Adiabatic Approximation#

The adiabatic approximation considers a time dependent Hamiltonian \(H(t)\) that preserves the adibataic theorem

Adiabatic Theorem : If the initial Hamiltonian \(H^{i}\) the system is in the \(n\)th eigenstate \(\psi^{i}_n\) then later on at some time \(t>0\) the Hamiltonian becomes \(H^{f}\) such that the eigenstate is still on the \(n\)th but of the current Hamiltonian \(\psi^{f}_n\).

While the adabatic process does not require a small perturbation what is required is a **slow change** in the time-dependent Hamiltonian $H(t)$

Given a system that abides the adiabatic theorem thus the adiabatic approximation, the total wavefunction is,

\[\begin{split} \boxed{\Psi(t) = \sum_n{c_n(0)\psi_n(t)}e^{i\left[ \theta_n(t) + \gamma_n(t) \right]}}\\ \boxed{\gamma_n(t) \equiv i\int_0^t{\braket{\psi_n(t')}{\dot\psi_n(t')}\d t'}}\\ \end{split}\]

  • \(\theta_n(t)\) is called the dynamic phase

  • \(\gamma_n(t)\) is called the geometric phase

Proof : For a time-dependent Hamiltonian \(H(t)\), the eigenfunction and eigenvalue changes with time

$$H(t)\psi_n(t) = E_n(t)\psi_n(t)$$

In general the total wavefunction $\Psi(t)$ is a linear combination of each eigenstate however this time the $E_n(t)$ is no longer constant over time,

$$
\boxed{\Psi(t) = \sum_n{c_n(t)\psi_n(t)e^{i\theta_n(t)}}}\\
\boxed{\theta_n(t) \equiv -\frac{1}{\hbar}\int_0^t{E_n(t')\d t}}
$$

Plugging $\Psi(t)$ into the TDSE and applying $\bra{\psi_m}$ gives,

$$
\dot c_m(t) = -\sum_n{c_n\braket{\psi_m}{\dot \psi_n}}e^{\theta_n - \theta_m}
$$

Differentiating the TISE you would find that

$$
\braket{\psi_m}{\dot\psi_n} = \frac{\bra{\psi_m}\dot H\ket{\psi_n}}{E_n - E_m}
$$

Plugging this into $\dot c_m(t)$ gives,

$$
\boxed{\dot c_m(t) = - c_m\braket{\psi_m}{\dot \psi_m} - \sum_{n\ne m}c_n \frac{\bra{\psi_m}\dot H\ket{\psi_n}}{E_n - E_m}\exp\left[\frac{-i}{\hbar}\int_0^t{E_n(t') - E_m(t')\d t'}\right]}
$$

This is the exact solution to the Time Dependent Hamiltonian problem (not an estimate like the TD Perturbation Theory).

Now let's apply the adibatic approximation. $\dot H$ is very small because the Hamiltonian changes slowly thus the whole second term droped in $\dot c_m(t)$

$$
    \dot c_m(t) = - c_m\braket{\psi_m}{\dot \psi_m}
$$

The solution to this differential equation is of course,

$$
\boxed{c_m(t) = c_m(0)e^{i\gamma_m(t)}}\\
\boxed{\gamma_m(t) \equiv i\int_0^t{\braket{\psi_m(t')}{\dot\psi_m(t')}\d t'}}\\
\boxed{\Psi_n(t) = \psi_n(t)e^{i\left[ \theta_n(t) + \gamma_n(t) \right]}}
$$

* Please note that the dot notation is a partial time derivative $\dot \psi_m(t) = \partial \psi_m / \partial t$

Berry’s Phase#

Previously we’ve seen the time dependent Hamiltonian that changes over time has the geometric phase:

\[ \gamma_n(t) \equiv i\int_0^t{\braket{\psi_n(t')}{\dot\psi_n(t')}\d t'} \]

Now let’s consider some parameter that changes over time which changes the Hamiltonian (like the size of the width in the infinite square well). Let’s denote this parameter as \(R(t)\) therefore the Hamiltonian is \(H(R(t))\). We can write the geometric phase as an integrate along \(R\),

\[\begin{split} \frac{\partial \psi}{\partial t} = \frac{\partial \psi_n}{\partial R}\frac{dR}{dt}\\ \gamma_n(R) = i\int_{R_i}^{R_f}{\braket{\psi_n(R)}{\psi_n'(R)}\d R} \end{split}\]

This is as expected and not very interesting since if \(R_f= R_i\), the integral goes to zero \(\gamma_n=0\)

Now consider multiple parameters such that \(\mathbf{R} = \mathbf{R_1} + \mathbf{R_2} + \dots \mathbf{R_N}\) then,

\[ \boxed{\gamma_n(R) = i\int_{\mathbf R_i}^{\mathbf R_f}{\braket{\psi_n}{\nabla_{\mathbf R}\psi_n}\d \mathbf R}} \]

Say after some time \(T\) the Hamiltonian return to it’s original form therefore travel around the closed loop around \(\mathbf R\),

\[ \boxed{\gamma_n(T) = i \oint{\braket{\psi_n}{\nabla_{\mathbf R}\psi_n}\d \mathbf R}} \]

From Stoke’s theorem we may write this as a surface integral,

\[ \boxed{\gamma_n(T) = i \int{\nabla_{\mathbf R} \times \braket{\psi_n}{\nabla_{\mathbf R}\psi_n}\d \mathbf a}} \]

Aharonov-Bohm Effect#

Consider an electron splitting dodging the solenoid magnetic field. Let’s center the coordinate system at the solenoid such that,

\[\begin{split} \mathbf B(r) = \begin{cases} B_0 \hat z & r < a\\ 0 & r > a \end{cases} \end{split}\]

The Aharonov-Bohm effect states there a geometric phase will occur to the wavefunction ofthe electron even if the electron does not travel through a nonzero magnetic field. The vector potential of the magnetic field is given by,

\[\begin{split} \mathbf B = \nabla \times \mathbf A \\ \mathbf A = \frac{\Phi}{2\pi r}\hat\phi \qquad (r>a) \end{split}\]

Clearly at regions where the magnetic field is zero (\(r > a\) in this experiment), \(\nabla \times \mathbf A = 0\). However this does not mean \(\mathbf A = 0\). The TDSE is given as,

\[ \left[ \frac{1}{2m}\left(\frac{h}{i}\nabla - q\mathbf A \right)^2 + V \right]\Psi = i\hbar \frac{\partial \Psi}{\partial t} \]

This has the solutions,

\[\begin{split} \Psi = e^{ig}\Psi'\\ g(\mathbf r) \equiv \frac{q}{\hbar}\int_{\mathcal{O}}^{\mathbf r}{\mathbf A(\mathbf r')\cdot \d \mathbf r'} = \pm \frac{q\Phi}{2\hbar} \end{split}\]