1
$\begingroup$

I am reading Marder (Condensed Matter Physics). He talks about the geometric phases and then we derives an expression in which the Berry connection and phase comes up.

The goal is to see how the geometrical phase of the wave function changes.
He uses the Schrodinger equation:

$$ -\frac{\hbar}{i} \frac{\partial|\Psi(t)\rangle}{\partial t}=\hat{\mathcal{H}}_{\vec{\lambda}(t)}|\Psi(t)\rangle $$

And in the adiabatic limit he considers the following solution:

$$ |\Psi(t)\rangle= e^{-(i / \hbar) \int_{0}^{t} d t^{\prime} \mathcal{E}_{\vec{\lambda}\left(t^{\prime}\right)}} e^{i \phi(t)}\left|\Psi_{\vec{\lambda}(t)}\right\rangle $$

Where $\left|\Psi_{\vec{\lambda}(t)}\right\rangle$ is an eigenstate of the Hamiltonian.

Using both expression he reaches,

$$ \begin{array}{l}{\left(\varepsilon_{\vec{\lambda}(t)}-\hbar \frac{\partial \phi}{\partial t}-\frac{\hbar}{i} \frac{\partial}{\partial t}\right)\left|\Psi_{\vec{\lambda}(t)}\right\rangle=\mathcal{E}_{\vec{\lambda}(t)}\left|\Psi_{\vec{\lambda}(t)}\right\rangle} \\ {\Rightarrow \frac{\partial \phi}{\partial t}=i \dot{\vec{\lambda}} \cdot\left\langle\Psi_{\vec{\lambda}}\left|\frac{\partial}{\partial \vec{\lambda}}\right| \Psi_{\vec{\lambda}}\right\rangle}\end{array} $$

First of all, why do the exponential terms vanish after doing the derivative? Secondly, how did he obtained the last expression? I think the $\dot{\vec{\lambda}}$ comes from the fact that it depends on $t$ and a chan rule is applied to have a derivative on $\vec{\lambda}$ but why do we have the bra $\left\langle\Psi_{\vec{\lambda}}\right|$?

$\endgroup$
3
$\begingroup$

The exponentials are a common factor of both sides of your penultimate equation, so he just cancels them. At the last step the ${\mathcal E}_\lambda$'s cancel, then he multiplies in the bra from the left so that he gets a number rather than a ket. The $|\Psi_\lambda\rangle$ are normalized so that the $\partial \phi/\partial t$ is just multiplied by unity.

$\endgroup$
  • $\begingroup$ "exponentials are a common factor of both sides of your penultimate equation, so he just cancels them." faceplam. I forgot about the right hand side exponential. "then he multiplies in the bra from the left so that he gets a number rather than a ket. The |Ψλ⟩ are normalized so that the ∂ϕ/∂t is just multiplied by unity." Thanks! $\endgroup$ – AA10 Nov 3 '19 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.