Two-Level Systems

Suppose we’ve got a physical system with only two linearly independent states.

(Classic example? LASER.)

Just go with me here — an electron has to be in one of these two: state 1, or state 2. That’s it. Those are the options.

So the state function $\Psi$ — let’s call it “some vector”:

$$|f\rangle$$

And we’ll let the basis vectors for state 1 and state 2 be:

$$|1\rangle=\begin{pmatrix}1\\0\end{pmatrix},\quad |2\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$

I’m just declaring this!!

(Not “this is what it IS!!!” — more like “let’s say it’s this.” Big difference.)

These two are orthogonal, so they’ll do the job nicely.

Then the thing that captures the state of the whole system (the alllll-encompassing total info)

$$|f\rangle$$

is gonna be a linear combination of the basis states it can possibly live in:

$$|f\rangle = a|1\rangle + b|2\rangle$$

And of course, $|f\rangle$ has to be normalizable, so:

$$a^{2} + b^{2} = 1$$

That condition rides along!!!

Oh, and also — it’s pretty obvious that $a^2$ and $b^2$ are the probabilities of being in each respective state!!!!

OK now the actual problem starts.

“If this physical system starts in state 1 at $t=0$, what state is it in at time $t$?????”

Alright, time to roll out the Schrödinger equation.

But — we need the time-dependent one! So:

$$i\hbar \frac{d}{dt}|f\rangle = \hat{H}|f\rangle$$

This has gotta be satisfied. And — since state 1 and state 2 are deeefiniiitely states with energy eigenvalues, let me first write down the Time Independent Schrödinger Equation (the time-free version):

$$\hat{H}|f\rangle = E|f\rangle$$

And the Hamiltonian operator $\hat{H}$!

Since we’ve got 2 states, it’s a 2×2 matrix. And since operators in QM are Hermitian matrices, let’s just assume the most general, dead-simple Hermitian operator we can get away with.

Hermitian means: take the transpose, take the complex conjugate, and you get yourself back. So:

$$\begin{pmatrix}h&g\\g&h\end{pmatrix}$$

Let’s roughly assume this and crank through it.

$$\begin{pmatrix}h&g\\g&h\end{pmatrix}|f\rangle = E|f\rangle$$$$\begin{pmatrix}h-E&g\\g&h-E\end{pmatrix}|f\rangle = 0$$

Which means: this matrix’s inverse must NOT exist.

$$\therefore \left|\begin{matrix}h-E&g\\g&h-E\end{matrix}\right| = 0$$$$(h-E)^{2} - g^{2} = 0$$$$(h-E+g)(h-E-g) = 0$$$$\therefore E = h+g \quad \text{or} \quad h-g$$

Got the eigenvalues!!!

Now let me grab the eigenvectors that give us those eigenvalues!!!

$$\begin{pmatrix}h&g\\g&h\end{pmatrix}X = (h\pm g)X$$$$\begin{pmatrix}h&g\\g&h\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix} = (h\pm g)\begin{pmatrix}\alpha\\\beta\end{pmatrix}$$$$\begin{cases}h\alpha + g\beta = (h\pm g)\alpha\\g\alpha + h\beta = (h\pm g)\beta\end{cases}$$$$\to \alpha = \pm\beta$$$$\therefore X = \begin{pmatrix}\alpha\\\beta\end{pmatrix} = \begin{pmatrix}\alpha\\\alpha\end{pmatrix} \text{ and } \begin{pmatrix}\alpha\\-\alpha\end{pmatrix} = \begin{pmatrix}1\\1\end{pmatrix} \text{ and } \begin{pmatrix}1\\-1\end{pmatrix}$$

So the eigenvectors of $\hat{H}$ come out to $(1,1)$ and $(1,-1)$ — but those don’t have magnitude 1.

Let’s normalize them so they’re nice unit vectors, all orthonormal and pretty:

$$\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix} \text{ and } \begin{pmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix}$$

There we go.

OK and we said the system starts in state 1 at $t=0$, so:

$$|f(0)\rangle = \begin{pmatrix}1\\0\end{pmatrix}$$

Expressed in terms of our energy eigenvectors:

$$\begin{pmatrix}1\\0\end{pmatrix} = \frac{\sqrt{2}}{2}\left(\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\end{pmatrix} + \begin{pmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix}\right)$$

Now slap on the time-dependent factor and boom — it becomes time-dependent!

$$|f(t)\rangle = \varphi(t)|f(0)\rangle = e^{-i\frac{E}{\hbar}t}\cdot \frac{\sqrt{2}}{2}\left(e_{1} + e_{2}\right)$$

<wrote it as $e_1$, $e_2$ ‘cause I was too lazy to spell the whole thing out… we good?>

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Originally written in Korean on my Naver blog (2015-08). Translated to English for gdpark.blog.