# Mutual information of shared state is larger than expectation values

Im trying to prove the following identity for a special case:

Alice and Bob share the Bell state \begin{align*} |\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). \end{align*} Consider the pair of observables \begin{align*} A = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{2} \end{pmatrix} , \qquad B = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{3} \end{pmatrix} . \end{align*}

Show the mutual information between Alice and Bob is larger than $$(\langle\psi | A \otimes B| \psi\rangle - \langle\psi |A|\psi \rangle \langle\psi |B|\psi \rangle)^2$$

I've make progress in obtaining the values for the mutual information using the following: $$I(\rho_A:\rho_B) = S(\rho_A) +S(\rho_B) - S(\rho_{AB}) = 1 + 1 - 0 = 2.$$

I would like to compute the expectation but I'm facing a problem in the case of $$\langle\psi |A|\psi \rangle$$ since the size of matrices in this multiplication do not match. namely, $$\langle\psi|$$ is of size $$1\times 4$$ and $$|\psi \rangle$$ is of size $$4\times 1$$ and the matrix $$A$$ is $$2 \times 2$$. I'm very new to the subject and I would greatly appreciate if I could have some guidance on how the computations for this expectation would be carried out.

additionally, I have computed the $$\langle\psi | A \otimes B| \psi\rangle$$ by first computing the tensor product of the two matrices $$A,B$$ and then taken the multiplication with the Bra and Ket of the state respectively deducing $$\langle\psi | A \otimes B| \psi\rangle = \frac{7}{12}\,.$$

I would appreciate any insight on this.

• I believe the statement is asking you to find $$\bigg(\langle\psi | A \otimes B| \psi\rangle - \langle\psi |A \otimes I |\psi \rangle \langle\psi |I \otimes B|\psi \rangle\bigg)^2$$ Commented Dec 1, 2023 at 9:27
• this seems to ask about an inequality between quantum mutual information and covariance with respect to a specific pair of observables. Out of curiosity, what's the context here?
– glS
Commented Dec 1, 2023 at 14:02