the entangled state (1)
from $n$ SPS qubits onto $n$ CS qubits. Note that since no measurement is
involved, i.e., all basic operations are unitary, the entangled state (1) can
be transferred back from $n$ CS qubits onto $n$ SPS qubits by performing
reverse operations, in the case when the dissipation of the system is
negligible.

The present proposal has the following features: (i) Because only one
superconducting flux qutrit is used, the circuit complexity is reduced; (ii)
The state transfer time is independent of the number of qubits, thus it does
not increase with the increase of the number of qubits; (iii) The higher-
energy level of the qutrit is not excited during the entire operation, thus
the effect of decoherence from this level is suppressed; and (iv) Since
measurement is not required, the entangled state can be transferred in a
deterministic manner when the dissipation is negligible. This proposal is
universal and can be applied to accomplish the same task, in a physical system
which consists of $2n$ microwave or optical cavities coupled to a natural or
artificial atom. To the best of our knowledge, how to transfer quantum
entangled states between SPS qubits and CS qubits has not been reported yet.

We should point out that this proposal works by applying the rotating-wave
approximation (RWA). As is well known, the RWA is often used in quantum optics
and quantum information processing. As an open question, it would be
interesting to generalize the protocol to a more general case without assuming
the RWA, which usually applies in the ultrastrong coupling regime (for a
review, see [29]).

This paper is organized as follows. In Sec. II, we introduce a few types of
interaction and state evolution, which are used for transferring entangled
states. In Sec. III, we show how to transfer the entangled states of $n$ SPS
qubits onto $n$ CS qubits. In Sec. IV, as an example, we investigate the
experimental feasibility for transferring a Bell state of two SPS qubits onto
two CS qubits in a circuit QED setup. A brief conclusion is given in Sec. V.

Figure 1: (color online) (a) Diagram of $2n$ microwave cavities coupled to a
superconducting flux qutrit (the circle $A$ in the middle). Each square
represents a cavity, which can be a one- or three-dimensional cavity. The
qutrit is capacitively or inductively coupled to each cavity. (b) Illustration
of $n-1$ cavities $(2,3,...,n)$ off resonant with the
$\left|g\right\rangle\leftrightarrow\left|f\right\rangle$ transition of the
qutrit, while $n$ cavities $(1^{\prime},2^{\prime},...,n^{\prime})$ off
resonant with the $\left|e\right\rangle\leftrightarrow\left|f\right\rangle$
transition of the qutrit. (c) Illustration of cavity $1^{\prime}$ and a
microwave pulse off resonant with the
$\left|g\right\rangle\leftrightarrow\left|e\right\rangle$ transition of the
qutrit.

##  II Type of interaction and state evolution

Consider $n$ single-mode cavities $(1,2,...,n)$ and another $n$ single-mode
cavities $(1^{\prime},2^{\prime},...,n^{\prime})$. The $2n$ cavities are
connected by a superconducting flux qutrit $A$ (Fig. 1a). The three levels of
the coupler qutrit are labelled as
$\left|g\right\rangle,\left|e\right\rangle,$and $\left|f\right\rangle$ (Fig.
1b). In this section, we will introduce a few types of interaction, the
corresponditing Hamiltonians, and the state evolutions, which will be used for
the entangled state transfer in next section.

###  II.1 Qutrit-cavity resonant interaction

Consider cavity $1$ on resonance with the $\left|g\right\rangle$
$\leftrightarrow$ $\left|e\right\rangle$ transition of the qutrit. In the
interaction picture and after making the rotating-wave approximation (RWA),
the Hamiltonian is given by (hereafter assuming $\hbar=1$)

$H_{1}=g_{r}\hat{a}_{1}\sigma_{eg}^{+}+\text{h.c.},$ (2)

where $\sigma_{eg}^{+}=\left|e\right\rangle\left\langle g\right|,$ $g_{r}$ is
the coupling strength between cavity $1$ and the $\left|g\right\rangle$
$\leftrightarrow$ $\left|e\right\rangle$ transition, and $\hat{a}_{1}$ is the
photon annihilation operator of cavity $1$. Under this Hamiltonian, it is easy
to get the following state evolution

$\displaystyle\left|e\right\rangle\left|0\right\rangle_{1}$
$\displaystyle\rightarrow$ $\displaystyle\cos
g_{r}t\left|e\right\rangle\left|0\right\rangle_{1}-i\sin
g_{r}t\left|g\right\rangle\left|1\right\rangle_{1},$
$\displaystyle\left|g\right\rangle\left|1\right\rangle_{1}$
$\displaystyle\rightarrow$ $\displaystyle\cos
g_{r}t\left|g\right\rangle\left|1\right\rangle_{1}-i\sin
g_{r}t\left|e\right\rangle\left|0\right\rangle_{1},$ (3)

while the state $\left|g\right\rangle\left|0\right\rangle_{1}$ remains
unchanged.

###  II.2 Off-resonant interaction between the coupler qutrit and the
$\left(\mathbf{2n-1}\right)$ cavities

Consider that the qutrit is decoupled from cavity $1$ but interacts with the
$\left(2n-1\right)$ cavities
($2,3,...,n,1^{\prime},2^{\prime},...,n^{\prime}$). Assume that cavity $j$ is
off resonant with the
$\left|g\right\rangle\leftrightarrow\left|f\right\rangle$ transition with
coupling constant $g_{j}$ and detuning $\Delta_{j}$ $(j=2,3,...,n),$ while
highly detuned (decoupled) from other intra-level transitions (Fig. 1b). In
addition, suppose that cavity $j^{\prime}$ is off resonant with the
$\left|e\right\rangle\leftrightarrow\left|f\right\rangle$ transition with
coupling constant $\mu_{j^{\prime}}$ and detuning $\Delta_{j^{\prime}}$
$(j^{\prime}=1^{\prime},2^{\prime},...,n^{\prime}),$ while highly detuned
(decoupled) from other intra-level transitions (Fig. 1b). After these
considerations, in the interaction picture and after the RWA, the Hamiltonian
is given by

$H_{2}=\sum_{j=2}^{n}g_{j}\left(e^{i\Delta_{j}t}\hat{a}_{j}\sigma_{fg}^{+}+\text{h.c.}\right)+\sum_{j^{\prime}=1^{\prime}}^{n^{\prime}}\mu_{j^{\prime}}\left(e^{i\Delta_{j^{\prime}}t}\hat{b}_{j^{\prime}}\sigma_{fe}^{+}+\text{h.c.}\right),$
(4)

where $\hat{a}_{j}$ ($\hat{b}_{j^{\prime}}$) is the photon annihilation
operator of cavity $j$ ($j^{\prime}$);
$\sigma_{fg}^{+}=\left|f\right\rangle\left\langle g\right|,$
$\sigma_{fe}^{+}=\left|f\right\rangle\left\langle e\right|,$
$\Delta_{j}=\omega_{fg}-\omega_{c_{j}}$ and
$\Delta_{j^{\prime}}=\omega_{fe}-\omega_{c_{j^{\prime}}}$. Here, $\omega_{fg}$
($\omega_{fe}$) is the
$\left|g\right\rangle\leftrightarrow\left|f\right\rangle$
($\left|e\right\rangle\leftrightarrow\left|f\right\rangle$) transition
frequency of the qutrit, while $\omega_{c_{j}}$ ($\omega_{c_{j^{\prime}}}$) is
the frequency of cavity $j$ (cavity $j^{\prime}$).

Under the large detunings $\left|\Delta_{j}\right|\gg g_{j}$
and$\;\left|\Delta_{j^{\prime}}\right|\gg\mu_{j^{\prime},}$ no energy exchange
is induced between the coupler qutrit and the cavities. When the following
conditions satisfy, i.e.,

$\frac{\left|\Delta_{j}-\Delta_{k}\right|}{\left|\Delta_{j}^{-1}\right|+\left|\Delta_{k}^{-1}\right|}\gg
g_{j}g_{k},\frac{\left|\Delta_{j^{\prime}}-\Delta_{k^{\prime}}\right|}{\left|\Delta_{j^{\prime}}^{-1}\right|+\left|\Delta_{k^{\prime}}^{-1}\right|}\gg\mu_{j^{\prime}}\mu_{k^{\prime}},\frac{\left|\Delta_{j}-\Delta_{k^{\prime}}\right|}{\left|\Delta_{j}^{-1}\right|+\left|\Delta_{k^{\prime}}^{-1}\right|}\gg
g_{j}\mu_{k^{\prime}},$ (5)

(where $j,k\in\\{2,3,...,n\\},$
$j^{\prime},k^{\prime}\in\\{1^{\prime},2^{\prime},...,n^{\prime}\\},$ $j\neq
k,$ $j^{\prime}\neq k^{\prime}$), no interaction between cavities
($2,3,...,n,1^{\prime},2^{\prime},...,n^{\prime}$) is induced by the coupler
qutrit. Thus, from the Hamiltonian (4), one can obtain the following effective
Hamiltonian [30,31]

$\displaystyle H_{\mathrm{e}}$ $\displaystyle=$
$\displaystyle-\sum_{j=2}^{n}\frac{g_{j}^{2}}{\Delta_{j}}\left(\sigma_{g}\hat{a}_{j}^{+}\hat{a}_{j}-\sigma_{f}\hat{a}_{j}\hat{a}_{j}^{+}\right)$
(6) $\displaystyle\
-\sum_{j^{\prime}=1^{\prime}}^{n^{\prime}}\frac{\mu_{j^{\prime}}^{2}}{\Delta_{j^{\prime}}}\left(\sigma_{e}\hat{b}_{j^{\prime}}^{+}\hat{b}_{j^{\prime}}\hat{-}\sigma_{f}\hat{b}_{j^{\prime}}\hat{b}_{j^{\prime}}^{+}\right),$

where $\sigma_{g}=\left|g\right\rangle\left\langle
g\right|,\sigma_{e}=\left|e\right\rangle\left\langle
e\right|,\sigma_{f}=\left|f\right\rangle\left\langle f\right|.$ In Eq. (6),
the terms in the first line represent the ac-Stark shifts of the levels
$\left|g\right\rangle$ and $\left|f\right\rangle$ induced by the cavities
$(2,3,...,n)$, while the terms in the second line are the ac-Stark shifts of
the levels $\left|e\right\rangle$ and $\left|f\right\rangle$ caused by the
cavities $(1^{\prime},2^{\prime},...,n^{\prime})$.

In the case of the level $\left|f\right\rangle$ being not occupied, the
Hamiltonian (6) reduces to

$H_{\mathrm{e}}=-\sum_{j=2}^{n}\lambda_{j}\sigma_{g}\hat{a}_{j}^{+}\hat{a}_{j}-\sum_{j^{\prime}=1^{\prime}}^{n^{\prime}}\lambda_{j^{\prime}}\sigma_{e}\hat{b}_{j^{\prime}}^{+}\hat{b}_{j^{\prime}},$
(7)

where $\lambda_{j}=g_{j}^{2}/\Delta_{j}$ and
$\lambda_{j^{\prime}}=\mu_{j^{\prime}}^{2}/\Delta_{j^{\prime}}.$ Under this
Hamiltonian (7), it is straightforward to obtain

$\displaystyle\left|g\right\rangle\left|0\right\rangle_{j}$
$\displaystyle\rightarrow$
$\displaystyle\left|g\right\rangle\left|0\right\rangle_{j},\text{
}\left|g\right\rangle\left|1\right\rangle_{j}\rightarrow
e^{i\lambda_{j}t}\left|g\right\rangle\left|1\right\rangle_{j},\text{\
}\left|g\right\rangle\left|\alpha\right\rangle_{j^{\prime}}\rightarrow\left|g\right\rangle\left|\alpha\right\rangle_{j^{\prime}},\text{\
\ }$ $\displaystyle\left|e\right\rangle\left|0\right\rangle_{j}$
$\displaystyle\rightarrow$
$\displaystyle\left|e\right\rangle\left|0\right\rangle_{j},\text{
}\left|e\right\rangle\left|1\right\rangle_{j}\rightarrow\left|e\right\rangle\left|1\right\rangle_{j},\text{
}\left|e\right\rangle\left|\alpha\right\rangle_{j^{\prime}}\rightarrow\left|e\right\rangle\left|e^{i\lambda_{j^{\prime}}t}\alpha\right\rangle_{j^{\prime}},$
(8)

where $j\in\left\\{2,3,...,n\right\\}$ and
$j^{\prime}\in\\{1^{\prime},2^{\prime},...,n^{\prime}\\}.$

###  II.3 Driving the qutrit conditional on the cavity’s state

Consider cavity $1^{\prime}$ to be off resonant with the
$\left|g\right\rangle$ $\leftrightarrow$ $\left|e\right\rangle$ transition of
the qutrit, with coupling strength $\widetilde{g}$ and detuning
$\widetilde{\Delta}=\omega_{eg}-\widetilde{\omega}_{c_{1^{\prime}}}$ (Fig.
1c). Here, $\widetilde{\omega}_{c_{1^{\prime}}}$ is the adjusted frequency of
cavity $1^{\prime}$. Meanwhile, assume that a classical pulse of frequency
$\omega_{p}$ and initial phase $\phi$ is applied to the qutrit.

Under the above assumptions, the Hamiltonian of the system in the interaction
picture and after the RWA is given by

$H_{3}=\widetilde{g}e^{i\widetilde{\Delta}t}\hat{a}_{1^{\prime}}\sigma_{eg}^{+}+\Omega_{p}e^{-i\left[\left(\omega_{p}-\omega_{eg}\right)t+\phi\right]}\sigma_{eg}^{+}+\text{h.c.,}$
(9)

where $\Omega_{p}$ is the Rabi frequency of the pulse.

Under the large detuning condition
$\left|\widetilde{\Delta}\right|\gg\tilde{g},$ and assuming
$\left|\widetilde{\Delta}\right|\gg\Omega_{p},$ we can obtain the following
effective Hamiltonian [22,31]

$H_{\mathrm{e}}=\frac{\widetilde{g}^{2}}{\tilde{\Delta}}\left(\hat{a}_{1^{\prime}}^{+}\hat{a}_{1^{\prime}}+\frac{1}{2}\right)\sigma_{z}+\left(\Omega_{p}e^{-i\left[\left(\omega_{p}-\omega_{eg}\right)t+\phi\right]}\sigma_{eg}^{+}+\text{h.c.}\right),$
(10)

where $\sigma_{z}=\left|e\right\rangle\left\langle
e\right|-\left|g\right\rangle\left\langle g\right|.$ Here, the stark-shift
effect of the qubit induced by the pulse is negligible because of
$\left|\widetilde{\Delta}\right|\gg\Omega_{p}.$ In a rotating frame under the
Hamiltonian
$H_{0}=\frac{\widetilde{g}^{2}}{\tilde{\Delta}}\left(\hat{a}_{1^{\prime}}^{+}\hat{a}_{1^{\prime}}+\frac{1}{2}\right)\sigma_{z},$
and for $\omega_{p}=\omega_{eg}+\widetilde{g}^{2}/\widetilde{\Delta},$ one
obtains

$H_{\mathrm{e}}=\Omega_{p}e^{-i\phi}e^{i4\widetilde{\omega}\hat{a}_{1^{\prime}}^{+}\hat{a}_{1^{\prime}}t}\sigma_{eg}^{+}+\text{h.c.,}$
(11)

where $\widetilde{\omega}=\widetilde{g}^{2}/\left(2\widetilde{\Delta}\right).$
When the cavity is in the vacuum state, one can easily find that this
Hamiltonian reduce to $H_{\mathrm{eff}}=\Omega_{p}e^{-i\phi}\sigma_{eg}^{+}+$
H.c., which rotates the qubit’s state as follows

$\displaystyle\left|g\right\rangle\left|0\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left(\cos\Omega_{p}t\left|g\right\rangle-
ie^{-i\phi}\sin\Omega_{p}t\left|e\right\rangle\right)\left|0\right\rangle_{1^{\prime}},$
$\displaystyle\left|e\right\rangle\left|0\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left(-ie^{i\phi}\sin\Omega_{p}t\left|g\right\rangle+\cos\Omega_{p}t\left|e\right\rangle\right)\left|0\right\rangle_{1^{\prime}}.$
(12)

On the other hand, when the cavity is in the coherent state
$\left|2\alpha\right\rangle,$ the Hamiltonian (11) becomes
$H_{\mathrm{eff}}=\Omega_{p}e^{-i\phi}e^{i4\widetilde{\omega}\overline{n}t}\sigma_{eg}^{+}+$h.c.,
with $\overline{n}=4\left|\alpha\right|^{2}.$ For a sufficient large $\alpha$,
such that $4\left|\widetilde{\omega}\right|\overline{n}\gg\Omega_{p}$ (large
detuning), it is easy to see that the qubit’s state is not changed by the
driving pulse, i.e.,

$\displaystyle\left|g\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left|g\right\rangle\left|2\alpha\right\rangle_{1^{\prime}},$
$\displaystyle\left|e\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left|e\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}.$
(13)

After returning to the original interaction picture by performing a unitary
operation $U=e^{-iH_{0}},$ we have from Eqs. (12) and (13)

$\displaystyle\left|g\right\rangle\left|0\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left(e^{i\widetilde{\omega}t}\cos\Omega_{p}t\left|g\right\rangle-
ie^{-i\phi}e^{-i\widetilde{\omega}t}\sin\Omega_{p}t\left|e\right\rangle\right)\left|0\right\rangle_{1^{\prime}},$
$\displaystyle\left|e\right\rangle\left|0\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$
$\displaystyle\left(-ie^{i\phi}e^{i\widetilde{\omega}t}\sin\Omega_{p}t\left|g\right\rangle+e^{-i\widetilde{\omega}t}\cos\Omega_{p}t\left|e\right\rangle\right)\left|0\right\rangle_{1^{\prime}},$
(14)

and

$\displaystyle\left|g\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$ $\displaystyle
e^{i\widetilde{\omega}t}\left|g\right\rangle\left|e^{i2\widetilde{\omega}t}2\alpha\right\rangle_{1^{\prime}},$
$\displaystyle\left|e\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}$
$\displaystyle\rightarrow$ $\displaystyle
e^{-i\widetilde{\omega}t}\left|e\right\rangle\left|e^{-i2\widetilde{\omega}t}2\alpha\right\rangle_{1^{\prime}}.$
(15)

In the next section, we will show how to use the results (3), (8), (14) and
(15) to transfer entangled states from $n$ PO qubits to $n$ WO qubits.

##  III Transfer of SPS-qubit entangled states onto CS-qubits

The setup is shown in Fig. 1(a). Initially, the coupler qutrit is in the
ground state $\left|g\right\rangle$ and decoupled from the cavity system. The
$n$ SPS qubits are initially in the entangled state

$c\left|0\right\rangle_{1}\left|0\right\rangle_{2}...\left|0\right\rangle_{n}+d\left|1\right\rangle_{1}\left|1\right\rangle_{2}...\left|1\right\rangle_{n},$
(16)

where $\left|0\right\rangle$ and $\left|1\right\rangle$ are the vacuum state
and the single-photon state.

Suppose that each of the $n$ cavities ($1^{\prime},2^{\prime},...,n^{\prime}$)
is in a coherent state $\left|\alpha\right\rangle$. Thus, the initial state of
the whole system is given by

$\left(c\left|0\right\rangle_{1}\left|0\right\rangle_{2}...\left|0\right\rangle_{n}+d\left|1\right\rangle_{1}\left|1\right\rangle_{2}...\left|1\right\rangle_{n}\right)\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}...\left|\alpha\right\rangle_{n^{\prime}}\left|g\right\rangle,$
(17)

where subscripts $1,2,...,n$ represent cavities $1,2,...,n$ and subscripts
$1^{\prime},2^{\prime},...,n^{\prime}$ represent cavities
$1^{\prime},2^{\prime},...,n^{\prime},$ respectively. The procedure for
transferring the entangled state is listed below:

Step (i): The purpose of this step is to convert the states
$\left|0\right\rangle$ and $\left|1\right\rangle$ of each of cavities
($2,3,...,n$) into $\left|+\right\rangle$ and $\left|-\right\rangle,$
respectively. Here,
$\left|\pm\right\rangle=\left(\left|0\right\rangle\pm\left|1\right\rangle\right)/\sqrt{2}.$
Note that this state convertion can be easily implemented by performing local
resonant operations on cavity $j$ and an auxilliary qubit $j$ (initially in
the ground state) placed in cavity $j,$ i.e., applying resonant interaction of
qubit $j$ with cavity $j$ and resonant interaction of qubit $j$ with a
classical pulse [32]. After this step of operation, the state (17) becomes

$\left(c\left|0\right\rangle_{1}\left|+\right\rangle_{2}\left|+\right\rangle_{3}...\left|+\right\rangle_{n}+d\left|1\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\right)\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}...\left|\alpha\right\rangle_{n^{\prime}}\left|g\right\rangle.$
(18)

Step (ii): Adjust the level spacings of the coupler qutrit to have cavity $1$
on resonance with the
$\left|g\right\rangle\leftrightarrow\left|e\right\rangle$ transition.
According to Eq. (3), one has
$\left|0\right\rangle_{1}\left|g\right\rangle\rightarrow\left|0\right\rangle_{1}\left|g\right\rangle$
and
$\left|1\right\rangle_{1}\left|g\right\rangle\rightarrow-i\left|0\right\rangle_{1}\left|e\right\rangle$
for an interaction time $t=\pi/\left(2g_{r}\right).$ Hence, the state (18)
becomes

$\left|0\right\rangle_{1}\left(c\left|g\right\rangle\left|+\right\rangle_{2}\left|+\right\rangle_{3}...\left|+\right\rangle_{n}-id\left|e\right\rangle\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\right)\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}...\left|\alpha\right\rangle_{n^{\prime}}.$
(19)

Then, the level spacings of the coupler qutrit is adjusted back such that the
qutrit is decoupled from cavity $1.$

Step (iii): Adjust the level spacings of the coupler qutrit so that the qutrit
is off resonant with the cavities
($2,3,...,n,1^{\prime},2^{\prime},...,n^{\prime}$) (Fig. 1b) to achieve an
effective Hamiltonian (7). According to Eq. (8), the state (19) changes to

$\displaystyle\left|0\right\rangle_{1}\left[c\left|g\right\rangle\left(\left|0\right\rangle_{2}+e^{i\lambda_{2}t}\left|1\right\rangle_{2}\right)\left(\left|0\right\rangle_{3}+e^{i\lambda_{3}t}\left|1\right\rangle_{3}\right)...\left(\left|0\right\rangle_{n}+e^{i\lambda_{n}t}\left|1\right\rangle_{n}\right)\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}\right.$
$\displaystyle-
id\left.\left|e\right\rangle\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\left|e^{i\lambda_{1^{\prime}}t}\alpha\right\rangle_{1^{\prime}}\left|e^{i\lambda_{2^{\prime}}t}\alpha\right\rangle_{2^{\prime}}..\left|e^{i\lambda_{n^{\prime}}t}\alpha\right\rangle_{n^{\prime}}\right].$
(20)

Now let us set

$\left|\lambda_{2}\right|=\left|\lambda_{3}\right|...=\left|\lambda_{n}\right|=\lambda,\text{
}\left|\lambda_{1^{\prime}}\right|=\left|\lambda_{2^{\prime}}\right|...=\left|\lambda_{n^{\prime}}\right|=\lambda.$
(21)

Thus, for an interaction time $t=\pi/\lambda,$ the state (20) becomes

$\left|0\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\left(c\left|g\right\rangle\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}-id\left|e\right\rangle\left|-\alpha\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}\right).$
(22)

Then, the level spacings of the qutrit is adjusted back such that the qutrit
is decoupled from the $2n$ cavities.

Step (iv): Apply a classical driving field on resonance with cavity
$1^{\prime}$ to obtain a displacement operator $D\left(\alpha\right)$ in phase
space, which results in the transformation
$\left|\alpha\right\rangle_{1^{\prime}}\rightarrow\left|2\alpha\right\rangle_{1^{\prime}}$
and
$\left|-\alpha\right\rangle_{1^{\prime}}\rightarrow\left|0\right\rangle_{1^{\prime}}$.
Thus, we have from Eq. (22)

$\left|0\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\left(c\left|g\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}-id\left|e\right\rangle\left|0\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}\right).$
(23)

Step (v): Adjust the frequency of cavity $1^{\prime}$ to have it off resonant
with the $\left|g\right\rangle\leftrightarrow\left|e\right\rangle$ transition
of the qutrit. Meanwhile apply a classical pulse (with initial phase
$\phi=\pi$ and frequency
$\omega_{p}=\omega_{eg}+\widetilde{g}^{2}/\widetilde{\Delta}$) to the qutrit.
According to Eqs. (14) and (15), for $t=\pi/\left(2\Omega_{p}\right),$ we have
$\left|e\right\rangle\left|0\right\rangle_{1^{\prime}}\rightarrow
ie^{i\widetilde{\omega}t}\left|g\right\rangle\left|0\right\rangle_{1^{\prime}}$
while the state $\left|g\right\rangle\left|2\alpha\right\rangle_{1^{\prime}}$
becomes
$e^{i\widetilde{\omega}t}\left|g\right\rangle\left|e^{i2\widetilde{\omega}t}2\alpha\right\rangle_{1^{\prime}}$.
Thus, we have from Eq. (23)

$e^{i\widetilde{\omega}t}\left|g\right\rangle\left|0\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\left\\{c\left|e^{i2\widetilde{\omega}t}2\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}+d\left|0\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}\right\\}.$
(24)

If the operation time of this step satisfies $2\widetilde{\omega}t=2m\pi$ ($m$
is an integer), we have $e^{i2\widetilde{\omega}t}=1.$ Thus, the state (24)
becomes

$\left|g\right\rangle\left|0\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\left(c\left|2\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}+d\left|0\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}\right),$
(25)

where we have dropped off the global phase $e^{i\widetilde{\omega}t}$. After
this step of operation, the frequency of cavity $1^{\prime}$ is adjusted such
that cavity $1^{\prime}$ is decoupled from the qutrit.

Step (vi): Apply a classical driving field on resonance with cavity
$1^{\prime},$ to obtain a displacement operator $D\left(-\alpha\right)$
resulting in the transformation
$\left|2\alpha\right\rangle_{1^{\prime}}\rightarrow\left|\alpha\right\rangle_{1^{\prime}}$
and
$\left|0\right\rangle_{1^{\prime}}\rightarrow\left|-\alpha\right\rangle_{1^{\prime}}$.
Hence, we have from Eq. (25)

$\left|g\right\rangle\left|0\right\rangle_{1}\left|-\right\rangle_{2}\left|-\right\rangle_{3}...\left|-\right\rangle_{n}\otimes\left(c\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}+d\left|-\alpha\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}\right),$
(26)

which shows that the $n$ CS qubits ($1^{\prime},2^{\prime},...,n^{\prime}$)
are in the entangled state
$c\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}+d\left|-\alpha\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}},$
i.e., the original entangled state (16) of the $n$ SPS qubits has been
transferred onto the $n$ CS qubits after the above operations. Here, note that
in Eq. (26), the two logic states of CS qubit $j^{\prime}$
($j^{\prime}=1^{\prime},2^{\prime},...,n^{\prime}$) are encoded with two
coherent states $\left|\alpha\right\rangle$ and $\left|-\alpha\right\rangle$
of cavity $j^{\prime}.$

To ensure that each step described above is performed by unitary operations,
the two wave vectors in each bracket of Eqs. (24-26) should be orthogonal.
Note that this requirement can be met for a large enough $\alpha.$ The reason
for this is as follows. Because of
$\left|\left\langle\alpha\right|\left.-\alpha\right\rangle\right|=$
$\left|\left\langle
2\alpha\right|\left.0\right\rangle\right|=e^{-2\left|\alpha\right|^{2}},$ the
two coherent states $\left|\alpha\right\rangle$ and
$\left|-\alpha\right\rangle$ or the two states $\left|0\right\rangle$ and
$\left|2\alpha\right\rangle$ involved in Eqs. (24-26) can be quasi-orthogonal
for a sufficiently large $\alpha,$ as mentioned in the introduction.

Several other points need to be made as follows:

(i) As shown above, there is no need of measurement for the state transfer,
i.e., all basic operations are unitary. Thus, by performing reverse unitary
operations, the entangled state
$c\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}..\left|\alpha\right\rangle_{n^{\prime}}+d\left|-\alpha\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}..\left|-\alpha\right\rangle_{n^{\prime}}$
of $n$ CS qubits can be converted back into the entangled state
$c\left|0\right\rangle_{1}\left|0\right\rangle_{2}...\left|0\right\rangle_{n}+d\left|1\right\rangle_{1}\left|1\right\rangle_{2}...\left|1\right\rangle_{n}$
of $n$ SPS qubits when the dissipation of the system is negligible.

(ii) The total operation time for the state transfer is

$t_{\mathrm{op}}=\pi/\lambda+\pi/\left(2g_{r}\right)+\pi/\left(2\Omega_{p}\right)+\tau_{p}+2\tau_{\alpha}+4\tau_{d}+2\tau_{c},$
(27)

which does not depend on the number of qubits involved in the entangled state.
Here, $\tau_{p}$ is the typical time for converting the states
$\left|0\right\rangle$ and $\left|1\right\rangle$ of each of cavities
($2,3,...,n$) into $\left|\pm\right\rangle$, $\tau_{\alpha}$ is the typical
time for performing a displacement operation on the cavity’s coherent state,
$\tau_{d}$ is the typical time required for adjusting the level spacings of
the coupler qutrit, and $\tau_{c}$ is the typical time for adjusting the
cavity frequency.

(iii) The state transfer requires adjusting the qutrit’s level spacings to
have the qutrit coupled with (or decoupled from) the cavities. For
superconducting devices, their level spacings can be rapidly (within 1–3 ns)
adjusted by varying external control parameters [33,34]. In addition, as shown
above, the frequency of cavity $1^{\prime}$ needs to be adjusted to have
cavity $1^{\prime}$ coupled with (or decoupled from) the qutrit. Note that for
superconducting microwave cavities, the cavity frequencies can be fast (within
a few nanoseconds) tuned in experiments [35,36].

(iv) Because of $\lambda_{j}=g_{j}^{2}/\Delta_{j}$ and
$\lambda_{j^{\prime}}=\mu_{j^{\prime}}^{2}/\Delta_{j^{\prime}},$ the condition
(21) turns out into

$\frac{g_{2}^{2}}{\left|\Delta_{2}\right|}=\frac{g_{3}^{2}}{\left|\Delta_{3}\right|}=...=\frac{g_{n}^{2}}{\left|\Delta_{n}\right|}=\frac{\mu_{1^{\prime}}^{2}}{\left|\Delta_{1^{\prime}}\right|}=\frac{\mu_{2^{\prime}}^{2}}{\left|\Delta_{2^{\prime}}\right|}=\frac{\mu_{3^{\prime}}^{2}}{\left|\Delta_{3^{\prime}}\right|}=...=\frac{\mu_{n^{\prime}}^{2}}{\left|\Delta_{n^{\prime}}\right|}.$
(28)

Due to $\Delta_{j}=\omega_{fg}-\omega_{c_{j}}$ $\left(2,3,...,n\right)$ and
$\Delta_{j^{\prime}}=\omega_{fe}-\omega_{c_{j^{\prime}}}$
$\left(1^{\prime},2^{\prime},...,n^{\prime}\right),$ the equality given in Eq.
(28) can be easily established by carefully selecting $\Delta_{j}$ (via
adjusting the frequency $\omega_{c_{j}}$ of cavity $j$) and carefully
selecting $\Delta_{j^{\prime}}$ (via adjusting the frequency
$\omega_{c_{j^{\prime}}}$ of cavity $j^{\prime}$).

(v) As described above, for step (v), the operation time should satify both
$t=\pi/\left(2\Omega_{p}\right)$ and $2\widetilde{\omega}t=2m\pi,$ i.e.,

$\Omega_{p}=\frac{\widetilde{g}^{2}}{4m\widetilde{\Delta}},$ (29)

which can be easily met by adjusting the Rabi frequency of the pulse.

The present proposal works out for either $c=d$ or $c\neq d$. Note that the
case of $c=d$ is specially interesting, because Bell states and GHZs are of
great importance in quantum communication and QIP. It should be mentioned that
how to directly prepare the initial entangled state (16) with $c=d$ was
previously reported [37]. Alternatively, the state (16) with $c=d$ can be
created by preparing $n$ natural or artificial atoms (distributed in $n$
cavities) in a maximally entangled state
$\left|g\right\rangle_{1}\left|g\right\rangle_{2}...\left|g\right\rangle_{n}+\left|e\right\rangle_{1}\left|e\right\rangle_{2}...\left|e\right\rangle_{n}$
[38-42] and then having each atom (in the excited state
$\left|e\right\rangle$) emitting a single photon into its cavity through local
operations on each atom and its cavity.

##  IV Possible experimental implementation

Figure 2: (color online) (a) Schematic circuit of four 3D microwave cavities
coupled to a superconducting flux qutrit. The qutrit consists of three
Josephson junctions and a superconducting loop. (b) Illustration of cavity $2$
off resonant with the
$\left|g\right\rangle\leftrightarrow\left|f\right\rangle$ transition of the
qutrit, while cavities $1^{\prime}$ and $2^{\prime}$ off resonant with the
$\left|e\right\rangle\leftrightarrow\left|f\right\rangle$ transition of the
qutrit. (c) Illustration of cavity $1^{\prime}$ and a microwave pulse off
resonant with the $\left|g\right\rangle\leftrightarrow\left|e\right\rangle$
transition of the qutrit. The unwanted couplings or interactions of the
qutrit’s energy-level transitions with the cavities or the pulse are not
illustrated in (b) and (c) for simplicity, which however are considered in our
numerical simulation, as described in the Hamiltonians (30) and (32).

This section is aimed at investigating the experimental feasibility for
transferring a Bell state of two SPS qubits $(1,2)$ onto two CS qubits
$\left(1^{\prime},2^{\prime}\right)$, by considering four 3D microwave
cavities coupled to a superconducting flux qutrit (Fig. 2a). In the following,
we will briefly discuss the fidelity for the Bell state transfer, which is
performed in a more realistic situation. Since steps (i), (ii), (iv) and (vi)
employ only the resonant interactions, these steps can be completed within a
very short time. In this sense, the influence of the qutrit decoherence, the
cavity decay, and the inter-cavity crosstalk is negligibly small for steps
(i), (ii), (iv) and (vi). Therefore, the imperformance of the Bell state
transfer is mainly caused by the system dissipation and the inter-cavity
crosstalk during the operation of steps (iii) and (v), because these steps
require reletively long operation times due to the use of the qutrit-cavity
off-resonant interaction.

After taking into account the unwanted couplings between the qutrit’s level
transitions and the cavities as well as the inter-cavity crosstalk, the
Hamiltonian $H_{2}$ in Eq. (4), with $n=2$ and $n^{\prime}=2^{\prime}$ for the
present case, is modified as

$\displaystyle H_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\left(g_{2}e^{i\Delta_{2}t}\hat{a}_{2}\sigma_{fg}^{+}+g_{2}^{\prime}e^{i\Delta_{2}^{\prime}t}\hat{a}_{2}\sigma_{fe}^{+}+g_{2}^{\prime\prime}e^{i\Delta_{2}^{\prime\prime}t}\hat{a}_{2}\sigma_{eg}^{+}+\text{h.c.}\right)$
(30)
$\displaystyle+\left(\mu_{1^{\prime}}e^{i\Delta_{1^{\prime}}t}\hat{b}_{1^{\prime}}\sigma_{fe}^{+}+\mu_{1^{\prime}}^{\prime}e^{i\Delta_{1^{\prime}}^{\prime}t}\hat{b}_{1^{\prime}}\sigma_{fg}^{+}+\mu_{1^{\prime}}^{\prime\prime}e^{i\Delta_{1^{\prime}}^{\prime\prime}t}\hat{b}_{1^{\prime}}\sigma_{eg}^{+}+\text{h.c.}\right)$
$\displaystyle+\left(\mu_{2^{\prime}}e^{i\Delta_{2^{\prime}}t}\hat{b}_{2^{\prime}}\sigma_{fe}^{+}+\mu_{2^{\prime}}^{\prime}e^{i\Delta_{2^{\prime}}^{\prime}t}\hat{b}_{2^{\prime}}\sigma_{fg}^{+}+\mu_{2^{\prime}}^{\prime\prime}e^{i\Delta_{2^{\prime}}^{\prime\prime}t}\hat{b}_{2^{\prime}}\sigma_{eg}^{+}+\text{h.c.}\right)$
$\displaystyle+\varepsilon,$

where the second (third) term in line one describes the unwanted coupling
between cavity 2 and the $\left|e\right\rangle\rightarrow\left|f\right\rangle$
$\left(\left|g\right\rangle\rightarrow\left|e\right\rangle\right)$ transition
with coupling constant $g_{2}^{\prime}$ ($g_{2}^{\prime\prime}$) and detuning
$\Delta_{2}^{\prime}=\omega_{fe}-\omega_{c_{2}}$
($\Delta_{2}^{\prime\prime}=\omega_{eg}-\omega_{c_{2}}$), the second (third)
term in line two describes the unwanted coupling between cavity $1^{\prime}$
and the $\left|g\right\rangle\rightarrow\left|f\right\rangle$
$\left(\left|g\right\rangle\rightarrow\left|e\right\rangle\right)$ transition
with coupling constant $\mu_{1^{\prime}}^{\prime}$
($\mu_{1^{\prime}}^{\prime\prime}$) and detuning
$\Delta_{1^{\prime}}^{\prime}=\omega_{fg}-\omega_{c_{1^{\prime}}}$
($\Delta_{1^{\prime}}^{\prime\prime}=\omega_{eg}-\omega_{c_{1^{\prime}}}$),
and the second (third) term in line three describes the unwanted coupling
between cavity $2^{\prime}$ and the
$\left|g\right\rangle\rightarrow\left|f\right\rangle$
$\left(\left|g\right\rangle\rightarrow\left|e\right\rangle\right)$ transition
with coupling constant $\mu_{2^{\prime}}^{\prime}$
($\mu_{2^{\prime}}^{\prime\prime}$) and detuning
$\Delta_{2^{\prime}}^{\prime}=\omega_{fg}-\omega_{c_{2^{\prime}}}$ (
$\Delta_{2^{\prime}}^{\prime\prime}=\omega_{eg}-\omega_{c_{2^{\prime}}}$). In
addition, $\varepsilon$ is the Hamiltonian describing the inter-cavity
crosstalk, which is given by

$\displaystyle\varepsilon$ $\displaystyle=$
$\displaystyle\left(g_{12}e^{i\Delta_{12}t}\hat{a}_{1}^{+}\hat{a}_{2}+\text{h.c.}\right)+\left(g_{1^{\prime}2^{\prime}}e^{i\Delta_{1^{\prime}2^{\prime}}t}\hat{b}_{1^{\prime}}^{+}\hat{b}_{2^{\prime}}+\text{h.c.}\right)$
(31)
$\displaystyle+\left(g_{11^{\prime}}e^{i\Delta_{11^{\prime}}t}\hat{a}_{1}^{+}\hat{b}_{1^{\prime}}+\text{h.c.}\right)+\left(g_{12^{\prime}}e^{i\Delta_{12^{\prime}}t}\hat{a}_{1}^{+}\hat{b}_{2^{\prime}}+\text{h.c.}\right)$
$\displaystyle+\left(g_{21^{\prime}}e^{i\Delta_{21^{\prime}}t}\hat{a}_{2}^{+}\hat{b}_{1^{\prime}}+\text{h.c.}\right)+\left(g_{22^{\prime}}e^{i\Delta_{22^{\prime}}t}\hat{a}_{2}^{+}\hat{b}_{2^{\prime}}+\text{h.c.}\right),$

where $g_{kl}$ is the crosstalk strength between the two cavities $k$ and $l$
while $\Delta_{kl}=\omega_{c_{k}}-\omega_{c_{l}}$ is the frequency difference
of the two cavities $k$ and $l$
$\left(k,l\in\\{1,2,1^{\prime},2^{\prime}\\};k\neq l\right)$.

On the other hand, when the unwanted couplings between the qutrit’s level
transitions and cavity $1^{\prime}$ are considered, the Hamiltonian $H_{3}$ in
Eq. (9) is modified as

$\displaystyle H_{3}^{\prime}$ $\displaystyle=$
$\displaystyle\left(\widetilde{g}e^{i\widetilde{\Delta}t}\hat{a}_{1^{\prime}}\sigma_{eg}^{+}+\widetilde{g}^{\prime}e^{i\widetilde{\Delta}^{\prime}t}\hat{a}_{1^{\prime}}\sigma_{fg}^{+}+\widetilde{g}^{\prime\prime}e^{i\widetilde{\Delta}^{\prime\prime}t}\hat{a}_{1^{\prime}}\sigma_{fe}^{+}+\text{h.c.}\right)$
$\displaystyle+\left(\Omega_{p}e^{-i\left[\left(\omega_{p}-\omega_{eg}\right)t+\phi\right]}\sigma_{eg}^{+}+\Omega_{p}^{\prime}e^{-i\left[\left(\omega_{p}-\omega_{fg}\right)t+\phi\right]}\sigma_{fg}^{+}+\Omega_{p}^{\prime\prime}e^{-i\left[\left(\omega_{p}-\omega_{fe}\right)t+\phi\right]}\sigma_{fe}^{+}+\text{h.c.}\right)\text{,}$

where the second (third) term in line one describes the unwanted coupling
between cavity $1^{\prime}$ and the
$\left|g\right\rangle\rightarrow\left|f\right\rangle$
$\left(\left|e\right\rangle\rightarrow\left|f\right\rangle\right)$ transition
with coupling constant $\tilde{g}^{\prime}$ ($\tilde{g}^{\prime\prime}$) and
detuning
$\widetilde{\Delta}^{\prime}=\omega_{fg}-\widetilde{\omega}_{c_{1^{\prime}}}$
($\widetilde{\Delta}^{\prime\prime}=\omega_{fe}-\widetilde{\omega}_{c_{1^{\prime}}}$),
while the second (third) term in line two describes the unwanted coupling
between the microwave pulse and the
$\left|g\right\rangle\rightarrow\left|f\right\rangle$
$\left(\left|e\right\rangle\rightarrow\left|f\right\rangle\right)$ transition
with Rabi frequency $\Omega_{p}^{\prime}$ ($\Omega_{p}^{\prime\prime}$). For
simplicity, we here assume that during the operation of step (v), all other
cavities are decoupled from the qutrit and the inter-cavity crosstalk is
negligible, which can be achieved by adjusting the cavity frequencies such
that the cavity-cavity frequency differences are sufficiently large.

During the operation of steps (iii) and (v), the dynamics of the lossy system
is determined by

$\displaystyle\frac{d\rho}{dt}$ $\displaystyle=$
$\displaystyle-i\left[H_{\mathrm{i}}^{\prime},\rho\right]+\sum_{j=1}^{2}\kappa_{j}\mathcal{L}\left[\hat{a}_{j}\right]+\sum_{j^{\prime}=1^{\prime}}^{2^{\prime}}\kappa_{j^{\prime}}\mathcal{L}\left[\hat{b}_{j^{\prime}}\right]$
(33)
$\displaystyle+\gamma_{eg}\mathcal{L}\left[\sigma_{eg}^{-}\right]+\gamma_{fe}\mathcal{L}\left[\sigma_{fe}^{-}\right]+\gamma_{fg}\mathcal{L}\left[\sigma_{fg}^{-}\right]$
$\displaystyle+\gamma_{e,\varphi}\left(\sigma_{e}\rho\sigma_{e}-\sigma_{e}\rho/2-\rho\sigma_{e}/2\right)$
$\displaystyle+\gamma_{f,\varphi}\left(\sigma_{f}\rho\sigma_{f}-\sigma_{f}\rho/2-\rho\sigma_{f}/2\right),$

where $H_{\mathrm{i}}^{\prime}$ is the modified Hamiltonian $H_{2}^{\prime}$
or $H_{3}^{\prime}$ given above,
$L\left[\Lambda\right]=\Lambda\rho\Lambda^{+}-\Lambda^{+}\Lambda\rho/2-\rho\Lambda^{+}\Lambda/2$
(with
$\Lambda=\hat{a}_{j},,\hat{b}_{j^{\prime}},\sigma_{eg}^{-},\sigma_{fe}^{-},\sigma_{fg}^{-})$,
$\sigma_{eg}^{-}=\left|g\right\rangle\left\langle e\right|,$
$\sigma_{fe}^{-}=\left|e\right\rangle\left\langle f\right|,$
$\sigma_{fg}^{-}=\left|g\right\rangle\left\langle f\right|,$
$\sigma_{e}=\left|e\right\rangle\left\langle e\right|$, and
$\sigma_{f}=\left|f\right\rangle\left\langle f\right|.$ In addition,
$\kappa_{j}$ ($\kappa_{j^{\prime}}$) is the decay rate of cavity $j$
($j^{\prime}$); $\gamma_{eg}$ is the energy relaxation rate for the level
$\left|e\right\rangle$ of the qutrit corresponding to the decay path
$\left|e\right\rangle\rightarrow\left|g\right\rangle$; $\gamma_{fe}$
($\gamma_{fg}$) is the relaxation rate for the level $\left|f\right\rangle$ of
the qutrit associated with the decay path
$\left|f\right\rangle\rightarrow\left|e\right\rangle$
($\left|f\right\rangle\rightarrow\left|g\right\rangle$); $\gamma_{e,\varphi}$
($\gamma_{f,\varphi}$) is the dephasing rate of the level
$\left|e\right\rangle$ ($\left|f\right\rangle$) of the qutrit.

For two SPS qubits, the Bell state is given by Eq. (16) with $c=d=1/\sqrt{2},$
$n=2$ and $n^{\prime}=2^{\prime},$ i.e., the Bell state is
$\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\right),$
where the subscripts 1 and 2 represent SPS qubits 1 and 2, respectively. The
initial state of the whole system is
$\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\right)\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}\left|g\right\rangle,$
where the subscripts $1^{\prime}$ and $2^{\prime}$ represent CS qubits
$1^{\prime}$ and $2^{\prime}$, respectively. According to Eq. (26), the ideal
output state of the whole system after the entire operation is

$\left|g\right\rangle\left|0\right\rangle_{1}\left|-\right\rangle_{2}\otimes\frac{1}{\sqrt{2}}\left(\left|\alpha\right\rangle_{1^{\prime}}\left|\alpha\right\rangle_{2^{\prime}}+\left|-\alpha\right\rangle_{1^{\prime}}\left|-\alpha\right\rangle_{2^{\prime}}\right).$
(34)

The fidelity of the entire operation is defined as
$F=\sqrt{\left\langle\psi_{id}\right|\rho\left|\psi_{id}\right\rangle},$ where
$\left|\psi_{id}\right\rangle$ is the ideal output state given by Eq. (34),
while $\rho$ is the final density matrix obtained by numerically solving the
master equation in a realistic situation.

For a flux qutrit, the typical transition frequency between neighboring levels
can be made as 1 to 20 GHz. As an example, consider $\omega_{eg}/2\pi=8.0$
GHz, $\omega_{fe}/2\pi=12.0$ GHz, and $\omega_{fg}/2\pi=20.0$ GHz. Choose
$\omega_{c_{1}}/2\pi=5.0$ GHz for cavity $1$. With a choice of
$\Delta_{2}/2\pi=-250$ MHz, $\Delta_{1^{\prime}}/2\pi=-125$ MHz, and
$\Delta_{2^{\prime}}/2\pi=125$ MHz, the cavity frequencies are
$\omega_{c_{2}}/2\pi=20.25$ GHz, $\omega_{c_{1^{\prime}}}/2\pi=12.125$ GHz,
and $\omega_{c_{2^{\prime}}}/2\pi=11.875$ GHz. Thus, we have
$\Delta_{2}^{\prime}/2\pi=-8.25$ GHz, $\Delta_{2}^{\prime\prime}/2\pi=-12.25$
GHz, $\Delta_{1^{\prime}}^{\prime}/2\pi=7.875$ GHz,
$\Delta_{1^{\prime}}^{\prime\prime}/2\pi=-4.125$ GHz,
$\Delta_{2^{\prime}}^{\prime}/2\pi=8.125$ GHz, and
$\Delta_{2^{\prime}}^{\prime\prime}/2\pi=-3.875$ GHz. For the values of the
cavity frequencies here, a simple calculation gives $\Delta_{12}/2\pi=-15.25$
GHz, $\Delta_{1^{\prime}2^{\prime}}/2\pi=0.25$ GHz,
$\Delta_{11^{\prime}}/2\pi=-7.125$ GHz, $\Delta_{12^{\prime}}/2\pi=-6.875$
GHz, $\Delta_{21^{\prime}}/2\pi=8.125$ GHz, and
$\Delta_{22^{\prime}}/2\pi=8.375$ GHz. In addition, we have
$\mu_{1^{\prime}}=$ $\mu_{2^{\prime}}=g_{2}/\sqrt{2}$ according to Eq. (28).

We choose $\widetilde{\Delta}/2\pi=160$ MHz, resulting in
$\widetilde{\omega}_{c_{1^{\prime}}}/2\pi=7.84$ GHz. For the value of
$\widetilde{\omega}_{c_{1^{\prime}}}$ here and the values of $\omega_{eg}$,
$\omega_{fe}$, $\omega_{fg}$ given above, we have
$\widetilde{\Delta}^{\prime}/2\pi=12.16$ GHz and
$\widetilde{\Delta}^{\prime\prime}/2\pi=4.16$ GHz. We choose
$\widetilde{g}/2\pi=24.1$ MHz, and thus $\Omega_{p}/2\pi\sim 0.454$ MHz
(calculated according to Eq. (29) for $m=1$ and $\alpha=1.86$). Because of
$\omega_{p}=\omega_{eg}+\widetilde{g}^{2}/\widetilde{\Delta},$ the driving
field frequency $\omega_{p}$ can be calculated given the values of
$\omega_{eg},$ $\tilde{g},$ and $\tilde{\Delta}$ here.

The dipole matrix element $\chi_{fg}$ between the two levels
$\left|g\right\rangle$ and $\left|f\right\rangle,$ the dipole matrix element
$\chi_{fe}$ between the two levels $\left|e\right\rangle$ and
$\left|f\right\rangle,$ and the dipole matrix element $\chi_{eg}$ between the
two levels $\left|g\right\rangle$ and $\left|e\right\rangle$ are assumed to
have the relationship of $\chi_{fg}\sim\chi_{fe}\sim 4\chi_{eg}$, which can be
made with appropriate design of the qutrit system [43]. Thus, we choose
$g_{2}^{\prime}\sim g_{2},$ $g_{2}^{\prime\prime}\sim 0.25g_{2};$
$\mu_{1^{\prime}}^{\prime}\sim\mu_{1^{\prime}},$
$\mu_{1^{\prime}}^{\prime\prime}\sim 0.25\mu_{1^{\prime}};$
$\mu_{2^{\prime}}^{\prime}\sim\mu_{2^{\prime}},$
$\mu_{2^{\prime}}^{\prime\prime}\sim 0.25\mu_{2^{\prime}}$;
$\widetilde{g}^{\prime}\sim 4\widetilde{g},$
$\widetilde{g}^{{}^{\prime\prime}}\sim 4\widetilde{g};$ and
$\Omega_{p}^{\prime}\sim 4\Omega_{p},$ $\Omega_{p}^{{}^{\prime\prime}}\sim
4\Omega_{p}.$

Other parameters used in the numerical simulation are: (i)
$\gamma_{eg}^{-1}=4T$ $\mu$s, $\gamma_{fe}^{-1}=2T$ $\mu$s,
$\gamma_{fg}^{-1}=T$ $\mu$s, (ii) $\gamma_{\phi e}^{-1}=\gamma_{\phi
f}^{-1}=T$ $\mu$s, (iii)
$\kappa_{1},\kappa_{2},\kappa_{1^{\prime}},\kappa_{2^{\prime}}=\kappa,$ and
(iv) $\alpha=1.86$. In addition, choose $g_{kl}=0.01g_{m}$ [14], with
$g_{m}=\max\\{g_{2},\mu_{1^{\prime}},\mu_{2^{\prime}}\\}.$

Figure 3: (color online) Fidelity versus $g_{2}$. The parameters used in the
numerical simulation are referred to the text. Figure 4: (color online)
Fidelity versus $\kappa^{-1}$. The parameters used in the numerical simulation
are referred to the text.

By numerically solving the master equation (33), Fig. 3 is plotted, which
illustrates the fidelity versus $g_{2}$ for $\kappa^{-1}=100$ $\mu$s and
$T=10$ $\mu$s, $15$ $\mu$s, $30$ $\mu$s. Note that all other coupling
constants can be determined for a given $g_{2}$, as described above. In Fig.
3, the optimal values of $g_{2}$, corresponding to the peaks of three curve
lines, are $2\pi\times 12.03$ MHz for $T=10$ $\mu$s, $15$ $\mu$s and $30$
$\mu$s. Next, we plot Fig. 4, which shows the fidelity versus $\kappa^{-1}$
for $g_{2}=2\pi\times 12.03$ MHz and $T=10$ $\mu$s, $15$ $\mu$s, $30$ $\mu$s.
From Fig. 4, one can see that the fidelity increases with $\kappa^{-1}$ and
$T$, and for $T\geq 15$ $\mu$s and $\kappa^{-1}\geq 100$ $\mu$s, the fidelity
exceeds $90.2\%$, impling the high-fidelity Bell-state transfer from two SPS
qubits onto two CS qubits. It should be mentioned that when taking into
account the qutrit decoherence, the cavity decay, and the inter-cavity
crosstalk in the operations of steps (i), (ii), (iv) and (vi), the fidelity
would be slightly decreased but will not be greatly decreased since these
steps of operation can be completed very fast due to the use of the resonant
interactions.

The coupling constants are readily achievable because a coupling constant
$\sim 2\pi\times 636$ MHz has been reported for a flux device coupled to a
microwave cavity [44]. For the parameters chosen above, the state transfer
time is estimated as $\sim 1.41$ $\mu$s, which is much shorter than the
decoherence times of the coupler qutrit ($10$ $\mu$s $-$ $120$ $\mu$s) and the
cavity decay times ($50$ $\mu$s $-$ $150$ $\mu$s) considered in Figs. 3 and 4.
For decoherence time of the flux qutrit, a rather conservative case is
considered in the numerical simulation, because decoherence time 70 $\mu$s to
1 ms for a superconducting flux device has been reported in experiments
[45,46]. For $\kappa^{-1}=100$ $\mu$s and the cavity frequencies
$\omega_{c_{1}},\omega_{c_{2}},\omega_{c_{1^{\prime}}},\omega_{c_{2^{\prime}}}$
given above, a simple calculation gives $Q_{1}\sim 3.14\times 10^{6}$ for
cavity $1,$ $Q_{2}\sim 1.27\times 10^{7}$ for cavity $2,$ $Q_{1^{\prime}}\sim
7.61\times 10^{6}$ for cavity $1^{\prime},$ and $Q_{2^{\prime}}\sim 7.46\times
10^{6}$ for cavity $2^{\prime}.$ Note that a high quality factor $Q=3.5\times
10^{7}$ of a 3D microwave cavity has been experimentally demonstrated [47].
Our analysis here shows that the high-fidelity transfer of the Bell state of
two SPS qubits onto two CS qubits is feasible within the current circuit QED
technology.

##  V Conclusion

We have explicitally shown how to transfer partially- or maximally-entangled
states (including Bell states and GHZ states) of $n$ single-photon-state (SPS)
qubits onto $n$ coherent-state (CS) qubits in a circuit-QED system. As shown
above, this proposal has these distinguishing features: (i) Only a coupler
qutrit is needed, hence the circuit hardware resources is significantly
reduced; (ii) The operation time does not increase with the increasing of the
number of qubits; (iii) Decoherence from the coupler qutrit is greatly
suppressed, because the higher energy level of the coupler qutrit is not
excited during the entire operation; and (iv) When the dissipation is
negligible, the state transfer is deterministic because of no measurement
being performed. Furthermore, we have numerically investigated the
experimental feasibility for transferring the Bell state of two SPS qubits
onto two CS qubits. The numerical simulation shows that high-fidelity transfer
of the Bell state, from two SPS qubits onto two CS qubits, can be achieved
with current circuit-QED technology.

Finally, it should be mentioned that since all basic operations are unitary
and measurement is not involved, the entangled states of SPS qubits can be
transferred back onto CS qubits by performing reverse operations when the
dissipation is negligible. This proposal can be applied to transfer maximally-
or partially-entangled states between $n$ SPS qubits and $n$ CS qubits in a
wide range of physical systems, such as $2n$ microwave or optical cavities
coupled to a natural or artificial atom.

## Acknowledgements

This work was partly supported by the National Natural Science Foundation of
China (NSFC) (11074062, 11374083, 11774076), the Key-Area Research and
Development Program of GuangDong province (2018B030326001), and the NKRDP of
China (2016YFA0301802).

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