Adiabatic equation related to the Berry phase for lambda with first order terms

Question

Consider the following derivation in David Vanderbilt's book "Berry Phases in Electronic Structure Theory - Electric Polarization, Orbital Magnetization and Topological Insulators" (2018, Cambridge University Press (page 100).

The wave function for the adiabatic approach is as follows: \begin{equation} |\psi(t)\rangle=e^{i\phi(\lambda(t))}e^{-i\gamma(t)}|n(t)\rangle \tag{1}\label{1} \end{equation}

where $e^{i\phi(\lambda(t))}$ is the geometric phase or Berry phase and $e^{-i\gamma(t)}$ is dynamic phase.

Berry phase has the following mathematical expression: \begin{equation} \phi(t)=\int_{\lambda(0)}^{\lambda(t)}A_n(\lambda)d\lambda \tag{2}\label{2} \end{equation} where $A_n$ is Berry connection.

Dynamic phase has the following mathematical expression: \begin{equation} \gamma(t)=\frac{1}{\hbar}\int_{0}^{t}E_n(t')dt' \tag{3}\label{3} \end{equation}

Expanding equation number (1) with first order $\lambda$ terms:

\begin{equation} |\psi(t)\rangle=e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\big[|n(\lambda)\rangle+\dot{\lambda}|\delta n(t)\rangle\big] \tag{4}\label{4} \end{equation}

Equation (4) solves the time-dependent Schrödinger equation to order zero in $\dot{\lambda}$, but we now require that it should also do so at first order.

For this purpose we can discard terms that scale like $\ddot{\lambda}$ or $\lambda^2$. You get: \begin{equation} (E_n-H_{\lambda})|\delta n\rangle=-i\hbar (\partial_{\lambda}+iA_n)|n\rangle. \tag{5}\label{5} \end{equation}

which can also be written like:

$$\begin{equation} (E_n-H_{\lambda})|\delta n\rangle=-i\hbar \mathcal{Q}_{n}|\partial_{\lambda}n\rangle \tag{6}\label{6} \end{equation} $$

where $\mathcal{Q}_n=1-|n\rangle\langle n|$.

How did they get equation (5) and (6) from equation (4)?

Answer 1

The TD-Schrodinger equation is $$H(\lambda)|\psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle$$

So, we just need to plug in \eqref{4} and remove all terms that aren't first-order in $\dot\lambda$. Let's start with the left-hand side, since its a little simpler.

$$\begin{align}H(\lambda)|\psi(t)\rangle&=e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\bigg[H(\lambda)|n(t)\rangle+\dot\lambda H(\lambda)|\delta n(t)\rangle\bigg]\\ &\overset{\text{First order}}{\to}e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\bigg[\dot\lambda H(\lambda)|\delta n(t)\rangle\bigg]\end{align}$$

Now for the right side, which is a bit more involved: $$\begin{align} i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle&=i\hbar\frac{\partial}{\partial t}e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\bigg[|n(t)\rangle+\dot\lambda|\delta n(t)\rangle\bigg]\\& =i\hbar \Bigg[iA_n(t)|\psi(t)\rangle-iE_n|\psi(t)\rangle+e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\bigg[\frac{\partial}{\partial t}|n(t)\rangle+\ddot\lambda|\delta n(t)\rangle+\dot\lambda\frac{\partial}{\partial t}|\delta n(t)\rangle\bigg]\Bigg]\\ &\overset{\text{First order}}{\to}i\hbar e^{i\phi(\lambda(t))}e^{-i\gamma(t)}\dot\lambda \Bigg[iA_n(\lambda)|n(t)\rangle-iE_n|\delta n(t)\rangle\rangle+\frac{\partial}{\partial \lambda}|n(\lambda)\rangle\Bigg] \end{align}$$

As mentioned in the text just before Eqn 3.53, we can make the substitution $\frac{\partial}{\partial t}|n(t)\rangle=\dot\lambda\frac{\partial}{\partial \lambda}|n(\lambda)\rangle$ using the chain rule to express the derivative in terms of $\dot\lambda$. The inconsistent switching between $|n(t)\rangle$ and $|n(\lambda(t))\rangle$ can make the derivation a bit confusing, but these expression are defined to be equivalent and you can verify that the equations are consistent with either expression for the functional form of $|n\rangle$.

Combining these expressions, removing the common factor of $\dot\lambda e^{i\phi(\lambda(t))}e^{-i\gamma(t)}$, and separating $|n\rangle$ and $|\delta n\rangle$ terms, we obtain:

$$(H(\lambda)-E_n)|\delta n\rangle=i\hbar\bigg(\frac{\partial}{\partial \lambda}+iA_n\bigg)|n\rangle$$

Which is just \eqref{5} with the sign of both sides flipped. To get \eqref{6} just requires using the definition $A_n(\lambda)=\langle n(\lambda)|i\frac{\partial}{\partial \lambda}|n(\lambda)\rangle$. We can then rearrange the right hand side to be in terms of $\frac{\partial}{\partial \lambda}|n\rangle$, which you can succinctly express by introducing $\mathcal{Q}_n=1-|n\rangle\langle n|$

$$(H(\lambda)-E_n)|\delta n\rangle=i\hbar\mathcal{Q}_n\frac{\partial}{\partial \lambda}|n\rangle$$