# Choosing a global phase in time-independent perturbation theory

I have been trying for quite some time now to find a satisfactory justification of a standard assumption we usually do in time-independent perturbation theory, but I am still puzzled.

Here is the problem: We consider a Hamiltonian of the type $$H=H_0+\lambda V$$, where $$V$$ is a perturbation and $$\lambda$$ is a small parameter. Let $$|\psi\rangle$$ be an eigenstate of $$H$$. We can write $$|\psi\rangle$$ as a series in powers of $$\lambda$$, $$|\psi\rangle=\sum_{q=0}^\infty\lambda^q|\psi_q\rangle$$ where we know that $$|{\psi_0}\rangle$$ is an eigenstate of $$H_0$$.

In most materials I have consulted, it is clearly stated that the global phase of $$|\psi\rangle$$ can be chosen such that $$\langle\psi_0|\psi\rangle$$ is real; see for example the Wikipedia page on perturbation theory, this (otherwise very clear) PDF (above equation (20)) or Chap. 11 of Cohen-Tannoudji, Diu and Laloë's famous book Quantum Mechanics (Vol. 2). This seems strange to me since, if you change the phase of $$|\psi\rangle$$, the phase of $$|\psi_0\rangle$$ will necessarily change as well. Namely, if $$|\psi'\rangle=\mathrm e^{\mathrm i\theta}|\psi\rangle$$, then $$|\psi'\rangle=\sum_{q=0}^\infty\lambda^q|\psi'_q\rangle=\sum_{q=0}^\infty\lambda^q\mathrm e^{\mathrm i\theta}|\psi_q\rangle$$ The uniqueness of the above expansion clearly imposes $$|\psi'_q\rangle=\mathrm e^{\mathrm i\theta}|\psi_q\rangle$$, such that $$\langle\psi'_0|\psi'\rangle=\langle\psi_0|\psi\rangle$$, so the phase transformation hasn't changed anything.

Does anyone has an idea on how to justify that $$\langle\psi_0|\psi\rangle$$ can be assumed to be real without loss of generality?

A phase change performed at the same time on all eigenstates $$|\psi_q\rangle \longrightarrow |\psi_q'\rangle = e^{i\theta}|\psi_q\rangle$$ won't alter probabilities computed with states spanned by {$$|\psi_q\rangle$$}: $$|\langle \phi|\sum_qc_q|\psi_q'\rangle|^2 = |e^{i\theta}\langle \phi|\sum_qc_q|\psi_q\rangle|^2 = |\langle \phi|\sum_qc_q|\psi_q\rangle|^2$$.
But, as you pointed out, we cannot perform different phase changes at each $$|\psi_q\rangle$$ individually: once we choose a phase for some $$|\psi_q\rangle$$, the phase of the remaining are fixed so that probabilities don't change.
So we are free to choose a global phase for {$$|\psi_q\rangle$$}, and we do so to make $$\langle\psi_0|\psi\rangle$$ real: let $$\langle\psi_0|\psi\rangle = e^{-i\theta}|\langle\psi_0|\psi\rangle|$$, so we fix a new phase ,$$|\psi_q\rangle \rightarrow |\psi_q'\rangle = e^{i\theta}|\psi_q\rangle$$, such that $$\langle\psi_0'|\psi\rangle = \langle\psi|\psi'_0\rangle$$.
By demanding $$\langle\psi_0|\psi\rangle = \langle\psi|\psi_0\rangle$$ we fixed a global phase for {$$|\psi_q\rangle$$}, so we cannot use this trick again to further assume that some other $$\langle\psi_k|\psi\rangle$$ is real without giving up on $$\langle\psi_0|\psi\rangle = \langle\psi|\psi_0\rangle$$.