The ground state of a time-reversal invariant system must has zero momentum?

[wdlang]

if the ground state is non-degenearate, this is easily understood

But what if the ground state is non-degenerate?

 

[Vanadium 50]

If you are arguing that the ground state is degenerate with eigenmomenta p and -p, I would argue this is not a state of zero momenta (even though the expectation value of the momentum operator is 0): if you measure p, you never get 0.

 

[Physics Monkey]

Hi wdlang,

The statement that the ground state of a time reversal invariant Hamiltonian must have zero momentum is simply not true.

The ground state may spontaneously break time reversal symmetry. However, if it does, then time reversal symmetry implies the existence of another perfectly good time reversed ground state. Thus the ground state is degenerate in this case.

I can give a simpler example than momentum that may help. Consider a ferromagnet at zero temperature. The ground state has all the spins aligned, but there is an equally good ground state with all spins reversed. The ground state is in fact degenerate.

For the case of momentum, here is a simple example that works. Consider the Hamiltonian H = E0 + (|p| - p0)^2/2m with E0, p0, and m constants . Such a Hamiltonian roughly describes rotons in He-4. The ground states of this hamiltonian have non-zero momentum and are degenerate.

Hope this helps.

 

[wdlang]

Hi wdlang,

The statement that the ground state of a time reversal invariant Hamiltonian must have zero momentum is simply not true.

The ground state may spontaneously break time reversal symmetry. However, if it does, then time reversal symmetry implies the existence of another perfectly good time reversed ground state. Thus the ground state is degenerate in this case.

I can give a simpler example than momentum that may help. Consider a ferromagnet at zero temperature. The ground state has all the spins aligned, but there is an equally good ground state with all spins reversed. The ground state is in fact degenerate.

For the case of momentum, here is a simple example that works. Consider the Hamiltonian H = E0 + (|p| - p0)^2/2m with E0, p0, and m constants . Such a Hamiltonian roughly describes rotons in He-4. The ground states of this hamiltonian have non-zero momentum and are degenerate.

Hope this helps.

thanks a lot!

yes, your example is good

 

[eaglelake]

When you say "ground state", I assume you mean ground state of the Hamiltonian, i.e. the energy ground state.

Generally speaking, because of the position dependent potential energy term in the Hamiltonian, the momentum operator does not commute with the Hamiltonian. In any energy eigenstate the momentum is uncertain; there is no ground state momentum. Thus, if we measure the momentum in the ground state, there is an entire eigenvalue spectrum of possible results, but there is no unique value .

The Hamiltonian eigenvalue equation is the Schrodinger time independent equation. (time independent??) So, I am confused! Could you please give me a reference where you saw this? Thank you.

 

[eaglelake]

Are you referring to a time-dependent Hamiltonian?

 

[wdlang]

When you say "ground state", I assume you mean ground state of the Hamiltonian, i.e. the energy ground state.

Generally speaking, because of the position dependent potential energy term in the Hamiltonian, the momentum operator does not commute with the Hamiltonian. In any energy eigenstate the momentum is uncertain; there is no ground state momentum. Thus, if we measure the momentum in the ground state, there is an entire eigenvalue spectrum of possible results, but there is no unique value .

The Hamiltonian eigenvalue equation is the Schrodinger time independent equation. (time independent??) So, I am confused! Could you please give me a reference where you saw this? Thank you.

i saw this statement in the RMP paper by I. Bloch et al.

RMP 80, 885 (2008)

on page 905, the paragraph under eq. 65.