Non-degenerate perturbation theory

Consider a Hamiltonian with a small perturbation $\delta H$ scaled by the parameter $\lambda \in [0,1]$ .

\begin{align*} H(\lambda) = H^{(0)} + \lambda \delta H \end{align*}

Often, $H^{(0)}$ has a known spectrum and $\delta H$ describes some further physics that we want to consider. In this case, we want to know the adjustment to the energy levels and eigenstates. We Taylor expand these in $\lambda$ and define

\begin{align*} E_n(\lambda) &= E\suppar0 + \lambda E\suppar1 + \lambda^2 E\suppar2 + \cdots \\ \ket n_\lambda &= \ket{n\suppar0} + \lambda \ket{n\suppar 1} + \lambda^2 \ket{n\suppar 2} + \cdots. \end{align*}

To find the energy corrections we write the Hamiltonian eigenvalue equation and apply the $\lambda$ expansions

\begin{align*} &(H\suppar0 + \lambda \delta H - E_n(\lambda)) \ket n_\lambda = 0, \\ &\left( (H\suppar 0 - E_n\suppar0) - \lambda(E_n\suppar 1 - \delta H) - \lambda^2 E_n\suppar2 - \cdots \right) \cdot \\ &\left( \ket{n\suppar 0} + \lambda \ket{n\suppar 1} + \lambda^2 \ket{n\suppar 2} + \cdots\right) = 0. \end{align*}

This equation must hold for all values of $\lambda$ , so the coefficient on each power $\lambda^n$ must be zero. We can then write the equations

\begin{align*} H\suppar 0 \ket{n\suppar 0} &= E_n\suppar 0 \ket{n\suppar 0} \tag{1} \\ (H\suppar 0 - E_n\suppar 0) \ket{n\suppar 1} &= (E_n\suppar 1 - \delta H) \ket{n \suppar 0} \tag{$\lambda$} \\ (H \suppar 0 - E_n\suppar0) \ket{n\suppar 2} &= (E_n\suppar 1 - \delta H)\ket{n\suppar 1} + E_n\suppar 2 \ket{n\suppar 0} \tag{$\lambda^2$}\\ (H\suppar 0 - E_n\suppar0) \ket{n\suppar3} &= (E_n\suppar 1 -\delta H) \ket{n\suppar 2} + E_n\suppar 2\ket{n\suppar 1} + E_n\suppar 3 \ket{n\suppar 0} \\ & \ \,\,\vdots \end{align*}

Acting from the left with $\bra{n\suppar0}$ gives the energy corrections. For example, the first order correction comes from the order $(\lambda)$ equation

\begin{align*} \braket{n\suppar 0 | H\suppar0 - E_n\suppar 0|n\suppar 1} &= \braket{n\suppar 0 | E_n\suppar 1 - \delta H | n\suppar 0} \\ E_n\suppar 1 &= \braket{n\suppar 0 | \delta H | n\suppar 0}. \end{align*}

This equation is the famous first-order result of non-degenerate perturbation theory. In general, the same procedure yields $E_n\suppar k = \braket{n\suppar 0| \delta H | n\suppar{k-1}}$ .

To find the state corrections, act from the left with $\bra{k\suppar0}$ for some $k \ne n$ . Assuming unperturbed states are orthogonal,

\begin{align*} \braket{k\suppar0 | H\suppar 0 -E_n\suppar 0 | n\suppar 1} &= \braket{k\suppar 0 | E_n\suppar 1 - \delta H | n\suppar 0} \\ (E_k\suppar 0 - E_n\suppar 0) \braket{k\suppar 0 | n \suppar 1} &= E_n\suppar 1 \braket{k\suppar 0 | n\suppar 0} - \braket{k\suppar 0 | \delta H | n\suppar0} \\ \braket{k\suppar0|n\suppar 1} &= - \frac{\braket{k\suppar 0|\delta H|n\suppar 0}}{E_k\suppar0-E_n\suppar0}. \end{align*}

$\braket{k\suppar0|\delta H|n\suppar0}$ is the $(k,n)$ matrix element of $\delta H$ . We will write it $\delta H_{kn}$ . Then by completeness

\begin{align*} \ket{n\suppar 1} = - \sum_{k \ne n} \frac{\delta H_{kn}}{E_k\suppar0-E_n\suppar0}. \end{align*}

We see that if the state being perturbed is degenerate, the denominator will vanish and the perturbed state will be undefined. To address this case we apply the more general degenerate perturbation theory.

Higher orders are more complicated. For second order we can take the same approach to find

\begin{align*} (H\suppar0 - E_n\suppar0)\ket{n\suppar2} &= (E_n\suppar1 - \delta H) \ket{n\suppar 1} + E_n\suppar 2\ket{n\suppar0} \\ (E_k\suppar0 - E_n\suppar 0) \braket{k\suppar0|n\suppar2} &= \bra{k\suppar0} (E_n\suppar 1 - \delta H) \left( - \sum_M \frac{\ket{m\suppar0}\delta H_{mn}}{E_m\suppar0 - E_n\suppar0} \right) \end{align*} \\ \begin{align*} \braket{k\suppar0|n\suppar2} &= \frac{1}{E_k\suppar0 - E_n\suppar0} \left[ \sum_m \frac{\delta H_{km} \delta H_{mn}}{E_m\suppar0 - E_n\suppar0} - \frac{\delta H_{nn}}{E_k\suppar0 - E_n\suppar0} \right] \\ \ket{n\suppar2} &= \sum_k \frac{1}{E_k\suppar0 - E_n\suppar0} \left[ \sum_m \frac{\delta H_{km} \delta H_{mn}}{E_m\suppar0 - E_n\suppar0} - \frac{\delta H_{nn}}{E_k\suppar0 - E_n\suppar0} \right]. \end{align*}