# More doubts in perturbation theory

• I
• Kashmir

#### 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 ?

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 ?

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.

• vanhees71
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'

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

• Kashmir
I like the treatment in Sakurai, Modern Quantum Mechanics.

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

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.