Entanglement concentration with different measurement in a 3-mode optomechanical system

TL;DR

This paper investigates how different phonon counting measurements in a 3-mode optomechanical system affect entanglement, showing that more precise measurements lead to greater entanglement enhancement.

Contribution

It compares the effects of imperfect, perfect, and on-off measurements on entanglement concentration in a 3-mode optomechanical system.

Findings

Perfect measurement yields the largest entanglement increase.

Imperfect measurement also enhances entanglement, but less than perfect.

On-off measurement shows less clear entanglement enhancement.

Abstract

In this work, we perform a series of phonon counting measurement with different methods in a 3-mode optomechanical system, and we compare the difference of the entanglement after measurement. In this article we focus on the two cases: imperfect measurement and on-off measurement. We find that whatever measurement you take, the entanglement will increase. The size of entanglement enhancement is the largest in perfect measurement, second in the imperfect measurement, and it is not obvious in the on-off measurement. We are sure that the more precise measurement information, the larger entanglement concentration.

Equations

$$\hat{H}={\omega }_{\mathbf{m}}{\hat{b}}^{†}\hat{b}+\sum _{j=1,2}\left[{\omega }_j{\hat{a}}_j^{†}{\hat{a}}_j+g_j\left({\hat{b}}^{†}+\hat{b}\right){\hat{a}}_j^{†}{\hat{a}}_j\right],$$

$${\hat{H}}_{\mathbf{int}}=\left(G_1{\hat{b}}^{†}{\hat{d}}_1+G_2\hat{b}{\hat{d}}_2\right)+h.c.$$

$$\frac{d}{dt}\hat{b}=-\frac{\gamma }{2}\hat{b}-\mathbf{i}(G_1{\hat{d}}_1+G_2{\hat{d}}_2^{†})-\sqrt{\gamma }{\hat{b}}^{\mathbf{in}}(t),$$

$$\frac{d}{dt}{\hat{d}}_1=-\frac{{\kappa }_1}{2}{\hat{d}}_1-\mathbf{i}{G}_1\hat{b}-\sqrt{{\kappa }_1}{\hat{d}}_1^{\mathbf{in}}(t),$$

$$\frac{d}{dt}{\hat{d}}_2^{†}=-\frac{{\kappa }_2}{2}{\hat{d}}_2^{†}+\mathbf{i}{G}_2\hat{b}-\sqrt{{\kappa }_2}{\hat{d}}_2^{\mathbf{in},†}(t),$$

$$|\Psi ⟩=\sum _{p,q}\frac{\sqrt{C_{p+q}^p}{N}_{\mathbf{m}}^{\frac{q}{2}}{N}_1^{\frac{p}{2}}}{{\left(1+N_2\right)}^{\left(p+q+1\right)/2}}|p,p+q,q⟩.$$

$$N_1=\frac{4C_1{C}_2}{{\left(1+C_1-C_2\right)}^2},$$

$$N_2=\frac{4C_2\left(C_1+1\right)}{{\left(1+C_1-C_2\right)}^2},$$

$$N_{\mathbf{m}}=\frac{4C_2}{{\left(1+C_1-C_2\right)}^2},$$

$$E_N=\ln \frac{{\left(1+C_1-C_2\right)}^2}{A+B+2C_2\left(1+2C_1\right)-4\sqrt{AB}},$$

$${\hat{M}}_1\left(q\right)=|q⟩⟨q|,$$

$${\hat{M}}_{\mu }\left(q\right)=\sum _{n=q}^{\infty }\sqrt{C_n^q}{\left(1-\mu \right)}^{\frac{n-q}{2}}{\mu }^{\frac{q}{2}}|n-q⟩⟨q|.$$

$${\hat{M}}_{\mathbf{off}}=|0⟩⟨0|,$$

$${\hat{M}}_{\mathbf{on}}=\mathbf{I}-|0⟩⟨0|=\sum _{k=1}^{\infty }|k⟩⟨k|.$$

$$|{\Psi }_q⟩=P_q^{-\frac{1}{2}}\widehat{M_1}\left(q\right)|\Psi ⟩=\sum _{p=0}^{\infty }\sqrt{f_p\left(q\right)}|p,p+q⟩.$$

$$f_p\left(q\right)=C_{p+q}^p{\zeta }^p{\left(1-\zeta \right)}^{1+q},$$

$$P_q=⟨\Psi |{\hat{M}}_1^{†}\left(q\right){\hat{M}}_1\left(q\right)|\Psi ⟩=\frac{N_{\mathbf{m}}^q}{{\left(1+N_{\mathbf{m}}\right)}^{1+q}}.$$

$$E_N\equiv \ln {\Vert {\hat{\rho }}^{\mathbf{T}}\Vert }_1,$$

$$E_N\left(q\right)=2\ln \sum _{p=0}^{\infty }\sqrt{f_p\left(q\right)}.$$

$$E_N\left(0\right)=\ln \frac{1+\sqrt{\zeta }}{1-\sqrt{\zeta }}.$$

$$f_p\left(q\right)\approx \frac{1}{\sqrt{2\pi }\sigma \left(q\right)}{e}^{-\frac{{\left[p-\kappa \left(q\right)\right]}^2}{2\sigma {\left(q\right)}^2}}$$

$$E_N\left(q\right)\approx \ln \frac{\sqrt{8\pi \zeta \left(1+q\right)}}{1-\zeta }.$$

$${\hat{\rho }}_q=\frac{1}{P_{\mu }\left(q\right)}T{r}_q\left[{\hat{M}}_{\mu }\left(q\right)\hat{\rho }{\hat{M}}_{\mu }^{†}\left(q\right)\right],$$

$$P_{\mu }\left(q\right)=\sum _{s=q}^{\infty }{P}_s{C}_s^q{\mu }^q{\left(1-\mu \right)}^{s-q}.$$

$$\begin{array}{rl}{\hat{\rho }}_q& =\sum _{s=q}^{\infty }\eta \left(s\right)|{\Psi }_s⟩⟨{\Psi }_s|\\ & =\sum _{p_1=0}^{\infty }\sum _{,p_2=0}^{\infty }\sum _{s=q}^{\infty }\sqrt{f_{p_1}\left(s\right)f_{p_2}\left(s\right)}\eta \left(s\right)\\ & \times |p_1,p_1+s⟩⟨p_2,p_2+s|,\end{array}$$

$$\eta \left(s\right)=\frac{N_{\mathbf{m}}^{s-q}{\left(1+N_{\mathbf{m}}\mu \right)}^{1+q}}{{\left(1+N_{\mathbf{m}}\right)}^{1+s}}{C}_s^q{\left(1-\mu \right)}^{s-q}.$$

$$\begin{array}{rl}{\hat{\rho }}_q^{\mathbf{T}}& =\sum _{p_1=0}^{\infty }\sum _{,p_2=0}^{\infty }\sum _{s=q}^{\infty }\sqrt{f_{p_1}\left(s\right)f_{p_2}\left(s\right)}\eta \left(s\right)\\ & \times |p_1,p_2+s⟩⟨p_2,p_1+s|.\end{array}$$

$$Q\left[q+1\right]=\left[\begin{array}[]{cc}\sqrt{f_{1}\left(q\right)f_{0}\left(q\right)}\eta\left(q\right)&0\\ f_{0}\left(q+1\right)\eta\left(q+1\right)&\sqrt{f_{0}\left(q\right)f_{1}\left(q\right)}\eta\left(q\right)\end{array}\right].$$

$$\begin{array}{rl}{\hat{\rho }}_q^{\mathbf{T}}& =\sum _{p_1=0}^{\infty }\sum _{,p_2=0}^{\infty }\sum _{s=q}^{q+1}\sqrt{f_{p_1}\left(s\right)f_{p_2}\left(s\right)}\eta \left(s\right)\\ & \times |p_1,p_2+s⟩⟨p_2,p_1+s|.\end{array}$$

$$\begin{array}{rl}{\hat{H}}_0& =\sum _{p_1=0}^{\infty }\sum _{,p_2=0}^{\infty }\sqrt{f_{p_1}\left(q\right)f_{p_2}\left(q\right)}\eta \left(q\right)\\ & \times |p_1,p_2+q⟩⟨p_2,p_1+q|,\end{array}$$

$$\begin{array}{rl}{\hat{H}}^{{}^{\prime }}& =\sum _{p_1=0}^{\infty }\sum _{,p_2=0}^{\infty }\sqrt{f_{p_1}\left(q+1\right)f_{p_2}\left(q+1\right)}\eta \left(q+1\right)\\ & \times |p_1,p_2+q+1⟩⟨p_2,p_1+q+1|.\end{array}$$

$$e_m^{(1)}=⟨m|{\hat{H}}^{{}^{\prime }}|m⟩,$$

$$e_m^{(2)}=\sum _{k\ne m}\frac{{\left[⟨m|{\hat{H}}^{{}^{\prime }}|k⟩\right]}^2}{e_m^{\left(0\right)}-e_k^{\left(0\right)}}.$$

Full text

I introductions

In recent years, quantum entanglement RH-RMP-2009 (1) has been regarded as a key source in the quantum information processing, for it can apply in terms of quantum cryptography AKE-PRL-1991 (2, 3), allow the realization of quantum teleportation DB-Nature-1997 (4, 5) and quantum dense coding CHB-PRL-1992 (6). A number of strategies to generate entanglement have been developed in different quantum systems, such as trapped ions KM-PRL-1999 (7), cold atoms JQY-PRB-2003 (8) and solid-state qubit YDW-PRB-2010 (9). The conventional methods for entanglement photons rely on nonlinear optical process like parameter amplification and second harmonic generation. However, the photons with vastly different frequencies, i.e., microwave photons and optical photons, can not be entangled directly in this way. Nevertheless, optomechanical system MA-RMP-2014 (10) provides probability to work out this difficulty and some related work have already been done YDW-PRL-2012 (11, 12, 13, 14, 15, 16, 17, 18). An optomechanical system suitable to this purpose is based on an optical cavity and a microwave cavity interacting with a mechanical element and such a 3-mode optomechanical system has been realized experimentally RA-Nature-2014 (19, 20).

Entanglement will be severely degraded by the channel noise due to it’s fragile nature. In order to overcome that decoherence effect, entanglement concentration or entanglement distillation will be utilized. The idea of the standard entanglement concentration is to exact a smaller number of elements with higher entanglement by distillation from a large number of elements with lower entanglement through local operations and classical communication. From 1996 when Bennett et al CHB-PRL-1996 (21) proposed entanglement concentration protocol firstly to the present, the various entanglement concentration protocol for discrete-variable DD-PRL-1996 (22, 23) and continuous-variable LMD-PRL-2000 (24, 25) quantum system have been developed as well as entanglement concentrations have been demonstrated experimentally ZZ-PRL-2003 (26, 27). However, distilling continuous-variable entanglement appears to be significantly harder to achieve than distilling discrete-variable, for one can not distill a Gaussian state by using only Gaussian operations JE-PRL-2002 (25, 28, 29). Thus non-Gaussian operations, in particular photon counting measurement AK-PRA-2006 (30), are indispensable for Gaussian entangled states distillation. The photon subtraction strategy, one of the available experimental operations beyond the Gaussian regime, is based on this idea. And the non-Gaussian operation with photon counting measurement can be implemented by beam splitters TO-PRA-2000 (31, 32, 33).

These ideas motivate us to explore an entanglement concentration protocol based on phonon counting measurement for 3-mode optomechanical system (Fig. 1). In this system, a genuine tripartite entanglement state, where the two cavity output mode and the mechanical output mode are entangled with each other, can be generated YDW-PRA-2015 (18). We perform the phonon counting measurement in the mechanical mode (indirectly through auxiliary photon counting) for the genuine tripartite entanglement state with different methods. In previous work WM-SCPMA-2015 (34), the perfect measurement, i.e., projective measurement, have been considered, but in practice it is difficult to find a measurement device which completely satisfy projective measurement. In this paper, we mainly focus on and get the general result with imperfect measurement and on-off measurement, which is available experimentally at present. While the amount of entanglement after measurement is measured in terms of logarithmic negativityGV-PRA-2002 (35). Numerical result and analytical result show that: 1, whatever measure you take, the entanglement will increase; 2, the entanglement enhancement is largest in perfect measurement, while smaller enhancement in imperfect measurement, and it is not obvious in on-off measurement. 3, we are sure that the more precise measurement information, the larger entanglement concentration.

The remainder of this paper is organized as follows. In Sec. II, we introduce the physical system and derive the amount of entanglement before concentration, along with the generating of a genuine tripartite entanglement state. In Section III, the the definition of logarithmic negativity is briefly summarized and the amount of entanglement after perfect measurement is introduced. Section IV is devoted to the entanglement distillation with imperfect measurement. We calculated the amount of entanglement after imperfect measurement perturbative order by order and compare the entanglement concentration effect analytically. In Sec. V, we discuss the on-off measurement and derive the average entanglement after on-off measurement. Finally, we conclude with a discussion and summary about three different measurement strategies numerically and analytically in Sec. VI.

II physical system and measurement operator

We consider a three-mode optomechanical system: two cavity modes ($\omega_{1}$ and $\omega_{2}$) are coupled to a single mechanical mode $\omega_{\mathbf{m}}$ (see Fig. 1). The cavities interact with the mechanics via the radiation pressure MA-RMP-2014 (10). The Hamiltonian of the system can be described by

[TABLE]

where $\hat{a}_{j}$ and $\hat{b}$ are the annihilation operator for cavity $j(j=1,2)$ and the mechanical mode respectively. The optomechanical coupling strengths are denoted by $g_{j}$. In order to generate steady state entanglement, we assume a strong coherent drive on each cavity, detuned to the red (blue) mechanical sideband for cavity 1(2), i.e., drive frequency $\omega_{\mathbf{d}1}=\omega_{1}-\omega_{\mathbf{m}},$ $\omega_{\mathbf{d}2}=\omega_{2}+\omega_{\mathbf{m}}$. We work in an interaction picture with respect to free Hamiltonian and split the cavity field into a classical cavity amplitude $\bar{a}_{j}$ and a small quantum fluctuation $\hat{d}_{j}$ with $\hat{d}_{j}=\hat{a}_{j}-\bar{a}_{j}$. Here $\bar{a}_{j}$ is the average number of photons for each cavity. After performing the rotating wave approximation and linearizing the Hamiltonian independently, We obtain:

[TABLE]

Here $G_{j}=g_{j}\bar{a}_{j}$ is the dressed coupling. In general, we take $g_{j},\bar{a}_{j}>0$. Taking the damping and the noise terns into account, we get the Langevin equation for the optical and mechanical modes operator CWG-QN-2004 (36):

[TABLE]

where $\kappa_{j}$ and $\gamma$ is the damping rate of the cavities and mechanics respectively and $\mathbf{i}$ is the imaginary unit. As discussed in CWG-QN-2004 (36), from the Langevin equations and the input-output relation, one can verify that the stationary output state in the Fock state basis $\left|n_{1},n_{2},n_{\mathbf{m}}\right\rangle$, can be expressed as:

[TABLE]

In Eq. (4), $C_{p+q}^{p}$ is the binomial coefficients and $N_{\mathbf{k}}(k=1,2,m)$ is the average photons or phonons number of each output mode with $N_{\mathbf{k}}=\left\langle\left(\hat{d}_{k}^{\mathbf{out}}\left[0\right]\right)^{\dagger}\hat{d}_{k}^{\mathbf{out}}\left[0\right]\right\rangle$ and they can be given as follow:

[TABLE]

with the cooperativity $C_{j}=4G_{j}^{2}/\gamma\kappa_{j}$. The result was derived under the assumptions of zero temperature and in the limit of narrow bandwidth around the bare frequencies $\omega_{j},\omega_{\mathbf{m}}$. Note that this is a Gaussian state (more specifically), it is a twice squeezed 3-mode vacuum state YDW-PRA-2015 (18), which is a genuine tripartite entangled state. By tracing out the mechanical mode, one obtains a 2-mode squeezed thermal state of the photon output fields, which has entanglement:

[TABLE]

with $A=C_{2}\left(C_{1}+C_{2}\right),B=\left(1+C_{1}\right)^{2}+C_{1}C_{2}$. The entanglement is maximized at the instability point $C_{1}\rightarrow C_{2}-1$ and remains finite ($E_{N}\rightarrow\ln\left(2C_{1}+1\right)$), at variance with the well-known divergence for a parametric amplifier. This is a natural result because the two modes are entangled with the mechanics. Indeed, the divergence is manifested only for the tripartite entanglement YDW-PRA-2015 (18). However, as we discuss in the following, a divergence of $E_{N}$ can be recovered by an ideal measurement. In this sense, the large entanglement of the three-body state is a physical resource which can be used to greatly enhance the bipartite entanglement of the emitted phonons.

In the practice, the mechanics can be connected to a strong damped auxiliary cavity (cf. Fig. 1) such that the mechanical output can be mapped to the optical output, $\hat{d}_{a,\mathbf{out}}=-\mathbf{i}\hat{b}_{\mathbf{out}}$ YDW-PRA-2015 (18). A recent experiment has demonstrated the readout of the phonon number through this mechanism JDC-Nature-2015 (37). So in the following text, the measurement of phonon of the mechanical mode is through the measurement of photon of the auxiliary cavity mode. We will simply refer to this method as "measurement of the phonon mode" and quantify its effect on the output entanglement of the two cavities.

In the measurement theory of quantum mechanics, projection operator is a perfect measurement operator

[TABLE]

but in experiment we often deal with imperfect measurement. A typical imperfect measurement is efficient measurement that the detect efficiency $\mu$ is considered CAH-PRA-1989 (38). For a single photon detector, the detect efficiency $\mu$ can be regarded as the probability for detecting one photon in time t from an one photon field. The explicit form of $\mu$ depends on the physical situation, here we just consider a constant value of $\mu$. According to ref. CAH-PRA-1989 (38), the operator of measurement with measure outcome $q$ is

[TABLE]

We find that the imperfect measurement become perfect measurement when $\mu=1$. Another measurement may be on-off measurement. The on case can be interpreted as: we detect photon, but we can’t identify the photon number. The off case is that we have not detected photon. In physics, they can be expressed as SLZ-PRA-2010 (39):

[TABLE]

III perfect measurement

We first consider the perfect measurement of the phonon number, described by projection operator $\hat{M_{1}}(q)$ where $q$ is the outcome of the phonon measurement. Such measurement increases the entanglement WM-SCPMA-2015 (34), as it can be computed straightforwardly from the state after measurement:

[TABLE]

Here we define

[TABLE]

where $\zeta=4C_{1}C_{2}(1+C_{1}+C_{2})^{-2}$, The normalization factor is

[TABLE]

Although Eq. (13) is not a Gaussian state, we can still quantify the entanglement directly from the definition of logarithmic negativity GV-PRA-2002 (35):

[TABLE]

where $\hat{\rho}$ is the density matrix of the state being evaluated, $\hat{\rho}^{\mathbf{T}}$ is the partial transpose with respect to one subsystem, $\left\|.\right\|_{\mathbf{1}}$ denotes trace norm. Also noticing that for a two mode entangled state written in a Schmidt decompostion as $\left|\varphi\right\rangle=\sum C_{n}\left|n_{\mathbf{A}},n_{\mathbf{B}}\right\rangle$, the entanglement is: $E_{N}=2\ln\sum_{n}\left|C_{n}\right|$. Thus the entanglement of the state in Eq. (13) can be written as:

[TABLE]

A special case is $q=0$, it means that no phonon has been detected. In this case,

[TABLE]

To get an analytical expression of the entanglement after measure, an approximation is necessary. Notice that $f_{p}(q)$ is normalized and it can be regarded as a Gaussian distribution for large $q$:

[TABLE]

with the mean and variance being $\kappa\left(q\right)=\zeta\frac{1+q}{1-\zeta}$, $\sigma\left(q\right)=\frac{\sqrt{\zeta\left(1+q\right)}}{1-\zeta}$. Then:

[TABLE]

The entanglement after measurement is found to increase logarithmically with the number of detected phonons ($q$), and it is larger than the entanglement before measurement, even the measurement outcome is zero (see Fig. 2 ). It can been found that measurement enhance entanglement. This is one of the most important conclusion in the perfect measurement.

IV imperfect measurement

Now we consider the imperfect measurement with the measurement operator $\hat{M}_{\mu}\left(q\right)$. According to ref. CAH-PRA-1989 (38), the state after measurement is given by

[TABLE]

here the trace is for detected mode, and $P_{\mu}(q)$ is the probability for detecting $q$ phonons from a field with the phonon number distribution $P_{s}$:

[TABLE]

For the given three mode entangled state $\left|\Psi\right\rangle$, when we consider the detect efficiency $\mu$, the state after the imperfect measurement with the detect outcome $q$ is:

[TABLE]

with

[TABLE]

Obvious, $\hat{\rho}_{q}$ is a mixed state. From the ref. GV-PRA-2002 (35), calculating logarithmic entanglement is to find the negativity $N\left(\hat{\rho}_{q}^{\mathbf{T}}\right)$ with

[TABLE]

The $\hat{\rho}_{q}^{\mathbf{T}}$ is a block diagonal matrix by identifying the different total photons $Q=p_{1}+p_{2}+s$ for each subblock. When $Q=q$, there is only one matrix element $Q\left[q\right]=f_{0}(q)\eta\left(q\right)$ in the subblock matrix. When $Q=q+1$, it is a $2\times 2$ square matrix with $Q\left[q+1\right]$:

[TABLE]

When $Q=q+n$, the sub block matrix $Q\left[q+n\right]$ is a $n\times n$ square matrix. Each sub block matrix of $\hat{\rho}_{q}^{\mathbf{T}}$ is a lower triangular matrix. To calculate $N\left(\hat{\rho}_{q}^{\mathbf{T}}\right)$, we make an approximation. By throwing away most elements and only remaining the main diagonal and the elements which are the nearest to the main diagonal for each sub blocks, then the expression of $\hat{\rho}_{q}^{\mathbf{T}}$ becomes

[TABLE]

Then calculating the entanglement after the imperfect measurement is calculating the eigenvalues of $\hat{\rho}_{q}^{\mathbf{T}}$. To calculate the eigenvalues of $\hat{\rho}_{q}^{\mathbf{T}}$, we thought about using perturbation theory. In quantum mechanics, the classical non degenerate stationary state perturbation theory is that for a given $\hat{H}$, we can devide $\hat{H}$ into two part: $\hat{H}=\hat{H}_{\mathbf{0}}+\hat{H}^{\mathbf{{}^{\prime}}}$($\hat{H}^{\mathbf{{}^{\prime}}}$ is a perturbation). Here, we just treat $\hat{\rho}_{q}^{\mathbf{T}}$ as $\hat{H}$ and let $\hat{\rho}_{q}^{\mathbf{T}}=\hat{H}=\hat{H}_{\mathbf{0}}+\hat{H}^{\mathbf{{}^{\prime}}}$ with

[TABLE]

[TABLE]

Also for a classical perturbation theory, the eigen equation of $\hat{H}_{\mathbf{0}}$ is $\hat{H}_{\mathbf{0}}\left|m\right\rangle=e_{m}^{(0)}\left|m\right\rangle$, where $e_{m}^{(0)}$ is eigenvalue in the eigenstate $\left|m\right\rangle$. The first-order approximate eigenvalue of $\hat{H}$ in a state $\left|\phi_{m}\right\rangle$ (which is close to $\left|m\right\rangle$):

[TABLE]

and the second-order approximate eigenvalue of $\hat{H}$ in a state $\left|\phi_{m}\right\rangle$:

[TABLE]

Then we have the approximate eigenvalue of $\hat{H}$ in a state $\left|\phi_{m}\right\rangle$

[TABLE]

For $\hat{\rho}_{q}^{\mathbf{T}}$, its eigenvectors and eigenvalues of $\hat{H}_{\mathbf{0}}$ is

[TABLE]

with $\left|\alpha\right\rangle=\mid p,p+q\rangle$, $\left|\beta_{\pm}\right\rangle=\frac{\mid p_{1},p_{2}+q\rangle\pm\mid p_{2},p_{1}+q\rangle}{\sqrt{2}}$. According to the definition of $N\left(\hat{\rho}_{q}^{\mathbf{T}}\right)$, only the eigenvalue of the state $\left|\beta_{-}\right\rangle$ is important. Comparing to the classic formula of perturbation theory, the entanglement after imperfect measurement is:

[TABLE]

with

[TABLE]

When only consider the first order, the entanglement can be expressed as:

[TABLE]

with $\varepsilon=\frac{\left(1-\mu\right)N_{\mathbf{m}}}{\left(1+N_{\mathbf{m}}\right)}$. Considering the second order, it is:

[TABLE]

with

[TABLE]

[TABLE]

In the Gaussian approximation, $\Omega$ can get a simple expression, and when $q$ is large, $\Omega$ will tend to $\frac{1}{2}$,

[TABLE]

Then we get

[TABLE]

In the numerical analysis of the entanglement after the imperfect measurement (see Fig. 2), we can see that the imperfect measurement will be close to the perfect measurement when $\mu$ is close to 1. In the analytical analysis, the perturbation approximation is effective when $\mu\rightarrow 1$ and the detected phonon number $q$ is small (see Fig. 4). It can be found that the entanglement after imperfect measurement is larger than the entanglement before measurement, but smaller than the entanglement after perfect measurement (see Fig. 3). That is to say: perfect measurement is more effective than imperfect measurement in the entanglement concentration, and no matter adopt what kind of measurement method, the entanglement is concentred.

V on-off measurement

Now we consider on-off measurement and the state after measure is

[TABLE]

For off case, the entanglement is the same with the perfect measurement when $q=0$.

[TABLE]

For on case, the logarithmic negativity theory still works. We can calculate the $N\left(\hat{\rho}_{\mathbf{on}}^{\mathbf{T}}\right)$. Then we get $E_{N}^{\mathbf{on}}=\ln\left(1+2N\left(\hat{\rho}_{\mathbf{on}}^{\mathbf{T}}\right)\right)$. In this way it is easy to get a precise numerical solution but difficult to find an analytical expression. Fortunately we find that when make the suitable parameters($C_{1}\gg C_{2}$), the average entanglement is close to the numerical on measurement entanglement (see Fig. 5)

[TABLE]

and the average entanglement is defined as:

[TABLE]

If use Gaussian approximation:

[TABLE]

We find even in the on-off measurement, the entanglement is concentred, but the effect is weakest comparing to the perfect measurement and the imperfect measurement (see Fig. 3).

VI conclusion

In quantum communication, we often need the maximum entangled state. How to achieve maximum entangled state is an important subject of quantum communication. Entanglement concentration is an important method in preparing the maximum entangled state. In the continuous-variable, for the Gaussian entangled state, only the non Gaussian operation is possible in the entanglement concentration. As a non Gaussian operation-quantum measurement, we use three different measurement operators(perfect measurement, imperfect measurement and on-off measurement) to concentrate a 3-mode Gaussian state. Perfect measurement is strongest in the entanglement concentration, imperfect measurement is second, and the on-off measurement is the weakest. But on the other hand, perfect measurement get the most precise measurement information. In this respect, the more precise measurement information, the larger entanglement concentration. But In the experiment the imperfect measurement is relatively easy to implementation, so making an efficient measuring instrument is very important. In the imperfect measurement, we use perturbation theory to calculate the entanglement of a mixed state, the numerical results and analytical results fit well. This has some reference for us to deal with the entanglement of other density matrices. No matter the perfect measurement, imperfect measurement and on-off measurement, they are strong measurement, the entanglement concentration based on weak measurement will be considered in the next work.

VII SUPPLEMENT

Supplement[A]: The 3-mode Entangled State

The 3-mode entangled state is

[TABLE]

with

[TABLE]

Obvious we have $N_{2}=N_{1}+N_{m}$. For the 3-mode entangled state, $p$ is one cavity mode, $p+q$ is another cavity mode, $q$ is mechanical mode. So $p,p+q$ represent the photon, $q$ represent the phonon. In order to make the following calculation process simple, we let

[TABLE]

The definition of $f_{p}\left(q\right)$ and $P_{q}$ are described in the following part. Then we can make a short expression for $\left|\Psi\right\rangle$:

[TABLE]

and the density matrix is

[TABLE]

also

[TABLE]

and the entanglement of the two cavity mode is

[TABLE]

with $A=C_{2}\left(C_{1}+C_{2}\right)$ , $B=\left(1+C_{1}\right)^{2}+C_{1}C_{2}$.

Supplement[B]: Phonon Measure with Perfect Measument

If we measure the mechanical mode and measure $q$ phonons , the measurement opreater is

[TABLE]

and the quantum state after measure is given by

[TABLE]

and the normalized state after measure is given by

[TABLE]

and the trace is calculated by the following

[TABLE]

After calculation we have

[TABLE]

we write as

[TABLE]

with

[TABLE]

the $\left|\Psi_{q}^{\prime}\right\rangle$ is not a normalized state. So we normalize the $\left|\Psi_{q}^{\prime}\right\rangle$

[TABLE]

with

[TABLE]

and

[TABLE]

[TABLE]

so we can define

[TABLE]

with $\left|\Psi_{q}\right\rangle$ is a normalized state of $\left|\Psi_{q}^{\prime}\right\rangle$

[TABLE]

and

[TABLE]

so we have the normalized $\left|\Psi_{q}\right\rangle$ and it can also be expressed as

[TABLE]

here we define

[TABLE]

[TABLE]

so we have the state after perfect measurement

[TABLE]

and the entanglement after perfect measurement is

[TABLE]

To get an analytical expression of the entanglement after measure, a approximation is necessary. Notice that $f_{p}\left(q\right)$ is normalized and it can be regarded as a Gaussian distribution for large $q$:

[TABLE]

Comparing a standard Gaussian fuction

[TABLE]

we have

[TABLE]

and

[TABLE]

[TABLE]

the mean and variance being is given by

[TABLE]

[TABLE]

So we have

[TABLE]

[TABLE]

[TABLE]

and the mean and variance is given by

[TABLE]

[TABLE]

Then after using Gaussian approximation, we have

[TABLE]

Supplement[C]: Phonon Measure with Imperfect Measument

If the phonon detect effiency is $\mu$, and the state is in the $\left|\Psi_{s}\right\rangle$, the probability for dectecting $q$ phonon is

[TABLE]

then calculate the $s$, get the probability for dectecting $q$ phonon in $\left|\Psi\right\rangle$

[TABLE]

Simplify it we have

[TABLE]

When we have measure mechanical mode with the measurement outcome $q$ , the condition of the two cavity mode after measure is given by

[TABLE]

where

[TABLE]

[TABLE]

then the main task is to calculate $Tr_{q}\left[\hat{M_{\mu}}\left(q\right)\rho\hat{M_{\mu}^{\dagger}}\left(q\right)\right]$ and it can be written as

[TABLE]

and because $\left|\Psi\right\rangle$

[TABLE]

and

[TABLE]

first, the $\hat{M_{\mu}}\left(q\right)\left|\Psi\right\rangle$ is

[TABLE]

obviously $n_{1}=q_{1}$, so

[TABLE]

second, the $\left\langle\Psi\right|\hat{M_{\mu}^{\dagger}}\left(q\right)$ is

[TABLE]

obviously $n_{2}=q_{2}$, so

[TABLE]

then, we have$\hat{M_{\mu}}\left(q\right)\rho\hat{M_{\mu}^{\dagger}}\left(q\right)$

[TABLE]

so we have

[TABLE]

and we can find $q_{3}=q_{1}-q,q_{3}=q_{2}-q,q_{1}=q_{2}$ and let $q_{1}=q_{2}=s$ finally, we have the conditional of state $p$ is

[TABLE]

define

[TABLE]

, so we have

[TABLE]

also we can write $\rho_{q}$ in another way, because:

[TABLE]

then

[TABLE]

then

[TABLE]

its entanglement is $E_{N}^{\mu}\left(q\right)=\ln\left[1+2N\left(\rho_{q}^{\mathbf{T}}\right)\right]$ ,where $N\left(\rho_{q}^{\mathbf{T}}\right)=\sum\left|\lambda_{i}\right|$, and $\lambda_{i}$ are the negativity eigenvalues of $\rho_{q}^{\mathbf{T}}$. $\rho_{q}^{\mathbf{T}}$ is the partial transpose of $\rho_{q}$ with

[TABLE]

Supplement[D]: Diagonalize the density matrix

the partial transpose of the density matrix is

[TABLE]

We let Q=p_{1}+p_{2}+s$$\left(Q\geq q\right), and define

[TABLE]

Then we get

when$Q=q$

[TABLE]

[TABLE]

when $Q=q+1$

[TABLE]

[TABLE]

when $Q=q+2$

[TABLE]

[TABLE]

so when $Q=q+n$

[TABLE]

[TABLE]

In this way we can diagonalize the density matrix with

[TABLE]

Supplement[E]: Limit Case of $\mu$

If $\mu=1$ only $s=q$ is valid, so $\eta\left(q\right)=1$ ,

[TABLE]

[TABLE]

So

[TABLE]

When $\mu\longrightarrow 1$, imperfect measurement will turn to be perfect measurement.

If $\mu\longrightarrow 0$ , $\eta\left(s\right)=\frac{N_{m}^{s}C_{s}^{q}}{N_{m}^{q}\left(1+N_{m}\right)^{1+s}}$,

[TABLE]

[TABLE]

from a physical standpoint it is necessary to make $q=0$.

[TABLE]

to make $\rho_{0}$ more clear, we may calculate this $Tr_{q}\left[\left|\Psi\right\rangle\left\langle\Psi\right|\right]$

[TABLE]

[TABLE]

[TABLE]

the condition is $s=q_{1}=q_{2}$

[TABLE]

Simplify it, we can get

[TABLE]

The $Tr_{q}\left[\left|\Psi\right\rangle\left\langle\Psi\right|\right]$ and $\rho_{0}$ are the same

[TABLE]

Supplement[F]: Entanglement Calculation after Imperfect Measuement

The state after measurement is

[TABLE]

with $\eta\left(s\right)=\frac{N_{m}^{s}}{\left(1+N_{m}\right)^{1+s}}\frac{\left(1+N_{m}\mu\right)^{1+q}}{N_{m}^{q}\mu^{q}}C_{s}^{q}\left(1-\mu\right)^{s-q}\mu^{q}$. The partial transpose is

[TABLE]

The sum of $s$ is from $q$ to $+\infty$ , it is difficult to calculate the negativity eigenvalues of $\rho_{q}^{\mathbf{T}}$ directly. Here we use perturbation theory, and we just consider two term $s=q$ , $s=q+1$ . So we rewrite it

[TABLE]

and define

[TABLE]

[TABLE]

so

[TABLE]

for $H_{0}$ , we have the eigenvalues and eigenvectors

[TABLE]

and we can calculate entanglement in this way

[TABLE]

with

[TABLE]

and in this article, because $E_{\beta_{-}}^{\left(0\right)}<0$ , $\left|E_{\beta_{-}}^{\left(0\right)}\right|\gg\left|E_{\beta_{-}}^{\left(1\right)}\right|$, $\left|E_{\beta_{-}}^{\left(0\right)}\right|\gg\left|E_{\beta_{-}}^{\left(2\right)}\right|$, so we have

[TABLE]

[TABLE]

[TABLE]

So the next mission is to calculate $\sum_{\beta_{-}}E_{\beta_{-}}^{\left(0\right)},\sum_{\beta_{-}}E_{\beta_{-}}^{\left(1\right)},\sum_{\beta_{-}}E_{\beta_{-}}^{\left(2\right)}$.

Supplement[G]: Zero Order and First Order Calculation

For the zero order calculation , it is easy.

[TABLE]

For the first order calculation, the following formulas may be useful

[TABLE]

[TABLE]

we calculate$E_{\beta_{-}}^{\left(1\right)}=\left\langle\beta_{-}\left|H_{1}\right|\beta_{-}\right\rangle$:

[TABLE]

[TABLE]

so

[TABLE]

If we just consider zero order and first order ,

[TABLE]

[TABLE]

So we get the expression of $E_{N}^{\mu}\left(q\right)$

[TABLE]

Use the accurate expression of $\eta\left(s\right)$ ,we get :

[TABLE]

[TABLE]

here we define:$\varepsilon=\frac{\left(1-\mu\right)N_{m}}{\left(1+N_{m}\right)}$ , and use the approximation formula $\left(1+x\right)^{n}\approx 1+nx$.

[TABLE]

We throw away some things such as $O\left[\left(1-\mu\right)^{2}\right]$, we can obtain

[TABLE]

We can get a simple expression of $E_{N}^{\mu}\left(q\right)$

[TABLE]

[TABLE]

make a further approximation

[TABLE]

with $\varepsilon=\frac{\left(1-\mu\right)N_{m}}{\left(1+N_{m}\right)}$ . In the perfect measurement, we have

[TABLE]

So we can get the entanglement of imperfcet measurement after first order perturbation approximation

[TABLE]

Supplement[H]: Second Order Calculation

The second order is$E_{\beta_{-}}^{\left(2\right)}=\sum_{n\neq\beta_{-}}\frac{\left(\left\langle\beta_{-}\left|H_{1}\right|n\right\rangle\right)^{2}}{E_{\beta_{-}}^{\left(0\right)}-E_{n}^{0}}$. $n$ is the eigenvectors of $H_{0}$ with the restricted condition $n\neq\beta_{-}$ . We just care $\sum_{\beta_{-}}E_{\beta_{-}}^{\left(2\right)}$ . First,

[TABLE]

and

[TABLE]

obvious

[TABLE]

so we only care

[TABLE]

also, define a expression to simplify calculation

[TABLE]

also because $f_{p}\left(q\right)=C_{p+q}^{p}\zeta^{p}\left(1-\zeta\right)^{1+q}$ we have a lot of different expressions of $g\left(p_{1},p_{2}\right)$ and we just write them for further use.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and we first care $\sum_{\beta_{-}}\sum_{\alpha}\frac{\left(\left\langle\beta_{-}\left|H_{1}\right|\alpha\right\rangle\right)^{2}}{E_{\beta_{-}}^{\left(0\right)}-E_{\alpha}^{0}}$ , and the following formulas may be useful

[TABLE]

because $p_{1}<p_{2}$

[TABLE]

[TABLE]

so

[TABLE]

so

[TABLE]

here we do a replace that is $p_{2}\rightarrow p_{3}+1$, then let $p_{3}\rightarrow p_{2}$ .

[TABLE]

Then calculate $\sum_{\beta_{-}}\sum_{\alpha_{+}}\frac{\left(\left\langle\beta_{-}\left|H_{1}\right|\alpha_{+}\right\rangle\right)^{2}}{E_{\beta_{-}}^{\left(0\right)}-E_{\alpha_{+}}^{0}}$. We do some preparation work firstly, the following four factor form may be useful

[TABLE]

and it easy to find that $AD=0,$ $BC=0$ that is

[TABLE]

[TABLE]

If $p_{1}\neq p_{2}$ ,we have $AC=0,$ $BD=0$ that is

[TABLE]

[TABLE]

the following factor form may be useful

[TABLE]

because $p_{1}\neq p_{2}$:

[TABLE]

if we consider $p_{1}<p_{2}$,and $a<b$,we have

[TABLE]

so we have divided the nonezero term of $\frac{\left(\left\langle\beta_{-}\left|H_{1}\right|\alpha_{+}\right\rangle\right)^{2}}{E_{\beta_{-}}^{\left(0\right)}-E_{\alpha_{+}}^{0}}$ into three parts, the first part is

[TABLE]

[TABLE]

so

[TABLE]

the second part is

[TABLE]

[TABLE]

so

[TABLE]

the third part is

[TABLE]

[TABLE]

[TABLE]

so

[TABLE]

and

[TABLE]

so we can get a simple expression of $\sum_{\beta_{-}}E_{\beta_{-}}^{\left(2\right)}$

[TABLE]

[TABLE]

If we consider second order, maybe we should calculate the contribution of the perturbation Hamiltonian $H_{2}$ when $s=q+2$ .

[TABLE]

and the following formulas may be useful

[TABLE]

[TABLE]

we calculate$\left\langle\beta_{-}\right|H_{2}\left|\beta_{-}\right\rangle$:

[TABLE]

[TABLE]

If we just consider zero order, first order and second order ,

[TABLE]

with

[TABLE]

so

[TABLE]

and we have calculated $\eta\left(q\right)$ and $\eta\left(q+1\right)$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

here $\varepsilon=\frac{\left(1-\mu\right)N_{m}}{1+N_{m}}$ , and use the approximation formula $\left(1+x\right)^{n}\approx 1+nx+\frac{n\left(n-1\right)}{2}x^{2}$.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

so

[TABLE]

and

[TABLE]

[TABLE]

Define

[TABLE]

[TABLE]

make a further approximation

[TABLE]

[TABLE]

[TABLE]

also $E_{N}\left(q\right)=2\ln\sum_{p=0}^{\infty}\sqrt{f_{p}(q)}$

[TABLE]

[TABLE]

In the Gaussian approximation and under large $q$ we have

[TABLE]

Then we get

[TABLE]

Supplement[I]: On-off Detection Measure

an on-off detection measure is given by

[TABLE]

[TABLE]

and the state after measure is given by

[TABLE]

[TABLE]

we first calculate the off detection measure,

[TABLE]

the trace is calculated by the following

[TABLE]

so

[TABLE]

and

[TABLE]

with

[TABLE]

the $\mid\psi^{\prime}\rangle_{off}$ is not a normalized state. So we normalize the $\left|\psi^{\prime}\right\rangle{}_{off}$ , and the normalization factor of $\left|\psi^{\prime}\right\rangle{}_{off}$ is

[TABLE]

and

[TABLE]

so the state after off detection measure is given by

[TABLE]

then normalize $\left|\psi^{\prime}\right\rangle{}_{off}$ we get

[TABLE]

[TABLE]

with

[TABLE]

and it’s entanglement is

[TABLE]

we now calculate the on detection measure,

[TABLE]

[TABLE]

so we get

[TABLE]

and $\left|\Psi_{q}\right\rangle=\sum_{p=0}^{\infty}\sqrt{f_{p}\left(q\right)}\left|p,p+q\right\rangle$

[TABLE]

so we have

[TABLE]

so

[TABLE]

then the normalized $\rho_{on}$ is

[TABLE]

and

[TABLE]

and the next problem is how to calculate the entanglement of $\rho_{on}$. In perfect measurement we have the average entanglement defined by

[TABLE]

then we can define the on average entanglement

[TABLE]

also we get

[TABLE]

Supplement[J]: Derivation of the Entanglement’s Equation

The entanglement is give by this equation

[TABLE]

for a simple density matrix the partial transpose is

[TABLE]

[TABLE]

or

[TABLE]

so

[TABLE]

then we just write $\rho^{\mathbf{T}}=\rho^{\mathbf{PT_{A}}}$. The trace form is

[TABLE]

[TABLE]

where $\left|\lambda_{i}\right|$ is the absolute eigenvalues of $\rho^{\mathbf{T}}$ and $N\left(\rho^{\mathbf{T}}\right)$ is the sum of the absolute value of the negative eigenvalues of $\rho^{\mathbf{T}}$ , $N\left(\rho^{\mathbf{T}}\right)=\sum_{\lambda_{i}<0}\left|\lambda_{i}\right|$,the relationship between $\sum_{i}\left|\lambda_{i}\right|$and $N\left(\rho^{\mathbf{T}}\right)$ are given by

[TABLE]

Consider a general two mode entangled state

[TABLE]

the normalization condition is $\sum_{n}C_{n}^{2}=1$

[TABLE]

and the partial transpose is

[TABLE]

to find the negative eigenvalues of $\rho^{\mathbf{T}}$,we rewrite $\rho^{\mathbf{T}}$ in this way

[TABLE]

Here we suppose $C_{n}$ and $C_{m}$ are real, so we have $C_{n}C_{m}^{*}=C_{m}C_{n}^{*}=C_{n}C_{m}$ .

[TABLE]

we use this formula

[TABLE]

$\left|n,m\right\rangle\left\langle m,n\right|+\left|m,n\right\rangle\left\langle n,m\right|$can be written as

[TABLE]

Define $\left|a\right\rangle=\frac{\left|n,m\right\rangle+\left|m,n\right\rangle}{\sqrt{2}}$, $\left|b\right\rangle=\frac{\left|n,m\right\rangle-\left|m,n\right\rangle}{\sqrt{2}}$, so we have

[TABLE]

In this way ,we give the diagonal form of $\rho^{\mathbf{T}}$

[TABLE]

use this relation

[TABLE]

So

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now it is easy to see the eigenvalues and eigenvectors of $\rho^{\mathbf{T}}$

[TABLE]

In the given $\rho^{\mathbf{T}}$, the $N\left(\rho^{\mathbf{T}}\right)$ is given by

[TABLE]

So we can get the expression of the entanglement

[TABLE]

with $\sum_{n}\left|C_{n}\right|^{2}=1$. So if we consider

[TABLE]

[TABLE]

The partial transpose is

[TABLE]

it is easy to see the eigenvalues and eigenvectors of $\rho^{\mathbf{T}}$

[TABLE]

the $N\left(\rho^{\mathbf{T}}\right)$ is given by

[TABLE]

So the entanglement of $\rho$ is

[TABLE]

Supplement[K]: Calculate of $\Omega$

we have calculate the entanglement

[TABLE]

with

[TABLE]

[TABLE]

and

[TABLE]

also we have a gaussian approximation for $f_{p}\left(q\right)$, that is

[TABLE]

with the mean and variance being $\kappa\left(q\right)=\zeta\frac{1+q}{1-\zeta}$ , $\sigma\left(q\right)=\frac{\sqrt{\zeta\left(1+q\right)}}{1-\zeta}$. If we want to make a simple expression of $E_{N}^{\mu}\left(q\right)$, we need know the expression of $\Omega$. In perfect measurement, undering the gaussian approximation:

[TABLE]

First we have a Simplification of $g\left(p_{1},p_{2}\right)$, and use $f_{p}\left(q\right)=C_{p+q}^{p}\zeta^{p}\left(1-\zeta\right)^{1+q}$ we get

[TABLE]

By using gaussian approximation for $f_{p}\left(q\right)\approx\frac{1}{\sqrt{2\pi}\sigma\left(q\right)}e^{-\frac{\left(p-\kappa\left(q\right)\right)^{2}}{2\sigma\left(q\right)^{2}}}$, we have

[TABLE]

Next we calculation of $\sum_{p_{1}=0}^{\infty}\sum_{p_{2}=0}^{\infty}g\left(p_{1},p_{2}\right)$. we may calculate $\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}g\left(p_{1},p_{2}\right)dp_{1}dp_{2}$ instead of

$\sum_{p_{1}=0}^{\infty}\sum_{p_{2}=0}^{\infty}g\left(p_{1},p_{2}\right)$. So

[TABLE]

and we do a taylor expansion for $\frac{\left(p_{1}+q+1\right)\left(p_{2}+q+1\right)}{\sqrt{1+\frac{q}{p_{1}+1}}+\sqrt{1+\frac{q}{p_{2}+1}}}$ :

[TABLE]

the $m$ is the taylor expansion center. Here we let $m=\kappa\left(q\right)$. We just calculate zero order with

[TABLE]

and beacuse $\int_{-\infty}^{+\infty}e^{-x^{2}}xdx=0$, the first order is

[TABLE]

Then we have

[TABLE]

and

[TABLE]

and q is large, we have

[TABLE]

Then we get

[TABLE]