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As far as I know "measurement" depends on the two variables : state and operator. So what exactly does measuring a qubit mean? is it implied that the operator is the $Z$ operator on that qubit? In that case does the "deferred measurement principle" apply to all operators or just the single $Z$?

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    $\begingroup$ Without any other context I would interpret an algorithm with instructions to "measure the qubits" to mean to measure in the computational $\{|0\rangle,|1\rangle\}$basis. But if there were instructions to "measure the qubits in the Hadamard basis" then that means either do a Hadamard first and then measure in the computational basis, or equivalently measure in the $\{|+\rangle,|-\rangle\}$ basis. The principle of deferred measurement applies regardless. $\endgroup$ May 31, 2022 at 16:57

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Measurement is defined with respect to a measurement basis. Given a state $\rho$, and a measurement basis $\{|u_k\rangle\}_k$, the measurement results in the outcome probabilities $p_k = \langle u_k|\rho |u_k\rangle$. More generally, measurement can be defined as a POVM $\{\mu_k\}_k$, in which case the outcome probabilities are given by $p_k=\operatorname{Tr}(\mu_k \rho)$.

This is always the case. When people say that they are "measuring an operator/observable", call it $A$, what they usually mean is that (1) the state is being measured in the eigenbasis of $A$, and (2) that the measurement probabilities are being (classically) post-processed to compute the expectation value $\langle A\rangle_\rho\equiv\operatorname{Tr}(A\rho)$.

In many-qubit circuits, unless otherwise specified, when just talking about "measuring the qubit" most sources will probably implicitly talk about measuring it in the computational basis (or if you prefer, the eigenbasis of the Pauli $Z$ operator).

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  • $\begingroup$ Is there a case when the measurement basis is not the eigenbasis of an operator $A$?...what's the advantage of using basis instead of operator? is it ordering of the basis? $\endgroup$
    – unknown
    Jun 1, 2022 at 15:21
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    $\begingroup$ an operator carries more information that its eigenbasis. For example, its eigenvalues are redundant information if all you're interested in are outcome probabilities. Operators are often used in the context of computing expectation values, in which case they describe computing specific linear functionals of the outcome probabilities in a given basis, in which case the eigenvalues are the associated weights. Note also that any (orthonormal) basis is a "measurement basis", and any orthonormal basis is the eigenbasis of some Hermitian operator. $\endgroup$
    – glS
    Jun 1, 2022 at 17:04
  • $\begingroup$ To me measuring an operator seems more "natural" than measuring in a basis or measuring a qubit...I'm sure that depends on what you learn first. At any rate I see the translation between the different variations now; thanks. $\endgroup$
    – unknown
    Jun 1, 2022 at 18:13
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I'm not sure if your question is about "what is a measurement", or what does "measuring a qubit" mean if we don't specify the operator or basis.

For the first answer, a generic measurement (called POVM: positive operator-valued measure) is defined by a set of positive semi-definite Hermitian matrices $\{\Pi_i\}_{i \in M}$ such that $\sum_i \Pi_i = I$. For instance, if you measure a qubit in the computational basis, then the operators are $\Pi_0 = |0\rangle \langle 0|$ and $\Pi_1 = |1\rangle \langle 1|$ (and you can trivially extend it to multi-qubit). Then, when you measure a quantum state $\rho$ (where $\rho$ is the density matrix of the state, that can be obtained for any pure state as $\rho = |\psi\rangle\langle\psi|$) using a POVM, you get two objects: a classical string $i \in M$ (the outcome of your measurement) and a post-measured state $\sigma$. More precisely, the probability of getting outcome $i$ is $Tr(\Pi_i\rho)$ and the post-measured state is $\sigma = \frac{\Pi_i \rho \Pi_i^\dagger}{Tr(\Pi_i\rho)}$ (the denominator is just used to renormalize the state). That's all.

For the second answer, typically, if you read "measuring a qubit" or "measuring a quantum state", you need more details to know exactly which measurement you need to apply on the qubit. Depending on the context, either you can guess the basis/POVM that is used, or it can also mean "for any POVM". Typically, however, when we say "measuring a qubit", it often meanss "measuring a qubit in the computational basis" (see above for the corresponding POVM).

Concerning the "deferred measurement principle", it applies for any measurement: you can always choose the postpone the measurement to the very end of your computation (unless, of course, if your operations depends on the result of the measurement… and still you can purify the state if you really want to postpone everything). This comes from the fact that $U \otimes V = (U \otimes I) (I \otimes V) = (I \otimes V) (U \otimes I)$, or, pictured:

enter image description here = enter image description here = enter image description here

(I used random gates as I was lazy to draw it for arbitrary gates, but this is true for any quantum process) This is directly linked to non-signaling (if the order of the measurement were important, then you could use it to transmit information, for instance by measuring later or sooner a quantum state).

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  • $\begingroup$ I was quoting what I see in many papers, presentations...no additional context is provided. Your answer seems to be providing the implied context. You use a set of projections instead of a single operator. To me a single operator seems more natural...you can build up the projections from it, for example $\Pi_0=(1+Z)/2, \Pi_1=(1-Z)/2$,... $\endgroup$
    – unknown
    May 31, 2022 at 17:42

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