Last day we explored a model of the ammonia molecule. Today we explore how such a molecule interacts with light. We will see an example of stimulated emission and stimulated absorption. We will be able to relate the absorption spectrum to the eigenvalues of the energy operator. This will give us an experimental way to measure energy eigenvalues, and helps motivate their study. [The other motivation, of course, is that one can use energy eigenvalues and eigenvectors to calculate dynamics.]

Overview

We are all familiar with absorption spectra: You shine light on gas, send it through a prism, and find black bands, corresponding to absorption at discrete frequencies. You have also seen emission spectra. You heat up a gas of sodium atoms, and a bright yellow light comes off -- which has essentially the same frequency as the missing light in the absorption spectrum. These discrete spectra were one of the main motivators for quantum mechanics.

The emission process is known as "spontaneous emission." To model it correctly one needs a quantum theory of light. The absorption process, however, can be understood through a model where a classical electromagnetic wave interacts with an atom or molecule. This is known as stimulated absorption. Remarkably, we will find that in modeling stimulated absorption, we will find additional process, stimulated emission. One observation of stimulated emission would be sending light through a properly prepared gas, followed by a prism, and finding that discrete frequencies are brighter. This phenomenon is the process behind the laser: light amplified by the stimulated emission of radiation.

Usually you do not think of a laser as an amplifier -- rather you think of it as a source. These are really the same thing. If you send noise through an amplifier, you get a narrow bandwidth source. To get an intense narrow bandwidth source, you make a feedback circuit, where the light is sent through the amplifier many times.

Ammonia

Recall, in the Ammonia atom, there is a nitrogen which can be in one of two possible places. Because of quantum tunneling this gives to closely spaced energy levels. It turns out that the energy spacing is in the microwave frequency range.
Not surprisingly, microwaves can therefore drive transitions between these levels. If the molecule transitions from the low energy state to the high energy state, it absorbs energy from the field. Thus for consistency the intensity of the field must diminish. This is stimulated absorption. The opposite process can also occur: The microwaves can cause the ammonia molecule to transition from the excited to ground state. By energy conservation this must increase the number of photons in the field. In this way you can make a microwave amplifier.

We begin with a simple two-state model for ammonia: $\psi_R(t)$ represents the amplitude for the nitrogen to be on the right, and $\psi_L(t)$ the amplitude to be on the left. In this truncated space, the Hamiltonian is
\begin{equation}
H \left( \begin{array}{c} \psi_L\\\psi_R\end{array}\right)=\left(
\begin{array}{cc}
a&b\\b&a
\end{array}\right)
\left( \begin{array}{c} \psi_L\\\psi_R\end{array}\right).
\end{equation}
As we saw last day, the constant $a$ plays no role in the dynamics, and we can save ourselves some writing by choosing $a=0$. This is not essential, but I am lazy and hate writing things that are unnecessary. The energy splitting between the two modes is $2b$, and $\pi/b$ is the period of oscillations. I haven't proven it to you, but $b<0$. Traditionally one write $b=-\Delta$. The Schrodinger equation then read
\begin{equation}
i\partial_t \left( \begin{array}{c} \psi_L\\\psi_R\end{array}\right) =
\left(
\begin{array}{cc}
0&-\Delta\\-\Delta&0
\end{array}\right)
\left( \begin{array}{c} \psi_L\\\psi_R\end{array}\right).
\end{equation}
We solved this equation last day.

Electric field

We now imagine placing this ammonia atom in an electric field. If the electric field is pointing to the right, then the energy of $\psi_R$ should be shifted relative to $\psi_L$: the molecule has a finite electric dipole moment. As a result we should find
\begin{equation}
i\partial_t \left(
\begin{array}{c}
\psi_L(t)\\
\psi_R(t)
\end{array}\right)
=\left(
\begin{array}{cc}
\epsilon&-\Delta\\
-\Delta&-\epsilon
\end{array}
\right)
\left(
\begin{array}{c}
\psi_L(t)\\
\psi_R(t)
\end{array}\right)
\end{equation}
where $\epsilon$ is proportional to $E$. Now the interesting thing is what happens if the electric field oscillates in time with frequency $\nu$. Clearly we then get
\begin{equation}
i\partial_t \left(
\begin{array}{c}
\psi_L(t)\\
\psi_R(t)
\end{array}\right)=\left(
\begin{array}{cc}
\epsilon\cos(\nu t)&-\Delta\\
-\Delta& -\epsilon \cos(\nu t)
\end{array}
\right)
\left(
\begin{array}{c}
\psi_L(t)\\
\psi_R(t)
\end{array}\right).
\end{equation}
Our goal is to solve this equation, and find the time dependence of the energy of the atom.

Symmetric/Antisymmetric Basis

Traditionally one does this calculation in a basis corresponding to the eigenstates in the absence of the magnetic field:
\begin{equation}
\psi= a(t) |+\rangle+ b(t)|-\rangle,
\end{equation}
where
\begin{eqnarray}
|+\rangle&=&\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\1\end{array}\right)\\
|-\rangle&=&\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\-1\end{array}\right)\\
\end{eqnarray}
Some simple algebra gives
\begin{equation}
\left(
\begin{array}{c}
\psi_L\\
\psi_R
\end{array}
\right)=
\frac{1}{\sqrt{2}}
\left(
\begin{array}{cc}
1&1\\
-1&1
\end{array}
\right)
\left(
\begin{array}{c}
a\\
b
\end{array}
\right).
\end{equation}
You might recognize this as a unitary basis change. If you go on to PHYS 4443, you will see a lot more of this. Some more straightforward algebra gives

\begin{equation}
i\partial_t
\left(
\begin{array}{c}
a\\
b
\end{array}
\right)
=
\left(
\begin{array}{cc}
-\Delta&+\epsilon\cos(\nu t)\\
\epsilon \cos(\nu t)&\Delta
\end{array}
\right)
\left(
\begin{array}{c}
a\\
b
\end{array}
\right).
\end{equation}

The oscillating electric field can either pump energy into this system, or take energy out. In the former case, the ammonia atom will be absorbing photons -- this is stimulated absorption. In the latter case, the ammonia atom will be emitting photons -- this is stimulated emission.
Our task in this lecture is to calculate this dynamics. Its pretty straightforward to numerically solve these equations, but we want to develop some approximations which will allow us to analytically understand the dynamics.

We will find that we get stimulated emission/absorption when the field is on-resonance -- that is $\nu=2\Delta$. Otherwise there is very little work being done by the electric field.

Short time

Suppose $\epsilon=0$. In that case we can readily integrate these equations to get
\begin{eqnarray}\label{noeps}
a(t)&=& e^{-i E_s t} a(0)\\
b(t)&=& e^{-i E_a t} b(0).
\end{eqnarray}
where $E_s=-\Delta$ and $E_a=\Delta=E_s+2\Delta$.

An idealized absorption experiment would start with the ammonia atoms in thermal equilibrium. If the temperature is low enough, that means they are all in the ground state ($a=1,b=0$). One drops them through a cavity, where they interact with a magnetic field.

Conversely, for a maser, one first uses a Stern-Gerlach apparatus to create a beam of ammonia atoms which are all in the antisymmetric state ($b=1,a=0$) -- we will model this alignment later. One then drops them through a cavity, where they interact with an electromagnetic field. Although it is not always true in experiments, the first approximation we can make to understand the system is to imagine that the atoms spend a very short time in the cavity.

In a short time, we will have just small corrections to Eq.~(\ref{noeps}). Formally we will do our book-keeping by using $\epsilon$ as an expansion parameter. The maser sounds cool, so lets specialize to the case $a(0)=0$ and $b(0)=1$, to first order we make the ansatz
\begin{eqnarray}
b(t)&=& e^{-i E_a t} + \epsilon \delta b(t)\\
a(t)&=& \epsilon \delta a(t).
\end{eqnarray}
Substituting this functional form into the second of our equations of motion, we have
\begin{equation}
i\partial_t [ \epsilon \delta a(t)]=\epsilon \cos(\nu t) [e^{-i E_a t}+ \epsilon \delta b(t)] + E_s [ \epsilon \delta a(t)].
\end{equation}
Throwing away terms which are quadratic or higher in $\epsilon$ yields
\begin{equation}\label{ode}
(i\partial_t -E_s) \delta a(t)= \cos(\nu t) e^{-i E_a t}
\end{equation}
with the boundary condition that $\delta a(t=0)=0$. Next day we will solve this equation.