Categories: Nonlinear optics, Optics, Perturbation, Physics, Quantum mechanics.

# Multi-photon absorption

Consider a quantum system where there are many eigenstates $\Ket{n}$, e.g. atomic orbitals, for an electron to occupy. Suppose an electromagnetic wave passes by, such that its Hamiltonian gets perturbed by $\hat{H}_1$, given in the electric dipole approximation by:

\begin{aligned} \hat{H}_1(t) = -\vu{p} \cdot \vb{E} \cos(\omega t) \approx -\vu{p} \cdot \vb{E} e^{-i \omega t} \end{aligned}

Where $\vb{E}$ is the electric field amplitude, and $\vu{p} \equiv q \vu{x}$ is the transition dipole moment operator. Here, we have made the rotating wave approximation to neglect the $e^{i \omega t}$ term, because it turns out to be irrelevant in this discussion.

We call the ground state $\Ket{0}$, but other than that, the other states need not be sorted by energy. However, we demand that the following holds for all even-numbered states $\Ket{e}$ and $\Ket{e'}$, and for all odd-numbered ($u$neven) states $\Ket{u}$ and $\Ket{u'}$:

\begin{aligned} \matrixel{e}{\hat{H}_1}{e'} = \matrixel{u}{\hat{H}_1}{u'} = 0 \qquad \quad \matrixel{e}{\hat{H}_1}{u} \neq 0 \end{aligned}

This is justified for atomic orbitals thanks to Laporte’s selection rule. Therefore, time-dependent perturbation theory says that the $N$th-order coefficient corrections are:

\begin{aligned} c_e^{(N)}(t) &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(N-1)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau} \\ c_u^{(N)}(t) &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(N-1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau} \end{aligned}

Where $\omega_{eu} = (E_e \!-\! E_u) / \hbar$. For simplicity, the electron starts in the lowest-energy state $\Ket{0}$:

\begin{aligned} c_0^{(0)} = 1 \qquad \qquad c_u^{(0)} = c_{e \neq 0}^{(0)} = 0 \end{aligned}

Finally, we prove the following useful relation for large $t$, involving a Dirac delta function $\delta$:

\begin{aligned} \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2 = 2 \pi \: \delta(x) \: t \end{aligned}

First, observe that we can rewrite the fraction using an integral:

\begin{aligned} \frac{e^{i x t} - 1}{x} = e^{i x t / 2} \frac{e^{i x t / 2} - e^{-i x t / 2}}{x} = i e^{i x t / 2} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau} \end{aligned}

By taking the limit $t \to \infty$, it can be turned into a nascent Dirac delta function:

\begin{aligned} \lim_{t \to \infty} \frac{e^{i x t} - 1}{x} = \lim_{t \to \infty} i e^{i x t / 2} \frac{2 \pi}{2 \pi} \int_{-\infty}^{\infty} e^{i x \tau} \dd{\tau} = \lim_{t \to \infty} i 2 \pi e^{i x t / 2} \: \delta(x) \end{aligned}

Consequently, the absolute value squared is as follows:

\begin{aligned} \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2 = 4 \pi^2 \delta^2(x) \end{aligned}

However, a squared delta function $\delta^2$ is not ideal, so we take a step back:

\begin{aligned} \delta^2(x) = \delta(x) \lim_{t \to \infty} \frac{1}{2 \pi} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau} = \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi} \end{aligned}

Where we have set $x = 0$ according to the first delta function. This gives the target:

\begin{aligned} \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2 = 4 \pi^2 \delta^2(x) = 2 \pi \: \delta(x) \: t \end{aligned}

## One-photon absorption

To warm up, we start at first-order perturbation theory. Thanks to our choice of initial condition, nothing at all happens to any of the even-numbered states $\Ket{e}$:

\begin{aligned} c_e^{(1)}(t) &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(0)} \: e^{i \omega_{eu} \tau} \dd{\tau} = 0 \end{aligned}

While the odd-numbered states $\Ket{u}$ have a nonzero correction $c_u^{(1)}$, where $\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$:

\begin{aligned} c_u^{(1)}(t) &= -\frac{i}{\hbar} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{0} \: c_0^{(0)} \: e^{i \omega_{u0} \tau} \dd{\tau} \\ &= i \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \int_0^t e^{i (\omega_{u0} - \omega) \tau} \dd{\tau} \\ &= i \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg[ \frac{e^{i (\omega_{u0} - \omega) \tau}}{i (\omega_{u0} - \omega)} \bigg]_0^t \end{aligned}

Consequently, the first-order correction (in the rotating wave approximation) is given by:

\begin{aligned} \boxed{ c_u^{(1)}(t) \approx \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega} } \end{aligned}

Since $\big| c_u^{(1)}(t) \big|^2$ is the probability of finding the electron in $\Ket{u}$, its transition rate $R_u^{(1)}(t)$ is as follows, averaged since the beginning $t = 0$:

\begin{aligned} R_u^{(1)}(t) = \frac{\big| c_u^{(1)}(t) \big|^2}{t} = \frac{1}{t} \bigg| \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg|^2 \cdot \bigg| \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega} \bigg|^2 \end{aligned}

For large $t \to \infty$, we can use the formula we proved earlier to get Fermi’s golden rule:

\begin{aligned} \boxed{ R_u^{(1)} = 2 \pi \bigg| \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg|^2 \delta(\omega_{u0} - \omega) } \end{aligned}

This well-known formula represents one-photon absorption: it peaks at $\omega_{u0} = \omega$, i.e. when one photon $\hbar \omega$ has the exact energy of the transition $\hbar \omega_{u0}$. Note that this transition is only possible when $\matrixel{u}{\vu{p}}{0} \neq 0$, i.e. for any odd-numbered final state $\Ket{u}$.

## Two-photon absorption

Next, we go to second-order perturbation theory. Based on the previous result, this time all odd-numbered states $\Ket{u}$ are unaffected:

\begin{aligned} c_u^{(2)}(t) &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau} = 0 \end{aligned}

While the even-numbered states $\Ket{e}$ have the following correction, using $\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$:

\begin{aligned} c_e^{(2)}(t) &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(1)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau} \\ &= i \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \int_0^t e^{i (\omega_{eu} + \omega_{u0} - 2 \omega) \tau} - e^{i (\omega_{eu} - \omega) \tau} \dd{\tau} \\ &= i \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \bigg[ \frac{e^{i (\omega_{e0} - 2 \omega) \tau}}{i (\omega_{e0} - 2 \omega)} - \frac{e^{i (\omega_{eu} - \omega) \tau}}{i (\omega_{eu} - \omega)} \bigg]_0^t \end{aligned}

The second term represents one-photon absorption between $\Ket{u}$ and $\Ket{e}$. We do not care about that, so we drop it, leaving only the first term:

\begin{aligned} \boxed{ c_e^{(2)}(t) \approx \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega} } \end{aligned}

As before, we can define a rate $R_e^{(2)}(t)$ for all transitions represented by this term:

\begin{aligned} R_e^{(2)}(t) = \frac{\big| c_e^{(2)}(t) \big|^2}{t} = \frac{1}{t} \bigg| \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \bigg|^2 \cdot \bigg| \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega} \bigg|^2 \end{aligned}

Which for $t \to \infty$ takes a similar form to Fermi’s golden rule, using the formula we proved:

\begin{aligned} \boxed{ R_e^{(2)} = 2 \pi \bigg| \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \bigg|^2 \delta(\omega_{e0} - 2 \omega) } \end{aligned}

This represents two-photon absorption, since it peaks at $\omega_{e0} = 2 \omega$: two identical photons $\hbar \omega$ are absorbed simultaneously to bridge the energy gap $\hbar \omega_{e0}$. Surprisingly, such a transition can only occur when $\matrixel{e}{\vu{p}}{0} = 0$, i.e. for any even-numbered final state $\Ket{e}$. Notice that the rate is proportional to $|\vb{E}|^4$, so this effect is only noticeable at high light intensities.

## Three-photon absorption

For third-order perturbation theory, all even-numbered states $\Ket{e}$ are unchanged:

\begin{aligned} c_e^{(3)}(t) &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(2)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau} = 0 \end{aligned}

And the odd-numbered states $\Ket{u}$ get the following third-order corrections:

\begin{aligned} c_u^{(3)}(t) &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(2)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau} \\ &= i \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}} \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}{\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \int_0^t e^{i (\omega_{ue} + \omega_{e0} - 3 \omega) \tau} - e^{i (\omega_{ue} - \omega) \tau} \dd{\tau} \\ &= i \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}} \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}{\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \bigg[ \frac{e^{i (\omega_{u0} - 3 \omega) \tau}}{i (\omega_{u0} - 3 \omega)} - \frac{e^{i (\omega_{ue} - \omega) \tau}}{i (\omega_{ue} - \omega)} \bigg]_0^t \end{aligned}

Once again, the second term is uninteresting, so we drop it and look at the first term only:

\begin{aligned} \boxed{ c_u^{(3)}(t) \approx \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}} \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})} {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \frac{e^{i (\omega_{u0} - 3 \omega) t} - 1}{\omega_{u0} - 3 \omega} } \end{aligned}

The resulting transition rate $R_u^{(3)}(t)$ is found to have the following familiar form:

\begin{aligned} R_u^{(3)}(t) = \frac{\big| c_u^{(3)}(t) \big|^2}{t} = \frac{1}{t} \bigg| \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}} \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})} {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \bigg|^2 \cdot \bigg| \frac{e^{i (\omega_{u0} - 3 \omega) t} - 1}{\omega_{u0} - 3 \omega} \bigg|^2 \end{aligned}

Applying our formula to this yields the following analogue of Fermi’s golden rule:

\begin{aligned} \boxed{ R_u^{(3)} = 2 \pi \bigg| \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}} \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})} {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \bigg|^2 \delta(\omega_{u0} - 3 \omega) } \end{aligned}

This represents three-photon absorption, since it peaks at $\omega_{u0} = 3 \omega$: three identical photons $\hbar \omega$ are absorbed simultaneously to bridge the energy gap $\hbar \omega_{u0}$. This process is similar to one-photon absorption, in the sense that it can only occur if $\matrixel{u}{\vu{p}}{0} \neq 0$. The rate is proportional to $|\vb{E}|^6$, so this effect only appears at extremely high light intensities.

## N-photon absorption

A pattern has appeared in these calculations: in $N$th-order perturbation theory, we get a term representing $N$-photon absorption, with a transition rate proportional to $|\vb{E}|^{2N}$. Indeed, we can derive infinitely many formulas in this way, although the results become increasingly unrealistic due to the dependence on $\vb{E}$.

If $N$ is odd, only odd-numbered destinations $\Ket{u}$ are allowed (assuming the electron starts in the ground state $\Ket{0}$), and if $N$ is even, only even-numbered destinations $\Ket{e}$. Note that nothing has been said about the energies of these states (other than $\Ket{0}$ being the minimum); everything is determined by the matrix elements $\matrixel{f}{\vu{p}}{i}$.

1. R.W. Boyd, Nonlinear optics, 4th edition, Academic Press.
2. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.