- #1
Malamala
- 299
- 27
Hello! Let's say we have 2 states of fixed parity ##| + \rangle## and ##| - \rangle## with energies ##E_+## and ##E_-## and we have a P-odd perturbing hamiltonian (on top of the original hamiltonian, ##H_0## whose eigenfunctions are the 2 above), ##V_P##. According to 1st order perturbation theory, the corrections to energy for both states is zero. So in order to get any difference in energy we need to go to second order and there the correction to, say, ##| + \rangle## state is:
$$E_+^2 = \frac{|\langle-|V_P|+\rangle|^2}{E_+-E_-}$$ Also the first order correction to the wavefunction implies that ##| + \rangle## becomes:
$$| + \rangle' = | + \rangle + \frac{\langle-|V_P|+\rangle}{E_+-E_-}| - \rangle$$ Now if I want to calculate the expectation value of the energy in this ##| + \rangle'## state, I would get
$$'\langle+|H_0+V_P| + \rangle' = \frac{\langle+|V_P| - \rangle\langle-|V_P|+\rangle}{E_+-E_-}+\frac{(\langle-|V_P| + \rangle)^\dagger\langle-|V_P|+\rangle}{E_+-E_-} + E_+ + \frac{|\langle-|V_P|+\rangle|^2}{(E_+-E_-)^2}E_-$$
So the correction to the energy would be:
$$\frac{\langle+|V_P| - \rangle\langle-|V_P|+\rangle}{E_+-E_-}+\frac{(\langle-|V_P| + \rangle)^\dagger\langle-|V_P|+\rangle}{E_+-E_-} + \frac{|\langle-|V_P|+\rangle|^2}{(E_+-E_-)^2}E_-$$
So there are a few things I am confused about (please let me know if I did any calculation mistakes):
1. Why isn't the expectation value I calculated in the last equation above, equal to the predicted shift in energy by the 2nd order correction? It seems like the expectation value is also second order in ##V_P##, so shouldn't they be the same?
2. If instead of ##V_P## I have a ##V_{PT}## i.e. a potential that is P,T-odd, what should I change in these calculations? In the case of parity it is clear when an expectation value is zero or not, based on the parities of the wavefunctions and the hamiltonian, but what should I do in case the potential is T-odd, too? The tricks from P-odd still applied (i.e. if a matrix element is zero for a P-odd, it is zero for a P,T-odd, too), but what further constraints do I get if I add the T-odd on top?
Thank you!
$$E_+^2 = \frac{|\langle-|V_P|+\rangle|^2}{E_+-E_-}$$ Also the first order correction to the wavefunction implies that ##| + \rangle## becomes:
$$| + \rangle' = | + \rangle + \frac{\langle-|V_P|+\rangle}{E_+-E_-}| - \rangle$$ Now if I want to calculate the expectation value of the energy in this ##| + \rangle'## state, I would get
$$'\langle+|H_0+V_P| + \rangle' = \frac{\langle+|V_P| - \rangle\langle-|V_P|+\rangle}{E_+-E_-}+\frac{(\langle-|V_P| + \rangle)^\dagger\langle-|V_P|+\rangle}{E_+-E_-} + E_+ + \frac{|\langle-|V_P|+\rangle|^2}{(E_+-E_-)^2}E_-$$
So the correction to the energy would be:
$$\frac{\langle+|V_P| - \rangle\langle-|V_P|+\rangle}{E_+-E_-}+\frac{(\langle-|V_P| + \rangle)^\dagger\langle-|V_P|+\rangle}{E_+-E_-} + \frac{|\langle-|V_P|+\rangle|^2}{(E_+-E_-)^2}E_-$$
So there are a few things I am confused about (please let me know if I did any calculation mistakes):
1. Why isn't the expectation value I calculated in the last equation above, equal to the predicted shift in energy by the 2nd order correction? It seems like the expectation value is also second order in ##V_P##, so shouldn't they be the same?
2. If instead of ##V_P## I have a ##V_{PT}## i.e. a potential that is P,T-odd, what should I change in these calculations? In the case of parity it is clear when an expectation value is zero or not, based on the parities of the wavefunctions and the hamiltonian, but what should I do in case the potential is T-odd, too? The tricks from P-odd still applied (i.e. if a matrix element is zero for a P-odd, it is zero for a P,T-odd, too), but what further constraints do I get if I add the T-odd on top?
Thank you!