Suppose I have a wavefunction which collapses to a certain eigenstate after a measurement of energy. In that state, I perform a calculation of position and obtain a certain position value, say $x_0$. After some time, if I perform a calculation of position again, will I obtain $x_0$ again, or will it be something different?

What I have learnt is that, if on performing a calculation of energy, the wavefunction collapses, then a calculation on the collapsed wavefunction will give the same energy, at any time later and for any observer. Does the same hold true for any observable?


3 Answers 3


Energy is a bit of a special case because the eigenfunctions of the Hamiltonian are time independent (assuming a time independent Hamiltonian). So when you make an energy measurement and collapse the system to an eigenfunction of the Hamiltonian it stays there.

However the position operator does not commute with the Hamiltonian so when you measure the position you will leave the system in some superposition of the energy eigenfunctions. This superposition is not time independant so subsequent measurements of position will not return the same value.

This applies to any operator that does not commute with the Hamiltonian.

  • $\begingroup$ Well, this point of view is in my opinion a little bit naïve. The OP is asking if the conditional expectation of measuring $x'$ immediately after having measured the position $x$ would be $\delta_{xx'}$. Of course the time evolution "messes things up" for any observable that do not commute with the Hamiltonian, but repeatability of a measurement is something that you would like to have in the sense above (immediately after), and not after the system has evolved. And even instantaneously, a repeatable measurement do not exist for observables with (partly) continuous spectrum. $\endgroup$
    – yuggib
    Nov 18, 2015 at 10:59
  • $\begingroup$ @yuggib: Perhaps I have misinterpreted what Tejas is asking or perhaps not. $\endgroup$ Nov 18, 2015 at 11:20
  • $\begingroup$ You are giving a sort of "superfluous motivation"; for it is not only the time evolution that makes the repeatability not possible, but already the continuous nature of the position observable. Take the momentum operator (for a free particle). It commutes with the Hamiltonian, nevertheless a repeated measurement of it is not possible. $\endgroup$
    – yuggib
    Nov 18, 2015 at 11:28

You can have a repeatable measurement process (i.e. a measurement process that, roughly speaking, gives the same result if done twice in a row) only for discrete observables.

A discrete observable is an observable whose spectrum is purely discrete. So with the Hamiltonian it is possible to have repeated measurements, provided it is a system with purely discrete energy spectrum (for example, the harmonic oscillator has a repeatable measure process; but the free Hamiltonian of an unconfined particle has no repeatable measure process).

The position operator, however, has always a continuous spectrum; therefore it is never possible to have repeatable measurements. You can have a measurement process of course, but even if performed twice in a row it will give two different results. The same is true for the momentum operator (of an unconfined particle) as well.


I believe that repeatable measurements of position (and momemtum) can never generate the same values simply because it violates the uncertainty principle:

$\delta _x \delta _p \geq \frac{\hbar}{2} $

The terms on the left side are the uncertainty for position and momemtum:

$\delta _x ^2 = <\hat{x}^2> - <\hat{x}>^2 $

If the wavefunction is an eigenstate of position, $\delta _x$ becomes 0 according to the equation above. Hence, the uncertainty relation (derived from the Cauchy-Schwarz inequality) does not hold.

For this reason, the wavefunction cannot be "updated" to an eigenstate after measurement.

You can use the same arguments above to conclude that wavefunction collapse for spin or photon polarization measurements does not violate the uncertainty relation:

$\delta^2 {_S{_x}} \delta^2 {_S{_y}} \geq \ | \frac{<[\hat{S}{_x},\hat{S}{_y}]>}{2i} |^2 $

The commutator on the right side is not a scalar, but rather another operator (in the z direction). The expectated value of $\hat{S}_z$ operating on an eigenstate of x and y is zero, thus the wavefunction can be "updated" as an eigenstate.


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.