In the question here, an answer is given as to why the wave function and energy can be written as follows

$\psi_n = \psi^0_n + \lambda\psi^1_n + \lambda^2 \psi^2_n ...$

$E_n = E_n^0 + \lambda E_n^1 + \lambda ^2 E_n^2 ...$

but in the answer it is said that

"Because $\lambda$ appears in the Hamiltonian,it should be pretty obvious that the energy eigenstates $\psi_n$ and the corresponding eigenvalues $E_n$ depend on $\lambda$ as well"

which is not obvious to me...

we say that

$$H\psi_n = E\psi_n$$

so if $H(\lambda)$, I do not see how it follows naturally that $\psi(\lambda)$. It only makes sense that the eigenvalue would change if the operator,$H$, changes (becomes a function of $\lambda$ in this instance). Why does $\psi_n$ become a function of lambda as well? why is :

$$H(\lambda)\psi_n = E(\lambda)\psi_n$$ an incomplete describtion, why does $\psi_n$ become a function of lambda as a result of the hamiltonian becoing a function of lambda?


2 Answers 2


$H(\lambda)$ and $H(\lambda')$ are different operators if $\lambda\neq \lambda'$. In the absence of additional information, why would they have the same eigenvectors? As a simple example, consider $$H(\lambda) = \pmatrix{1&\lambda \\ \lambda & -1}$$ whose eigenvectors are, to first order in $\lambda$, $$\psi_+ \approx \pmatrix{1\\\lambda/2}\qquad \psi_- \approx \pmatrix{\lambda/2\\1}$$

It is of course possible that the operator might change in such a way that its eigenvectors don't depend on $\lambda$ - an example would be $$H(\lambda) = \pmatrix{1+\lambda & 0 \\ 0 & -(1+\lambda)}$$ But this is a special case, so there's no reason to expect it to be true in general. And even here, there's no harm in assuming the eigenvectors to be functions of $\lambda$, because all we'd find is that they are constant functions of $\lambda$ (i.e. $\psi(\lambda) = \psi(0)$).

  • $\begingroup$ If $\psi_n$ is an eigenvector of H and we act the operator on it, $H(\lambda)\psi_n$, then by definition of $\psi_n$ being an eigenvector then it should stay the same, unless you are saying that we have to change our original $\psi_n$ to be a function of $\lambda$ so it can be an eigenvector of $H(\lambda)$ since H and H(\lambda) do not have the same eigenvectors? $\endgroup$ Jul 2, 2022 at 1:21
  • 1
    $\begingroup$ @realanswers "unless you are saying ..." That is precisely what I'm saying. In the first example I gave in my answer, $\pmatrix{1\\ 0}$ is an eigenvector of $H(0)$, but clearly not an eigenvector of $H(\lambda)$ for $\lambda\neq 0$, right? $\endgroup$
    – J. Murray
    Jul 2, 2022 at 1:22
  • $\begingroup$ That makes a lot of sense, thank you for your help $\endgroup$ Jul 2, 2022 at 1:23

$\psi_n$ is a function of $\lambda$: $$ \psi_n = \psi^0_n + \lambda\psi^1_n + \lambda^2 \psi^2_n \ldots $$

The point is: you fix a basis set to be the eigenstates of the unperturbed eigenstates, and you expand the perturbed solution in terms of the unperturbed one.

If you accept that the eigenvalues must depend on $\lambda$, then finding the eigenvector will also depend on $\lambda$ since you need to solve

\begin{align} H\psi_n&=E(\lambda)\psi_n\,,\\ &=(E_n^0+\lambda E_n^1+\ldots)\psi_n \end{align} using the unperturbed eigenstates.

  • $\begingroup$ The question is asking WHY $\psi_n$ becomes a function of $\lambda$ because the Hamiltonian is a function of $\lambda$ $\endgroup$ Jul 2, 2022 at 1:14
  • $\begingroup$ sorry your comment makes no sense…. How is $H$ a function of $n$? $\endgroup$ Jul 2, 2022 at 1:15
  • $\begingroup$ edited, sorry meant $\lambda$ $\endgroup$ Jul 2, 2022 at 1:17

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