Categories: Physics, Quantum mechanics.

Berry phase

Consider a Hamiltonian H^\hat{H} that does not explicitly depend on time, but does depend on a given parameter R\vb{R}. The Schrödinger equations then read:

iddtΨn(t)=H^(R)Ψn(t)H^(R)ψn(R)=En(R)ψn(R)\begin{aligned} i \hbar \dv{}{t}\Ket{\Psi_n(t)} &= \hat{H}(\vb{R}) \Ket{\Psi_n(t)} \\ \hat{H}(\vb{R}) \Ket{\psi_n(\vb{R})} &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})} \end{aligned}

The general full solution Ψn\Ket{\Psi_n} has the following form, where we allow R\vb{R} to evolve in time, and we have abbreviated the traditional phase of the “wiggle factor” as LnL_n:

Ψn(t)=exp(iγn(t))exp(iLn(t)/)ψn(R(t))Ln(t)0tEn(R(t))dt\begin{aligned} \Ket{\Psi_n(t)} = \exp(i \gamma_n(t)) \exp(-i L_n(t) / \hbar) \: \Ket{\psi_n(\vb{R}(t))} \qquad L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'} \end{aligned}

The geometric phase γn(t)\gamma_n(t) is more interesting. It is not included in ψn\Ket{\psi_n}, because it depends on the path R(t)\vb{R}(t) rather than only the present R\vb{R} and tt. Its dynamics can be found by inserting the above Ψn\Ket{\Psi_n} into the time-dependent Schrödinger equation:

ddtΨn=idγndtΨnidLndtΨn+exp(iγn)exp(iLn/)ddtψn=idγndtΨn+1iEnΨn+exp(iγn)exp(iLn/)RψndRdt=idγndtΨn+1iH^Ψn+exp(iγn)exp(iLn/)RψndRdt\begin{aligned} \dv{}{t}\Ket{\Psi_n} &= i \dv{\gamma_n}{t} \Ket{\Psi_n} - \frac{i}{\hbar} \dv{L_n}{t} \Ket{\Psi_n} + \exp(i \gamma_n) \exp(-i L_n / \hbar) \dv{}{t}\Ket{\psi_n} \\ &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} E_n \Ket{\Psi_n} + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \\ &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} \hat{H} \Ket{\Psi_n} + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}

Here we recognize the Schrödinger equation, so those terms cancel. We are then left with:

idγndtΨn=exp(iγn)exp(iLn/)RψndRdt\begin{aligned} - i \dv{\gamma_n}{t} \Ket{\Psi_n} &= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}

Front-multiplying by iΨni \Bra{\Psi_n} gives us the equation of motion of the geometric phase γn\gamma_n:

dγndt=An(R)dRdt\begin{aligned} \boxed{ \dv{\gamma_n}{t} = - \vb{A}_n(\vb{R}) \cdot \dv{\vb{R}}{t} } \end{aligned}

Where we have defined the so-called Berry connection An\vb{A}_n as follows:

An(R)iψn(R)|Rψn(R)\begin{aligned} \boxed{ \vb{A}_n(\vb{R}) \equiv -i \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} } \end{aligned}

Importantly, note that An\vb{A}_n is real, provided that ψn\Ket{\psi_n} is always normalized for all R\vb{R}. To prove this, we start from the fact that R1=0\nabla_\vb{R} 1 = 0:

0=Rψn|ψn=Rψn|ψn+ψn|Rψn=ψn|Rψn+ψn|Rψn=2Re{iAn}=2Im{An}\begin{aligned} 0 &= \nabla_\vb{R} \Inprod{\psi_n}{\psi_n} = \Inprod{\nabla_\vb{R} \psi_n}{\psi_n} + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}^* + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} = 2 \Real\{ - i \vb{A}_n \} = 2 \Imag\{ \vb{A}_n \} \end{aligned}

Consequently, An=Imψn|Rψn\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} is always real, because ψn|Rψn\Inprod{\psi_n}{\nabla_\vb{R} \psi_n} is imaginary.

Suppose now that the parameter R(t)\vb{R}(t) is changed adiabatically (i.e. so slow that the system stays in the same eigenstate) for t[0,T]t \in [0, T], along a circuit CC with R(0) ⁣= ⁣R(T)\vb{R}(0) \!=\! \vb{R}(T). Integrating the phase γn(t)\gamma_n(t) over this contour CC then yields the Berry phase γn(C)\gamma_n(C):

γn(C)=CAn(R)dR\begin{aligned} \boxed{ \gamma_n(C) = - \oint_C \vb{A}_n(\vb{R}) \cdot \dd{\vb{R}} } \end{aligned}

But we have a problem: An\vb{A}_n is not unique! Due to the Schrödinger equation’s gauge invariance, any function f(R(t))f(\vb{R}(t)) can be added to γn(t)\gamma_n(t) without making an immediate physical difference to the state. Consider the following general gauge transformation:

ψ~n(R)exp(if(R))ψn(R)\begin{aligned} \ket{\tilde{\psi}_n(\vb{R})} \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})} \end{aligned}

To find An\vb{A}_n for a particular choice of ff, we need to evaluate the inner product ψ~nRψ~n\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}:

ψ~nRψ~n=exp(if)(iRfψ~nψn+ψ~nRψn)=iRfψnψn+ψnRψn=iRf+ψnRψn\begin{aligned} \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n} &= \exp(i f) \Big( i \nabla_\vb{R} f \: \inprod{\tilde{\psi}_n}{\psi_n} + \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big) \\ &= i \nabla_\vb{R} f \: \inprod{\psi_n}{\psi_n} + \inprod{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n} \end{aligned}

Unfortunately, ff does not vanish as we would have liked, so An\vb{A}_n depends on our choice of ff.

However, the curl of a gradient is always zero, so although An\vb{A}_n is not unique, its curl R×An\nabla_\vb{R} \cross \vb{A}_n is guaranteed to be. Conveniently, we can introduce a curl in the definition of γn(C)\gamma_n(C) by applying Stokes’ theorem, under the assumption that An\vb{A}_n has no singularities in the area enclosed by CC (fortunately, An\vb{A}_n can always be chosen to satisfy this):

γn(C)=S(C)Bn(R)dS\begin{aligned} \boxed{ \gamma_n(C) = - \iint_{S(C)} \vb{B}_n(\vb{R}) \cdot \dd{\vb{S}} } \end{aligned}

Where we defined Bn\vb{B}_n as the curl of An\vb{A}_n. Now γn(C)\gamma_n(C) is guaranteed to be unique. Note that Bn\vb{B}_n is analogous to a magnetic field, and An\vb{A}_n to a magnetic vector potential:

Bn(R)R×An(R)=Im ⁣{R×ψn(R)|Rψn(R)}\begin{aligned} \vb{B}_n(\vb{R}) \equiv \nabla_\vb{R} \cross \vb{A}_n(\vb{R}) = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\} \end{aligned}

Unfortunately, Rψn\nabla_\vb{R} \psi_n is difficult to evaluate explicitly, so we would like to rewrite Bn\vb{B}_n such that it does not enter. We do this as follows, inserting 1=mψmψm1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m} along the way:

iBn=R×ψn|Rψn=ψn|R×Rψn+Rψn×Rψn=mRψn|ψm×ψm|Rψn\begin{aligned} i \vb{B}_n = \nabla_\vb{R} \cross \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} &= \Inprod{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \Bra{\nabla_\vb{R} \psi_n} \cross \Ket{\nabla_\vb{R} \psi_n} \\ &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}

The fact that ψn|Rψn\Inprod{\psi_n}{\nabla_\vb{R} \psi_n} is imaginary means it is parallel to its complex conjugate, and thus the cross product vanishes, so we exclude nn from the sum:

Bn=mnRψn|ψm×ψm|Rψn\begin{aligned} \vb{B}_n &= \sum_{m \neq n} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}

From the Hellmann-Feynman theorem, we know that the inner products can be rewritten:

ψm|Rψn=ψnRH^ψmEnEm\begin{aligned} \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} = \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m}}{E_n - E_m} \end{aligned}

Where we have assumed that there is no degeneracy. This leads to the following result:

Bn=ImmnψnRH^ψm×ψmRH^ψn(EnEm)2\begin{aligned} \boxed{ \vb{B}_n = \Imag \sum_{m \neq n} \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m} \cross \matrixel{\psi_m}{\nabla_\vb{R} \hat{H}}{\psi_n}}{(E_n - E_m)^2} } \end{aligned}

Which only involves RH^\nabla_\vb{R} \hat{H}, and is therefore easier to evaluate than any Rψn\Ket{\nabla_\vb{R} \psi_n}.


  1. M.V. Berry, Quantal phase factors accompanying adiabatic changes, 1984, Royal Society.
  2. G. Grosso, G.P. Parravicini, Solid state physics, 2nd edition, Elsevier.