Categories: Optics, Physics, Quantum mechanics.

# Rabi oscillation

In quantum mechanics, from the derivation of time-dependent perturbation theory, we know that a time-dependent term $$\hat{H}_1$$ in the Hamiltonian affects the state as follows, where $$c_n(t)$$ are the coefficients of the linear combination of basis states $$\ket{n} \exp\!(-i E_n t / \hbar)$$:

\begin{aligned} i \hbar \dv{c_m}{t} = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp\!(i \omega_{mn} t) \end{aligned}

Where $$\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$$ for energies $$E_m$$ and $$E_n$$. Note that this equation is exact, despite being used for deriving perturbation theory. Consider a two-level system where $$n \in \{a, b\}$$, in which case the above equation can be expanded to the following:

\begin{aligned} \dv{c_a}{t} &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp\!(- i \omega_0 t) \: c_b - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{a} \: c_a \\ \dv{c_b}{t} &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp\!(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b \end{aligned}

Where $$\omega_0 \equiv \omega_{ba}$$ is positive. We assume that $$\hat{H}_1$$ has odd spatial parity, in which case Laporte’s selection rule states that the diagonal matrix elements vanish, leaving:

\begin{aligned} \dv{c_a}{t} &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp\!(- i \omega_0 t) \: c_b \\ \dv{c_b}{t} &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp\!(i \omega_0 t) \: c_a \end{aligned}

We now choose $$\hat{H}_1$$ to be as follows, sinusoidally oscillating with a spatially odd $$V(\vec{r})$$:

\begin{aligned} \hat{H}_1(t) = V \cos\!(\omega t) = \frac{V}{2} \Big( \exp\!(i \omega t) + \exp\!(-i \omega t) \Big) \end{aligned}

We insert this into the equations for $$c_a$$ and $$c_b$$, and define $$V_{ab} \equiv \matrixel{a}{V}{b}$$, leading us to:

\begin{aligned} \dv{c_a}{t} &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!-\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!+\! \omega_0) t\big) \Big) \: c_b \\ \dv{c_b}{t} &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!+\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t\big) \Big) \: c_a \end{aligned}

Here, we make the rotating wave approximation: assuming we are close to resonance $$\omega \approx \omega_0$$, we decide that $$\exp\!(i (\omega \!+\! \omega_0) t)$$ oscillates so much faster than $$\exp\!(i (\omega \!-\! \omega_0) t)$$, that its effect turns out negligible when the system is observed over a reasonable time interval.

In other words, over this reasonably-sized time interval, $$\exp\!(i (\omega \!+\! \omega_0) t)$$ averages to zero, while $$\exp\!(i (\omega \!-\! \omega_0) t)$$ does not. Dropping the respective terms thus leaves us with:

\begin{aligned} \boxed{ \begin{aligned} \dv{c_a}{t} &= - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b \\ \dv{c_b}{t} &= - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a \end{aligned} } \end{aligned}

Now we can solve this system of coupled equations exactly. We differentiate the first equation with respect to $$t$$, and then substitute $$\dv*{c_b}{t}$$ for the second equation:

\begin{aligned} \dv{c_a}{t} &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + \dv{c_b}{t} \bigg) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \\ &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a \bigg) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \\ &= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a \end{aligned}

In the first term, we recognize $$\dv*{c_a}{t}$$, which we insert to arrive at an equation for $$c_a(t)$$:

\begin{aligned} 0 = \dv{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a \end{aligned}

To solve this, we make the ansatz $$c_a(t) = \exp\!(\lambda t)$$, which, upon insertion, gives us:

\begin{aligned} 0 = \lambda^2 - i (\omega - \omega_0) \lambda + \frac{|V_{ab}|^2}{(2 \hbar)^2} \end{aligned}

This quadratic equation has two complex roots $$\lambda_1$$ and $$\lambda_2$$, which are found to be:

\begin{aligned} \lambda_1 = i \frac{\omega - \omega_0 + \tilde{\Omega}}{2} \qquad \quad \lambda_2 = i \frac{\omega - \omega_0 - \tilde{\Omega}}{2} \end{aligned}

Where we have defined the generalized Rabi frequency $$\tilde{\Omega}$$ to be given by:

\begin{aligned} \boxed{ \tilde{\Omega} \equiv \sqrt{(\omega - \omega_0)^2 + \frac{|V_{ab}|^2}{\hbar^2}} } \end{aligned}

So that the general solution $$c_a(t)$$ is as follows, where $$A$$ and $$B$$ are arbitrary constants, to be determined from initial conditions (and normalization):

\begin{aligned} \boxed{ c_a(t) = \Big( A \sin\!(\tilde{\Omega} t / 2) + B \cos\!(\tilde{\Omega} t / 2) \Big) \exp\!\big(i (\omega \!-\! \omega_0) t / 2 \big) } \end{aligned}

And then the corresponding $$c_b(t)$$ can be found from the coupled equation we started at, or, if we only care about the probability density $$|c_a|^2$$, we can use $$|c_b|^2 = 1 - |c_a|^2$$. For example, if $$A = 0$$ and $$B = 1$$, we get the following probabilities

\begin{aligned} |c_a(t)|^2 &= \cos^2(\tilde{\Omega} t / 2) = \frac{1}{2} \Big( 1 + \cos\!(\tilde{\Omega} t) \Big) \\ |c_b(t)|^2 &= \sin^2(\tilde{\Omega} t / 2) = \frac{1}{2} \Big( 1 - \cos\!(\tilde{\Omega} t) \Big) \end{aligned}

Note that the period was halved by squaring. This periodic “flopping” of the particle between $$\ket{a}$$ and $$\ket{b}$$ is known as Rabi oscillation, Rabi flopping or the Rabi cycle. This is a more accurate treatment of the flopping found from first-order perturbation theory.

The name generalized Rabi frequency suggests that there is a non-general version. Indeed, the Rabi frequency $$\Omega$$ is based on the special case of exact resonance $$\omega = \omega_0$$:

\begin{aligned} \Omega \equiv \frac{V_{ba}}{\hbar} \end{aligned}

As an example, Rabi oscillation arises in the electric dipole approximation, where $$\hat{H}_1$$ is:

\begin{aligned} \hat{H}_1(t) = - q \vec{r} \cdot \vec{E}_0 \cos\!(\omega t) \end{aligned}

After making the rotating wave approximation, the resulting Rabi frequency is given by:

\begin{aligned} \Omega = - \frac{\vec{d} \cdot \vec{E}_0}{\hbar} \end{aligned}

Where $$\vec{E}_0$$ is the electric field amplitude, and $$\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$$ is the transition dipole moment of the electron between orbitals $$\ket{a}$$ and $$\ket{b}$$. Apparently, some authors define $$\vec{d}$$ with the opposite sign, thereby departing from its classical interpretation.

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

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