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

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: $$$$|\psi(t)\rangle=e^{i\phi(\lambda(t))}e^{-i\gamma(t)}|n(t)\rangle \tag{1}\label{1}$$$$

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: $$$$\phi(t)=\int_{\lambda(0)}^{\lambda(t)}A_n(\lambda)d\lambda \tag{2}\label{2}$$$$ where $$A_n$$ is Berry connection.

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

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

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

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: $$$$(E_n-H_{\lambda})|\delta n\rangle=-i\hbar (\partial_{\lambda}+iA_n)|n\rangle. \tag{5}\label{5}$$$$

which can also be written like:

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

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

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

• I don't know what all the symbols mean here, and don't have that book, so can't look up what they mean easily. But Eq. 6 seems to be somewhat common: I've been working today on describing p-DMRG for my types of DMRG question, and you can see that Eq. 9 in this: arxiv.org/abs/1803.07150 is very similar to Eq. 6 in your question. Eq. 5 is similar. Maybe we just need to know what all these symbols mean? Jul 29, 2020 at 20:17
• Eq. 5 is a direct application of the expression for first-order correction from perturbation theory. I don't totally understand eq. 6, but it is somehow leveraging the completeness of the basis. In particular, I just don't know what the notation $|\partial_{\lambda}n\rangle$ means. Sep 1, 2020 at 20:59
• @jheindel: We have a policy not to ask 2 questions in one post, and combined with the fact that the notation was not made clear, the question was closed. However, I have deleted the question about Eq. 6, so if you can write an answer about Eq. 5 only, it would be very appreciated as it would help clean up our unanswered queue for the upcoming new year! Dec 31, 2020 at 16:53
• Hi @jheindel ! Since this is now one of our longest standing unanswered questions, and has been left up for over 6 months, I wonder if you might be able to spare a couple minutes to answer the part about Eq. 5? The part about Eq. 6 was deleted. I'm trying to clear up the unanswered queue! Mar 19, 2021 at 15:56

## 1 Answer

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

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

• Wow! What a complete answer! Thanks for the effort Tyberius ♦. Jul 13, 2021 at 16:07