Categories: Perturbation, Physics, Quantum mechanics.

Time-dependent perturbation theory

In quantum mechanics, time-dependent perturbation theory exists to deal with time-varying perturbations to the Schrödinger equation. This is in contrast to time-independent perturbation theory, where the perturbation is is stationary.

Let $$\hat{H}_0$$ be the base time-independent Hamiltonian, and $$\hat{H}_1$$ be a time-varying perturbation, with “bookkeeping” parameter $$\lambda$$:

\begin{aligned} \hat{H}(t) = \hat{H}_0 + \lambda \hat{H}_1(t) \end{aligned}

We assume that the unperturbed time-independent problem $$\hat{H}_0 \ket{n} = E_n \ket{n}$$ has already been solved, such that the full solution is:

\begin{aligned} \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp(- i E_n t / \hbar) \end{aligned}

Since these $$\ket{n}$$ form a complete basis, the perturbed wave function can be written in the same form, but with time-dependent coefficients $$c_n(t)$$:

\begin{aligned} \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp(- i E_n t / \hbar) \end{aligned}

We insert this ansatz in the time-dependent Schrödinger equation, and reduce it using the known unperturbed time-independent problem:

\begin{aligned} 0 &= \hat{H}_0 \ket{\Psi(t)} + \lambda \hat{H}_1 \ket{\Psi(t)} - i \hbar \dv{t} \ket{\Psi(t)} \\ &= \sum_{n} \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar) \\ &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar) \end{aligned}

We then take the inner product with an arbitrary stationary basis state $$\ket{m}$$:

\begin{aligned} 0 &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp(- i E_n t / \hbar) \end{aligned}

Thanks to orthonormality, this removes the latter term from the summation:

\begin{aligned} i \hbar \frac{d c_m}{dt} \exp(- i E_m t / \hbar) &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar) \end{aligned}

We divide by the left-hand exponential and define $$\omega_{mn} = (E_m - E_n) / \hbar$$ to get:

\begin{aligned} \boxed{ i \hbar \frac{d c_m}{dt} = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t) } \end{aligned}

So far, we have not invoked any approximation, so we can analytically find $$c_n(t)$$ for some simple systems. Furthermore, it is useful to write this equation in integral form instead:

\begin{aligned} c_m(t) = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} \end{aligned}

If this cannot be solved exactly, we must approximate it. We expand $$c_m(t)$$ in the usual way, with the initial condition $$c_m^{(j)}(0) = 0$$ for $$j > 0$$:

\begin{aligned} c_m(t) = c_m^{(0)} + \lambda c_m^{(1)}(t) + \lambda^2 c_m^{(2)}(t) + ... \end{aligned}

We then insert this into the integral and collect the non-zero orders of $$\lambda$$:

\begin{aligned} c_m^{(1)}(t) &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} \\ c_m^{(2)}(t) &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} \\ c_m^{(3)}(t) &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} \end{aligned}

And so forth. The pattern here is clear: we can calculate the $$(j\!+\!1)$$th correction using only our previous result for the $$j$$th correction. We cannot go any further than this without considering a specific perturbation $$\hat{H}_1(t)$$.

References

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.