The Time-Independent Schrödinger Equation, Part 2

OK so — when we see an equation, what do we do? We hunt for the $x$ that makes both sides equal, right?

And that $x$ we find — we call it a root, or a solution.

So what about a differential equation? Same deal — we’re hunting for a function $f(x)$ that makes both sides equal. And that $f(x)$ we find? Also called a solution.

Alright. Now we’ve run into the Schrödinger equation. Same game.

Let’s go find the wave function $\psi(x,t)$ that satisfies it!!!

…we’ll actually do that later.

For now, let’s make one big assumption: “the wave function is separable — not tangled up in time and space together.”

$$\psi (x,t) = \psi (x)\Phi (t)$$

This is what we’re assuming. Then we’ll go find a solution that satisfies the Schrödinger equation.

(We also need to… assume… that the potential energy $V$ is time-independent too… heh heh, because we’re still beginners here.)

This is basically the whole point of Chapter 2!

But honestly — this is the exact same thing we did back in E&M Chapter 3 with electric potential. Remember? We had

$$V(x,y) = X(x)Y(y)$$

We “assumed” the potential splits like that and cranked through Laplace’s equation. Literally almost identical calculation. Hella familiar.

OK, let’s actually do it. Plug the assumption into the Schrödinger equation:

$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} + V\psi$$

Substitute in

$$\psi (x,t) = \psi (x)\Phi (t)$$

and you get

$$\psi (x)\,i\hbar \frac{\partial \Phi (t)}{\partial t} = -\Phi (t)\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi (x)}{\partial x^{2}} + V(x)\psi (x)\Phi (t)$$

Now group the $x$ stuff on one side, the $t$ stuff on the other:

$$i\hbar \frac{1}{\Phi}\frac{\partial \Phi}{\partial t} = -\frac{\hbar^{2}}{2m}\frac{1}{\psi}\frac{\partial^{2}\psi}{\partial x^{2}} + V$$

Left side = something that only depends on $t$. Right side = something that only depends on $x$.

But these two sides are supposed to be equal. Equaaaaaaal. For all $x$ and all $t$.

The only way that works? Both sides have to be constants. If they weren’t constants, there’s no way they’d stay equal as $x$ and $t$ wiggle around independently. (Exact same argument from E&M, by the way.)

$$(\text{for reference})\\ At = B\\ Ct = Dx\\ E = Fx\\ G = H\\ \text{For both sides to always be equal, for all } x\text{, all } t!!!\\ \text{Both sides must be constants}!!!$$

OK so let’s say each side equals some constant. Call it $E$. (Why $E$? Totally random choice. No reason at all. Definitely not going to get confused later with “energy”…)

$$\text{Left side)}\\ i\hbar \frac{1}{\Phi}\frac{\partial \Phi}{\partial t} = E\\ \frac{\partial \Phi}{\partial t} = -\frac{i}{\hbar}E\cdot \Phi$$$$\text{Right side)}\\ -\frac{\hbar^{2}}{2m}\frac{1}{\psi}\frac{\partial^{2}\psi}{\partial x^{2}} + V = E\\ -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} + V\psi = E\psi$$

The first equation (the $t$ one) gives us $\Phi(t)$. The second (the $x$ one) gives us $\psi(x)$.

So… what’s the point of all this? -_- Why did we go through all that?

$$\psi (x,t) = (\text{function of } x) \times (\text{function of } t)$$

We assumed it splits like this (and also that $V$ only depends on $x$!), plugged it into the Schrödinger equation, and we can now solve for the “function of $x$” and “function of $t$” separately.

Meaning: multiply the two together and boom — we’ve got $\psi(x,t)$!!!!!!!!!!!!!!!!!!!!!

$$\frac{\partial \Phi}{\partial t} = -\frac{i}{\hbar}E\cdot \Phi$$$$\Phi (t) = Ce^{-\frac{i}{\hbar}E\cdot t}$$

First-order ODE — piece of cake!!!

Then

$$\psi (x,t) = \psi (x)\left( Ce^{-\frac{i}{\hbar}E\cdot t} \right)$$

And since $\psi(x)$ can also absorb a constant in front of it, let’s just lump all the constants together and roll them into $\psi(x)$ later. For $\Phi$, we’ll just set the constant to 1 and call it a day:

$$\Phi (t) = e^{-\frac{i}{\hbar}E\cdot t}$$

OK. The $t$ part — done.

From here on out, the real work is this:

$$\text{Right side)}\\ -\frac{\hbar^{2}}{2m}\frac{1}{\psi}\frac{\partial^{2}\psi}{\partial x^{2}} + V = E\\ -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} + V\psi = E\psi$$

And this — this is the time-independent Schrödinger equation!!!!!!

(Yeah, we totally saw this in the last post~ heh.)

I want to go solve it properly, but we kinda need to know what $V(x)$ actually is first — remember, we said it only depends on $x$. Without knowing $V$, we’re stuck.

So the plan going forward is basically: split into cases based on $V(x)$.

For this kind of $V(x)$, we get this $\psi(x)$.

For that kind of $V(x)$, we get that $\psi(x)$.

· · ·

Honest confession though — every time I do separation of variables, I feel a rock on my chest. ;_; It feels like maybe I shouldn’t be doing this?

Because there really are cases where separation of variables just… doesn’t work. And we’re only looking at the somewhat-special cases where it does. And there aren’t that many of even those. lol

BUT — the book has a section that’s basically like “no no, separation isn’t meaningless!!!!!!!” with three reasons. Let me transcribe those here. (Two are physical interpretations, one is mathematical.)

1. Separable solutions represent “stationary states.”

The wave function itself obviously changes with time, yeah. But the probability density — what the wave function actually represents physically — doesn’t change with time.

$$\psi (x,t) = \psi (x)e^{-\frac{i}{\hbar}E\cdot t}$$

and

$$\psi^{*}(x,t) = \psi^{*}(x)e^{\frac{i}{\hbar}E\cdot t}$$$$\left| \psi (x,t) \right|^{2} = \left( \psi (x)e^{-\frac{i}{\hbar}E\cdot t} \right) \cdot \left( \psi^{*}(x)e^{\frac{i}{\hbar}E\cdot t} \right) = \psi(x)\psi^{*}(x)$$

The $\Phi(t)$ just… cancels. Clean. And the same thing happens when you compute expectation values — the $\Phi(t)$ bit dies off the same way. So when you’re writing down the wave function, you can basically get away with just omitting $\Phi(t)$.

That means $\langle x \rangle$ comes out as a constant. $\langle p \rangle$ comes out as $0$.

In other words — stationary state. Nothing moves. That’s what it represents.

2. Separable solutions are states where the total energy has a definite, measured value.

(Flashback to classical mechanics: total energy $E$ — aka mechanical energy? — sometimes gets written as the Hamiltonian. $H(x,p) = p^2/2m + V(x)$.)

But in quantum mechanics, $p$ is an operator, so

$$H = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}} + V(x)$$

which means the time-independent Schrödinger equation can be written as

$$H\psi = E\psi$$

(Ohhhh. Now I see why we “happened to” call the constant $E$. Sneeeakily foreshadowed.)

And this is where the linear algebra post pays off. Skipping the full linear algebra background: think of the operator $H$ as a linear map, and we’re applying it to the function $\psi(x)$. What do we get?

linear map $\times$ $\psi(x)$ = constant $\times$ $\psi(x)$.

In linear algebra, when you see $Lx = \lambda x$ for a linear map $L$ — $x$ is the eigenvector and $\lambda$ is the eigenvalue.

Same exact deal here. In quantum mechanics, with the same setup, $E$ is the eigenvalue and $\psi(x)$ is the eigenvector. (Or actually, since it’s a function, let’s call it an eigenfunction.)

3. When you solve the time-independent Schrödinger equation, you actually get infinitely many solutions.

Which means the general solution is a linear combination of all of them:

$$\psi (x,t) = \sum C_{n}\psi_{n}e^{-i\frac{E_{n}}{\hbar}t}$$

We’ll build this up slowly over the next few posts. And since the whole thing is literally the same as E&M, there’s nothing scary about it… heh heh heh heh heh.

From the next post on, we’ll keep rolling — we’ll look at what $\psi(x)$ looks like for different choices of $V(x)$, and try to grab the physical meaning on the way.

Let’s go let’s go!!!

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