# Deriving the Berry phase from the Schroedinger equation

Let $$|n(\mathbf{R})\rangle$$ be eigenstates of the snapshot Hamiltonian $$H|\mathbf{R} \rangle$$, of eigenvalues $$E_n(\mathbf{R})$$. The vector $$\mathbf{R}$$ contains the parameters upon which the system depends. The objective is to solve the Schroedinger equation $$i | \dot{\psi}(t) \rangle = H(\mathbf{R}(t)) |\psi(t)\rangle$$ when the parameter vector $$\mathbf{R}(t)$$ varies slowly with time. Consider the trial solution \begin{align*} |\psi(t)\rangle = e^{i\gamma_n(t)} e^{-i\phi_n(t)} |n(\mathbf{R}(t))\rangle \end{align*}

with $$\phi_n(t) = \displaystyle{\int}_{0}^t E_n(\mathbf{R}(t')) dt'$$. I obtain (omitting functional dependencies for clarity) \begin{align*} |\dot{\psi}\rangle &= e^{i\gamma_n } e^{-i\phi_n}(i\dot{\gamma}_n |n\rangle - i \dot{\phi}_n |n\rangle+ |\dot{n}\rangle ) \\ &= e^{i\gamma_n} e^{-i\phi_n }(i\dot{\gamma}_n |n\rangle - i E_n |n\rangle+ |\dot{n}\rangle ) \end{align*} so putting $$|\dot{\psi}\rangle = -iH |\psi \rangle = -iE_n e^{i\gamma_n} e^{-i\phi_n} |n\rangle$$ gives \begin{align*} |\dot{n}\rangle = -i\dot{\gamma}_n |n\rangle \end{align*} Now I'm trying to figure out how to solve for $$\gamma_n$$, which is supposed to be: \begin{align*} \gamma_n(\mathcal{C}) = \int_{\mathcal{C}} i \langle n(\mathbf{R}) | \nabla_{\mathbf{R}} n(\mathbf{R}) \rangle d\mathbf{R} \ \ \ (\dagger) \end{align*} I assume that $$\mathcal{C}$$ is a path in $$\mathbf{R}$$-space, parameterised by time $$t$$. That is to say, \begin{align*} i\gamma_n(t) &= -\int_0^t \langle n(\mathbf{R}(t') | \nabla_{\mathbf{R}(t')} n(\mathbf{R}(t')) \rangle \dot{\mathbf{R}}(t') dt' \\ i\dot{\gamma}_n(t) &= - \langle n(\mathbf{R}(t) | \nabla_{\mathbf{R}(t)} n(\mathbf{R}(t)) \rangle \dot{\mathbf{R}}(t) \end{align*} From here (again omitting functional depencies for clarity), I'm not totally sure how to show that $$i\dot{\gamma}_n |n \rangle = -|\dot{n}\rangle$$, so that $$(\dagger)$$ is indeed a solution of the differential equation. I had the idea to re-write \begin{align*} | \nabla_{\mathbf{R}} n \rangle \dot{\mathbf{R}} = |\dot{n} \rangle \end{align*} so that \begin{align*} i\dot{\gamma}_n |n \rangle &= - \langle n |\dot{n} \rangle | n \rangle = -| n \rangle \langle n |\dot{n} \rangle \end{align*} but the RHS doesn't look quite like $$|\dot{n}\rangle$$, since the identity operator is rather a sum $$\displaystyle{\sum_n} | n \rangle \langle n |$$ over $$n$$. Where did I go wrong? For reference, I am following these notes.

$$\def\g{\gamma} \def\d{\delta} \def\R{{\bf R}} \newcommand\ket[1]{|#1\rangle} \newcommand\bra[1]{\langle #1|} \newcommand\braket[2]{\langle #1|#2\rangle}$$The condition that \begin{align*} \ket{\dot n} &= -i\dot\g_n\ket{n}\tag{1} \end{align*} follows as a requirement on our trial solution. This condition can be satisfied by solving the differential equation (1) for $$\g_n$$. Condition (1) implies \begin{align*} \braket{m}{\dot n} &= -i\dot\g_n\braket{m}{n} \\ \braket{m}{\dot n} &= -i\dot\g_n\d_{mn} \\ \dot\g_n &= i\braket{n}{\dot n}. \tag{2} \end{align*} Thus, for example, $$i\dot\g_n\ket{n} = -\ket{n}\braket{n}{\dot n}$$ as claimed. (Note that $$\dot\g_n$$ is just a $$c$$-number.) Multiplying by $$\ket{n}$$ has not gotten us closer to finding $$\g_n$$, so we leave this.
From (2) we have \begin{align*} \dot\g_n &= i\braket{n}{\dot n} \\ &= i\braket{n(\R(t)}{\frac{d}{dt}n(\R(t))} \\ &= i\braket{n(\R(t))}{\nabla_{\R(t)}n(\R(t))\cdot \dot\R(t)} & \textrm{mv chain rule} \\ &= i\braket{n(\R(t))}{\nabla_{\R(t)}n(\R(t))}\cdot\dot\R(t). & \textrm{\R a vect of c-nums} \end{align*} Thus, \begin{align*} \g_n(t) &= \int_0^t i\braket{n(\R(t))}{\nabla_{\R(t)}n(\R(t))}\cdot\dot\R(t) dt \\ &= \int_{\mathcal{C}} i\braket{n(\R)}{\nabla_{\R}n(\R)}\cdot d\R. \end{align*}
• Got it, thanks! I hadn't realised that the eigenstates $|n\rangle$ are normalised :) Nov 14, 2021 at 22:58