Does the Heisenberg uncertainty principle apply to the free particle? If it does or doesn't how would one go about proving this mathematically?

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    $\begingroup$ The answer is yes.The "proof" is the success of quantum mechanics. $\endgroup$ Jan 24 '17 at 1:07
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    $\begingroup$ It does. And you don't prove things about physics mathematically, except in so far as you say "If this mathematics is a good representation of the system then I expect that is also true." But if experiment doesn't show that what you conclude is that this isn't actually a good model of the system. $\endgroup$ Jan 24 '17 at 1:07
  • $\begingroup$ It amounts to a certain inequality between characteristics (second absolute moments) of a wave function (position basis) and its Fourier transform (momentum basis), see an explanation in lame terms on Quora, more formally in Pascuzzo's paper. Of course, the physics part is in the fact that the position and momentum bases can be interpreted as related by a Fourier transform. $\endgroup$
    – Conifold
    Jan 24 '17 at 1:22
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    $\begingroup$ The proof is the same for a free particle as for an un-free particle. What step in the proof did you think might break down? $\endgroup$
    – WillO
    Jan 24 '17 at 1:33
  • $\begingroup$ For some explicit formulas how this works, see e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Jan 24 '17 at 5:51

The Heisenberg uncertainty principle follows from the axioms/postulates of quantum mechanics. This means that it is a property of the theory as a whole, regardless of the situation. Therefore, (quite trivially) the Heisenberg uncertainty principle applies to the free particle...in the quantum mechanical picture of the world.

The derivation of the Heisenberg uncertainty principle is pretty standard and can be found in most undergrad QM textbooks. (See Griffiths, Shankar, or Sakurai.)


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