Relation between Mutual information and Expectation Values

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The discussion focuses on calculating mutual information and expectation values in quantum mechanics. The mutual information between Alice and Bob has been computed as I(ρ_A:ρ_B) = 2. A challenge arises in computing the expectation value ⟨ψ|O_A|ψ⟩ due to mismatched matrix sizes, with the user seeking guidance on resolving this issue. The user successfully calculated ⟨ψ|O_A ⊗ O_B|ψ⟩ as 7/12 by using the tensor product of the matrices. Suggestions were made to adjust the calculations by incorporating identity matrices to align the dimensions for the expectation values.
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Homework Statement
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*}
\mathcal{O}_A =
\begin{pmatrix}
1 & 0 \\ 0 & \frac{1}{2}
\end{pmatrix}
, \qquad \mathcal{O}_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 | \mathcal{O}_A \otimes \mathcal{O}_B| \psi\rangle - \langle\psi |\mathcal{O}_A|\psi \rangle \langle\psi |\mathcal{O}_B|\psi \rangle)^2 $
Relevant Equations
\begin{align*}
|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).
\end{align*}
Consider the pair of observables
\begin{align*}
\mathcal{O}_A =
\begin{pmatrix}
1 & 0 \\ 0 & \frac{1}{2}
\end{pmatrix}
, \qquad \mathcal{O}_B =
\begin{pmatrix}
1 & 0 \\ 0 & \frac{1}{3}
\end{pmatrix}
.
\end{align*}
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 $\langle\psi |\mathcal{O}_A|\psi \rangle$ 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 $\mathcal{O}_A$ 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 $\langle\psi | \mathcal{O}_A \otimes \mathcal{O}_B| \psi\rangle$ 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
$$\langle\psi | \mathcal{O}_A \otimes \mathcal{O}_B| \psi\rangle = \frac{7}{12}$$.

I would appreciate any insight on this.
 
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Here it is edited:
Show the mutual information between Alice and Bob is larger than ##(\langle\psi | \mathcal{O}_A \otimes \mathcal{O}_B| \psi\rangle - \langle\psi |\mathcal{O}_A|\psi \rangle \langle\psi |\mathcal{O}_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 |\mathcal{O}_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 ##\mathcal{O}_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 | \mathcal{O}_A \otimes \mathcal{O}_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...

I think that for ##\langle\psi |\mathcal{O}_A|\psi \rangle## you calculate ##\langle\psi | \mathcal{O}_A \otimes \mathcal I| \psi\rangle##, and for ##\langle\psi |\mathcal{O}_B|\psi \rangle##, it is ##\langle\psi |\mathcal I \otimes \mathcal{O}_B | \psi\rangle##.

(But I am new to it, too.)
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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