4
$\begingroup$

In Quantum Computing quantum measurements are described by a collection of measurement operators $\{M_n\}$ such that the probability of the outcome $m$ is given by $p(m)=\langle\psi|M_m^\dagger M_m|\psi\rangle$ and the state of the system after measurement is $\frac{M_m|\psi\rangle}{\sqrt{\langle\psi|M_m^\dagger M_m|\psi\rangle}}$. In addition these measurement operators satisfy $\sum_m M_m^\dagger M_m$. Now let $P$ and $Q$ be two hermitian matrices that commute and let $|\psi\rangle$ be a pure state in some Hilbert space. My question then is what does it mean to measure $A$ and $B$ simultaneously on $|\psi\rangle$?

$\endgroup$
1
2
$\begingroup$

Firstly, it probably helps to switch over to specifying a measurement by observables. An observable is a matrix, $A$, that defines the measurement. In effect, you take its spectral decomposition, $$ A=\sum_i\lambda_iP_i $$ where the $\lambda_i$ are the eigenvalues and $P_i$ are projectors onto the corresponding eigenspaces. Then your measurement corresponds to using the projectors $\{P_i\}$.

Now if you talk about measuring $A$ and $B$ (two observables) simultaneously well, it kind of means what it says - that you can perform both measurements at the same time. There are some simple cases (e.g. for a pair of qubits) where it's obvious that you can do this. For instance, if $A=X\otimes I$ and $B=I\otimes Z$ then you can measure both qubits at the same time.

However, the terminology tends to more get used in the sense "well, you could measure them at the same time, but practically we don't". Instead, you measure one after the other, but the act of measuring one observable does not affect the outcome of measuring the other observable$^*$. Mathematically, this is conveyed by the condition that the two observables commute, $AB-BA=0$. You can see that this is trivially true in my two-qubit example.

That said, you could theoretically take this further. The fact that two observables commute mean that they are simultaneously diagonalizable. So if you took that common eigenbasis, that would define a single set of projectors that you could measure in one go.

$^*$ The ideal language for describing this fails me at this point, as it so often does when talking about quantum. The case of measuring each of the qubits in a two-qubit state is clear - I can do that in any order (including at the same time). But the outcome of one measurement does affect the outcome of the other, just think of perfectly correlated outcomes of measurements on a Bell state. So that wording is inadequate. Really, it's the expected value of the outcomes which is unchanged by ordering of the two measurements.

$\endgroup$
2
  • $\begingroup$ Thankyou! Mathematically speaking what does it mean to make a simultaneous measurement? Does it mean that we take for instance $P,Q$ hermitian matrices and we simultaneously diagonalize them with a basis and use that basis for measurement? $\endgroup$ – Musk Nation Mar 30 at 16:14
  • 1
    $\begingroup$ @MuskNation two observables are simultaneously diagonalizable if and only if they commute. There is no definition of "simultaneous measurement", the point is that if two observables are compatible, in the sense that they commute, then the order in which we measure them won't affect the outcome. $\endgroup$ – Condo Mar 30 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.