# What is the form of the $n$-th order term of the perturbation series of an eigenvalue?

Suppose I have a matrix given by a sum $A=D+\epsilon B$, where $D$ is diagonal and $\epsilon$ is small, and I want the eigenvalues of $A$ as power series in $\epsilon$. The leading order is just the eigenvalues of $D$, the first corrections are the diagonal elements of $B$, the second order is also well known.

I would like to know what is the particular form of the $n$-th order term in the eigenvalue perturbation series. Apparently it can be written as a sum over partitions, but I can't find this anywhere.

• Symmetric matrices or not? – user154997 Oct 25 '17 at 13:28
• In any case: migrate to math? – user154997 Oct 25 '17 at 13:28
• $B$ is symmetric, yes – thedude Oct 25 '17 at 13:34

## 1 Answer

The answer can be found in Kato's book Perturbation theory for linear operators. I will use Kato's notation. In fact I will answer a more general question where you have an operator which depends analytically on a parameter $x$ (your $\epsilon$). Let such operator be $T(x)$ and let

$$T(x) = \sum_{n=0}^{\infty} x ^n T^{(n)}$$

such that the series converges in a neighborhood of $x=0$. I also call $T=T^{(0)}$. In your case you simply have $T^{(n)}=0$ for $n\ge 2$. We seek the perturbation series of an eigenvalue $\lambda$ of $T$. This means that there exist an eigen-projector of $T$, $P$ such that

$$TP = \lambda P +D,$$

where $D$ is a nilpotent term that may arise from the Jordan decomposition. If $m = \mathrm{dim} P$ is the dimension of the range of $P$, $D^m=0$. Note that for non-degenerate eigenvalue ($m=1$) we have necessarily $D=0$. Define also $Q=1-P$ and the reduced resolvent

$$S = \lim_{z\to \lambda} = Q (T - z)^{-1} Q$$

lastly let's define

$$S^{(0)} = -P, \ \ S^{(n)} = S^n, \ \ S^{(-n)} = - D^n, \ \mathrm{for}\ n\ge 1.$$

Let $P(x)$ be the eigenprojector of $T(x)$ analytically connected to $P$. Then one has the following series:

$$(T(x) - \lambda) P(x) = D + \sum_{n=1}^{\infty} x^n \tilde{T}^{(n)}$$

with

$$\tilde{T}^{(n)} = - \sum_{p=1}^{\infty} (-1)^p \sum_{\mathcal{A}} S^{(k_1)} T^{(n_1)} S^{(k_2)} \cdots S^{(k_p)} T^{(n_p)} S^{(k_{p+1})},$$

where $\mathcal{A}$ corresponds to the indices satisfying the following constraint

$$\mathcal{A} = \left \{ \sum_{i=1}^p n_i = n ; \sum_{j=1}^{p+1} k_j = p; n_j \ge 1; k_j \ge -m+1 \right \}.$$

In the non-degenerate case ($m=1$) this provides the final answer, i.e.

\begin{eqnarray} \lambda(x) &=& \lambda + \sum_{n=1}^{\infty} x^n \lambda^{(n)} \\ \lambda^{(n)} &=& \mathrm{Tr} \tilde{T}^{(n)}. \end{eqnarray}

Note that in this case one must have $D=0$. Moreover taking the trace already kills many terms because of the cyclic property of the trace and noting that $SP = PS = 0$.

To make contact with possibly more familiar expressions, note that for a self-adjoint unperturbed operator $T$, the reduced resolvent should look familiar:

$$S = \sum_{\lambda_j \neq \lambda} \frac{ |j\rangle \langle j|}{ \lambda_j - \lambda},$$

where I called here $\lambda_j$ and $|j\rangle$ the eigenvalues and eigenvector of $T$.