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Re: [Phys-L] ground-state energies



Looks like others have discussed this problem. For example, the chemistry stack exchange proposes that HUP has an additional term when considering angle because it is 2 pi periodic, which again has to do with the BCs or equivalently the symmetry:

https://chemistry.stackexchange.com/questions/68880/uncertainty-principle-in-the-ground-state-of-a-rigid-rotor <https://chemistry.stackexchange.com/questions/68880/uncertainty-principle-in-the-ground-state-of-a-rigid-rotor>

On Nov 2, 2018, at 11:13 AM, John Denker via Phys-l <phys-l@mail.phys-l.org <mailto:phys-l@mail.phys-l.org>> wrote:

On 11/2/18 7:39 AM, Carl Mungan wrote:

In intro quantum mechanics, we explain that the translational
(particle in a box) or vibrational (simple harmonic oscillator)
ground-state energy cannot be zero because of Heisenberg’s
Uncertainty Principle (HUP). This is also reinforced by choosing the
ground-state quantum number to be 1 not 0.

We, Kemosabe?

The uncertainty principle doesn't actually say any of those
things.

As a separate matter, FWIW, the ground state quantum number
for a harmonic oscillator is conventionally N=0, not N=1.
The energy is proportional to N+½.

However, the same argument doesn’t hold for the rotational (rigid
rotor) ground-state energy which is zero and which is labeled with
quantum number 0.

An alert student might ask: Why doesn’t HUP apply? If the energy were
zero, wouldn’t I simultaneously know the exact angular position and
angular speed?

The angular speed is zero, but you don't know anything
whatsoever about the angle. There is no violation of
the uncertainty principle.

When I think further about what the ground-state wavefunctions
actually look like, it seems to me that a key difference is that in
the translational or vibrational case, the wavefunction has to go to
zero at the two ends (either the two walls for an infinite wall, or
at +/-infinity for a finite well or oscillator), whereas we impose
periodic boundary conditions (BCs) instead for the rotational case.

That's the right way to approach it.

The wavefunction has to be nonzero somewhere.
It has to go to zero at the wall (for a particle in a box)
or eventually (for a harmonic oscillator).
Therefore it has some curvature.
Therefore it has some nonzero kinetic energy.
This proves that the ground-state energy is nonzero,
without mentioning the uncertainty principle.

But how does this difference in BCs factor into HUP?

It would be simpler to ask how the BCs factor into the assertion
of nonzero energy; this question is answered above.

As for using such scenarios to shed light on the uncertainty
principle, I'm not sure that's possible. It might be, but I'd
have to think about it some more. Hmmmmm.
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