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.


1 Answer 1


$\def\g{\gamma} \def\d{\delta} \def\R{{\bf R}} \newcommand\ket[1]{|#1\rangle} \newcommand\bra[1]{\langle #1|} \newcommand\braket[2]{\langle #1|#2\rangle}$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*}

  • $\begingroup$ Got it, thanks! I hadn't realised that the eigenstates $|n\rangle$ are normalised :) $\endgroup$
    – James W.
    Nov 14, 2021 at 22:58
  • $\begingroup$ @JamesW.: Great! I am glad to help. $\endgroup$
    – user26872
    Nov 14, 2021 at 23:05

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