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*From*: Carl Mungan <mungan@usna.edu>*Date*: Fri, 2 Nov 2018 10:39:35 -0400

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.

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?

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. But how does this difference in BCs factor into HUP?

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Carl E. Mungan, Professor of Physics 410-293-6680 (O) -3729 (F)

Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363

mailto:mungan@usna.edu http://usna.edu/Users/physics/mungan/

**Follow-Ups**:**Re: [Phys-L] ground-state energies***From:*John Denker <jsd@av8n.com>

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