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.

  • $\begingroup$ 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! $\endgroup$
    – Mithrandir24601
    Oct 22, 2019 at 21:33

1 Answer 1


In the following I'm assuming you're talking about a two-qubit system. You don't explicitly say, but I infer from the calculation you're showing. If not, this can all be generalised easily enough...

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 $U$.


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