2
$\begingroup$

In a non-relativistic quantum mechanical system in an infinite potential well. I try to measure the energy and the position of the system simultaneously. Since, the respective operators do commute according to Heisenberg's uncertainty relation I should be able to measure them both with infinite precision. Now, since I know that there is no potential energy in the well I can use $ E=\frac{p^2}{2m} $ since the potential energy is 0 and determine it's momentum provided I know it's mass. But I shouldn't be able know the momentum and position simultaneously with infinite precision! So where am I going wrong?

$\endgroup$
  • $\begingroup$ What do you mean "There is no energy in the well"?! That the well has it's base at an arbitrary $V_0$ does not mean that the particle in it has to have that $V_0$ as its energy! $\endgroup$ – ACuriousMind Aug 15 '14 at 14:50
  • $\begingroup$ By no energy in the well I mean the potential energy is V=0 ... Yes the particle can have kinetic energy T and T=V. $\endgroup$ – drewdles Aug 15 '14 at 14:55
  • $\begingroup$ The notion of "measure A and B simutaneously" where A and B do not commute, is ill defined. $\endgroup$ – zzz Aug 15 '14 at 16:48
  • $\begingroup$ As clarified in the comments to an answer, the question is based from the wrong understanding that the Energy operator is $E=i\frac{\partial}{\partial t}$, which is not even an operator on the Hilbert space of the particle. $\endgroup$ – fqq Aug 15 '14 at 18:30
2
$\begingroup$

If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $ H = \frac{p^2}{2m} + V(x) $ where $ V(x) $ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $ \frac{p^2}{2m} $ does not commute with $ x $, how are you saying that the energy and position operators commute? Besides, if that were true than we could place a particle in any point inside the well and the particle would stay there forever.

$\endgroup$
  • $\begingroup$ $ E = i \hbar \frac{d}{dt} $ and $ p = -i \hbar \frac{d}{dx} $ where these are partial derivatives hence they do commute $\endgroup$ – drewdles Aug 15 '14 at 15:22
  • $\begingroup$ Nevermind I understood where I was going wrong u were right ... Thanks $\endgroup$ – drewdles Aug 15 '14 at 15:29
  • 1
    $\begingroup$ That's not the energy operator, it's not even an operator on the Hilbert space. $\endgroup$ – fqq Aug 15 '14 at 18:34
1
$\begingroup$

At the first look the question seemed very interesting, but later I found the mistake. You said you are measuring the position of the particle precisely. But how? You can tell that the particle is inside the well but you can not know the exact position of the particle. For more info read http://physicspages.com/2012/07/10/infinite-square-well-uncertainty-principle/

$\endgroup$
1
$\begingroup$

I try to measure the energy and the position of the system simultaneously

The states with definite energy are not states with definite position so there is no particle state of both definite energy and definite position.

$\endgroup$

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.