# How can we prove that the covariance satisfies $\mathrm{Cov}_\rho(X,Y)=\mathrm{Cov}_\rho(Y,X)$?

While attempting to prove the Cauchy Schwarz Inequality I came across this problem. First of all, if we are given a $$\rho$$ density matrix and two matrix of obserables $$X,Y$$, after defining the Covariance as: $$COV_{\rho}(X,Y) = \mathbb{E}_{\rho}[(X-\mu_XI)^{\dagger}(Y-\mu_YI)]$$ where $$\mathbb{E}_{\rho}(X)= tr(\rho^{\dagger}X) = \sum_{ij}\rho_{ij}X_{ij}$$ and $$\mu_X = \mathbb{E}_{\rho}(X)$$ and the same for Y.

How can we formally prove that: $$COV_{\rho}(X,Y)= COV_{\rho}(Y,X)$$

I was considering that since $$X,Y$$ are both observable then a meaningful conclusion can be taken by that, since $$X^{\dagger} = X$$ but what if $$X$$ is not necessarily like this? Would it still hold?

• What does this have to do with CS? – Norbert Schuch Apr 10 at 21:47
• for $\rho$ pure this covariance is an inner product, thus at least connected to CS (which is one way to get to the uncertainty relations). I'm not sure if this still holds for a general $\rho$ though – glS Apr 11 at 8:21

## 1 Answer

I don't believe this holds even for Hermitian matrices. The main reason being is that $$\mathrm{tr}[\rho X Y ] \neq \mathrm{tr}[\rho Y X]$$ in general. This is a problem because we see that $$COV_{\rho}(X,Y) = \mathrm{tr}[\rho X Y] - \mu_X \mathrm{tr}[\rho Y] - \mu_Y \mathrm{tr}[\rho X] + \mu_X \mu_Y$$ and $$COV_{\rho}(Y,X) = \mathrm{tr}[\rho Y X] - \mu_X \mathrm{tr}[\rho Y] - \mu_Y \mathrm{tr}[\rho X] + \mu_X \mu_Y.$$ And so $$COV_{\rho}(X,Y) = COV_{\rho}(Y,X)$$ iff $$\mathrm{tr}[\rho X Y] = \mathrm{tr}[\rho Y X]$$. But this doesn't hold in general. For instance take $$\rho = |0\rangle \langle 0 |$$, $$X = \sigma_x$$ and $$Y = \sigma_Y$$. Then $$\mathrm{tr}[\rho X Y] = \mathrm{tr}[\rho i \sigma_z] = i$$ but $$\mathrm{tr}[\rho Y X] = \mathrm{tr}[\rho (-i \sigma_z)] = - i.$$

Result holds in absolute value

However we will always have $$|COV_{\rho}(X,Y)| = |COV_{\rho}(Y,X) |$$ to see this note that all terms above apart from $$\mathrm{tr}[\rho X Y]$$ and $$\mathrm{tr}[\rho Y X]$$ will be real. But for these two terms we have $$\mathrm{tr}[\rho X Y] = \mathrm{tr}[X Y \rho] = \mathrm{tr}[(\rho Y X)^{\dagger}] = \overline{\mathrm{tr}[\rho Y X]}$$ where $$\overline{z}$$ denotes complex conjugate. Thus $$COV_{\rho}(X,Y) = \overline{COV_{\rho}(Y,X)}$$ and therefore $$|COV_{\rho}(X,Y)| = |COV_{\rho}(Y,X) |$$.

Result holds in absolute value for general $$X$$ and $$Y$$

For the same reasons as above the result will hold even when $$X$$ and $$Y$$ are not Hermitian: $$\mathrm{tr}[\rho X^\dagger Y] = \mathrm{tr}[X^\dagger Y \rho] = \mathrm{tr}[(\rho Y^\dagger X)^{\dagger}] = \overline{\mathrm{tr}[\rho Y^\dagger X]}$$

• I will look into your claim. However, it is definitely strange because I expected the Cauchy schwarz inequality to hold – simonegiancola09 Apr 10 at 20:58
• Actually, maybe it holds as a modulus operation, because your claim is working even if I do not completely get the meaning of sigma in the notation. It might be that $$|Cov_{\rho}(X,Y)| = |Cov_{\rho}(Y,X)|$$ – simonegiancola09 Apr 10 at 21:01
• Sorry the sigma notation is the Pauli matrices, e.g., $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. – Rammus Apr 10 at 21:37
• And yes, it will hold in absolute value because $\mathrm{tr}[\rho X Y] = \mathrm{tr}[X Y \rho] = \mathrm{tr}[(\rho Y X)^{\dagger}] = \overline{\mathrm{tr}[\rho Y X]}$. Where $\overline{z}$ denotes complex conjugate. – Rammus Apr 10 at 21:44
• Do you have any idea if this applies also to matrices which are not hermitian? – simonegiancola09 Apr 10 at 22:05