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?


1 Answer 1


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 $\langle\psi_0|\psi\rangle = \langle\psi|\psi_0\rangle$.


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