Time dependent perturbation theory (Berry phase)

[dRic2]

Homework Statement

An Hamiltonian $H$ is characterized by a parameter $\lambda(t)$ which is varied with time. In the adiabatic approximation the solution is given by
$$\ket{\psi(t)} = e^{i\phi}e^{-i\gamma}\ket{n(t)}$$
with $\phi$ the Berry phase and $\gamma$ the dynamical phase.

Carrying the the adiabatic perturbation theory to the next order in $\lambda$, that is $\dot \lambda$, assume that the solution is given by:
$$\ket{\psi(t)} = e^{i\phi}e^{-i\gamma}[\ket{n(t)} + \dot \lambda \ket{\delta n}]$$
where $\ket{\delta n}$ is to be determined. We arlerady know that the solution satisfy the Schrodinger equation to the order zero in $\dot \lambda$. Show that:
$$(E_n - H_{\lambda})\ket{\delta n} = -i \hbar (\partial_{\lambda} + i A_n) \ket{n}$$.
$A_n$ is the Berry connection $\bra{n} i \partial_{\lambda} \ket{n}$.

Relevant Equations

Berry phase $$\phi = \int_{t_0}^{t} d\tau A_n(\tau)$$
Dynamical phase $$\gamma = \int_{t_0}^{t} d\tau E_n(\tau)$$

If I plug the solution into the Schrodinger equation I get
$$(i \hbar \partial_t - H)\ket{\psi} = 0$$
Since I know that the zeroth-order expansion is lambda is already a solution I think this is equal to
$$(i \hbar \partial_t - H)e^{i\phi} e^{-i\gamma}\ket{\delta n} = 0$$
If now I carry on with the differentiation I get the solution, except for the fact that I have $\delta n$ everywhere, while on the right-hand side the should be a $\ket{n}$. So, either my assumption is wrong (why?), or I am missing something. I've been thinking for 2 days...

(the $\ddot \lambda$ term is dropped)

Thanks,
Ric

 

[dRic2]

Solved it. For anyone interested:

Since I know that the zeroth-order expansion is lambda is already a solution I think this is equal to

This is plain wrong and it makes no sense. It would correspond to adding 0 to my previous solution which is true for the zeroth-order, but then it will remain true for the zeroth-order and not for the first order. I was confused by the sentence "We already know that the solution satisfies the Schrodinger equation to the order zero".

The problem is actually a very simple one: just performing all the required derivatives in $(i\hbar \partial_t -H) \ket{\psi} = 0$, simplifying the expression with simple algebric manipulations, and remembering that $\frac 1 {\dot \lambda} \partial_t = \partial_\lambda$, one gets the desired result.