2
$\begingroup$

If I have a large number of identical systems in identical quantum state $\Psi$ and an observable $A$ whose eigenstates are $\alpha_n$:

$$ \Psi = \sum_n c_n \alpha_n $$

I can get absolute values of $|c_n|^2$ by measuring $A$, but with which experiment I can get relative phases of $c_n$?

$\endgroup$
  • $\begingroup$ arxiv.org/pdf/1410.0916v1.pdf $\endgroup$ – valerio Jun 9 '16 at 13:30
  • $\begingroup$ @valerio92 could you maybe summarize the article? I'm not a professional physicist. $\endgroup$ – xaxa Jun 9 '16 at 16:17
5
$\begingroup$

The phase ambiguity is a bit worse than you think. There is a global phase ambiguity in $|\Psi⟩$, for sure, but if the state $$ |\Psi⟩=\sum_n c_n |\alpha_n⟩ \tag 1 $$ is all you have around, then there is also a phase ambiguity in the phase of each individual $c_n$. This is because if you change $|\alpha_n⟩$ to $|\alpha'_n⟩=e^{i\theta_n}|\alpha_n⟩$, the transformed state is also an eigenstate of $A$ with eigenstate $|\alpha_n⟩$, so the two have exactly equal standing. If your universe only ever contains $|\Psi⟩$ (and a measurement device for $A$), then you only ever care about the absolute values $|c_n|^2$ of the coefficients.

The phase of these coefficients comes into play if you have two or more states, i.e. if you introduce some second state $$ |\Phi⟩=\sum_n b_n |\alpha_n⟩ \tag 2 $$ into the mix. Here transforming the eigenstates by some phase, as before, will still change the phases of the individual coefficients, but it will leave their differences $\arg(c_n)-\arg(b_n)$ unchanged. These are the phase differences that can be detected by experiment.

So, to come to the question, how do you detect them? There are a bunch of ways, but they all involve interference in some way. As a simple example, take your two states $|\Phi⟩$ and $|\Psi⟩$, and a superposition of them, $$ \frac{|\Phi⟩+|\Psi⟩}{\sqrt 2} = \sum_n \frac{c_n+b_n}{\sqrt 2} |\alpha_n⟩,\tag 3 $$ and measure $A$. Then you will measure the eigenvalue $\alpha_n$ with a probability $$P_n=\frac12 |c_n+b_n|^2$$ which is sensitive to the relative phase of the two coefficients.

$\endgroup$
  • $\begingroup$ That's a nice catch about phases of eigenstates... Now I'm a bit confused: to which one of them does a system collapse after measurement according to QM? Of a set of systems in identical state, will each one collapse to the same $e^{i\theta_n}\alpha_n$ or $\theta_n$ may be different? $\endgroup$ – xaxa Jun 9 '16 at 15:24
  • $\begingroup$ They're the same physical state, because they only differ by a phase. The phase would matter if this was one branch of an interference experiment, but because you've done a projective measurement, all the coherence gets destroyed and you won't see any interference. $\endgroup$ – Emilio Pisanty Jun 9 '16 at 15:50
  • $\begingroup$ Seems like I'm missing something... Can't they be used as inputs in interference experiment later - after I've done the measurement? $\endgroup$ – xaxa Jun 9 '16 at 16:13
  • $\begingroup$ Interferometers always measure phase differences, and those are always reset during a projective measurement. For more details, consult your favourite QM textbook. $\endgroup$ – Emilio Pisanty Jun 9 '16 at 16:50

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.