How can there be an Uncertain Momentum at a Specific Energy?

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  • #1
BHL 20
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It's been a few years since I failed my physics degree but I still really want to reach an understanding of QM, and I'm currently going through a QM textbook.

One thing I cannot understand no matter how much I think about it, is momentum uncertainty. In classical mechanics a specific kinetic energy implies a specific momentum. And solving the time-independent Schrödinger equation gives specific energy levels for the system.

Now for systems with a mix of potential and kinetic energy there is of course no reason for the Hamiltonian eigenvalues to imply a specific momentum. But if you take a system like the particle in a potential well, where the Hamiltonian eigenvalues are essentially kinetic energies, and you force that system to collapse to a specific eigenvalue, how can there be any uncertainty in momentum at all? And in a system like the particle in a well there's always some amount of certainty in position so the momentum should never be 100% certain, right?

Also I'm always thrown off by comments like "the ground state energy can not be zero due to the uncertainty principle". How would collapsing to a zero energy ground state be any different from collapsing to any other eigenstate, in terms of reducing certainty?

I hope my question is clear, thank you.
 

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  • #2
hilbert2
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The kinetic energy doesn't tell anything about the direction of momentum. There's the uncertainty you're looking for
 
  • #3
andresB
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Also I'm always thrown off by comments like "the ground state energy can not be zero due to the uncertainty principle". How would collapsing to a zero energy ground state be any different from collapsing to any other eigenstate, in terms of reducing certainty?

I hope my question is clear, thank you.
About this one. You can always redefine energy by adding a constant, just as in classical mechanics. So, the ground state can be any number you want it to be. I think you meant momentum.
 
  • #4
hutchphd
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About this one. You can always redefine energy by adding a constant, just as in classical mechanics. So, the ground state can be any number you want it to be. I think you meant momentum.
The act of building a box around a particle raises its energy. These are not usually arbitrary numbers, and the context makes it clear.
 
  • #5
andresB
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The act of building a box around a particle raises its energy. These are not usually arbitrary numbers, and the context makes it clear.
Not sure what you mean. You can add the same number to all the energy levels of a particle inside a box and nothing bad happens. Do you disagree with that?
 
  • #6
dextercioby
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The uncertainty in momentum can very well be 0. And the HUP needs adjustment for an infinitely deep well.

But if you take a system like a particle in a potential well, where the Hamiltonian eigenvalues are essentially kinetic energies, and you force that system to collapse to a specific eigenvalue, how can there be any uncertainty in momentum at all? And in a system like a particle in a well, there's always some amount of certainty in position so the momentum should never be 100% certain, right?
 
  • #7
hutchphd
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Not sure what you mean. You can add the same number to all the energy levels of a particle inside a box and nothing bad happens. Do you disagree with that?
Yes of course but the context usually makes it clear what the best choice of zero should be. For a particle in a box we would compare the ground state KE of the bound particle to that of a free particle (in ground state). What you say is simply semantics: does it contain physics?
 
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hilbert2
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If you construct an operator for the absolute value of momentum (or its norm in the 2D or 3D case), then it's most certainly hermitian and has common eigenstates with kinetic energy. But even that operator can't have common eigenstates with the position operator. Is someone able to calculate the uncertainty relation for ##\hat{|\mathbb{p}|}## and ##\hat{\mathbb{x}}## ?.
 
  • #9
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In classical mechanics a specific kinetic energy implies a specific momentum
This isn’t correct even classically. Momentum is a vector. A specific amount of classical KE constrains only the magnitude of that vector, not it’s direction.
 
  • #10
andresB
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Yes of course but the context usually makes it clear what the best choice of zero should be. For a particle in a box we would compare the ground state KE of the bound particle to that of a free particle (in ground state). What you say is simply semantics: does it contain physics?
Of course it has physical meaning. It means many relevant things can only depend on the difference between energy levels.
 
  • #11
hutchphd
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Of course it has physical meaning.
I mean what is the relevance to this thread?? I think it is not useful. The Hamiltonian for any system defines the conventional zero for energy. The OP is about that "zero point".
 
  • #12
andresB
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I was pointing out that there is no such thing as "the ground state energy can not be zero due to the uncertainty principle"

Though maybe the OP meant "the ground state kinetic energy can not be zero due to the uncertainty principle"
 
  • #13
BHL 20
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This isn’t correct even classically. Momentum is a vector. A specific amount of classical KE constrains only the magnitude of that vector, not it’s direction.
Yes I may have not finished my degree but I certainly know the difference between vectors and scalars 🤣. I just didn't think that uncertainity in direction would count as satisfying the uncertainty principle. I was thinking of uncertainty as having to be more like a continuous probability distribution rather than a discrete one. Brain fart.
 
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  • #14
vanhees71
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The uncertainty in momentum can very well be 0. And the HUP needs adjustment for an infinitely deep well.

But if you take a system like a particle in a potential well, where the Hamiltonian eigenvalues are essentially kinetic energies, and you force that system to collapse to a specific eigenvalue, how can there be any uncertainty in momentum at all? And in a system like a particle in a well, there's always some amount of certainty in position so the momentum should never be 100% certain, right?
Don't underestimate the subtleties of the apparently simply "infinitely deep well" ;-)). You cannot define a momentum operator as a self-adjoint operator and thus a momentum observable is a very tricky business for this example, which is of course unphysical to begin with.

It's just taken as a simple example for solving an energy eigenproblem, and indeed the Hamiltonian ##\hat{H}=-\hbar/(2m) \partial_x^2## is a well-defined self-adjoint operator on the Hilbert space ##\mathrm{L}^2([0,L])##.

So this is a bad example to illustrate the question whether momentum can be uncertain if a particle is prepared in an energy eigenvalue, because in this example there doesn't exist a well-defined momentum observable to begin with ;-)).

The question itself is very easy to answer even for a free particle. Since in this case ##\hat{H}=\hat{p}^2/(2m)## with a properly defined self-adjoint momentum operator, there's a simultaneous set of eigenvectors of ##\hat{p}## and ##\hat{H}##, but the eigenvalues of ##\hat{H}## are all two-fold degenerate except the ground state (which is the momentum eigenstate to the eigenvalue ##p=0##), because the momentum eigenstates ##|p \rangle## and ##|-p \rangle## are both eigenstates of the Hamiltonian with the same eigenvalue ##p^2/(2m)##. Thus any non-trivial superposition ##\lambda_1 |p \rangle + \lambda_2 |p \rangle## is also an eigenvector of ##\hat{H}## with the eigenvalue ##p^2/(2m)## but not an eigenvector of ##\hat{p}##.
 
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