What is a proof for the principle of deferred measurement?

Question

Does anyone know of a proof for the 'principle of deferred measurement'?

Answer 1

I'll assume the "principle of deferred measurement" refers to the equality shown in the figure in the relevant Wikipedia page: the possibility of writing a "classically conditioned" operation as a standard conditional operation followed by a measurement (where measurement basis and basis used for the conditional operation are presumably the same).

(Measure and then conditional operation) Consider a physical situation where one register is measured, and a channel is applied to another register conditionally to the measurement outcome. We can actually model this situation in full generality for quantum channels and POVMs. If you only care about pure states and unitary evolutions, you can replace in the following POVMs with standard projective measurements (i.e. $\mu(b)=|b\rangle\!\langle b|$), channels with unitary gates (i.e. $\Phi_b(X)= U_b X U_b^\dagger$), and generic operators/density matrices with pure states (i.e. $X=|\psi\rangle\!\langle\psi|$). In fact, if only pure states are of interest, you can outright skip this section and jump the next one, as I'll specialise the formalism to pure states there anyway.

Assume a bipartite space $\newcommand{\Tr}{\operatorname{Tr}}{\cal X\otimes Y}$ for some finite dimensional vector spaces $\mathcal X,\mathcal Y$. Denote with $\operatorname{Pos}(\mathcal X)$ the set of positive semidefinite operators on $\mathcal X$, with $\Sigma$ some finite set labeling measurement outcomes, and with $\mathrm C(\mathcal X,\mathcal Y)$ the set of channels from $\mathcal X$ to $\mathcal Y$.

Suppose you measure on ${\cal X}$ with some POVM $\{\mu(b):b\in\Sigma\}\subset\operatorname{Pos}({\cal X})$, and the overall input state is some $\rho\in\operatorname{Pos}(\mathcal X\otimes\mathcal Y)$. If the measurement gives the outcome $b$, which happens with probability $p_b\equiv \operatorname{tr}[(\mu_b\otimes I)\rho]$, the residual state on the other register "collapses" to $$\rho_{\cal Y}(b)=\frac{\operatorname{tr}_{\cal X}[(\mu_b\otimes I_{\cal Y})\rho]}{p_b}.$$ If a channel $\Phi_b\in\mathrm C(\mathcal Y)$ is applied to the second register conditionally to observing the outcome $b$ on the first register, then the overall final state, disregarding the measurement outcome, is going to be $$\sum_b p_b \Phi_b(\rho_{\cal Y}(b)) = \sum_b \Phi_b(\operatorname{tr}_{\cal X}[(\mu_b\otimes I_{\cal Y})\rho]).$$ A concise way to rewrite this is as $$\sum_b (\mu_b^\star\otimes \Phi_b)(\rho),$$ having defined the quantum map $\mu_b^\star\in\mathrm C(\mathcal X,\mathbb{C})$ as $\mu_b^\star(\sigma)\equiv \langle\mu_b,\sigma\rangle$. This is but a formal trick to get a concise expression for the channel $\Psi$ describing the overall measurement+channel evolution, which can therefore be concisely written as $\Psi=\sum_b \mu_b^\star\otimes\Phi_b$.

If on the other hand the observed measurement outcome is remembered, then we can write the final state as $$\sum_b p_b \mathbb{P}_b\otimes\Phi_b(\rho_{\cal Y}(b)) = \sum_b (M_b\otimes\Phi_b)(\rho), \qquad \mathbb{P}_b\equiv |b\rangle\!\langle b|,$$ where I defined the quantum maps $M_b(\sigma)\equiv \mathbb{P}_b \langle\mu_b,\sigma\rangle$. Note that $M\equiv \sum_b M_b$ would be the (entanglement breaking) measurement channel describing the act of measuring with the POVM $\mu$ and remembering the result in a classical register.

(Conditional operation and then measure) Our goal would be now to show that we can equivalently perform some operation on the state and only afterwards measure the first register, and obtain the same identical final state.

Working in the general formalism of quantum channels, however, actually makes this a bit of a trivial question. The reason is that a channel might itself model an evolution involving a measurement, which would make the question moot. In fact, we could simply define the "conditioned channel" $\tilde\Phi\equiv\sum_k |k\rangle\!\langle k|\otimes\Phi_k$, and trivially find that performing a computational basis measurement after applying this channel is identical to what we obtained before measuring and then applying the channel. But this $\tilde\Phi$ corresponds in practice to applying $\Phi_k$ conditioned to a measurement outcome on the first register, and so this solution is useless in this context.

To get a better idea of how and where exactly measurements are involved, let's specialise our formalism above to pure states, unitary evolutions, and projective measurements. Let $|\Psi\rangle$ be the input state. Say we measure in the computational basis the first register (we don't lose any generality assuming computational basis measurement: it amounts to a specific choice of basis to describe the states). Then the possible post-measurement states on the second register are $$|\Psi_b\rangle\equiv \frac{1}{p_b} (\langle b|\otimes I)|\Psi\rangle, \qquad p_b\equiv \|(\langle b|\otimes I)|\Psi\rangle\|^2,$$ and $|\Psi_b\rangle$ is found with probability $p_b$. If the unitary $U_b$ is applied conditionally to observing the $b$-th outcome, the final state is $$U_b |\Psi_b\rangle\equiv \frac{1}{p_b}(\langle b|\otimes U_b)|\Psi\rangle.$$

Suppose we instead deferred measurement to the end, and replaced the classically-conditioned unitary operation with the controlled-unitary operation $U\equiv \sum_b |b\rangle\!\langle b|\otimes U_b$. This unitary gives the evolution $$|\Psi\rangle\to U|\Psi\rangle = \sum_b (|b\rangle\!\langle b|\otimes U_b)|\Psi\rangle = \sum_b p_b |b\rangle \otimes U_b |\Psi_b\rangle.$$ This last expression makes the conclusion immediate: if we now measure the first register of $U|\Psi\rangle$ in the computational basis, we'll get the $b$-th outcome with probability $p_b$, and the corresponding post-measurement state on the second register is going to be $U_b |\Psi_b\rangle$, which is exactly what we got doing the measurement and then applying the unitary $U_b$ conditionally to the classical outcome.