I and my friend were disputing about some weird behaviour of the act of measuring some observables quantities e.g. Energy, position.

But I still don't think what he said is strictly true.

He said" each observable has its own Eigenstate, and when you measure it, the wave function will collapse to that eigenstate , giving its eigenvalue for that particle observable quantity. However ,I think some observables do share same eigenstate. i.e. a free particle with zero potential, you can prove this by Solving the TISE. When you now measure and obtained an exact momentum of a particle of a system, by uncertainty principle, the uncertainty of $x$ will be infinite, then the wave function will be spread everywhere,therefore it has a well define wavelength, thus you have well defined $k$ for wavenumber, hence a well defined energy by $p^2 / 2m$.

To conclude, He said: is it true that every time you measure a quantity, it only produces a eigenstate for that particular quantity? I.e. measure position gives position eigenstate, momentum for momentum eigenstate.

My argument shown above claimed the above is not strictly true: So my argument is energy and momentum do share same eigenfunction, when potential is zero. Who is correct?

I also mentioned something called "Conjugate variables" e.g. position and momentum, you can only know one at a time. But I think energy and momentum are not, hence resulting in the reasoning i wrote above!

  • $\begingroup$ Maybe it's just me and I'm already too tired at the moment, but could you maybe completely rephrase your question? What is it you are stating, what is exactly your question? I only get an incoherent picture, e.g. your punctuation lacks coherence, e.g. there is only one quotation mark. However, from what I think I grasped, maybe you should have a look at the difference between point spectrum and continuous spectrum of bounded Hermitian operators. $\endgroup$
    – Martin
    Commented Oct 24, 2013 at 19:41
  • $\begingroup$ Sorry for me not being clear on my question, let me try to edit it $\endgroup$ Commented Oct 24, 2013 at 20:26

1 Answer 1


In the case of a free particle, $ [H, p] = [\frac{p^2}{2m} , p] = 0$. So $H$ and $p$ have the same eigenfunctions. Although this doesn't mean that they necessarily have the same eigenvalue, but it does say that you can measure both simultaneously. In the general case, if two operators commute, they share the same eigenfunctions.
BUT, while solving the TISE for a free particle, it's evident that if you take the wave function to be $e^{\pm ikx}$ (or some super position of these), then you can't normalize it. So $e^{\pm ikx}$ doesn't describe a physical state, and this is why the wave packet descripption is used.

  • $\begingroup$ In the case of a bound state, we have boundary condition to restrict the value of k. But for a unbound/scattering state, we don't have BC , hence Fourier analysis was used to construct a wavepacket to make the wavefunction normalizable $\endgroup$ Commented Oct 24, 2013 at 22:17
  • $\begingroup$ so is it true that for a non-free particle case, [H,p] happens not commuting? What would happen to the wave function when you try to measure the energy of the system? I'm wondering how could one measure the "ENERGY" without refering to momentum and its potential? $\endgroup$ Commented Oct 24, 2013 at 22:20
  • $\begingroup$ In the case of a non-free particle, $[H,p]$ depends on the commutation of the potential ($V(x)$) with $p$. If $[V(x),p]=0$ then $[H,p]=0$ also. But in the case that $H$ doesn't commute with $p$, the eigenfunctions of the operators aren't the same. So, when you measure the energy for example, the wave function will collapse into one of the energy eigenstates, then if you measure the momentum afterwards, the system will collapse into one of the eigenstates of $p$, and you'll lose the information about the energy. Therefor you can't have the measurements of energy and momentum at the same time. $\endgroup$
    – QnoP
    Commented Oct 25, 2013 at 10:11
  • $\begingroup$ In these cases, the potential and kinetic energies can't be separated by measurement, this means measuring the momentum can't give you the kinetic portion of the energy you measure. $\endgroup$
    – QnoP
    Commented Oct 25, 2013 at 10:24
  • $\begingroup$ I wanted to add that the only situation in which $V(x)$ commutes with momentum is where $V(x) = constant$, which doesn't really have a physical significance. Unless $V$ has different functional dependencies other than $x$ (like t for instance). $\endgroup$
    – QnoP
    Commented Oct 25, 2013 at 12:14

Your Answer

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