2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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].

$\endgroup$
10
  • $\begingroup$ 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. $\endgroup$
    – Hafez
    Oct 15 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. $\endgroup$
    – Hafez
    Oct 15 at 6:29
  • 1
    $\begingroup$ 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 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 at 15:06
  • 1
    $\begingroup$ @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 at 15:09

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.