Parity Operator


by tshafer
Tags: operator, parity
tshafer
tshafer is offline
#1
Jan15-10, 03:28 PM
P: 42
We're working on the parity operator in my second semester quantum mechanics class and there is one point I am confused about, either in the definition of degeneracy or in the parity operator itself. We talked about a theorem whereby the parity operator and the Hamiltonian cannot share simultaneous eigenkets (or, alternatively, wave functions) if there is a degeneracy in the Hamiltonian, regardless of whether or not parity and the Hamiltonian commute.

However, I thought that the Hydrogen wave functions have a definite parity (going as whether [tex]\ell[/tex] is even or odd), even though the Hydrogen spectrum is highly degenerate ignoring corrections. What am I missing?

Thanks!
Tom
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blechman
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#2
Jan15-10, 04:02 PM
Sci Advisor
P: 779
that theorem is false! at least as you quoted it (your own example correctly contradicts it!). If H and P commute then they have simultaneous eigenstates. So there must be something missing... Do you have a reference? Does this theorem have a name?

What can happen is that if you have degenerate states in the Hamiltonian with different parities, that degeneracy can be destroyed by a perturbation that does not respect parity. For example, the Stark effect.
tshafer
tshafer is offline
#3
Jan15-10, 04:24 PM
P: 42
The statement is along the lines of "For commuting H and P, if H is degenerate its eigenkets do not have definite parity."

i.e. there is room for wiggling out of this due to degeneracy. We also talked about the double-well potential in the context of symmetry breaking... but I don't claim to fully grasp that yet.

blechman
blechman is offline
#4
Jan15-10, 05:03 PM
Sci Advisor
P: 779

Parity Operator


If [itex]|\phi\rangle[/itex] is a normalized eigenket of H, then so is [itex]P|\phi\rangle[/itex] if H and P commute. Furthermore, these states are degenerate. Therefore, any linear combination of these states are also eigenstates of H.

In particular:

[tex]|\pm\rangle=\frac{1}{\sqrt{2}}\left(|\phi\rangle\pm P|\phi\rangle\right)[/tex]

are degenerate eigenstates with definite parity. So you should always be able to chose eigenstates that have definite parity.

There must be more to this theorem...
tshafer
tshafer is offline
#5
Jan15-10, 05:06 PM
P: 42
Thanks — I thought so, too, but this is my first serious exposure to parity or discrete symmetries at all. I'm making an attempt at trying to actually understand it rather than just floating along.


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