---
title: "Multi-photon absorption"
date: 2022-01-30
categories:
- Physics
- Optics
- Quantum mechanics
- Nonlinear optics
- Perturbation
layout: "concept"
---

Consider a quantum system where there are many eigenstates $\Ket{n}$,
e.g. atomic orbitals, for an electron to occupy.
Suppose an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
passes by, such that its Hamiltonian gets perturbed by $\hat{H}_1$, given in the
[electric dipole approximation](/know/concept/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](/know/concept/electric-field/) amplitude,
and $\vu{p} \equiv q \vu{x}$ is the transition dipole moment operator.
Here, we have made the
[rotating wave approximation](/know/concept/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](/know/concept/selection-rules/).
Therefore, [time-dependent perturbation theory](/know/concept/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](/know/concept/dirac-delta-function/) $\delta$:

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

<div class="accordion">
<input type="checkbox" id="proof-sinc"/>
<label for="proof-sinc">Proof</label>
<div class="hidden">
<label for="proof-sinc">Proof.</label>
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:

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


## 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](/know/concept/fermis-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 $\matrixel{f}{\vu{p}}{i}$.



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