Time Perturbation Theory

TDSE

If the transitions between different energy levels are allowed, we must use time-dependent wave function. Thus the TDSE \begin{equation} H\Psi = i\hbar\frac{\partial \Psi}{\partial t} \end{equation} can not be solved by variable seperation. But if the time-dependent part of Hamiltonian is small, we can consider is as perturbation.

Two Energy-level System

We start with the Hamiltonian that have only two eigen states: \begin{equation} H^0 \psi_a=E_a\psi_a \end{equation} \begin{equation} H^0 \psi_b=E_b \psi_b \end{equation} which are orthornormalized \(\langle \psi_a | \psi_b\rangle = \delta_{ab}\). Thus any state at any time can be expressed as: \begin{equation} \label{Psi} \Psi(t)=c_a\psi_a e^{-iE_a t/\hbar}+c_b\psi_b e^{-iE_b t/\hbar} \end{equation} where \(c_a\) and \(c_b\) are the initial constant.

Add Perturbation

If we add a time-dependent Hamiltonian pertrbation \(H'(t)\), now \(c_a\) and \(c_b\) would become time-dependent. Assuming that \(c_a(0) = 1\) as initial condition, as \(\Psi(t)\) obeys the TDSE, we can solve the constant into: \begin{equation} \label{9.13} \dot{c_a}=-\frac{i}{\hbar}H’_{ab} e^{-i\omega_0 t}c_b \end{equation}

\begin{equation} \dot{c_b}=-\frac{i}{\hbar}H’_{ba} e^{i\omega_0 t}c_a \end{equation}

where \(\omega_0 = \frac{E_b - E_a}{\hbar}\) and \(H'_{ab} = \langle \psi_a |H'|\psi_b\rangle\)

Time-Dependent Perturbation Theory

We can solve the equations by iteration, substituting the right hand of \eqref{9.13} with the lower order approximation value.
Assume \(c_a(0)=1\).

0th Order

\begin{equation} c_a^{(0)}(t) = 1 \end{equation}

\begin{equation} c_b^{(0)}(t) = 0 \end{equation}

1st Order

\begin{equation} c_a^{(1)}(t)=1 \end{equation}

\begin{equation} \dot{c_b^{(1)}}(t)=-\frac{i}{\hbar} H_{ba}’ e^{i\omega_0 t} \to c_b^{(1)}(t)=-\frac{i}{\hbar}\int_{0}^{t}H_{ba}’ (t’)e^{i\omega_0 t’}dt’ \end{equation}

2nd Order

\begin{equation} c_a^{(2)}(t)=1-\frac{1}{\hbar^2} \int_0^t H_{ab}’(t’)e^{-i\omega_0 t’}[\int_0^{t’}H_{ba}’(t^”)e^{i\omega_0 t^”}dt^”]dt’ \end{equation} \begin{equation} c_b^{(2)}(t)=c_b^{(1)}(t) \end{equation}

Sinusoidal Perturbation

If we consider the sinusoidal perturbation \begin{equation} \prime{H}(\vec r,t) = V(\vec r)\cos (\omega t) \end{equation} to the 1st order approximation, we obtain \begin{equation} \label{9.25} c_b(t)\approx -\frac{i}{\hbar}V_{ab} \int_0^t \cos (\omega t’)e^{i\omega_0 t} dt’=-\frac{V_{ab}}{2\hbar}[\frac{e^{i(\omega_0+\omega)t }-1}{\omega_0+\omega}+\frac{e^{i(\omega_0-\omega)t}-1}{\omega_0-\omega}] \end{equation} If we only consider the situation that drive frequency is close to transition frequency (\(\omega_0+\omega >> |\omega_0-\omega|\)), we can ignore the first term in \eqref{9.25}. So, \begin{equation} c_b(t)\approx -i\frac{V_{ba}}{\hbar}\frac{\sin [(\omega_0-\omega)t/2]}{\omega_0-\omega}e^{i(\omega_0-\omega)t/2} \end{equation} The transition probability at time \(t\) is: \begin{equation} \label{9.28} P_{a\to b}(t) = |c_b(t)|^2\approx \frac{|V_{ab}|^2}{\hbar^2}\frac{\sin^2 [(\omega_0-\omega)t/2]}{(\omega_0-\omega)^2} \end{equation} which is in sinusoidal form.

Emission and Absorption of Radiation

EM Wave

We can consider an atom in the EM wave as under a sinusoidal electric field \begin{equation} \vec E = E_0 \cos(\omega t)\hat k \end{equation} The Hamiltonian perturbation is \begin{equation} H’=-qE_0 z\cos(\omega t) \end{equation} We have \begin{equation} H’_{ba} = -R E_0 z\cos(\omega t) \text{, where} R=q\langle \psi_b |z|\psi_a \rangle \end{equation} So the transition probability can be written from equation \eqref{9.28} with \(V_{ba} = -RE_0\)

Absorption, Stimulated Emission and Spontaneous Emission

Temperarily skiped.

Incoherence Perturbation

The energy density of EM wave: \begin{equation} u=\frac{\epsilon_0}{2}E_0^2 \end{equation} Thus transition probability is directly proportional to energy density. But this is only correct to single frequency. In practice, the \(u \to \rho(\omega)d\omega\). We can substitute \(\rho(\omega)\) with \(\rho(\omega_0)\), making the result in integral form: \begin{equation} P_{b\to a}(t)=\frac{2}{\epsilon_0 \hbar^2}|R|^2 \rho(\omega_0) \int_0^{\infty} \lbrace \frac{\sin^2 [(\omega_0-\omega)t/2]}{(\omega_0-\omega)^2} \rbrace d\omega \end{equation} After substitution and expansion of integral range, we obetan \begin{equation} P_{b\to a}(t) \approx \frac{\pi |R|^2}{\epsilon_0 \hbar^2}\rho(\omega_0) t. \end{equation} If we consider incident waves from all directions and contribute equally to \(\rho(\omega)\), we need to substitute \(|R|^2\) with \(|\vec R \cdot \hat n|^2_{avg}=\frac{1}{3}|R|^2\) We can obtain the average value \(|\vec R \cdot \hat n|^2_{avg}\) by spherical coordinate integral. So, under incoherence light from all directions, the stimulated transition speed from b to a is: \begin{equation} \label{9.47} R_{b\to a}(t)=\frac{\pi}{3\epsilon_0 \hbar^2}|\vec R|^2 \rho(\omega_0) \end{equation}

Spontaneous Emission

Einstein’s Constants of Emission and Absorption

As the stimulated emission speed and absorption speed are proportional to EM field energy density, we can get the result with Boltzman distribution and Planck distribution: \begin{equation} B_{ab}=B_{ba} \end{equation} \begin{equation} A=\frac{\omega_0^3 \hbar}{\pi^2 c^3}B_{ba} \end{equation} where \(B_{ab}\) and \(B_{ba}\) are absoprtion and stimulated emission constant, and \(A\) is spontaneous emission constant. From equation\eqref{9.47} we know: \begin{equation} B_{ba} = \frac{\pi}{3\epsilon_0\hbar^2}|\vec R|^2 \end{equation} So the spontaneous emission speed is \begin{equation} A=\frac{\omega_0^3|\vec R|^2}{3\pi\epsilon_0 \hbar c^3}. \end{equation}

Life Time of Stimulated State

The number of stimulated particle would decrease by exponential function: \begin{equation} N_b(t)=N_b(0)e^{-At} \end{equation} The time constant: \begin{equation} \tau = \frac{1}{A} \end{equation} is the life time of state.

Selection Rules