Time-Dependent Perturbation Theory

Time-Dependent Perturbation Theory#

The Time Dependent Perturbation Theory (TDPT) solve for the perturbed hamiltonian as a function of time \(H'(t)\) such that the full wavefunction \(\Psi(x,t)\) (or simply \(\Psi(t)\)) now has a time-dependent probability coefficient,

\[ \begin{equation} \Psi(t) = \sum_k{c_k(t)\psi_k e^{\frac{-iE_k t}{\hbar}}} \end{equation} \]

For some given Hamiltonian \(H = H^0 + H'\). \(\Psi(t)\) should satisfies the TDSE where,

\[ H\Psi = i\hbar \frac{\partial \Psi}{\partial t} \]

Note that the goal of TDPT is to solve for \(c_k(t)\) for each wavefunction which determines the probability of measuring the wavefunction at some given time \(t\).

System of Two Eigenvalues#

For a system of two eigenvalues \(H^0\psi_a = E_a\) and \(H^0\psi_b = E_b\), we can write a nice solution for \(c_k(t)\) which is,

\[\begin{split} \begin{gather} \boxed{\dot c_a = -\frac{i}{\hbar}\left[c_aH'_{aa} + H'_{ab}e^{i\omega_{ba} t}c_b\right], \qquad \dot c_b = -\frac{i}{\hbar}\left[c_bH'_{bb} + H'_{ba}e^{i\omega_{ba} t}c_a\right]}\\ H'_{ij} = \bra{\psi_i}H'\ket{\psi_j} \\ \omega_{ba} \equiv \frac{E_b - E_a}{\hbar} \end{gather} \end{split}\]

  • In many cases the diagonal matrices are zero \(H_{aa}' = H_{bb}' = 0\)

Proof#

Plugging in the wavefunction \(\Psi(t)\) into the TDSE we find that,

\[ \begin{equation} \boxed{\dot c_a^n = -\frac{i}{\hbar}H'_{ab}e^{i\omega_{ba} t}c_b^{(n-1)}, \qquad \dot c_b^n = -\frac{i}{\hbar}H'_{ba}e^{i\omega_{ba} t}c_a^{(n-1)}} \end{equation} \]

Unfortunately we cannot simply integrate by time to get \(c_a\) or \(c_b\) noticing that both formula is dependent on each other. To solve this consider the intial state that the system in definitely in state \(\psi_a\) but not \(\psi_b\) this means:

\[ c_a(0) = 1 \qquad c_b(0) = 0 \]

We apply this to the zeroth order to find \(c_a^0(t)\), \(c_b^0(t)\) and then plug that into the first order integrate to find \(c_a^1(t)\), \(c_b^1(t)\) and so on. This is a recursive process that is generalized like so:

\[ \begin{equation} \boxed{\dot c_a^n = -\frac{i}{\hbar}H'_{ab}e^{i\omega_{ba} t}c_b^{(n-1)}, \qquad \dot c_b^n = -\frac{i}{\hbar}H'_{ba}e^{i\omega_{ba} t}c_a^{(n-1)}} \end{equation} \]

This states that the \(n\)th order approximation is dependent on the \((n-1)\)th order thus a recursive process.

Zeroth Order : For the zeroth order we consider \(\dot c_a(t) = 0\) and \(\dot c_b(t) = 0\) therefore the coefficients are constant throughout time,

\[ \begin{equation} \boxed{c_a^0(t) = 1 \qquad c_b^0(t) = 0} \end{equation} \]

First Order : For the first order, we plug in the zeroth order coefficient,

\[\begin{split} \begin{gather} \frac{d c_a^1}{dt} = -\frac{i}{h}\left[c_a^0H'_{aa} + H'_{ab}e^{i\omega_{ba} t}c_b^0\right] = 0\\ \boxed{c_a^1(t) = 1 -\frac{i}{\hbar}\int_0^{t}{H'_{aa}}\d t'}\\ \frac{d c_b^1}{dt} = -\frac{i}{h}H'_{ba}e^{i\omega_{ba} t}c_a^0 = -\frac{i}{h}H'_{ba}e^{i\omega_{ba} t}\\ \boxed{c_b^1(t) = \frac{-i}{\hbar}\int_0^{t}{H'_{ba}e^{i\omega_{ba} t'}\d t'}} \end{gather} \end{split}\]

The probability of measuring \(\psi_b\) in this case is actually interpretted as the probability of the transition from \(\psi_a\) to \(\psi_b\),

\[ \boxed{P^1_{a\rightarrow b} = \abs{c_b^1}^2} \tag{First Order}\]

Therefore this generalizes to any initial state \(\psi_k\) with \(c_k(0)=0\) transitioning to another state \(\psi_b\).

Example - Sinusoidal Perturbations#

Consider the perturbed Hamiltonian,

\[ \begin{equation} H'(\boldsymbol{r},t) = V(\boldsymbol{r})\cos(\omega_{ba} t) \end{equation} \]

Immediately we can see that the perturbed Hamiltonian has a separable characteristic where:

\[\begin{split} \begin{gather} H'_{ba} \equiv V_{ba}\cos(\omega_{ba} t)\\ V_{ba} \equiv \bra{\psi_a} V \ket{\psi_b} \end{gather} \end{split}\]

Once again we use the initial condition that \(c_a(0)=1\) and \(c_b(0)=0\) therefore the first order coefficient \(c_b(t)\) is,

\[ \begin{gather} c_b^1 = \int{H'_{ba}(t)e^{i\omega_{ba} t}dt} = -\frac{V_{ba}}{2\hbar}\left[\frac{e^{i(\omega_{ba} + \omega)t} - 1}{\omega_{ba} + \omega} + \frac{e^{i(\omega_{ba} - \omega)t} - 1}{\omega_{ba} - \omega} \right] \end{gather} \]
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