Let $\hat{H}(t)$ be the Hamiltonian of a quantum system depending on time $t$ through $k$ parameters $R(t) = (R_1(t), R_2(t), \dots, R_k(t))$: $$ \hat{H}(t) = \hat{H}(R_1(t),R_2(t),\dots,R_k(t)). $$ Suppose that the Hamiltonian changes very gradually. The adiabatic theorem tells us that the system which starts out from the $n$th eigenstate will remain in the $n$th eigenstate even as the eigenfunctions itself evolves: $$ \Psi_n(x,t) = e^{i \gamma_n(t)} \psi_n(x,t) e^{-\frac{i}{\hbar}\int_0^t E_n(\tau) d\tau}. $$ Here, the phase $\gamma_n(t)$ is called the geometric phase, and $\psi_n(x,t)$ is the $n$th instant eigenstate of the Hamiltonian, $\hat{H}(t)\psi_n(x,t) = E_n(t)\psi_n(t)$. If the Hamiltonian returns to its original form after a time $T$, the net geometric phase is $$ \gamma_n(T) = \oint \langle \psi_n|\nabla_R \psi_n \rangle d R, $$ where the line integral is taken around a closed loop in the $k$-dimensional parametric space. When proving this formula, it is usually assumed that the eigenstates $\psi_n(x,t)$ are normalized.

I am interested in the following questions.

  1. Does Berry's phase make sense for non-normalized states?
  2. If "yes", how could we derive Berry's phase for non-normalized eigenstates?
  3. Are there non-trivial examples of Berry's phase for spinless non-relativistic particles in external magnetic fields?

1 Answer 1


The Berry phase for non-normalized states makes perfect sense, however you do not gain any additional information from using a non-normalized state. To put simply, when you consider non-normalized states you are just using a non-unitary gauge choice. You could then derive the non-normalized version of the gauge field by standard formulas.

In particular, you can write $\psi=\psi'/||\psi'||$, so that $\psi'=g\psi$, with $g=||\psi'||$. Using the usual transformation rule for gauge fields you find

$A'=\langle \psi|d\psi\rangle +g^{-1}dg=\frac{\langle\psi'|d \psi'\rangle}{||\psi'||^2} +||\psi'|| d\left(\frac{1}{||\psi'||}\right) +d\log(||\psi'||)=\frac{\langle\psi'|d \psi'\rangle}{||\psi'||^2}$.

So in particular, if you have a loop of states $[0,1]\ni t\mapsto \gamma(t)=|\psi(t)\rangle \langle \psi(t)|=\frac{|\psi'(t)\rangle \langle \psi'(t)|}{||\psi'(t)||^2}$ (the first representation using a unitary gauge and the second a non-unitary, non-normalized, gauge), parallel transport using the gauge $\psi'$ would yield the complex number

$\exp(-\int_{\gamma}A')=\exp\left(-\int_{0}^{1}dt \frac{\langle \psi'(t)|\frac{d}{dt}|\psi'(t)\rangle }{||\psi'(t)||^2}\right)=g(1)\exp(-\int_{\gamma}A) g^{-1}(0)=\frac{||\psi'(1)||}{||\psi'(0)||} \exp(-\int_{0}^1\langle \psi(t)|\frac{d}{dt}|\psi(t)\rangle)$.

To answer your third question, I will assume that you are referring to the canonical example of a spin $S$ particle in a magnetic field. In that case, an irreducible representation of angular momentum gives you $2S+1$ states corresponding to the projection of angular momentum in an axis, namely $m\in \{-S,...,S\}$. These same numbers label the eigenstates of the family of Hamiltonians $H(B)= -B\cdot S$, where $B$ is the magnetic field and $S$ are the generators of angular momentum. For each family of eigenstates, the Berry curvature can be seen as the magnetic field, in $B$-space, produced by a monopole of topological charge $2m$. For a spin $S=0$ representation there is only one eigenstate with $m=0$ -- hence, every Berry phase you can consider will be trivial.

I hope I answered your questions in a satisfactory way.

  • 1
    $\begingroup$ Perhaps in a non-Hermitian system the Berry curvature can be complex, meaning we can't gauge away the "normalization part" of the connection? It might be interesting to look at. $\endgroup$ Nov 22, 2020 at 22:57
  • 1
    $\begingroup$ I believe that, unfortunately, that won't be the case. The Berry connection preserves the inner product. In the non-Hermitian case, the connection appearing in the adiabatic theorem may be different because, assuming the Hamiltonian is diagonalizable, the spectral projectors are not necessarily orthogonal. However, one can use a modification of the inner product (w.r.t. which the spectral projectors are orthogonal) so that the connection preserves this twisted inner product. In terms of this inner product, the parallel transport is unitary. $\endgroup$
    – B. Mera
    Nov 23, 2020 at 9:02
  • $\begingroup$ thanks that's a good point! $\endgroup$ Nov 23, 2020 at 22:22
  • $\begingroup$ You are very welcome. Glad it was useful! $\endgroup$
    – B. Mera
    Nov 23, 2020 at 23:19

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