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 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 \dv{c_n}{t} \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 \dv{c_m}{t} \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 $$\omega_{mn} \equiv (E_m - E_n) / \hbar$$ to get:

\begin{aligned} \boxed{ i \hbar \dv{c_m}{t} = \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 $$\hat{H}_1(t)$$.

## Sinusoidal perturbation

Arguably the most important perturbation is a sinusoidally-varying potential, which represents e.g. incoming electromagnetic waves, or an AC voltage being applied to the system. In this case, $$\hat{H}_1$$ has the following form:

\begin{aligned} \hat{H}_1(\vec{r}, t) \equiv V(\vec{r}) \sin\!(\omega t) = \frac{1}{2 i} V(\vec{r}) \: \big( \exp\!(i \omega t) - \exp\!(-i \omega t) \big) \end{aligned}

We abbreviate $$V_{mn} = \matrixel{m}{V}{n}$$, and take the first-order correction formula:

\begin{aligned} c_m^{(1)}(t) &= - \frac{1}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)} \int_0^t \exp\!\big(i \tau (\omega_{mn} \!+\! \omega)\big) - \exp\big(i \tau (\omega_{mn} \!-\! \omega)\big) \dd{\tau} \\ &= \frac{i}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)} \bigg( \frac{\exp\!\big(i t (\omega_{mn} \!+\! \omega) \big) - 1}{\omega_{mn} + \omega} + \frac{\exp\!\big(i t (\omega_{mn} \!-\! \omega) \big) - 1}{\omega_{mn} - \omega} \bigg) \end{aligned}

For simplicity, we let the system start in a known state $$\ket{a}$$, such that $$c_n^{(0)} = \delta_{na}$$, and we assume that the driving frequency is close to resonance $$\omega \approx \omega_{ma}$$, such that the second term dominates the first, which can then be neglected. We thus get:

\begin{aligned} c_m^{(1)}(t) &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) \big) - 1}{\omega_{ma} - \omega} \\ &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) - \exp\!\big(\!-\! i t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega} \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) \\ &= - \frac{V_{ma}}{\hbar} \frac{\sin\!\big( t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega} \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) \end{aligned}

Taking the norm squared yields the transition probability: the probability that a particle that started in state $$\ket{a}$$ will be found in $$\ket{m}$$ at time $$t$$:

\begin{aligned} \boxed{ P_{a \to m} = |c_m^{(1)}(t)|^2 = \frac{|V_{ma}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ma} - \omega) t / 2 \big)}{(\omega_{ma} - \omega)^2} } \end{aligned}

The result would be the same if $$\hat{H}_1 \equiv V \cos\!(\omega t)$$. However, if instead $$\hat{H}_1 \equiv V \exp\!(- i \omega t)$$, the result is larger by a factor of $$4$$, which can cause confusion when comparing literature.

In any case, the probability oscillates as a function of $$t$$ with period $$T = 2 \pi / (\omega_{ma} \!-\! \omega)$$, so after one period the particle is back in $$\ket{a}$$, and after $$T/2$$ the particle is in $$\ket{b}$$. See Rabi oscillation for a more accurate treatment of this “flopping” behaviour.

However, when regarded as a function of $$\omega$$, the probability takes the form of a sinc-function centred around $$(\omega_{ma} \!-\! \omega)$$, so it is highest for transitions with energy $$\hbar \omega = E_m \!-\! E_a$$.

Also note that the sinc-distribution becomes narrower over time, which roughly means that it takes some time for the system to “notice” that it is being driven periodically. In other words, there is some “inertia” to it.

## References

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
2. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.