# Time-independent perturbation theory: why i'th order perturbations are orthogonal to base state?

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 $$\alpha$$!

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