Let $b_{1,2}$ be bosonic operators that satisfy $[b_i, b_i^\dagger]=1$. Two boson operators give the Hilbert space $\mathcal H_1 \otimes \mathcal H_2$ in the tensor product form.

Consider the Gaussian state $$|\Phi\rangle = e^{\lambda (b_1^\dagger b_2^\dagger - b_1b_2)} |0\rangle,$$ where $|0\rangle$ is the vaccum and $\lambda\in\mathbb R$. Taking partial trace to subsystem labeled by 2, we obtain $\rho_1 = \operatorname{Tr}_2(|\Phi\rangle \langle \Phi|)$.

Question: What is the purity $\operatorname{Tr}(\rho_1^2)$?

Even though the state is Gaussian, it seems that the purity computation is not easy. I tried to expand $$|\Phi\rangle \langle \Phi| = \sum_{n,m\geq 0} \frac{(-1)^m}{n!m!} \lambda^{n+m} (b_1^\dagger b_2^\dagger -b_1b_2)^n |0\rangle \langle 0| (b_1^\dagger b_2^\dagger -b_1b_2)^m.$$ Since there are too many terms, taking partial trace seems not easy.

I am also interested in to what extent we can compute the entanglement measures explicitly, for Gaussian states.

  • 1
    $\begingroup$ Since it is Gaussian, you might find it much easier to do computations with its 2-point correlator matrix. Remember that wicks theorem guarantees this is all you need! Having said that, another straight forward approach in your case is to first diagonalize the exponent by doing a change of basis, i.e. define a new set of $c$ and $c^\dagger$. $\endgroup$ Commented Mar 14, 2023 at 8:18
  • $\begingroup$ If you use the straightforward approach, it is easier to start with evaluating $\langle n,m|\Phi\rangle$. Though this is likely not the easiest one. $\endgroup$
    – Roger V.
    Commented Mar 14, 2023 at 8:59
  • $\begingroup$ @flippiefanus even after tracing out $2$? $\endgroup$
    – Roger V.
    Commented Mar 14, 2023 at 10:24

1 Answer 1


This problem has the symmetry $1 \leftrightarrow 2$ and the operator in the exponent creates and annihilates only $12$ pairs . Therefore it admits a straightforward solution. Let's consider the state as a function of $\lambda$: $$ |\lambda \rangle = \exp\left(\lambda(b_1^\dagger b_2^\dagger - b_2 b_1)\right) |0\rangle. $$ This function

  • satisfies equation $$ \frac{d}{d \lambda} |\lambda\rangle = (b_1^\dagger b_2^\dagger - b_2 b_1)|\lambda\rangle \qquad(1) $$

  • satisfies initial condition $$ |\lambda\rangle\Biggr|_{\lambda = 0} = |0\rangle\qquad\qquad\qquad\qquad(2) $$

  • being a Gaussian state with the symmetry of the problem, it can be written as $$ |\lambda\rangle = A(\lambda)\exp\left(\phi(\lambda)b_1^\dagger b_2^\dagger \right)|0\rangle \qquad(3) $$ From (1-3) follow equations and initial conditions for $A(\lambda)$ and $\phi(\lambda)$: $$ \frac{d}{d\lambda} \phi = (1-\phi^2),\qquad \frac{d}{d\lambda} A = -\phi A, \qquad(4) $$ $$ \phi(0) = 0, \qquad A(0) = 1. \qquad(5) $$ The solution of the problem (4, 5) is $$ \phi(\lambda) = \tanh(\lambda),\qquad A(\lambda) = \frac1{\cosh(\lambda)}. \qquad (6) $$

Finally, substitution of (6) to (3) give $$ |\lambda\rangle = \frac1{\cosh(\lambda)}\sum_{n=0}^\infty \frac{\tanh^n(\lambda)}{n!} (b_1^\dagger)^n (b_2^\dagger)^n |0\rangle $$ It is now straightforward to obtain $$ \rho_1 = \frac1{\cosh^2(\lambda)}\sum_{n=0}^\infty \frac{\tanh^{2n}(\lambda)}{n!} (b_1^\dagger)^n |0\rangle\langle 0| (b_1)^n = \frac1{\cosh^2(\lambda)}\sum_{n=0}^\infty \tanh^{2n}(\lambda) |n\rangle\langle n| $$ and $$ \mbox{Tr} \rho_1^2 = \frac1{\cosh^4(\lambda)}\frac1{1-\tanh^4(\lambda)}. $$

  • $\begingroup$ Dear Gec, would you mind to elaborate the following statement: "being a Gaussian state with the symmetry of the problem, it can be written as..."? $\endgroup$ Commented Mar 14, 2023 at 9:26
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    $\begingroup$ @TobiasFünke Alas, I am not an expert in the field of Gaussian states, and my approach is rather intuitive. Isn't it known that Gaussian states of two bosons can be written in the form $\exp(a_1(b_1^\dagger)^2 + a_2(b_2^\dagger)^2 + cb_1^\dagger b_2^\dagger)|0\rangle$? If this fact is incorrect, my answer can be changed as follows: (3) is a natural ansatz for finding a solution to the problem (1,2) , $\endgroup$
    – Gec
    Commented Mar 14, 2023 at 9:43
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    $\begingroup$ @TobiasFünke long time ago I studied Berezin's book link From this book I've accepted an idea that any state $\exp(Q)|0\rangle$, with $Q$ being any quadratic operator (with restrictions in bosonic case), can be written in the form $A\exp(B)|0\rangle$, where $A$ is a constant and $B$ is quadratic operator constructed from creations operators only. $\endgroup$
    – Gec
    Commented Mar 14, 2023 at 9:58
  • $\begingroup$ Interesting; I have to think about it more. Thanks for the reference! $\endgroup$ Commented Mar 14, 2023 at 10:15
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    $\begingroup$ @eigenvalue, equation (3) is specific for this particular problem. In general $1\leftrightarrow 2$ symmetric case you will need to change (3) to $A\exp(\phi b_1^\dagger b_2^\dagger + \varphi(b_1^\dagger b_1^\dagger + b_2^\dagger b_2^\dagger))|0\rangle$. In the case without $1\leftrightarrow 2$ symmetry you will need to use the most general form $A\exp(\phi b_1^\dagger b_2^\dagger + \varphi_1 b_1^\dagger b_1^\dagger + \varphi_2 b_2^\dagger b_2^\dagger)|0\rangle$. $\endgroup$
    – Gec
    Commented Mar 15, 2023 at 6:29

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