I More doubts in perturbation theory

[Kashmir]

Townsend, quantum mechanics
" In our earlier derivation we assumed that each unperturbed eigenstate $\left|\varphi_{n}^{(0)}\right\rangle$ turns smoothly into the exact eigenstate $\left|\psi_{n}\right\rangle$ as we turn on the perturbing Hamiltonian. However, if there are $N$ states
$\left|\varphi_{n, i}^{(0)}\right\rangle \quad i=1,2, \ldots, N$ all with the same energy, it isn't clear which are the right linear combinations of the unperturbed states that become the exact eigenstates. For example, in the case of two-fold degeneracy, is it
$\left|\varphi_{n, 1}^{(0)}\right\rangle \text { and }\left|\varphi_{n, 2}^{(0)}\right\rangle$
or
$\frac{1}{\sqrt{2}}\left(\left|\varphi_{n, 1}^{(0)}\right\rangle+\left|\varphi_{n, 2}^{(0)}\right\rangle\right) \quad \text { and } \frac{1}{\sqrt{2}}\left(\left|\varphi_{n, 1}^{(0)}\right\rangle-\left|\varphi_{n, 2}^{(0)}\right\rangle\right)$
or some other of the infinite number of linear combinations that we can construct from these two states? If we choose the wrong linear combination of unperturbed states as a starting point, even the small change in the Hamiltonian generated by turning on the perturbation with an infinitesimal $\lambda$ must produce a large change in the state
1) We find the eigenstate of the total Hamiltonian using the below series $\begin{aligned}\left|\psi_{n}\right\rangle &=\left|\varphi_{n}^{(0)}\right\rangle+\lambda\left|\varphi_{n}^{(1)}\right\rangle+\lambda^{2}\left|\varphi_{n}^{(2)}\right\rangle+\cdots \\ E_{n} &=E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda^{2} E_{n}^{(2)}+\cdots \end{aligned}$ What happens if I use the above series expansion for the wrong states? 2) Is it that the states which change abruptly aren't "the exact eigenstates" of the total perturbed Hamilton ?

 

[HomogenousCow]

1) You end up with meaningless expressions since the $(E_1 - E_2)^{-1}$ factors diverge.
2) No, the necessity to choose the "right" basis only arises in perturbation theory because we're looking for a power series in terms of $\lambda$ which cannot approximate jump-discontinuities.

 

[Kashmir]

1) You end up with meaningless expressions since the $(E_1 - E_2)^{-1}$ factors diverge.
2) No, the necessity to choose the "right" basis only arises in perturbation theory because we're looking for a power series in terms of $\lambda$ which cannot approximate jump-discontinuities.

So it is not possible to write the series expansion using"wrong" states ?

 

[DrDu]

This case is known as "degenerate perturbation theory" and is discussed in any textbook. Specifically, you first have to diagonalize the perturbation V in the set of degenerate zeroth order states.

 

[Kashmir]

This case is known as "degenerate perturbation theory" and is discussed in any textbook. Specifically, you first have to diagonalize the perturbation V in the set of degenerate zeroth order states.

I'm having confusion reading my textbook 'McIntyre'

 

[DrDu]

So try to find another book which fits your way of thinking better :-)

 

[vanhees71]

I like the treatment in Sakurai, Modern Quantum Mechanics.

 

[Kashmir]

So it is not possible to write the series expansion using"wrong" states ?

Anyone??

 

[DrDu]

Maybe you can think of it like this. This degeneracy can be lifted in several ways. For example, in a spin 1/2 problem, you can add either a sigma_z or a sigma_x term to lift the degeneracy. Assume you take the degenerate case as the limit of a small sigma_z potential tending to 0. Now if you add a sigma_x potential as a real perturbation, the convergence of the series will break down as soon as the term containing sigma_x becomes larger than the term containing sigma_z. In the limit of vanishing sigma_z, the perturbation series will not converge at all.