2
$\begingroup$

There is a very interesting answer given by Peter Shor in this website here. However, I admit I don't fully understand it.

In particular, I don't understand:

  1. If we have a probability density μv on quantum states v, we can predict any experimental outcome from the density operator

    What is the formula given? Is it $\operatorname{trace}{\left(\rho A\right)}$ for an observable $A$?

  2. a probability distribution on quantum states is an overly specified distribution, and it is quite cumbersome to work with

    How can we work with the density operator if, from an experimental point of view, all we can measure is the probability distribution?

$\endgroup$
  • 5
    $\begingroup$ Welcome to SE.Physics! Could you think of a more descriptive title for this question? Ideally, something that someone seeing this question on the list can understand the physical subject matter from reading. $\endgroup$ – Nat Oct 28 '18 at 2:21
  • $\begingroup$ Shor is not a physicist, and he is using totally nonstandard terminology there. $\endgroup$ – Buzz Oct 28 '18 at 3:07
  • 2
    $\begingroup$ @Buzz: Note that the OP for the previous question was a mathematician, and not a physicist. But I'm curious—what terminology do you think I should have used? $\endgroup$ – Peter Shor Oct 28 '18 at 23:48
  • $\begingroup$ @PeterShor You make repeated references to "probability distributions on quantum states," defined in a way that is never used. Moreover, you call such distributions "overly specified," which makes it sound like the "probability distributions on quantum states" formalism is adequate but unnecessarily complicated. However, that is not the case. The "probability distributions on quantum states" as you write them down are not overly specified but underspecified—I would even say wrongly specified—because they leave no room for phase information; and that is why they are never used. $\endgroup$ – Buzz Oct 28 '18 at 23:51
  • 1
    $\begingroup$ @Buzz Definitely the phase information is there inside the density matrix. Or the probability is just a deduction of the geometrical structure of quantum mechanics, the probability is just a distance, right? $\endgroup$ – XXDD Oct 29 '18 at 2:30
4
$\begingroup$

From your question, it sounds like you've learned about observables but maybe not von Neumann measurements and POVMs. In my previous answer, when I mentioned "experimental outcome", I was referring to these types of measurement. If you want to understand my previous answer completely, you probably should learn about them.

For your first question, the expectation valuable of an observable $A$ on a density matrix $\rho$ is indeed $\mathrm{Tr}\, (\rho A)$.

For the second question, if you can prepare a mixed quantum state (represented by a density matrix) repeatedly, you can measure the density matrix by doing quantum state tomography on it. If you only have one instance of a quantum state, you can't even get a probability distribution by measuring it ... you just get one measurement outcome.

The probability distribution I referred to in my previous answer is a probability distribution that the OP was asking about. This is related to how the state was prepared, and you can't measure this probability distribution even if you can prepare arbitrarily many copies of the quantum state.

$\endgroup$

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.