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.

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

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:

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


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

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 $T$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.