I have been learning about time independent perturbation theory (non-degenerate for the moment), and am not satisfied about a particular point: the justification for setting $\langle n^i|n^0\rangle = 0$ for $i>0$ where $i$ denotes the order of the approximation.

I should think that we can set this to zero iff it has no observable consequences. And since the only quantities that we observe/measure are the eigenvalues, this is iff the eigenvalues remain unchanged. Now it is far from obvious to me as to how we can make a claim that the eigenvalues of ANY operator we apply to the subsequent kets will be unchanged, and therefore that we really can never know by ANY measurement whether the $i>0$ order corrections to the $n$th eigenvalue, $|n\rangle$, really are entirely orthogonal to $|n^0\rangle$. I would appreciate a comment on this, but even just considering the Hamiltonian operator it seems not to be true.

If we have $(H^0+\lambda H^1)(|n^0\rangle + \lambda |n^1\rangle + \lambda ^2|n^2\rangle + ...) = (E_n^0 +\lambda E_n^1 +\lambda ^2 E_n^2+...)(|n^0\rangle + \lambda |n^1\rangle + \lambda ^2|n^2\rangle + ...)$

and from the first order approximation

$E_n^1= \langle n^0 | H^1 | n^0\rangle$

which is independent on $|n^1\rangle$.

One can easily find the coefficients $\langle m^0|n^1 \rangle = \frac{\langle m^0|H^1|n^0\rangle}{E_n^0-E_m^0}$ for $m \neq n$ such that

$|n^1\rangle = \alpha |n^0\rangle + \Sigma _{m\neq n} |m^0\rangle \frac{\langle m^0|H^1|n^0\rangle}{E_n^0-E_m^0}$

And that is about as much as one can say from considering the first order terms. I appreciate that the independence of $E_n^1$ on $|n^1\rangle$ means that, working to first order, and if the Hamiltonian's spectrum are the only quantities we care about, we can set $\alpha = 0$. And in fact, it is convenient that way because then the $|n\rangle = |n^0\rangle+\lambda |n^1\rangle$ is normalised to $O(\lambda ^2)$.

However, the value of $\alpha$ does have second order consequences (as demonstrated below), which is why I do not understand why authors use $\langle n^1|n^0\rangle = 0$ in finding second-order uentities, i.e. $E_n^2$.

To second order:

$H^1|n^1\rangle + H^0|n^2\rangle = E_n^0|n^2\rangle + E_n^1|n^1\rangle + E_n^2|n^0\rangle$

and then taking the inner product with $\langle n^0|$, one gets

$E_n^2 = \langle n^0|H^1|n^1 \rangle - E_n^1\langle n_0|n^1\rangle$

and substituting in for $E_n^1$ and $|n^1\rangle$, and after some rearranging, one finds

$E_n^2 = \langle n^0|H^1 [|n^1\rangle - \alpha |n^0\rangle]$

which does depend on $\alpha$!

  • $\begingroup$ what is this $\alpha$ you refer to? $\endgroup$ – ZeroTheHero Oct 13 '18 at 12:30
  • $\begingroup$ @ZeroTheHero It is the component of $|n^1\rangle$ parallel to $|n^0\rangle$ $\endgroup$ – 21joanna12 Oct 13 '18 at 14:40

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