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

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    $\begingroup$ Related: this answer of mine works through two important cases. $\endgroup$ Jul 1, 2021 at 19:06
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    $\begingroup$ Can you take the hints from problem 5 of O'Donnell's homework set? I always intuited it as a consequence of the no-communication theorem; a violation of the principle of deferred measurement may allow Alice to take some of qubits, and make a measurement of her qubits whose results were contingent on whether or not Bob waited until the end to measure his qubits, or instead measured his qubits in-situ during the running of the circuit. $\endgroup$ Jul 1, 2021 at 23:57

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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.

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  • $\begingroup$ +1 It's cool to know the principle applies to channels and POVMs, too. However, I think it's more general than the ability to commute measurements through controls. It says that for any circuit $C$ with intermediate measurements there is another circuit $C'$ where all measurements are terminal and that agrees with $C$ on all inputs, e.g. measurement sandwiched between two unitaries can be made terminal too (by adding ancilla, see 2nd example here). $\endgroup$ Jul 4, 2021 at 6:39
  • $\begingroup$ @AdamZalcman I see. I suppose "circuit with intermediate measurements" could be modeled in general as a channel with Kraus operators corresponding to the different possible measurement outcomes (or more precisely, the different possible combinations of measurement outcomes). Then a possible corresponding matching unitary evolution with measurements only at the end should be the isometry obtained putting all the Kraus operators one above the other, with the measurement outcomes "selecting" a corresponding block in this isometry. $\endgroup$
    – glS
    Jul 4, 2021 at 20:49
  • $\begingroup$ I'm not sure whether one can also control the gate decomposition with this type of argument though. That is to say, the isometry/unitary thus obtained might not admit a gate decomposition resembling the original one. Though I guess that might make sense, after all, removing intermediate measurements from a circuit potentially makes for a completely different type of dynamics $\endgroup$
    – glS
    Jul 4, 2021 at 20:51
  • $\begingroup$ I'm sorry I haven't missed the detailed steps of the above mathematical derivation. First not quite sure what each symbol, and parameter means. Second how to derive the next step, I can't figure out yet. Third, the significance of delayed measurement and what the advantages are cannot be seen from the two equations above. $\endgroup$ Aug 22, 2023 at 11:38
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    $\begingroup$ @R-XZhao sorry, can you rephrase that? I don't really understand what you're trying to say. Anyway, rereading this I do think there were some issues in the answer. I reworked most of the derivation, so it should be solid now $\endgroup$
    – glS
    Aug 23, 2023 at 18:57

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