path: root/content/know/concept/berry-phase/index.pdc

---
title: "Berry phase"
firstLetter: "B"
publishDate: 2021-11-29
categories:
- Physics
- Quantum mechanics

date: 2021-11-25T20:42:45+01:00
draft: false
markup: pandoc
---

# Berry phase

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:

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

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 $\nabla_\vb{R} 1 = 0$:

$$\begin{aligned}
    0
    &= \nabla_\vb{R} \braket{\psi_n}{\psi_n}
    = \braket{\nabla_\vb{R} \psi_n}{\psi_n} + \braket{\psi_n}{\nabla_\vb{R} \psi_n}
    \\
    &= \braket{\psi_n}{\nabla_\vb{R} \psi_n}^* + \braket{\psi_n}{\nabla_\vb{R} \psi_n}
    = 2 \Re\{ - i \vb{A}_n \}
    = 2 \Im\{ \vb{A}_n \}
\end{aligned}$$

Consequently, $\vb{A}_n = \Im \braket{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real,
because $\braket{\psi_n}{\nabla_\vb{R} \psi_n}$ 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
$\braket*{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:

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

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:

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

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:

$$\begin{aligned}
    i \vb{B}_n
    = \nabla_\vb{R} \cross \braket{\psi_n}{\nabla_\vb{R} \psi_n}
    &= \braket{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \bra{\nabla_\vb{R} \psi_n} \cross \ket{\nabla_\vb{R} \psi_n}
    \\
    &= \sum_{m} \braket{\nabla_\vb{R} \psi_n}{\psi_m} \cross \braket{\psi_m}{\nabla_\vb{R} \psi_n}
\end{aligned}$$

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

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

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

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

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

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

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.