I am going to repeat the argument from last day, then solve the resulting equation.
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 $\delta a(t=0)=0$. Remembering your differential equations course,
the solution of this equation is of the form
\begin{equation}
\delta a(t) = A [e^{i (\nu-E_a ) t}-e^{-i E_s t}] + B[ e^{-i (\nu+E_a) t}-e^{-i E_s t}].
\end{equation}
Substituting this into Eq.~(\ref{ode}) yields
\begin{equation}
(E_a-\nu-E_s) A e^{i(\nu-E_a)t} + (E_a+\nu-E_s) B e^{-i(\nu+E_a)t} =\frac{e^{i(\nu-E_a)t}}{2}+\frac{e^{-i(\nu+E_a)t}}{2}.
\end{equation}
For this to be true at all times, we must have
\begin{eqnarray}
A&=& \frac{1}{2 (E_a-\nu-E_s) }\\
B&=& \frac{1}{2 (E_a+\nu-E_s) }
\end{eqnarray}
If $\nu\approx E_s-E_a$, then $B$ is much bigger than $A$, and we can neglect $A$.
We then have
\begin{equation}
a(t)=\frac{\epsilon}{ 2\delta} e^{-i(\nu +E_s+E_a)t/2} [e^{-i\delta t/2}-e^{i\delta t/2}]
\end{equation}
where $\delta=E_a+\nu-E_s$ is the ``detuning" from resonance. The probability that the ammonia atom added a photon to the cavity mode is then
\begin{equation}\label{prob1}
P= 4 \epsilon^2 t^2 \frac{\sin^2 (\delta t/2)}{(\delta t/2)^2}.
\end{equation}
Our approximations are of course only consistent if $P\ll1$. Regardless, we see we have a very narrow band-width amplifier, whose bandwidth is set by the amount of time the atoms spends in the cavity.
Absorption of light
The same argument, in reverse tells us what happens if we start in the ground state, and shine microwaves on the molecules, and the expression ends up being the same. Since Eq.~(\ref{prob1}) is sharply peaked, we get only narrow absorption bands -- and the frequencies tell us the differences between the energies of the eigenstates.
If we have a source which is broad band compared to the width of the peak, then the total probability of absorbing photons of any frequency is
\begin{eqnarray}
P_{\rm total} &=& 4 \epsilon^2 t^2 \int \frac{\sin^2 (\delta t/2)}{(\delta t/2)^2} d\omega\\
&=& 8\pi\epsilon^2 t.
\end{eqnarray}
This has the reasonable result that the absorption probability is proportional to the amount of time you shine photons on the system. In other words, there is a constant rate. This result (with appropriate bells and whistles added) is known as "Fermi's Golden Rule." The tricky part in a real atomic calculation is figuring out how to relate $\epsilon$ to the field strength and the atomic parameters.
Lasers
The key step in making a maser is the Stern-Gerlach step: you need to set the initial conditions so that $|b|^2>|a|^2$. This is known as "population inversion". Although we have not explained yet how to do this Stern-Gerlach filtering, I can tell you there is no optical analog. Thus the tricky part of a laser is producing the population inversion. The most popular scheme uses a 3-level atom with a structure where the top level rapidly decays to the middle, but the middle only slowly decays to the ground. You pump from the ground state to the most excited state, and atoms accumulate in the middle state. The frequency of the amplified light corresponds to the difference between the energy of the middle and ground states.
Familiar laser pointers are "injection lasers" or "diode lasers." They involve having a p-type semiconductor in contact with an n-type semiconductor. Electrons an holes are injected into the device (this is the "excited state"). They then recombine, emitting light (decay to the ground state). This yields a "light emitting diode". Such a diode is a fairly large bandwidth optical amplifier (so fits the definition of the acronym LASER).
What we usually call a laser, adds a cavity, so that there is one mode which gets preferentially amplified. You then spontaneously generate copious amounts of light at that frequency.
