Time Reversal Invariance Of Hamiltonian

[casdan1]

Homework Statement

Suppose that the Hamiltonian is invariant under time reversal: [H,T] = 0. Show that, nevertheless, an eigenvalue of T is not a conserved quantity.

Homework Equations

The Attempt at a Solution

Using Kramer's Theorem.

Consider the energy eigenvalue equation, $$H|\Psi\rangle = E|\Psi\rangle$$ for a time-reversal-invariant Hamiltonian, TH = HT. Therefore
$$HT|\Psi\rangle = TH|\Psi\rangle = ET|\Psi\rangle$$, so both $$|\Psi\rangle$$ and $$T|\Psi\rangle$$ are eigenvectors with energy eigenvalue E.
This implies two possibilities.

1. $$|\Psi\rangle$$ and $$T|\Psi\rangle$$ are linearly dependent, and so describe the same state, or

2. They are linearly independent, and so describe two degenerate states.

It can further be shown that case 1 is not possible in certain circumstances.

How can I show that there is no conserved quantity?

 

[Jhero]

To show that there is no conserved quantity, we can look at the expectation value of the time-reversal operator, T, for the state |\Psi\rangle. If T|\Psi\rangle = |\Psi\rangle, then the expectation value of T for the state will be +1. If T|\Psi\rangle = -|\Psi\rangle, then the expectation value of T for the state will be -1.

Since T is an anti-unitary operator, its eigenvalues are always ±1. This means that for any state |\Psi\rangle, the expectation value of T will always be either +1 or -1, depending on the eigenvalue of T for that state. Therefore, the expectation value of T is not a constant and cannot be considered a conserved quantity.

Furthermore, since T is an anti-unitary operator, it does not commute with the Hamiltonian, [H,T] ≠ 0. This means that the time-reversal operator does not commute with the Hamiltonian and therefore, the eigenvalues of T are not conserved quantities in a time-reversal invariant system.

In conclusion, even though the Hamiltonian is invariant under time reversal, the eigenvalues of the time-reversal operator are not conserved quantities. This is because T is an anti-unitary operator and its eigenvalues are not constant and do not commute with the Hamiltonian.