1
$\begingroup$

I would like to ask how to properly calculate the off-diagonal terms in covariance matrix for the entangled Gaussian state?

E.g. from https://arxiv.org/abs/0810.0534v1 we have a coherent Gaussian state in the following form $$ |\psi\rangle = \sum_{n=0}^{\infty}\sqrt{\frac{N^n}{(N+1)^{n+1}}}|n\rangle_A|n\rangle_B $$ and the covariance matrix $$ V=\frac{1}{4} \left( \begin{array}{cccc} S & 0 & C & 0 \\ 0 & S & 0 &-C \\ C & 0 & S & 0 \\ 0 &-C & 0 & S \\ \end{array} \right) $$ where $S=2N+1$ and $C=2\sqrt{N(N+1)}$.

Using the definition of the covariance matrix $$ V_{ij}=\frac{1}{2}Tr[\hat \rho\{\hat q_i;\hat q_j\}] $$ (assuming zero displacement) where $\hat \rho$ is the appropriate density operator and vector $\hat q=(\hat X_A, \hat P_A, \hat X_B, \hat P_B)$ can be expressed using kvadrature, i.e. $\hat X = \hat a + \hat a^\dagger$, $\hat P = \hat a^\dagger - \hat a$. It is clear to me how I get the diagonal terms, but not the off-diagonal.

For example $$ V_{13}=\frac{1}{2}Tr[\hat \rho(\hat X_A\hat X_B + \hat X_B\hat X_A ) ] $$ since operators $A$ and $B$ commute $$ V_{13}=Tr[\hat \rho(\hat X_A\hat X_B ) ]\\ =Tr[\hat \rho(\hat a_A + \hat a_A^\dagger )(\hat a_B + \hat a_B^\dagger ) ] $$ but this combination of creation and annihilation operators change the states but never "return" and, therefore, trace will be zero.

Probably I do something trivially wrong, but I'm blind. Thanks.


Edit 1: in more details: My density matrix reads $$ \hat \rho = \sum_{n=0}^{\infty}\frac{N^n}{(N+1)^{n+1}} |n\rangle_A\langle n| |n\rangle_B \langle n| $$ Then the above described term 13 of the covariance matrix is $$ V_{13}=Tr\left[\sum_{n=0}^{\infty}\frac{N^n}{(N+1)^{n+1}} _B\langle n|_A\langle n| \hat a_A\hat a_B + \hat a_A^\dagger \hat a_B^\dagger + \hat a_A^\dagger\hat a_B + \hat a_A^\dagger\hat a_B |n\rangle_A|n\rangle_B \right]\\ Tr\left[\sum_{n=0}^{\infty}\frac{N^n}{(N+1)^{n+1}} _B\langle n|_A\langle n| \left( n|n-1\rangle_A|n-1\rangle_B + (n+1)|n+1\rangle_A|n+1\rangle_B + \sqrt{n(n+1)}|n+1\rangle_A|n-1\rangle_B + \sqrt{n(n+1)}|n-1\rangle_A|n+1\rangle_B \right) \right] $$ then the trace gives zero. Where I do a mistake? What is wrong?

$\endgroup$
  • $\begingroup$ Your state is a superposition of states with all particle numbers. Why would you get zero? (E.g., you could remove one A and one B particle.) $\endgroup$ – Norbert Schuch Apr 27 '18 at 15:57
  • $\begingroup$ Thanks for answer. But I still do not see it. I will write it in more details. $\endgroup$ – Naake May 3 '18 at 18:31
0
$\begingroup$

You need to take two independent summation indices for the ket and the bra vector in the density matrix (and thus in the computation of expectation values).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.