I am defining a density matrix via

$\rho = \frac{X^\dagger X}{\textrm{tr}(X^\dagger X)}$, where $X$ belongs to the Ginibre ensemble. This results in an induced distribution on the set of density matrices; let us call it $\mu$.

I am wondering if we can infer properties of $\mu$ given the underlying Ginibre ensemble...

for e.g. the average state $\int \textrm{d} \mu(\rho) \rho$, and more generally the average over a function of $\rho$ such as fidelity between $\rho$ and some fixed state $\sigma$, $\int \textrm{d} \mu(\rho) F(\rho, \sigma)$...

  • $\begingroup$ If I'm not mistaken, $X$ is a vector and the Ginibre ensemble is that of vectors whose components are standard centered normal variables. If this is the case, then it's a Haar-random matrix and many properties of such matrices have been discussed on this site: the average state is the maximally mixed one for instance, that is $\frac1dI_d$ $\endgroup$
    – Tristan Nemoz
    Commented Aug 28, 2023 at 15:29
  • $\begingroup$ I'm afraid not. The Ginibre ensemble is a set of random matrices such that each entry of a matrix is a complex number, with both the real and imaginary part sampled from the standard normal distribution. . $\endgroup$ Commented Aug 29, 2023 at 4:54

2 Answers 2


Let $X=A+\mathrm{i}B$, with $A,B\in\mathbb{R}^{N\times M}$ whose entries are standard normal centered gaussian variables.

We thus have: $$X^\dagger X=(A^\top-\mathrm{i}B^\top)(A+\mathrm{i}B)=A^\top A+B^\top B+\mathrm{i}(A^\top B-B^\top A)$$ We're interested in: $$\begin{align}\mathbb{E}\left[\frac{X^{\dagger}X}{\mathrm{tr}\left(X^\dagger X\right)}\right]&=\mathbb{E}\left[\frac{A^\top A+B^\top B+\mathrm{i}(A^\top B-B^\top A)}{\mathrm{tr}\left(A^\top A\right)+\mathrm{tr}\left(B^\top B\right)+\mathrm{i}\underbrace{\left(\mathrm{tr}\left(A^\top B\right)-\mathrm{tr}\left(B^\top A\right)\right)}_{0}}\right] \end{align}$$ Let us perform an analysis on a per-component basis. On the diagonal, say for the $x^{\text{th}}$ line and column, the coefficient in this density matrix is the expected value of: $$\frac{1}{\sum\limits_{j=1}^M\sum\limits_{i=1}^N\left(A_{i,j}^2+B_{i,j}^2\right)}\left[\sum_{i=1}^{N}A_{i,x}^2+\sum_{i=1}^{N}B_{i,x}^2+\mathrm{i}\underbrace{\left(\sum_{i=1}^NA_{i,x}B_{i, x}-\sum_{i=1}^NB_{i,x}A_{i, x}\right)}_0\right]$$ where all the $A_{i,j}$ and $B_{i,j}$ are centered standard normal random variables. We're thus interested in: $$\mathbb{E}\left[\frac{\sum\limits_{i=1}^NA_{i,x}^2+\sum\limits_{i=1}^NB_{i,x}^2}{\sum\limits_{j=1}^M\sum\limits_{i=1}^NA_{i,j}^2+B_{i,j}^2}\right]$$

We can write it as: $$\frac{S_1}{S_1+S_2}$$ where $S_1\sim\chi^2(2N)$ and $S_2\sim\chi^2(2MN-2N)$. Because of the link between the $\chi^2$ and $\Gamma$ distribution, we have $S_1\sim\Gamma(N, 2)$ and $S_2\sim\Gamma(N(M-1), 2)$, which means that $\frac{S_1}{S_1+S_2}\sim\beta(N, N(M-1))$, whose expected value is $\frac{N}{N(M-1)+N}=\frac{1}{M}$. Thus, every diagonal coefficient in this density matrix is equal to $\frac1M$, which was expected: there's no reason to prefer a computational state to another.

If we now take the coefficient at line $x$ and column $y$, the expression of this coefficient is: $$\frac{1}{\sum\limits_{j=1}^M\sum\limits_{i=1}^N\left(A_{i,j}^2+B_{i,j}^2\right)}\left[\sum_{i=1}^{N}A_{i,x}A_{i,y}+\sum_{i=1}^{N}B_{i,x}B_{i,y}+\mathrm{i}\left(\sum_{i=1}^NA_{i,x}B_{i, y}-\sum_{i=1}^NB_{i,x}A_{i, y}\right)\right]$$ We thus want to compute: $$\mathbb{E}\left[\frac{A_{i,x}A_{i,y}}{\sum\limits_{j=1}^M\sum\limits_{i=1}^NA_{i,j}^2+B_{i,j}^2}\right]=\mathbb{E}\left[\frac{XY}{X^2+Y^2+S}\right]$$ since this will give us the expectation we're looking for by linearity. Here, $X$ and $Y$ are independent and sampled from a centered standard gaussian, while $S\sim\chi^2(NM-2)$ is also independent from $X$ and $Y$. What is important here is that $X$ (and $Y$, though it doesn't matter because of the independence) is centered around $0$. Indeed, computing this expectation yields, using the fact that $f_{X,Y,S}(x,y,s)=f_X(x)f_Y(y)f_S(s)$: $$\mathbb{E}\left[\frac{XY}{X^2+Y^2+S}\right]=\int_{0}^{+\infty}\left[\int_{-\infty}^{+\infty}\left(y\int_{-\infty}^{+\infty}\frac{x}{x^2+y^2+s}f_X(x)\,\mathrm{d}x\right)f_Y(y)\,\mathrm{d}y\right]f_S(s)\,\mathrm{d}s$$ Since $f_X$ is even, clearly the middle integrand is odd, which means that the integral is nil. All in all, this expectaion is equal to $0$.

Thus, the final density matrix is the maximally mixed state $\frac1MI_M$.

Now that we know that $\rho$ is the maximally mixed state, it is easy to compute $F(\rho, \sigma)$: $$F(\rho, \sigma)=\frac1M\mathrm{tr}(\sqrt{\sigma})=\frac1M\sum_{i=1}^M\sqrt{\lambda_i}$$ where the $\lambda_i$ are the eigenvalues of $\sigma$.

For the average Fidelity between $\sigma$ and some $\rho$ taken from $\mu$, I'm not sure much can be said (maybe someone will add to this answer). If $\sigma=|\psi\rangle\langle\psi|$ is a pure state, this will give $\int\langle\psi|\rho|\psi\rangle\,\mathrm{d}\mu(\rho)=\langle\psi|\int\rho\,\mathrm{d}\mu(\rho)|\psi\rangle=\frac1M$, which corresponds to the previous case, but I'm not sure this is still true for general mixed states.

  • $\begingroup$ So how is the induced measure $d\mu(\rho) = d\mu(A+iB)$? $\rho$ is non-linear in X and hence A and B. Think transformation of probability measure under a function. $\endgroup$ Commented Sep 1, 2023 at 4:34
  • $\begingroup$ @Ghost-of-PPPF it's just an abuse of notation, it should really be $\mathrm{d}\mu\left(\left(A^\top-\mathrm{i}B^\top\right)(A+\mathrm{i}B)\right)$. We simply decomposed $X$ into its real and imaginary parts. The gist of it is that we're taking an average of some function of centered standard normal random variables. You can just write the equations with $\mathbb{E}$ instead of the integral and it should work just fine $\endgroup$
    – Tristan Nemoz
    Commented Sep 1, 2023 at 9:16
  • $\begingroup$ No there is also a denominator tr (X^\dagger X). In any case the induced measure is not a uniform distribution i.e. NOT a centered Gaussian with mean zero due to this non-linear change of variables. $\endgroup$ Commented Sep 2, 2023 at 4:53
  • $\begingroup$ You have done the calculation on the imaginary part assuming a linear transform of variables which is simply not true. $\endgroup$ Commented Sep 2, 2023 at 5:07
  • $\begingroup$ @Ghost-of-PPPF I did not simply use the linearity of the expectation, but the fact that the density was symmetric around 0, which ensured a nil expectation. I've added more details, can you tell me whether that solves your concern? $\endgroup$
    – Tristan Nemoz
    Commented Sep 2, 2023 at 14:15

Turns out this induced measure is the so called the Hilbert-Schmidt measure: https://arxiv.org/pdf/quant-ph/0012101.pdf


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