The expansion of a function in powers of a parameter

Question

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.

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.