How do I properly express adding perturbed states to unperturbed states?
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I have a problem set due tomorrow, and the last problem is driving me nuts. Been combing through griffiths trying to find similar examples to no avail, so it'd be greatly appreciated if stackexchange could help me out =)
Suppose that a quantum system has 4 independent states. The unperturbed energy of state |n1⟩|n1⟩ is E1E1 while the states |n2⟩|n2⟩, |n3⟩|n3⟩, and |n4⟩|n4⟩ have energy E2E2. Now a perturbation H′H′ acts on the system such that
H′|n1⟩H′|n2⟩H′|n3⟩H′|n4⟩=−k|n1⟩=k|n2⟩=k2–√|n4⟩=k2–√|n3⟩+k|n4⟩H′|n1⟩=−k|n1⟩H′|n2⟩=k|n2⟩H′|n3⟩=k2|n4⟩H′|n4⟩=k2|n3⟩+k|n4⟩
So the way I understand it, I have two unperturbed energy levels here, E2E2 and E1E1, the former of which is triply degenerate. When the system is perturbed, E2E2 splits off into 3 new distinct energy levels. The first order correction would typically be given by
En=E0+δEn,En=E0+δEn,
where EnEn is the unperturbed Hamiltonian and δEnδEn is the correction due to the perturbation. What I would have to do is add the perturbed states to the unperturbed states, but how do I work with the notation to express that mathematically? My base intuition is that I can treat the kets as eigenfunctions of H0H0 and treat the kk coefficients as eigenvalues, but I'm not really sure if that would be going on the right track.
Suppose that a quantum system has 4 independent states. The unperturbed energy of state |n1⟩|n1⟩ is E1E1 while the states |n2⟩|n2⟩, |n3⟩|n3⟩, and |n4⟩|n4⟩ have energy E2E2. Now a perturbation H′H′ acts on the system such that
H′|n1⟩H′|n2⟩H′|n3⟩H′|n4⟩=−k|n1⟩=k|n2⟩=k2–√|n4⟩=k2–√|n3⟩+k|n4⟩H′|n1⟩=−k|n1⟩H′|n2⟩=k|n2⟩H′|n3⟩=k2|n4⟩H′|n4⟩=k2|n3⟩+k|n4⟩
So the way I understand it, I have two unperturbed energy levels here, E2E2 and E1E1, the former of which is triply degenerate. When the system is perturbed, E2E2 splits off into 3 new distinct energy levels. The first order correction would typically be given by
En=E0+δEn,En=E0+δEn,
where EnEn is the unperturbed Hamiltonian and δEnδEn is the correction due to the perturbation. What I would have to do is add the perturbed states to the unperturbed states, but how do I work with the notation to express that mathematically? My base intuition is that I can treat the kets as eigenfunctions of H0H0 and treat the kk coefficients as eigenvalues, but I'm not really sure if that would be going on the right track.
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Hey dear here is your answer
Suppose that the Hamiltonian is written as
H=H(0)+VH=H(0)+V
It is assumed that both H(0)H(0), the unperturbed hamiltonian, and HH, the perturbed hamiltonian, are self-adjoint. In particular, there is an orthonormal basis ψ(0)mψm(0) for the Hilbert space consisting of eigenvectors of H(0)H(0), and a an orthonormal basis ψmψm consisting of eigenvectors of HH.
One can choose to write any state ψψ in either basis. In particular, one can choose to write the first order perturbed state in the basis ψ(0)mψm(0) of H(0)H(0) and then use this to derive lots of useful things.
Addendum. One thing I forgot to mention is why one would want to write the perturbed state in the unperturbed basis. The simple, essential answer is that perturbation theory is so useful precisely because the unperturbed Hamiltonian can be easily diagonalized; namely it's eigenvectors and eigenvalues are known exactly, whereas those of the unperturbed Hamiltonian are not.
Hope it's help you
Suppose that the Hamiltonian is written as
H=H(0)+VH=H(0)+V
It is assumed that both H(0)H(0), the unperturbed hamiltonian, and HH, the perturbed hamiltonian, are self-adjoint. In particular, there is an orthonormal basis ψ(0)mψm(0) for the Hilbert space consisting of eigenvectors of H(0)H(0), and a an orthonormal basis ψmψm consisting of eigenvectors of HH.
One can choose to write any state ψψ in either basis. In particular, one can choose to write the first order perturbed state in the basis ψ(0)mψm(0) of H(0)H(0) and then use this to derive lots of useful things.
Addendum. One thing I forgot to mention is why one would want to write the perturbed state in the unperturbed basis. The simple, essential answer is that perturbation theory is so useful precisely because the unperturbed Hamiltonian can be easily diagonalized; namely it's eigenvectors and eigenvalues are known exactly, whereas those of the unperturbed Hamiltonian are not.
Hope it's help you
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