# Kramers Theorem.

1. Nov 6, 2009

### MathematicalPhysicist

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Dec 3, 2009

### physcoast

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.