Physics, asked by powersantuk8095, 1 year ago

QM perturbation theory : When do I have to use degenerate/non-degenerate perturbation theory?

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Answered by vrrunda
0
I am considering a perturbation theory problem in quantum mechanics. The unperturbed hamiltonian is

H0=A1BS1z+A2BS2z.H0=A1BS1z+A2BS2z.

The eigenstates of the unperturbed hamiltonien are ∣M1M2⟩∣M1M2⟩, where −S1,2≤M1,2≤S1,2−S1,2≤M1,2≤S1,2 (SS's are spin operators). Here, A1≪A2A1≪A2. I want to know if this system is degenerate to apply the perturbation theory. The energy of each state is :

EM1M2=ℏ(A1BM1+A2BM2).EM1M2=ℏ(A1BM1+A2BM2).

(Am I correct ?) But because A1A1 is way smaller than A2A2, it means that the energy will have some kind of structure : separated energy levels of order A2BM2A2BM2, and then close to that, small variations around this value owing to A1BM1A1BM1.

In my problem, it would be simpler to do non-degenerate perturbation theory. But I have the feeling that it may be incorrect because of the small difference of energy added by the first operator (like : ∣M1,M2⟩∣M1,M2⟩ and ∣M1+1,M2⟩∣M1+1,M2⟩have almost the same energy, but not ∣M1M2⟩∣M1M2⟩and ∣M1M2+1⟩∣M1M2+1⟩), even if technically each state has a different energy.

Answered by Anonymous
0
Hey mate... ^_^

 A1 is way smaller than A2, it means that the energy will have some kind of structure : separated energy levels of order A2BM2, and then close to that, small variations around this value owing to A1BM1....

#Be Brainly❤️
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