# Degenerate perturbation theory, vector derivative

1. Feb 5, 2014

### jostpuur

Stuff starts with this:

$$\mathbb{R}\to\mathbb{C}^{N\times N},\quad t\mapsto A(t)$$

$$A(t)^{\dagger} = A(t)$$

$$A(t)v_n(t) = \lambda_n(t)v_n(t)$$

$$v_n(t)^{\dagger}v_{n'}(t) = \delta_{nn'}$$

$$\dot{A}(t)v_n(t) + A(t)\dot{v}_n(t) = \dot{\lambda}_n(t)v_n(t) + \lambda_n(t)\dot{v}_n(t)$$

Suppose that $K>1$ is the dimension of some eigenspace corresponding to some eigenvalue $\lambda_n(0)$. We can assume the components arranged so that

$$\lambda_n(0)=\lambda_{n+1}(0)=\cdots = \lambda_{n+K}(0)$$

I have not understood how to solve $\dot{v}_n(0)$.

Isn't the main aim to solve the coefficients $v_{n'}(0)^{\dagger}\dot{v}_n(0)$, like in the non-degenerate case too? If these are known for all $n'$, then the whole vector can be written as

$$\dot{v}_n(0) = \sum_{n'=1}^N \big(v_{n'}(0)^{\dagger}\dot{v}_n(0)\big)v_{n'}(0)$$

I understand that for those $n'$ such that $\lambda_{n'}(0)\neq\lambda_n(0)$ we have

$$v_{n'}(0)^{\dagger}\dot{v}_n(0) = \frac{v_{n'}(0)^{\dagger}\dot{A}(0)v_n(0)}{\lambda_n(0) - \lambda_{n'}(0)}$$

I also understand that we can assume

$$v_n(0)^{\dagger}\dot{v}_n(0) = 0$$

The real part will be zero due to the assumption $\|v_n(t)\|=1$, and the imaginary part can be chosen zero by choosing right complex phase.

What I have not understood is how to obtain the coefficients for such $n'$ that $n'\neq n$ and $\lambda_{n'}(0)=\lambda_n(0)$.

I have Quantum Mechanics Second Edition by Bransden and Joachain. After discussing how to obtain $v_{n'}(0)$ (by some rotation in the eigenspace) they state

Well to me it seems that I cannot obtain all the coefficients in a way "similar" to the non-degenerate case.