path: root/source/know/concept/berry-phase/index.md

---
title: "Berry phase"
sort_title: "Berry phase"
date: 2021-11-29
categories:
- Physics
- Quantum mechanics
layout: "concept"
---

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

$$\begin{aligned}
    i \hbar \dv{}{t}\Ket{\Psi_n(t)}
    &= \hat{H}(\vb{R}) \Ket{\Psi_n(t)}
    \\
    \hat{H}(\vb{R}) \Ket{\psi_n(\vb{R})}
    &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})}
\end{aligned}$$

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

$$\begin{aligned}
    \Ket{\Psi_n(t)}
    = \exp(i \gamma_n(t)) \exp(-i L_n(t) / \hbar) \: \Ket{\psi_n(\vb{R}(t))}
    \qquad
    L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
\end{aligned}$$

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

$$\begin{aligned}
    \dv{}{t}\Ket{\Psi_n}
    &= i \dv{\gamma_n}{t} \Ket{\Psi_n} - \frac{i}{\hbar} \dv{L_n}{t} \Ket{\Psi_n}
    + \exp(i \gamma_n) \exp(-i L_n / \hbar) \dv{}{t}\Ket{\psi_n}
    \\
    &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} E_n \Ket{\Psi_n}
    + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
    \\
    &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} \hat{H} \Ket{\Psi_n}
    + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
\end{aligned}$$

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

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

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

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

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

$$\begin{aligned}
    \boxed{
        \vb{A}_n(\vb{R})
        \equiv -i \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})}
    }
\end{aligned}$$

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

$$\begin{aligned}
    0
    &= \nabla_\vb{R} \Inprod{\psi_n}{\psi_n}
    = \Inprod{\nabla_\vb{R} \psi_n}{\psi_n} + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
    \\
    &= \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}^* + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
    = 2 \Real\{ - i \vb{A}_n \}
    = 2 \Imag\{ \vb{A}_n \}
\end{aligned}$$

Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real,
because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary.

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

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

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

$$\begin{aligned}
    \ket{\tilde{\psi}_n(\vb{R})}
    \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})}
\end{aligned}$$

To find $$\vb{A}_n$$ for a particular choice of $$f$$,
we need to evaluate the inner product
$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$:

$$\begin{aligned}
    \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
    &= \exp(i f) \Big( i \nabla_\vb{R} f \: \inprod{\tilde{\psi}_n}{\psi_n} + \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big)
    \\
    &= i \nabla_\vb{R} f \: \inprod{\psi_n}{\psi_n} + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
    \\
    &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
\end{aligned}$$

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

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

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

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

$$\begin{aligned}
    \vb{B}_n(\vb{R})
    \equiv \nabla_\vb{R} \cross \vb{A}_n(\vb{R})
    = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
\end{aligned}$$

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

$$\begin{aligned}
    i \vb{B}_n
    = \nabla_\vb{R} \cross \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
    &= \Inprod{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \Bra{\nabla_\vb{R} \psi_n} \cross \Ket{\nabla_\vb{R} \psi_n}
    \\
    &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
\end{aligned}$$

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

$$\begin{aligned}
    \vb{B}_n
    &= \sum_{m \neq n} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
\end{aligned}$$

From the [Hellmann-Feynman theorem](/know/concept/hellmann-feynman-theorem/),
we know that the inner products can be rewritten:

$$\begin{aligned}
    \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
    = \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m}}{E_n - E_m}
\end{aligned}$$

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

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

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



## References
1.  M.V. Berry,
    [Quantal phase factors accompanying adiabatic changes](https://doi.org/10.1098/rspa.1984.0023),
    1984, Royal Society.
2.  G. Grosso, G.P. Parravicini,
    *Solid state physics*,
    2nd edition, Elsevier.