1
$\begingroup$

What does it mean when we say that the outcome of a quantum measurement is a random variable (or a quantum ensemble) such as in renes notes (page 51) or in this paper (page 4 B-1) ? Does it mean that we have as many copies of the state , on which we can perform the POVMs and obtain the random variable by counting frequencies ? Because when I hear measuring with a POVM, I understand using one element of the list of POVMs , getting one outcome and altering the initial state, so we should not be allowed to use a second element from the list of the POVMs because we lost the initial state. Can someone help me clarify this confusion ?

$\endgroup$
1
  • $\begingroup$ does quantumcomputing.stackexchange.com/a/16532/55 answer your question? The gist is that yes, every time you measure a state you "consume it" (it's possible you still have some state after the measurement, but it will be different than the one before the measurement, except for cases where the measurement extracts no information from the state), obtaining a single measurement outcome. Which outcome you get is, of course, stochastic, hence why the measurement outcome is effectively/should be described as a random variable $\endgroup$
    – glS
    Dec 12, 2023 at 16:53

2 Answers 2

1
$\begingroup$

A random variable can be interpretted with ensembles as you say, but it can also be intrinsic: you measure something, and it can actually produce any of the outcomes, just each with different probabilities

$\endgroup$
4
  • $\begingroup$ So one measure produces only one outcome , not as many outcome as there are elements in the POVM used , right ? If I measure with the POVM {$P_0, P_1, P_2$} I only get one outcome that depend on the $P_i$ I chose , isn't ? If so why Do the references I linked in my first questions talk about the outcome being a Random variable $\endgroup$
    – yosh
    Dec 10, 2023 at 17:41
  • $\begingroup$ @yosh I'm not quite sure what the confusion is. Random variables have a list of possible outcomes but also only yield one outcome when measured. The outcome of a coin flip is a random variable that yields either heads or tails, for example, and the process could be described by a POVM with one element corresponding to measurement result heads and one element corresponding to measurement result tails: $\{P_{h},P_t\}$ $\endgroup$ Dec 10, 2023 at 23:48
  • $\begingroup$ Consider the state $\rho = \frac{1}{2}(|0\rangle\langle0|+|1\rangle\langle1|)$, and the POVM {$P_0 =|0\rangle\langle0|,P_1= |1\rangle\langle1|$}. If we measure with this POVM, we get the outcome 0 with proba 1/2 and post measurement state $\frac{1}{2}(|0\rangle\langle0|$ or outcome 1 with other post measurement state. But we can't measure a second time (we will get the same outcome due to the collapse of the wave function) so after one measurement we only get one real number (the outcome) . So why do the references I linked talk about the outcome being a Random variable $\endgroup$
    – yosh
    Dec 11, 2023 at 12:25
  • $\begingroup$ @yosh that's what a random variable is! Something that you measure once and could take a variety of outcomes, but before you measure it you cannot say for certain which outcome it will take. Maybe you'll like the linked Wikipedia page's sentence "The term 'random variable' can be misleading as its mathematical definition is not actually random nor a variable" $\endgroup$ Dec 11, 2023 at 14:44
0
$\begingroup$

When we speak about the mean value of an observable in a quantum state, we are referring to the average of many measurements of that observable. This means we assume the original state has been recreated many times, not measured in a state that has been altered by a previous measurement. If a measurement is realized previous on a system, subsequent measurements will all yield the same result. So, the answer for the following question is YES, IT DOES:

Does it mean that we have as many copies of the state , on which we can perform the POVMs and obtain the random variable by counting frequencies ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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