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Kashmir

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*Townsend, quantum mechanic*s

" 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 ?