Is it possible to show that the 6 states generated are not all unique?

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SUMMARY

The discussion centers on Kramers theorem, which asserts that in systems with Hamiltonians invariant under time reversal, energy levels must exhibit at least double degeneracy. The user seeks to demonstrate that threefold degeneracy is impossible by analyzing eigenstates |A1>, |A2>, and |A3> with a common eigenvalue A. The conclusion drawn is that assuming three eigenstates leads to contradictions when applying the time reversal operator T, particularly when considering the implications of Kramers theorem on systems with odd numbers of particles, such as three electrons.

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Homework Statement


Kramers theorem states that if the Hamiltonian of a system is invariant under time reversal, and if T^2|\psi>=-|\psi> where T is the time reversal operator, then the energy levels must be at least doubly degenerate. In fact the degree of degenercay must be even. Show explicitly that threefold degeneracy is not possible.


Homework Equations





The Attempt at a Solution


So I have assumed that there are three eigenstates with the same eigenvalue, let's write them as |A1>,|A2>,|A3> and the eigenvalue is A, so if H is the Hamiltonian then:
H|A1>=A|A1>
H|A2>=A|A2>
H|A3>=A|A3>
A is a real valued number.
Now, HT|A1>=AT|A1>
so T|A1> is one of the three above eigenstates, either A1,A2 or A3.
Now I need to show that either way we get a contradiction.
Now if T|A1>=|A1>, then -|A1>=T^2|A1>=T|A1>=|A1> which is impossible.
Now i am left with two options, either T|A1>=|A2> or T|A1>=|A3>, here is where I am stuck I didn't succeed in showing a contradiction.

Any hints?
 
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I have this same problem. I think you need to use the fact that Kramer's Theorem only applies to odd numbers of particles. If you use 3 electrons (a, b, c) then you can create 6 states from the combinations. You can also have 3 pairwise exchange operators P12, P13, P23.

Then you can show that for any T |abc> choice, you have to get T and T^2 equivalent to one of the pairwise operators, which is false since P^2=1.
 

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