How do we experimentally find the probability distribution of quantum particles, i.e the one obtained by mod squaring the solution of the Schrödinger equation? For example, to study the case of an infinite potential well, how do we proceed in the lab?


We cannot, in a strict sense, measure the wave function $\psi(x)$ or its square $\lvert \psi(x)\rvert^2$. This is because one cannot infer the "true" probabilitity distribution of a random variable by making finitely many measurements on it. The only thing you can do is place your hope in the law of large numbers and expect that, if you prepare many identical quantum states all with the same state/wavefunction and then measure their position, you can then fit a probability distribution to the results that approximates the "true" wavefunction. However, it is unclear for what purpose this would be of any use. Note also that you cannot do these repeated measurement on a single particle/state, since the first measurement will already change it.

If you are willing to relax your notion of "measuring the wavefunction" to "measuring the state", then this is the active field of quantum tomography. The basic idea is that we don't attempt to measure $\psi(x)$, but a complete set of commuting observables, whose values uniquely determine the state - and crucially, this measurement can be carried out with finitely many measurement, and the state is not changed further after the first measurement.

  • $\begingroup$ Actually i want to know if we cant,then how can the schrodinger equation could be experimentally verified? $\endgroup$ – user157588 Oct 20 '17 at 11:04
  • $\begingroup$ @user157588 Note that although we cannot actually measure the wavefunction, the test of the probabilities for a large ensemble I talk about in the first paragraph does constitute an experimental test (i.e. it could, in principle, falsify the predicted form of the wavefunction). If you think experiments should "verify" hypotheses, verificationism hasn't been the mode of operation of science for a long time. $\endgroup$ – ACuriousMind Oct 20 '17 at 11:28
  • $\begingroup$ @ACuriousMind Thanks for this answer. Does it also matter to mention that after all the wavefunction is a complex valued function hence already non-measurable in an experimental sense? or that isn't necessarily a limitation? I also wanted to ask about the 2nd paragraph: What do we mean by measuring the state and not the wavefunction? I mean, isn't measuring in QM always assumed to be for an observable of the system? Many thanks for any further clarifications. $\endgroup$ – user929304 Oct 23 '17 at 16:40
  • $\begingroup$ @user929304 I don't like saying "we can't meaure complex numbers" because a complex number is described by two real numbers which we could measure. Also, I'm talking about a "state" and not a "wavefunction" in the second paragraph because tomography is often done for effectively finite-dimensional systems that have no position operator, hence no wavefunction. The notion of a quantum state is more general than that of a wavefunction. $\endgroup$ – ACuriousMind Oct 24 '17 at 1:21
  • $\begingroup$ Thanks for the clarification. I see, very cool to learn about tomography. I admit I am still confused by the "state" and "wavefunction" difference. I mean we always learn that state of a system in QM is encoded by a wavefunction or wave-vector in Hilbert space. So is the more correct terming that, "wavefunctions" are reserved for the full ket representing the system being projected to the subspace of position eigenstates in Hilbert space? Or is there an issue because there are not really eigenstates for position operator? Sorry for all this... really confused $\endgroup$ – user929304 Oct 24 '17 at 12:56

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.