First order state correction for time independent perturbation theory Following the derivation on the wikipedia page, https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)_ , $|n_{0}\rangle$ is an eigenstate of the unperturbed Hamiltonian $\hat{H}_{0}$, the perturbed state is given by
$$
|n\rangle=|n_{0}\rangle+\lambda|n_{1}\rangle+\lambda^2|n_{2}\rangle+....
$$
and the perturbed Hamiltonian is given by
$$
\hat{H}=\hat{H}_{0}+\lambda\hat{V}.
$$
In their derivation, they state that $\langle n_{0}|n_{1}\rangle$ must be real because "the overall phase is not determined in quantum mechanics". If this is the case, it is possible to show that $\langle n_{0}|n_{1}\rangle=0$, however I can't fully understand why we can deduce that $\langle n_{0}|n_{1}\rangle$ is real. Any ideas?
 A: Let's see what happens if $\langle n^{(0)}|n^{(1)}\rangle$ is not real. Take $\hat{H}_0 \psi^{(0)}(x)=E^{(0)}\psi^{(0)}(x)$ and (to the linear order in $\lambda$) $$(\hat{H}_0+\lambda\hat{V})(\psi^{(0)}(x)+\lambda \psi^{(1)}(x)) = (E^{(0)}+\lambda E^{(1)})(\psi^{(0)}(x)+\lambda \psi^{(1)}(x))$$. Equating the terms linear in $\lambda$ we obtain
$$\hat{V}\psi^{(0)}+\hat{H_0}\psi^{(1)}=E^{(1)}\psi^{(0)}+ E^{(0)}\psi^{(1)}$$
Since, as said, the overall phase of the wave function is unobservable, we can choose that $\psi^0(x)$ is real (remember, we always assume the eigenstates are non-degenerate). Now the normalization requires that $\langle n^{(0)}|n^{(1)}\rangle + \langle n^{(1)}|n^{(0)}\rangle=0$, hence $\int (\psi^{(0)})^{*}\psi^{(1)}(x)dx$ can only be imaginary. The real part of $\psi^{(1)}(x)$ is taken care of in the textbooks; let's look at the imaginary part. Take $\psi^{(1)}(x)\equiv i g(x)$. Then have a look what we have:
$$\hat{V}\psi^{(0)}(x)+i\hat{H_0}g(x)=E^{(1)}\psi^{(0)}(x)+ iE^{(0)}g(x)$$
but since both $\hat{H}_0$ and $\hat{V}$ are hermitian (and hence $E^{(0)}$ and $E^{(1)}$ real), it also true by complex conjugation that
$$\hat{V}\psi^{(0)}(x)-i\hat{H_0}g(x)=E^{(1)}\psi^{(0)}(x)- iE^{(0)}g(x)$$
now take a difference and see that $\hat{H_0}g(x)=E^{(0)}g(x)$, that is $g(x)=\psi^{(0)}(x)$ (all states are non-degenerate). This means that the imaginary first-order correction to $\psi^{(0)}(x)$ is proportional to itself. There is nothing badly wrong about this: $|(1+i\lambda)|^2=1+\lambda^2\approx 1$, but in a sense such "correction" is trivial as it is nothing but a phase shift. And since it is independent of perturbation one can always safely ignore it.
A: If $\langle n_0|n_1\rangle$ is imaginary, it can be written as $re^{i\theta}$. If $|n_0\rangle$ solves $\hat H_0|n_0\rangle=E_0|n_0\rangle$, then so does $e^{i\theta}|n_0\rangle$. Choosing instead this as our $|n_0\rangle$ (as we are free to do) , we have that $\langle n_0|n_1\rangle = r$. (It does not $have$ to be real, but it can be taken to be real wolog).
