Absorption, Emission, and Stimulated Emission

Alright, diving right in.

Sinusoidal Perturbation

Time to make the perturbation Hamiltonian concrete.

Let’s throw in a cosine and actually put the theory from last time to use:

$$H'(t) = V(r)\cos wt$$

That’s the setup.

(You can basically think of this kind of $H'$ as light, by the way. Foreshadowing.)

So:

$$H'_{ab} = \langle \psi_a | H' | \psi_b \rangle = V_{ab}\cos wt \quad \left( V_{ab} = \langle \psi_a | V(r) | \psi_b \rangle \right) \\ H'_{ba} = \langle \psi_b | H' | \psi_a \rangle = V_{ba}\cos wt \quad \left( V_{ba} = \langle \psi_b | V(r) | \psi_a \rangle \right)$$

Agreed? Good.

And before we turned the perturbation on, $C_a = 1$, $C_b = 0$ — i.e. the system was just siiitting there in state $a$ the whole time.

Plug it in:

$$\dot{C_a}(t) = 0 \\ \dot{C_b}(t) = -\frac{i}{\hbar} V_{ba} (\cos wt) e^{iw_0 t}$$$$\text{So,} \quad \begin{cases} C_a : \text{constant} \\ \\ \dfrac{dC_b}{dt} = -\dfrac{i}{\hbar} V_{ba} (\cos wt) e^{iw_0 t} \end{cases} \\ \to \quad C_b = -\frac{i}{\hbar} V_{ba} \int_0^t \left( \cos wt' \right) e^{iw_0 t'} dt' \\ = -\frac{i}{\hbar} V_{ba} \int_0^t \left( \frac{e^{iwt} + e^{-iwt}}{2} \right) e^{iw_0 t'} dt' \\ = -\frac{iV_{ba}}{2\hbar} \int_0^t \left( e^{i(w_0+w)t'} + e^{i(w_0-w)t'} \right) dt' \\ = -\frac{iV_{ba}}{2\hbar} \left[ \frac{1}{i(w_0+w)} e^{i(w_0+w)t'} + \frac{1}{i(w_0-w)} e^{i(w_0-w)t'} \right]_0^t$$$$\therefore \quad C_b(t) = -\frac{V_{ba}}{2\hbar} \left( \frac{e^{i(w_0+w)t} - 1}{w_0+w} + \frac{e^{i(w_0-w)t'} - 1}{w_0-w} \right)$$

* Heads up — $w$ is a variable here!

OK, now let’s actually try to read what this $C_b(t)$ is telling us.

If we hit a two-level system with a perturbation of $V(r)\cos wt$,

$C_a$, the amplitude for being in state $a$, $C_b$, the amplitude for being in state $b$,

evolve in time like the expressions above.

<The thing where $C_a$ is constant but $C_b$ keeps changing —>

<

$$|C_a|^2 + |C_b|^2 = 1$$

doesn’t that break this condition?? you might be yelling at the screen right now.>

(Yeah, so there’s something I didn’t really emphasize. What we just computed is only the 1st-order correction.)

So the above line shouldn’t actually be written with equality — it should be

$\cong$

Strictly speaking.

The correction to $C_a$ is zero at 1st order, but it is not zero at 2nd order!!

OK so — now, when the frequency of the perturbation is close to $w_0$, versus when it isn’t…

those two cases give wildly different values of $C_b$. Night and day.

Because that second term blows up when $w \sim w_0$.

So let’s zoom in on exactly that case — when the driving frequency $w$ of the cosine satisfies

$$w \cong w_0$$

and look at $C_b$ while ignoring the first term (it’s small compared to the second).

$$\therefore \quad C_b(t) = -\frac{V_{ba}}{2\hbar} \left( \frac{e^{i(w_0-w)t} - 1}{w_0-w} \right) = -\frac{V_{ba}}{2\hbar} \left( \frac{e^{i(w_0-w)t} - 1}{w_0-w} \right) \frac{e^{-i(w_0-w)\frac{t}{2}}}{e^{-i(w_0-w)\frac{t}{2}}} \\ = -\frac{V_{ba}}{2\hbar} \left( \frac{e^{i(w_0-w)\frac{t}{2}} - e^{-i(w_0-w)\frac{t}{2}}}{(w_0-w)\, e^{i(w_0-w)\frac{t}{2}}} \right) = -\frac{V_{ba}}{2\hbar(w_0-w)} e^{-i\frac{(w_0-w)}{2}t} \left( \frac{e^{i(w_0-w)\frac{t}{2}} - e^{-i(w_0-w)\frac{t}{2}}}{1} \right) \\ = -\frac{V_{ba}}{2\hbar(w_0-w)} e^{-i\frac{(w_0-w)}{2}t} \left( \frac{e^{i(w_0-w)\frac{t}{2}} - e^{-i(w_0-w)\frac{t}{2}}}{2i} \right) 2i \\ = -\frac{2iV_{ba}}{2\hbar(w_0-w)} e^{-i\frac{(w_0-w)}{2}t} \sin\!\left(\frac{w_0-w}{2}t\right) \\ = \frac{V_{ba}}{i\hbar(w_0-w)} e^{-i\frac{(w_0-w)}{2}t} \sin\!\left(\frac{w_0-w}{2}t\right)$$

This thing — this right here — $|C_b(t)|^2$ is the probability of finding the system in state $b$ after $t$ seconds!!!

Oh! At $t=0$ we were in state $a$!!!

Then $t$ seconds later, there’s a chance we’ve hopped over to $b$.

So the square of that guy? That’s exactly the Transition Probability, right?!

$$|C_b(t)|^2 : \text{Transition Probability} \\ = P_{a \to b}(t) = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\left(\dfrac{w_0-w}{2}t\right)}{(w_0-w)^2}$$

* The exponential out front dies once we multiply by the complex conjugate. Gone.

Graph it:

$$\text{Width:} \quad \frac{4\pi}{t} \quad \left( \because \text{it's a } \sin^2 \right) \\ \text{Height:} \quad \left(\frac{V_{ab}t}{2\hbar}\right)^2$$

But that’s for some finite $t$.

As $t$ gets bigger, and bigger, and bigger ($t\to\infty$),

"

$$\text{transition only happens at } w = w_0.$$

"

$$\text{The atom absorbs only what satisfies } w = w_0, \text{ and that's when the transition happens!}$$

We’ll do this more rigorously later — this is the Fermi Golden Rule.

* Wait!!!!!

Even if it’s not exactly two levels, for a system that has nice discrete ’lines’ like this, think gas.

That is — picture a gas absorbing light.

Solids don’t have sharp levels like this; they have energy bands, so the story there is a bit different.

Emission & Absorption of Radiation

Now let’s take what we just did and crank it up a level.

Well — not really “up a level” so much as: let’s get more concrete about what’s actually going on.

Concrete about what, exactly?

Before, we were kind of hand-wavy with $V_{ab}$, $V_{ba}$ as just “some matrix elements.”

Now we’re going to actually spell out what $V(r)$ is, and see absorption and emission fall out.

Time to admit it out loud: $H'(t)$ is light.

Yep. We’re shooting light at a two-level system!!!

Light is an electromagnetic wave.

EM waves carry both $E$ and $B$ together, sure, but — remember in E&M class? The prof was basically like, “come on, stop dragging $B$ around, just write $E$, it’s fine.”

Just picture $B$ tagging along beside $E$. That’s the vibe.

So the claim was: writing only the electric field is enough.

OK, so only the electric field:

The electric field that some ‘atom’ feels is

$$\vec{E} = \bigl(E(r)\cos wt\bigr)\hat{k}$$

Makes sense — the atom’s going to interact with the EM wave.

But if the wavelength is waaaay bigger than the atom itself ($\lambda \gg a$),

then from the wave’s perspective, the atom is basically a point — might as well not be there in terms of spatial variation.

So we can just say

$$\vec{E} = E_0 \cos wt\, \hat{k}$$

(This is assuming no interaction between the atom and the EM wave — i.e. we’re just writing down the field the atom sees.)

Now, sneak alert: another assumption just slipped in.

Monochromatic. Single frequency! One $w$, nothing else.

And let me throw in one more while we’re at it: polarization along $z$.

OK now we can write down our potential $V$ concretely.

$$V = -\int \vec{F}\cdot d\hat{r} = -\int qE_0\cos wt\,\hat{k}\cdot d\hat{z} = qE_0\cos wt \int_0^z dz = qE_0 z\cos wt$$

Feed this $V$ into the two-level system machine:

$$H'_{ba} = \langle \psi_b | H' | \psi_a \rangle = -qE_0\cos wt\, \langle \psi_b | z | \psi_a \rangle \\ = -pE_0\cos wt \\ \quad \left( p \equiv q\langle \psi_b | z | \psi_a \rangle \right)$$$$\text{So effectively, } V_{ba} = qE_0.$$

And hey — now we can actually compute this:

$$P_{a \to b}(t) = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\left(\dfrac{w_0-w}{2}t\right)}{(w_0-w)^2} = \frac{|pE_0|^2}{\hbar^2} \frac{\sin^2\!\left(\dfrac{w_0-w}{2}t\right)}{(w_0-w)^2}$$

Still the transition probability.

(And yeah, it only absorbs the frequency $w_0$.)

* Reference)

The wavefunctions of a two-level system

$$\psi_a, \quad \psi_b$$

turn out to be either even or odd functions!!!!!

Which is why

$$H'_{ii} = 0$$

is legit.

And also why

$$V_{ab} = V_{ba}$$

holds!!!

OK so here’s what I really wanted to get to. If instead we set

$$C_a(0) = 0 \\ C_b(0) = 1$$

and go compute

$$P_{b \to a}(t)$$

!!!! (the emission probability)

it’s exactly the same as

$$P_{a \to b}.$$

But since this is light coming out because we shone light in, it’s called stimulated emission (a.k.a. induced emission)!!!!!!!!!!!

Was that not emphatic enough?

The induced transition probability and the induced emission probability are the same.

The thing that exploits this principle of stimulated emission? The laser. Yep, that laser.

And LASER is an acronym — take the first letter of each of the following:

LASER : Light Amplification by Stimulated Emission of Radiation

Quick note on spontaneous emission too.

So there’s not just stimulated emission — as I sort of hinted — there’s also this other thing called “spontaneous emission.”

The idea: the system is just chillin’… and then at some random moment, on its own, it drops down and spits out a photon.

That’s spontaneous emission.

Why does that even happen?

The usual story is quantum fluctuation.

Because of quantum fluctuation, there’s energy floating around even at 0 K,

and — some people run with this — the universe itself popped into existence from nothing, so the fact that even at 0 K you still have some tiny bit of energy hanging around is kinda natural… lol lol lol lol lol lol lol

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Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.