I want to know the analytical solution of $\mathbb{E}_{\psi}\frac{\langle \psi |A|\psi\rangle}{\langle \psi |A^2|\psi\rangle}$. I see similar questions before approximate average, but it does not provide a very specific calculation method. So I want to know if there is any method to calculate this expectation, and it would be best if some simple examples being given, such as when A is a density operator. If not, is there anything similar in classical probability theory? Please advise.


1 Answer 1


This is only a partial answer since it only considers diagonalizable matrices. Maybe the fact that those are dense in the space of matrices could be used to extend the result to all matrices?

Let $A$ and $B$ be two commuting diagonalizable matrices. Since they commute, they are jointly diagonalizable, which means that they share their eigenvectors. Let us denote $\lambda_i$ the eigenvalues of $A$, $\mu_i$ those of $B$ and $\left|\varphi_i\right\rangle$ the associated eigenvectors.

Note that the expectation we want to compute is the same if we consider Haar-random $|\psi\rangle$ that aren't unit vectors. It means that we can write: $$|\psi\rangle=\sum_j\left(X_j+\mathrm{i}Y_j\right)\left|\varphi_j\right\rangle$$ with the $X_j$ and $Y_j$ being independently sampled from a $\mathcal{N}(0, 1)$ distribution. We then have: $$\begin{align*} \mathbb{E}\left[\frac{\langle\psi|A|\psi\rangle}{\langle\psi|B|\psi\rangle}\right] &= \mathbb{E}\left[\frac{\sum\limits_j\lambda_j\left(X_j^2+Y_j^2\right)}{\sum\limits_k\mu_k\left(X_k^2+Y_k^2\right)}\right]\\ &= \sum_j\lambda_j\mathbb{E}\left[\frac{X_j^2+Y_j^2}{\sum\limits_{k}\mu_k\left(X_k^2+Y_k^2\right)}\right]. \end{align*}$$ Let us denote $E_j$ the inner expectation. First of all, note that we have $E_j=E_k$ for all $j$ and $k$. Let us denote $E$ this value, that is $E_j=E$ for all $j$. Furthermore, we have: $$\sum_j\mu_jE_j=\mathbb{E}\left[\frac{\sum\limits_{j}\mu_j\left(X_j^2+Y_j^2\right)}{\sum\limits_{k}\mu_k\left(X_k^2+Y_k^2\right)}\right]=1.$$ Thus: $$E\sum_j\mu_j=1$$ which gives us $E=\frac{1}{\mathrm{tr}(B)}$. Thus: $$\mathbb{E}\left[\frac{\langle\psi|A|\psi\rangle}{\langle\psi|B|\psi\rangle}\right]=\sum_j\lambda_j\frac{1}{\mathrm{tr}(B)}=\frac{\mathrm{tr}(A)}{\mathrm{tr}(B)}.$$ In particular, the expectation you're looking for is equal to $\frac{\mathrm{tr}(A)}{\mathrm{tr}\left(A^2\right)}$ if $A$ is diagonalizable. Note that this includes density operators.


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