thanks for your answer,but i don't get it well
Is the Hamiltonian commutable with momentum operator in the 1- dimensional finite well?
can you see [p^2/2m +v , P]=0 is to formed, for V=constant?
why in this case, we can't find degeneracy?
why eigenstate of momenta are not exsist when...
[H,P]=0 , where P is momentum operator.
Hamiltonian is commutable with momentum operator. so H and p have
wave function simultaniously, but in 1-dimensional potential well degeneracy
not exist.
what is the reason?
:cry: could someone explain the following sentence to me?
"First-order perturbation correction is often
precluded because of a symmetry principle operating
for the state under consideration."
"In second-order, on the other hand, the perturbation
has to connect the given state...