In the perturbation theory for non-degenerate levels, the energy $E_n(\lambda)$ of an eigenstate $|\psi_n(\lambda)\rangle$ of the hamiltonian $\mathcal{H}=\mathcal{H}_0+\lambda \mathcal{H}_1$ (where $\mathcal{H}_0$ is the unperturbed hamiltonian with eigenstates $E_n^0$ and $\lambda\mathcal{H}_1$ is the perturbation) is described by the equation

$$E_n(\lambda)=E_n^0+\dfrac{\lambda\langle\phi_n|\mathcal{H}_1|\phi_n\rangle+\lambda^2\langle\phi_n|\mathcal{H}_1|\psi_n^1\rangle+\lambda^3\langle\phi_n|\mathcal{H}_1|\psi_n^2\rangle+\dots}{1+\lambda a_n^{(1)}+\lambda^2 a_n^{(2)}+\dots}.$$

Here, $|\phi_n\rangle$ is the $n$'th eigenstate of the unperturbed hamiltonian, and $a_n^{(p)}=\langle\phi_n|\psi_n^p\rangle$, with $|\psi_n^p\rangle$ being the $p$'th correction to the $n$'th eigenstate. The zeroth correction of course equals the eigenstate of the unperturbed hamiltonian: $|\psi_n^0\rangle=|\phi_n\rangle$.

My question is the following: how does one go from the above equation to this? $$\begin{array}{r l} E_n(\lambda)&=E_n^0+\lambda\langle\phi_n|\mathcal{H}_1|\phi_n\rangle\\ &+\lambda^2\left[\langle\phi_n|\mathcal{H}_1|\psi_n^1\rangle-\langle\phi_n|\mathcal{H}_1|\phi_n\rangle a_n^{(1)}\right]\\ &+\lambda^3\left[\langle\phi_n|\mathcal{H}_1|\psi_n^2\rangle-\langle\phi_n|\mathcal{H}_1|\psi_n^1\rangle a_n^{(1)}-\langle\phi_n|\mathcal{H}_1|\phi_n\rangle a_n^{(2)}\right]+\dots \end{array}$$

The book I'm working with tells me this is done by "Expanding this expression in powers of $\lambda$", but I can't recognize the procedure. Of course the first term $E_n^0$ is of order $\lambda^0$, and the second term $\lambda\langle\phi_n|\mathcal{H}_1|\phi_n\rangle$ is of order $\lambda^1$, and these can be taken from the original equation right away, but I can't find out how the subtractions in the third and fourth term got there.

  • $\begingroup$ Are you certain that the order $\lambda^3$ coefficient in the equation after "My question is the following" was transcribed correctly? Also, which book is this from? $\endgroup$ Commented Dec 15, 2013 at 21:38
  • $\begingroup$ Just checked it, can't find any difference between the equation here and the one in the book. Looking at the answer by David Z however, I guess the squared coefficient was left out because it is even an order smaller for small $a_n^{(1)}$. This is from "Atomic Physics 1 & 2", a book (or collection of lecture notes) that our teacher Atom Physics has written during the last 10 years of teaching this subject at the University of Amsterdam (not published as far as I know). $\endgroup$
    – Betohaku
    Commented Dec 16, 2013 at 8:27
  • 1
    $\begingroup$ I see ok; the issue pointed out by David was precisely my point of confusion; I had written a response, but decided not to post because of the extra term. $\endgroup$ Commented Jan 3, 2014 at 9:56

1 Answer 1


You can work it out from the Taylor series

$$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots$$

where $x = \lambda a_n^{(1)} + \lambda^2 a_n^{(2)} + \cdots$. Each term can then be expanded in a power series in $\lambda$:

$$\begin{align} -x &= -\lambda a_n^{(1)} - \lambda^2 a_n^{(2)} - \lambda^3 a_n^{(3)} - \cdots \\ x^2 &= \lambda^2 \bigl(a_n^{(1)}\bigr)^2 + 2\lambda^3 a_n^{(1)}a_n^{(2)} + \cdots \\ -x^3 &= -\lambda^3 \bigl(a_n^{(1)}\bigr)^3 - \cdots \end{align}$$

You can then add these up term-by-term in $\lambda$:

$$\begin{align}\frac{1}{1 + \lambda a_n^{(1)} + \lambda^2 a_n^{(2)} + \cdots} \\ = 1 - \lambda a_n^{(1)} + & \lambda^2\Bigl[-a_n^{(2)} + \bigl(a_n^{(1)}\bigr)^2\Bigr] + \lambda^3\Bigl[-a_n^{(3)} + 2a_n^{(1)}a_n^{(2)} - \bigl(a_n^{(1)}\bigr)^3\Bigr] + \cdots\end{align}$$

and then multiply this by the series in the numerator ($\lambda\langle\phi_n\rvert\mathcal{H}_1\lvert\phi_n\rangle + \cdots$)to get

$$\begin{align} &\lambda\langle\phi_n\rvert\mathcal{H}_1\lvert\phi_n\rangle \\ + &\lambda^2\Bigl[\langle\phi_n\rvert\mathcal{H}_1\lvert\psi_n^1\rangle - \langle\phi_n\rvert\mathcal{H}_1\lvert\phi_n\rangle a_n^{(1)}\Bigr] \\ + &\lambda^3\biggl(\langle\phi_n\rvert\mathcal{H}_1\lvert\psi_n^2\rangle - \langle\phi_n\rvert\mathcal{H}_1\lvert\psi_n^1\rangle a_n^{(1)} + \langle\phi_n\rvert\mathcal{H}_1\lvert\phi_n\rangle\Bigl[-a_n^{(2)} + \bigl(a_n^{(1)}\bigr)^2\Bigr]\biggr) \\ + &\cdots \end{align}$$

Note that in the expression as you wrote it in the question, there is a term missing in the coefficient of $\lambda^3$.

I'm sure you can tell that calculating additional higher-order terms becomes increasingly tedious, but it's just straightforward series multiplication. There's nothing particularly complicated about it.


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