The celebrated adiabatic theorem states that for a system initially in the eigenstate $|\psi(0)\rangle = |n(0)\rangle$ for $t=0$, it will stay in that state afterward under adiabatic evolution: $$ |\psi(t)\rangle = e^{i \gamma_n(t)}e^{-\frac{i}{\hbar}\int_0^t \varepsilon_n(t^\prime)dt^\prime}|n(t)\rangle. $$ where $|n(t)\rangle $is the instantaneous eigenstate of $H(t)$ and $\gamma_n$ is the Berry phase.

However, in this paper it states that up to first order, the adiabatic evolution of the state is: $$ |\psi(t)\rangle =e^{-\frac{i}{\hbar}\int_0^t \varepsilon_n(t^\prime)dt^\prime} (|n(t) \rangle+i\hbar \sum_{m\neq n} |m(t)\rangle \frac{\langle m(t) |\frac{d}{dt}|n(t)\rangle}{E_m - E_n}). $$ My question is, why does the Berry phase term disappear? The derivation in Shen's book is confusing, can anyone give a clear derivation?

  • $\begingroup$ The paper that you refer to is quite old (1983). The Berry phase wasn't properly described until 1984. Perhaps the authors glanced over its importance. Moreover, from my understanding, if the evolution is non-cyclical then the geometrical phase can be made to vanish. $\endgroup$ Jun 29, 2021 at 10:09
  • $\begingroup$ Wiki has a nice derivation of the adiabatic theorem. Introduction to quantum mechanics by David J. Griffiths also have a very nice description and proof of the theorem. $\endgroup$ Jun 29, 2021 at 10:13
  • $\begingroup$ Thank you for replying. But I need the derivation up to the first order. $\endgroup$
    – Hao
    Jun 29, 2021 at 10:59

1 Answer 1


I think that if you express the wavefunction as an infinite series of some coefficients $c_n(t)$, your question is easier to answer. I made some notes on the adiabatic theorem that I will attach here for you to check if it answers your doubt (see below).

Equation (10) is equivalent to the eq. of the paper that you are referring to. Here you can see that apart from the term of the temporal variation of the Hamiltonian $\dot{H}$, which in the adiabatic approximation is assumed to be very small compared with the other terms, the term that defines the Berry Phase is still present: $c_m(t)<\psi_m|\dot{\psi}_m>$.

Proof of the Adiabatic Theorem

At the end of the day what really matters to have a non-zero Berry phase is not the velocity the system have in a process along the evolution path. The Berry Phase only depends on the path taken, for this reason is usually denoted as a topological phase and is also present in classical mechanics.

More information in: https://en.wikipedia.org/wiki/Berry_connection_and_curvature

I hope this to be useful to you.


  • $\begingroup$ Thank you for answering, but it seems (10) is not identical to what I refer to. You only derive a differential equation, but you do not solve it. Can you give the solution to (10) up to first order? $\endgroup$
    – Hao
    Jun 29, 2021 at 10:57
  • $\begingroup$ I guess you need to use perturbation theory to reexpress the sumation and substract a global $c_m(t)$. Then, you can separate variables and integrate. Nevertheless, the term of the Berry phase is going to be present because in eq 10. is actually isolated. It will be something like: $c_m(t) = e^{i\gamma_m} + \mathcal{F}(t, E_m)$, where $\mathcal{F}$ is zero in the adiabatic approximation. $\endgroup$
    – T. ssP
    Jun 29, 2021 at 11:40

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