Berry's phase for non-normalized wave functions

Question

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?

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.