Proving $\langle j_2|\langle j_1| U(|0\rangle\!\langle0|\otimes\rho)U^\dagger|j_2\rangle|j_1\rangle =\operatorname{Tr}(M_j\rho)$

I'm trying to prove that:

$$\langle j_2|\langle j_1| U(|0\rangle\!\langle0|\otimes\rho)U^\dagger|j_2\rangle|j_1\rangle =\operatorname{Tr}(M_j\rho)$$

where $$\rho$$ is the density operator, $$M_j=\frac{1}{2}|\psi_j\rangle\!\langle\psi_j|$$, and $$U$$ is unitary.

Assuming $$\rho = a|0\rangle\langle 0|+b|1\rangle \langle 1|$$

and $$M = \frac{1}{2}|0\rangle\langle 0|+ \frac{1}{2}|1\rangle\langle 1|$$

I got :

$$\mathrm{Tr}(M_j\rho) = \mathrm{Tr}(\frac{1}{2} (a|0\rangle\langle 0|+b|1\rangle \langle 1|)) = \frac{1}{2}ab$$

I calculated $$(|0⟩⟨0|\otimes \rho)$$ as well (a matrix of 4x4)

$$\vert 0 \rangle \langle 0 \vert \otimes \rho = \begin{pmatrix} 1/2 & 0 & 0 & 0\\ 0 & 1/2 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &0\\ \end{pmatrix}$$

Not sure though how to continue from here and apply $$U$$ to the matrix.

• I'm a tad confused by what $\left|j_1\right>$, $\left|j_2\right>$ and $\left|\psi_j\right>$ are - would you be able to give the definitions of these? Thanks! – Mithrandir24601 Oct 22 '19 at 21:33

I would start by writing $$\langle j_2|\langle j_1| U(|0\rangle\!\langle0|\otimes\rho)U^\dagger|j_2\rangle|j_1\rangle=\text{Tr}\left(U(|0\rangle\!\langle0|\otimes\rho)U^\dagger|j_2\rangle\langle j_2|\otimes|j_1\rangle\langle j_1|\right).$$ If it's not clear to you why that's correct, work backwards - when you take the trace you can sum over any orthonormal basis. Just pick any basis such that $$|j_2\rangle|j_1\rangle$$ is a member of that basis, so all other elements are orthogonal.
Next, I can use the cyclic properties of the trace to move the $$U$$. $$\text{Tr}\left((|0\rangle\!\langle0|\otimes\rho)U^\dagger|j_2\rangle\langle j_2|\otimes|j_1\rangle\langle j_1|U\right)$$ Next, define $$|\Psi\rangle=U^\dagger|j_2\rangle|j_1\rangle$$, so this is the same as $$\text{Tr}\left((|0\rangle\!\langle0|\otimes\rho)|\Psi\rangle\langle \Psi|\right)$$
Now, I can perform the trace by first applying the partial trace on the first qubit, and then the trace on the second qubit: $$\text{Tr}\text{Tr}_1\left((|0\rangle\!\langle0|\otimes\rho)|\Psi\rangle\langle \Psi|\right)=\text{Tr}\left(\rho\cdot\text{Tr}_1\left((|0\rangle\!\langle0|\otimes I)|\Psi\rangle\langle \Psi|\right)\right).$$ So, if we define $$M=\text{Tr}_1\left((|0\rangle\!\langle0|\otimes I)|\Psi\rangle\langle \Psi|\right),$$ then the answer is expressed as $$\text{Tr}(M\rho)$$.
Finally, we can certainly write (without knowing anything more about it) that $$|\Psi\rangle=\alpha|0\rangle|\phi_0\rangle+\beta|1\rangle|\phi_1\rangle,$$ for an arbitrary pair of single-qubit states $$|\phi_0\rangle,|\phi_1\rangle$$ and $$|\alpha|^2+|\beta|^2=1$$, which would mean that $$M=|\alpha|^2|\phi_0\rangle\langle\phi_0|.$$ If you want it to be the more specific form given in the question ($$|\alpha|^2=\frac12$$), then you need more information about $$U$$.