2
$\begingroup$

Suppose I have two non-commuting operators, $U_1$ and $U_2$ with eigenvalues $\lambda_{1,1}, \lambda_{1,2}$ and $\lambda_{2,1}, \lambda_{2,2}$, respectively. In order to determine how measuring one value for $U_1$ (say $\lambda_{1,1}$) affects the probability of measuring a particular value for $U_2$ I would want to write the particle in terms of a new basis given by the eigenvectors to get the probability of each outcome, right?

I realize this question might not make perfect sense. I am self-teaching quantum computing from a linear algebra perspective.

$\endgroup$
1
$\begingroup$

In order to determine how measuring one value for $U_1$ (say $\lambda_{1,1}$) affects the probability of measuring a particular value for $U_2$ I would want to write the particle in terms of a new basis given by the eigenvectors to get the probability of each outcome, right?

Pretty much, yes. Given two observables (Hermitian operators) $\newcommand{\calO}{{\mathcal O}}\calO_1,\calO_2$, "measuring $\calO_1$" results in collapsing the state you are measuring, call it $|\psi\rangle$, in one of the eigenvectors of $\calO_1$, call this $|m_1\rangle$.

If you then immediately afterwards measure $\calO_2$, to find the probabilities of finding each one of the eigenvectors of $\calO_2$ you compute the squared overlap between $|m_1\rangle$ and said eigenvectors.

About the general question of how measurements affect each other, it's hard to answer in full generality because it will depend a lot on the context and on what exactly you mean with "how".

For example, if you are measuring a single-qubit state and the observables considered are two Pauli matrices, say $\calO_1=X$ and $\calO_2=Y$ (observe how they do not commute), then after the first measurement you will have maximal uncertainty about the results of the second measurement (in other words, the outcome probabilities for the second measurement are always $50/50$). This happens because any pair of eigenvectors of $X$ and $Y$ have the same squared overlap (we say that the corresponding bases are mutually unbiased).

The canonical answer about why the commutator between the observables is relevant for their being compatible is the uncertainty principle. One can show that, given any pair of observables $A$ and $B$, you have $$\sigma_A \sigma_B \ge \lvert\langle AB\rangle - \langle A\rangle\langle B\rangle\rvert,$$ where $\sigma_A^2\equiv \langle A^2\rangle-\langle A\rangle^2$ are the variances. The right-hand side is lower bounded by the (expectation value of) the commutator $[A,B]$, which is the standard form of Heisenberg's uncertainty principle.

$\endgroup$
6
  • $\begingroup$ this might be a dumb question, but when we have two operators I was thinking they would act on a 2 qubit state, but it seems in your Pauli matrix example they are acting on a single qubit state. Am I thinking about that wrong? $\endgroup$
    – Indy500
    Mar 17 at 20:46
  • $\begingroup$ @Indy500 well, it depends on what you are doing. I interpreted the question as about measuring the same state in different ways, which corresonds to using different observables to measure the same, in this case single-qubit, state. Not that you need to restrict to the single-qubit case, mind you. For example, the uncertainty principle equation above works for arbitrary observables defined on a state in arbitrary dimensions $\endgroup$
    – glS
    Mar 17 at 21:38
  • $\begingroup$ to be clear, in the example I'm thinking of sequential measurements: you measure in one basis, then you measure the resulting state in a different basis. You can also consider a different scenario where you start with the same initial state $|\psi\rangle$ and measure $\mathcal O_1$ repeatedly (resetting the state each time) to collect the associated statistics, and then measure repeatedly $\mathcal O_2$, again resetting the state at each experiment. In this case you need "two qubits" (well, you'll need more than that to actually get the statistics you need to compare the probabilities) $\endgroup$
    – glS
    Mar 17 at 21:43
  • $\begingroup$ But if you measure one qubit, it collapses to a measurable state, so how could you measure it again in a different basis? $\endgroup$
    – Indy500
    Mar 18 at 21:20
  • $\begingroup$ @Indy500 that's a good question, and there are a couple of possible answers I think, depending on the context. If you consider a measurement which still results in a post-measurement state, then clearly you can measure the post-measurement state. The post-measurement state will be, of course, an eigenstate of the first operator that was measured. Usually you don't actually have a post-measurement state in practice. In these cases, you can understand these statements along the lines of "if I measure $\mathcal O_2$ on a given eigenstate of $\mathcal O_1$, what will I find?" $\endgroup$
    – glS
    Mar 19 at 9:51

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.