Assume having a one-mode quantum Gaussian state with quadrature observable vector $\hat r = [\hat q , \hat p ] $ and covariance matrix $\sigma$. According to definition [1]: \begin{equation} \sigma = \text{tr}\left(\begin{bmatrix} \hat q^2 & \frac{1}{2}\{\hat q, \hat p\}\\ \frac{1}{2}\{\hat p, \hat q\} & \hat p^2 \end{bmatrix} \rho \right) \end{equation} My question is how can we show the covariance matrix as a function of the average photon number $N = \text{tr}(\hat a^\dagger \hat a \rho)$? I have found an answer in [2] section III.B. (gauge-invariant states) which states the covariance matrix as: \begin{equation} \alpha = \begin{bmatrix} \text{Re}N + I/2 & -\text{Im}N \\ \text{Im}N & \text{Re}N + I/2 \end{bmatrix} \end{equation} But it is confusing to me as these two cannot be equal to each other as the off-diagonal elements in the second one have opposite signs while the off-diagonal elements of the first one are the same.

EDIT: Would you please also explain about the feasibility and meaning of $\text{Im}N$? I thought $N$ is physical observable thus it can just have real values representing the average number of photons.

Any help or reference is highly appreciated. Thanks.

[1] C. Weedbrook et al., “Gaussian quantum information,” Rev. Mod. Phys., vol. 84, no. 2, pp. 621–669, May 2012, doi: 10.1103/RevModPhys.84.621.

[2] A. Holevo and R. Werner, “Evaluating capacities of bosonic Gaussian channels,” Phys. Rev. A, vol. 63, no. 3, p. 032312, Feb. 2001, doi: 10.1103/PhysRevA.63.032312.


1 Answer 1


Recall that $a=(q+ip)/\sqrt{2}$ in some dimensionless units (Weedbrook might change the units because I think they like $\hbar=2$; I'm using $[q,p]=i$ and $[a,a^\dagger]=1$). We can thus find the equality $$a^\dagger a=(q-ip)(q+ip)/2=[q^2+p^2+i(qp-pq)]/2=\frac{q^2+p^2-1}{2}.$$ This relates the total number of photons to the trace of your matrix $\sigma$.

Your definition of $N$, which is standard, must necessarily be real. Still, it cannot fully characterize the state, because there are multiple Gaussian states with the same average photon number $N$. These possibilities are arranged in the relative sizes of the diagonal elements of $\sigma$ and the magnitude and phase of its two off-diagonal elements (the two off-diagonal elements are equal, because $\{A,B\}=\{B,A\}$). I presume the latter source is using some other definition to encode the other parameters in the "phase" of some other variable $N$. Now that I check, it indeed does; it assumes $a$ and $a^\dagger$ are vectors of operators; when they are scalar, their definition of $N$ should coincide with yours. When we use vectors, it is indeed possible that $N_{ij}=\langle a_i^\dagger a_j\rangle$ be complex for $i\neq j$.

The two resources use different definitions for their covariance matrices but [2] doesn't seem to realize it, so I would rely on [1] or trace through the mistakes in [2]. Both define the same vector $$x^{[1]}=R^{[2]}=(q_1,p_1,\cdots,q_n,p_n)^\top,$$ from which they both have $$\sigma_{ij}^{[1]}=\alpha_{ij}^{[2]}=\frac{1}{2}\langle \{R_i-\langle R_i\rangle,R_j-\langle R_j\rangle\}\rangle,$$ so the two definitions should be the same. However, [2] quotes their Ref. [13] to define $\alpha$, and in that reference they arrange their parameters in a differente order: $$R^{[13]}=(q_1,\cdots,q_n,p_1,\cdots,p_n)^\top.$$ This means that for $n>1$ the definitions will not match up.

Okay, so what about the covariance matrix not being symmetric? In the case of $n=1$ we don't have to worry, because $\mathrm{Im}N=0$ and the two formulas match up because Ref. [2] is specifically looking at states with $\langle qp+pq\rangle=\langle a^2 - a^{\dagger 2}\rangle/2i=0$. In the case of $n>1$, all of the elements of $V$ are real, because they all correspond to expectation values of Hermitian operators, and $V$ is explicitly symmetric. Why isn't $\alpha$ symmetric? If we go back to Ref. [2]'s Ref. [13], they indeed define this asymmetric $\alpha$ with some $-\mathrm{Im}N$, but then they go on to treat $\alpha$ as being symmetric, saying "For arbitrary real symmetric matrix $\alpha$" and giving the $n=1$ example with $$\alpha=\begin{pmatrix}\alpha^{qq}&\alpha^{qp}\\\alpha^{qp}&\alpha^{pp}\end{pmatrix};$$ notably, they do not say $\alpha^{qp}=-\alpha^{pq}$, so I'm even less inclined to trust the details of [2].

  • $\begingroup$ Thanks a lot for your reply. Indeed the traces of the two matrices are equal as you correctly mentioned. However, I don't think that vectorizing $a, a\dagger$ would have anything to do with the off-diagonal elements being different. In [2] he just puts the ladder operators of the different modes into a big ladder vector for simpler notation and vector algebra. As we are discussing the one-mode case, the operators will automatically be scalar and 2x2 matrices. But still, the problem that off-diagonal elements of the covariance matrix of [2] having different signs is not solved for me. $\endgroup$ Oct 15, 2021 at 6:17
  • $\begingroup$ My confusion comes from the fact that in [1] as you said because $\{A,B\}= \{B,A\}$ the off-diagonals are equal. But in [2], the ImN and -ImN are different unless N is real (i.e. $p, q$ anti-commute). And finding the rigorous proof for this equality is very important for me because in [2] he makes a correspondence mapping from the 2x2 matrix with that form into complex scalar numbers and continues with them, which means the off-diagonal elements are important to consider. $\endgroup$ Oct 15, 2021 at 6:29
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    $\begingroup$ When we vectorize, we have off-diagonal components like $N_{ij}=\langle a_i^\dagger a_j\rangle$, which are not guaranteed to be real $\endgroup$ Oct 15, 2021 at 15:04
  • $\begingroup$ Also, the states in [2] are not the most general Gaussian states: they are invariant under $a\to a \exp(i \phi)$ $\endgroup$ Oct 15, 2021 at 15:06
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    $\begingroup$ @Hafez you are correct, the two covariances should be the same in their original definitions, but it seems as though [2] misquotes a source (by the same authors) that might itself have mistakes in it $\endgroup$ Oct 19, 2021 at 15:09

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