path: root/source/know/concept/multi-photon-absorption/index.md

---
title: "Multi-photon absorption"
sort_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/) $$\delta$$:

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


{% include proof/start.html id="proof-relation" -%}
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:

$$\begin{aligned}
    \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
    = 4 \pi^2 \delta^2(x)
    = 2 \pi \: \delta(x) \: t
\end{aligned}$$
{% include proof/end.html id="proof-relation" %}



## 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.