Is a "Bell measurement" equivalent to a projective measurement in a different basis?

Question

Let \begin{align} H = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix}\in M_2(\mathbb C), \quad \mathrm{CNOT} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{bmatrix} \in M_4(\mathbb C), \quad U = (H\otimes I_2)\mathrm{CNOT}.\end{align} Then the Bell state measurement is the unitary channel given by $\Phi:\rho \mapsto U\rho U^*$.

The Bell states are \begin{align}\beta_{i,j} = \frac{|0,j\rangle + (-1)^i|1, j\oplus 1\rangle}{\sqrt{2}}, \qquad i,j\in\{0,1\}. \end{align} Let $P_{i,j}$ be the orthogonal projection onto the subspace spanned by $\beta_{i,j}$. Then $\{P_{i,j}:i,j\in\{0,1\}\}$ forms a projective measurement.

I want to understand if the Bell measurement $\Phi$ is the same as the projective measurement $\{P_{i,j}\}$, in the sense whether one can be derived from the other? I thought in this case, the Bell states should be the eigenvectors of $U$, but they are not.

In a more general sense, given a unitary channel, is it possible to construct a projective measurement from it, and vice versa?

Answer 1

The process of measuring $\rho$ in a basis $\{|u_k\rangle\}$ after evolution through a unitary $U$ results in the outcome probabilities: $$p_k = \langle u_k|U\rho U^\dagger |u_k\rangle= \operatorname{Tr}[\mathbb{P}(|u_k\rangle) U\rho U^\dagger], \qquad \mathbb{P}(|u\rangle)\equiv |u\rangle\!\langle u|.$$ You can notice that these probabilities are identical to those resulting from directly measuring $\rho$ in the basis $\{U^\dagger|u_k\rangle\}$: $$\operatorname{Tr}(\mathbb{P}(|u_k\rangle) U\rho U^\dagger) = \operatorname{Tr}[\mathbb{P}(U^\dagger |u_k\rangle) \rho].$$ You can think this as a special case of the more general statement that measuring $\rho$ with a POVM $\{\mu_k\}$ after evolution through a channel $\Phi$ is equivalent to directly measuring $\rho$ with the POVM $\{\Phi^\dagger(\mu_k)\}$, where $\Phi^\dagger$ is the adjoint channel of $\Phi$, and one can show that $\{\Phi^\dagger(\mu_k)\}$ is indeed always a POVM. This amounts to the observation that the outcome probabilities read $$p_k = \langle \mu_k,\Phi(\rho)\rangle = \langle\Phi^\dagger(\mu_k),\rho\rangle,$$ where $\langle \cdot,\cdot\rangle$ is here the L2 inner product of matrices: $\langle A,B\rangle\equiv \operatorname{Tr}(A^\dagger B)$.

You can also understand these statements as thinking in the Schrodinger picture vs Heisenberg's one: you either think of states as evolving and then being measured, or as the states being fixed while the measurements "back-evolve".

Answer 2

If I am reading you right, you are asking how the measurement performed in the Bell Measurement to determine the bell state is related to the projective measurments onto the subspaces those states span.

Their equivalence comes from the equivalence between the set of product projectors performed on the two qubits after the above unitary disentangles them, and the non-local projector performed on the state if the unitary hasn't taken place.

ie, given $$\beta_{i,j} = \frac{|0,j\rangle + (-1)^i|1, j\oplus 1\rangle}{\sqrt{2}}$$

the state after the unitary interaction will be $$\beta_{i,j} = \frac{|i\rangle \otimes|j\rangle}{\sqrt{2}}$$

The first projective measurement, performed before the interaction, and the second, performed after, can be related to eachother in that they obtain the same statistics before and after the measurement is taken.

As for deriving the latter from the former, you have to do it by performing the unitary interaction on the state that the projector is the density operator of.