Normalization

So the thing I just pulled off a second ago —

$$\int_{-\infty}^{\infty}{\rho(x)\,dx} = 1$$

pinning down that constant $A$ that makes this hold…

OK so here’s the deal. Once you realize that

$$|\psi|^2$$

is doing duty as a probability density function, well — for it to actually behave like one, we need

$$\int_{-\infty}^{\infty}{|\psi|^2\,dx} = 1$$

to be true. And the way you make that happen? Just tune a constant out front so the total area comes out to 1. That whole “squeeze it with a constant until the magnitude is 1” move is what people call normalization.

The book literally spells it out: “Without this condition, statistical interpretation is meaningless.”

OK cool. So, broadly — we know what normalization is.

But the professor kept hammering this in lecture, and the book says the same thing: the process here is the thing you should actually sit and think about.

So… what are we supposed to be thinking about?

Let’s see. $\psi$ is a function of position $x$ and time $t$. Say we normalize at $t = 0$. Fine.

But — hold on. A little time ticks by. Do we have to go back and normalize again?

Wait wait wait wait.

And the answer is — nope. Nuh-uh. Nononono absolutely not.

Why???????????

The secret is hiding inside the Schrödinger equation.

Just like in classical mechanics every secret is stashed inside $F = ma$, in quantum mechanics every secret lives inside the Schrödinger equation. That’s the whole game.

OK let’s actually check it. Say $|\psi|^2$ has already been normalized. Then

$$\int_{-\infty}^{\infty}{|\psi|^2\,dx} = 1$$

holds. The question is: as time rolls on, does

$$\int_{-\infty}^{\infty}{|\psi|^2\,dx} = 1$$

juuuust keep on being true? Forever?

Way to check: compute

$$\frac{d}{dt}\left(\int_{-\infty}^{\infty}{|\psi|^2\,dx}\right)$$

and if this thing equals 0, we can say yes, we’re good.

Let’s go.

$$\begin{aligned} \frac{d}{dt}\!\left(\int_{-\infty}^{\infty}{|\psi|^2\,dx}\right) &= \int_{-\infty}^{\infty}{\frac{\partial}{\partial t}\!\left(|\psi|^2\right)dx} \\[4pt] &= \int_{-\infty}^{\infty}{\frac{\partial}{\partial t}\!\left(\psi^*\psi\right)dx} \\[4pt] &= \int_{-\infty}^{\infty}{\left(\frac{\partial\psi^*}{\partial t}\psi + \psi^*\frac{\partial\psi}{\partial t}\right)dx} \end{aligned}$$

Now reach for Schrödinger:

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

solve for $\partial\psi/\partial t$:

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

And then, by conjugating, automatically:

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

Plug both of these back in:

$$\begin{aligned} \frac{d}{dt}\!\left(\int_{-\infty}^{\infty}{|\psi|^2\,dx}\right) &= \int_{-\infty}^{\infty}{\left(\frac{\partial\psi^*}{\partial t}\psi + \psi^*\frac{\partial\psi}{\partial t}\right)dx} \\[4pt] &= \int_{-\infty}^{\infty}{\left(\left(-\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{i}{\hbar}V\psi^*\right)\psi + \psi^*\left(\frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{i}{\hbar}V\psi\right)\right)dx} \\[4pt] &= \int_{-\infty}^{\infty}{\left(-\psi\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{i}{\hbar}V\psi^*\psi + \frac{i\hbar}{2m}\psi^*\frac{\partial^2\psi}{\partial x^2} - \frac{i}{\hbar}V\psi\psi^*\right)dx} \\[4pt] &= \int_{-\infty}^{\infty}{\left(-\psi\frac{i\hbar}{2m}\frac{\partial^2\psi^*}{\partial x^2} + \frac{i\hbar}{2m}\psi^*\frac{\partial^2\psi}{\partial x^2}\right)dx} \\[4pt] &= \frac{i\hbar}{2m}\int_{-\infty}^{\infty}{\left(\frac{\partial^2\psi}{\partial x^2}\psi^* - \psi\frac{\partial^2\psi^*}{\partial x^2}\right)dx} \\[4pt] &= \frac{i\hbar}{2m}\int_{-\infty}^{\infty}{\frac{d}{dx}\!\left(\frac{\partial\psi}{\partial x}\psi^* - \psi\frac{\partial\psi^*}{\partial x}\right)dx} \\[4pt] &= \frac{i\hbar}{2m}\left[\frac{\partial\psi}{\partial x}\psi^* - \psi\frac{\partial\psi^*}{\partial x}\right]_{-\infty}^{\infty} \end{aligned}$$

And now the finishing move: the wave functions we actually care about always die off to 0 out at infinity. So that bracket evaluated at $\pm\infty$? Both ends zero.

$$= 0$$

How clean is that?!

So — once you’ve normalized, anytime, anywhere — from there on out, as time marches forward, the thing just stays normalized. No do-over needed.

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