I'm currently taking a class in QM and we came across the topic of non-degenerate perturbation theory. Let us for further discussions assume that $H_0$ is the unperturbed Hamiltonian with solutions of the form $$H_0\phi_n=\epsilon_n\phi_n,$$ for $n\in\mathbb{N}$. Since we are looking at the non-degenerate case we know that $\epsilon_n\neq \epsilon_m$ for all $m\neq n$. Let $H'$ be the perturbation and let $$\psi = \sum_{i=0}^{\infty}\lambda^i\psi_i,\hspace{0.5cm} E = \sum_{i=0}^{\infty}\lambda^iE_i.$$ The new problem that we now want to solve is $$H\psi = E\psi,$$ where $H\equiv H_0+H'$. We can now show that the following relationship holds $$H_0\psi_i+H'\psi_{i-1} = \sum_{j=0}^iE_j\psi_{i-j},$$ with the convention $\psi_{-1}=0$. For $i=0$ we get $\psi_0=\phi_n$, and for the case $i=1$ we find $$(H_0-E_0)\psi_1=(E_1-H')\psi_0.$$ In my lecture notes it now says that any linear combination of $\psi_1$ and $\psi_0$ would solve this equation and we therefore need another condition, which we eventually choose to be $\langle \phi_n\vert\psi_k\rangle =\delta_0^k$, to have a unique solution.

I'm having a hard time seeing the need for such an extra condition. I've tried to look it up but most of the time the normalization is just assumed (for example here, or in this answer) to hold anyways and not much of an explanation is given.

TL;DR: Why do we need such an extra condition and how can I see that?

  • $\begingroup$ I think you've messed up your notation a bit. You state that $H_0\phi_i+H'\phi_{i-1} = \sum_{j=0}^iE_j\phi_{i-j}$, but you define $\phi_n$ as the $n$th eigenstate of $H_0$, which would imply that the corrections to $\phi_0$ depends on $\phi_1$. What you really want is the first order correction to the $n$th eigenstate. $\endgroup$
    – cxx
    Dec 15, 2018 at 16:56
  • $\begingroup$ @HantingZhang thank you very much for pointing this out! Could you maybe take a look at it now and see if it works? $\endgroup$
    – Sito
    Dec 15, 2018 at 17:16
  • 1
    $\begingroup$ Yes, I'm writing a answer rn $\endgroup$
    – cxx
    Dec 15, 2018 at 17:17

1 Answer 1


In my lecture notes it now says that any linear combination of $\psi_1$ and $\psi_0$ would solve this equation [...]

Indeed, if $(H^0-E^0_n)\psi_n^1=(E^1_n-H')\psi_n^0$, then we can see that if $\psi_n^1$ satisfies the equation above, then $\psi_n^1 + k\psi_n^0$ will too: $$(H_0-E_n^0)(\psi_n^1 + k\psi_n^0) = (H_0-E_n^0)\psi_n^1 + k(H_0-E_n^0)\psi_0$$ $$ = (H_0-E_n^0)\psi_n^1 + k(E_n^0-E_n^0)\psi_n^0 = (H_0-E_n^0)\psi_n^1.$$ We've added nothing to our equation.

Since our unperturbed eigenfunctions constitute of complete, orthonormal set, we can write $$\psi_n^1 = \sum_{m} c^{(n)}_m\psi_m^0,$$ where the fancy $(n)$ in $c^{(n)}_m$ in just bookkeeping for which $\psi_n^1$ we're talking about. Now, from above, we have the freedom to subtract off the $c^{(n)}_n\psi^0_n$ term in the sum, turning it into: $$\psi_n^1 = \sum_{m \neq n} c^{(n)}_m\psi_m^0.$$ Taking the inner product with $\langle\psi_n^0\vert$ on both sides,

$$\langle\psi_n^0\vert\psi_n^1\rangle = \langle\psi_n^0\vert\sum_{m \neq n} c^{(n)}_m\vert\psi_m^0\rangle = \sum_{m \neq n} c^{(n)}_m\langle\psi_n^0\vert\psi_m^0\rangle = 0.$$

That's how we get the normalization.

Why do we need such an extra condition and how can I see that?

Sometimes, we are interested in the perturbations to the wavefunction $\psi_n^1$, for which we can solve for with $$(H^0-E^0_n)\psi_n^1=(E^1_n-H')\psi_n^0.$$ Since $$\psi_n^1 = \sum_{m \neq n} c^{(n)}_m\psi_m^0,$$ the problem reduces to finding the coeifficents $c^{(n)}_m$. Plugging in $\psi_n^1$, we have $$\sum_{m \neq n} (E^0_m - E^0_n)c^{(n)}_m\psi_m^0 = (E^1_n-H')\psi_n^0.$$ Then, taking the inner product with $\psi_k^0$,

$$\sum_{m \neq n} (E^0_m - E^0_n)c^{(n)}_m\langle\psi_k^0\vert\psi_m^0\rangle = E^1_n\langle\psi_k^0\vert\psi_n^0\rangle-\langle\psi_k^0\vert H'\vert \psi_n^0\rangle.$$

If $k = n$, then the entire LHS is $0$, and we just get the equation for the first order perturbations to the energy. If $k \neq n$, then $$ (E^0_k - E^0_n)c^{(n)}_m = -\langle\psi_k^0\vert H'\vert \psi_n^0\rangle.$$ Hence $$c^{(n)}_m = \frac{\langle\psi_m^0\vert H'\vert \psi_n^0\rangle}{E^0_n - E^0_m} \rightarrow \psi_n^1 = \sum_{m \neq n} \frac{\langle\psi_m^0\vert H'\vert \psi_n^0\rangle}{E^0_n - E^0_m}\psi^0_m.$$

Notice that the demonimator is safe, because we pulled out the $n = m$ term right from the beginning. By choosing the extra condition, we can avoid some serious trouble.

  • Note: this is only for non-degenerate perturbation theory.

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