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I calculated that the product of the uncertainty in position $\sigma_x$ for the ground state of an infinite square well of width $L$ with the uncertainty in the momentum $\sigma_p$ for the same state, and it does not change with respect to the width of the well $L$. Intuitively, why does this happen?

(I would also like to know what the physical reasons are for the decrease in the momentum uncertainty, but I have asked that separately.)

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  • $\begingroup$ Why would you expect it to? Neither the position nor the momentum operator know anything about the potential. Actually, why should the uncertainty of a single one of these operators depend on the potential? It depends on the states you evaluate them on, not on the potential. $\endgroup$ – ACuriousMind Feb 26 '16 at 22:06
  • $\begingroup$ @ACuriousMind yes, you are right! That's one way to see it. Thank you. Although i also want to see an answer that addresses the fact that with a wider well, we in general can superimpose more eigenstates. $\endgroup$ – TheQuantumMan Feb 26 '16 at 22:10
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    $\begingroup$ The state actually does change in this case. It becomes wider (in space). The uncertainty will depend on the potential, because the state will depend on the potential... All this is, of course, due to the assumption of ground state, which is not stated (no pun intended) in the question. $\endgroup$ – safkan Feb 26 '16 at 22:11
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    $\begingroup$ @safkan I am talking about the classic infinite well problem with ground state equal to (πhbar/L)^2/(2m) where L is the width and m is the mass. $\endgroup$ – TheQuantumMan Feb 26 '16 at 22:13
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    $\begingroup$ Nice. This is now a much more accessible problem. +1. $\endgroup$ – DanielSank Feb 29 '16 at 17:09
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Honestly, there really isn't all that much going on here beyond the fact that you're scaling the entire system, and the fact that the momentum is a derivative, $$p=-i\hbar\frac d{dx},$$ or, alternatively, that the commutator $$[x,p]=i\hbar$$ is fixed, means that if you rescale $x$ by some amount $\lambda$ then you need to rescale $p$ by the reciprocal of that, $1/\lambda$. The uncertainties in both operators are tied to the operators themselves in terms of scaling, so their product must remain constant under such a scaling transformation.

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  • $\begingroup$ So this works for any potential? If I take a 1D Schrodinger equation and scale the $x$ axis does $\sigma_x \sigma_p$ always stay constant? $\endgroup$ – DanielSank Feb 29 '16 at 17:53
  • $\begingroup$ For the ground state, yes. (Also for any definite eigenstate. But you can't just specify the transformation of the potential without saying what happens to the wavefunction.) $\endgroup$ – Emilio Pisanty Feb 29 '16 at 17:57
  • $\begingroup$ The explanation is correct, but again, it's kind of mathematical. It offers no explanation based on intuition. $\endgroup$ – TheQuantumMan Feb 29 '16 at 18:12
  • $\begingroup$ That is the intuition. Momentum scales like a wavevector because of de Broglie's relation, and that completely constrains its behaviour under scaling. You're doing very little to the system (heck, from a passive-transformation viewpoint you're doing nothing at all beyond changing your unit system) so you cannot expect the physics to change. I don't know what you were expecting but there's seriously nothing much there. $\endgroup$ – Emilio Pisanty Feb 29 '16 at 19:48
  • $\begingroup$ @EmilioPisanty i have found an explanation that goes like this: the narrower the well, the more confined the particle is, so it bounces off the walls of the well constantly, thus having big uncertainty. So, i guess i was looking something like this as an explanation. I agree with you that your explanation is intuitive and informative, but i was looking for a more layman approach to it. Also, could you please provide me with any sources for transformations and things like the arguments that you provided in the answer? Because i like this approach, it is very powerful $\endgroup$ – TheQuantumMan Feb 29 '16 at 19:56

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