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+---
+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.