what does it mean to "measure a qubit" if measurement is defined on operators

Question

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 $Z$?

Answer 1

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).

Answer 2

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:

= =

(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).