The Free Particle

OK so this time: the free particle. $V(x) = 0$ everywhere. Not trapped in a box, not sitting in some potential well — just… loose, flying around in empty space.

Ready for another round of existential crisis? heh

Let’s go.

We’re still grinding through the Schrödinger equation case by case, split up by what $V(x)$ looks like, and this is case #3.

The case where $V(x) = 0$ for all $x$ — the free particle.

So plug $V = 0$ into the (time-independent) Schrödinger equation:

$$-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}} = E\psi \\ \frac{d^{2}\psi}{dx^{2}} = -\frac{2mE}{\hbar^{2}}\psi$$

$E$ is a positive constant, so let’s rename that whole blob in front real quick:

$$\frac{d^{2}\psi}{dx^{2}} = -k^{2}\psi \quad \left( k \equiv \sqrt{\frac{2mE}{\hbar^{2}}} \right)$$

Wait — this ODE is kind of easy?!

$$\psi(x) = Ae^{ikx} + Be^{-ikx}$$

Second derivative has to give us a negative sign back, so the exponent has to be imaginary — hence the $i$. heh.

heh heh.

But… hmm. Something’s off.

$\psi(x)$ is supposed to go to zero at $x = \infty$ and $x = -\infty$. (Every wave function we’ve been messing with has to do that, right?!)

But… there is literally no choice of $A, B$ that makes $Ae^{ikx} + Be^{-ikx}$ vanish at infinity. (T_T)

And there’s a second problem too, and to get there we need to push the discussion a little further than we normally would.

Remember — for the time-independent Schrödinger equation, we had

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

so we can just multiply by the $\Phi(t)$ we already found and get the full time-dependent thing. Let’s actually do that:

$$\begin{aligned} \psi(x,t) &= \psi(x)\Phi(t) \\ &= \left(Ae^{ikx} + Be^{-ikx}\right)\left(e^{-i\frac{E}{\hbar}t}\right) \\ &= Ae^{i\left(kx - \frac{E}{\hbar}t\right)} + Be^{-i\left(kx + \frac{E}{\hbar}t\right)} \end{aligned}$$

Now to see the wave interpretation more nakedly, let’s stop thinking of $k$ as “that constant from earlier.” In wave-land, $k$ is the wave number:

$$\frac{2\pi}{\lambda}$$

That is — “how many wavelengths of $\lambda$ meters fit inside $2\pi$ meters?” That kind of thing.

OK. Going back to

$$\psi(x,t) = Ae^{i\left(kx - \frac{E}{\hbar}t\right)} + Be^{-i\left(kx + \frac{E}{\hbar}t\right)}$$

let’s also rewrite those $E/\hbar$ guys in terms of $k$:

$$\begin{aligned} k &= \frac{\sqrt{2mE}}{\hbar} \\ k^{2} &= \frac{2mE}{\hbar^{2}} \\ \frac{E}{\hbar} &= \frac{k^{2}\hbar}{2m} \end{aligned}$$$$\psi(x,t) = Ae^{i\left(kx - \frac{k^{2}\hbar}{2m}t\right)} + Be^{-i\left(kx + \frac{k^{2}\hbar}{2m}t\right)} \\ = Ae^{ik\left(x - \frac{k\hbar}{2m}t\right)} + Be^{-ik\left(x + \frac{k\hbar}{2m}t\right)}$$

Now — look at that. $\psi(x,t)$ is a function of

$$\left(x \pm \text{something} \cdot t\right)$$

That exact form.

So yeah — $\psi$ really is a wave function!! (Remember from E&M 2, electromagnetic waves — when we derived the classical wave equation from a rope wave, that was exactly the form we got!)

And that “something” is the speed of the wave.

So the wave is moving at speed $\frac{k\hbar}{2m}$. Cool!

But wait —

$$\begin{aligned} k^{2} &= \frac{2mE}{\hbar^{2}} \\ E &= \frac{\hbar^{2}k^{2}}{2m} = \frac{1}{2}mv^{2} \\ \therefore \quad v &= \frac{\hbar k}{m} \end{aligned}$$

If we back out the velocity from the energy using classical mechanics…

It comes out to be exactly twice the wave’s speed. Hold on. That’s half. The wave speed is half the classical particle speed. What??

OK so the time-independent Schrödinger equation solution has given us two things that don’t line up:

1. Can’t normalize it.
2. Velocity mismatch.

At this point we have no choice but to conclude: the assumption we made in setting up the equation is wrong.

Specifically — “independent of time” is wrong.

I had been told “independent of time” means a stationary state. So you could also read this as: a free particle has no stationary state. (…what am I even saying, is the mental breakdown kicking in)

OK OK OK OK. Time-independence is out. So what does the general solution of the time-dependent Schrödinger equation look like?

“The general solution of the time-dependent Schrödinger equation is a linear combination of the solutions of the time-independent Schrödinger equation.” …yeah.

A wave built out of superposing a whole bunch of waves with different wave numbers!!

(And then the book drops this on us)

$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{i\left(kx - \frac{\hbar k^{2}}{2m}t\right)}dk$$

The $\frac{1}{\sqrt{2\pi}}$ out front is just there to make the normalization bookkeeping nicer later.

$\Phi(k)$ is the density of states.

(Before it was $c_n$, but here it depends on $k$ rather than on some quantum number $n$?!

But it’s the same thing. Because $k$ determines the energy of the wave, right?? (actually it’s $n$ that determines $k$, heh heh heh)

Anyway — that’s why $k$ is the important variable in a wave.)

“OK but what IS $\Phi(k)$, concretely?!?!”

Answer: you find it by normalizing against the initial condition.

(Way back when we first learned normalization, the idea was “normalization shouldn’t change over time — meaning, you only have to do it once. That’s how it’s consistent with nature.” Right?)

So — normalize once, at the initial condition.

Plug in $t=0$:

$$\psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{ikx}dk$$

And the $\Phi(k)$ that’s consistent with this initial condition… boom, Fourier transform:

$$\Phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{-ikx}dx$$

Done. That’s how you pin it down.

Let’s run an example using the logic above.

Ex. 2.6: A free particle whose initial position is restricted to $-a < x < a$ has the following wave function at $t=0$:

$$\psi(x,0) = \begin{cases} A & (-a < x < a) \\ 0 & (\text{otherwise}) \end{cases}$$

$A$ and $a$ are positive real numbers. Find $\psi(x,t)$.

• First, normalize using $\psi(x,0)$:

$$\int_{-a}^{a}A^{2}dx = 2aA^{2} = 1 \\ \therefore \quad A = \frac{1}{\sqrt{2a}}$$

Now grind out $\Phi(k)$:

$$\begin{aligned} \Phi(k) &= \frac{1}{\sqrt{2\pi}}\int_{-a}^{a}\psi(x,0)e^{-ikx}dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{-a}^{a}\frac{1}{\sqrt{2a}}e^{-ikx}dx \\ &= \frac{1}{2\sqrt{\pi a}}\left[-\frac{1}{ik}e^{-ikx}\right]_{-a}^{a} \\ &= \frac{1}{2\sqrt{\pi a}}\frac{1}{ik}\left[-e^{-ika} + e^{ika}\right] \\ &= \frac{1}{k\sqrt{\pi a}}\frac{e^{ika} - e^{-ika}}{2i} \\ &= \frac{1}{k\sqrt{\pi a}}\sin(ka) \end{aligned}$$$$\Phi(k) = \frac{1}{k\sqrt{\pi a}}\sin(ka)$$

Got it.

Now plug back into

$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{i\left(kx - \frac{\hbar k^{2}}{2m}t\right)}dk$$

and:

$$\psi(x,t) = \frac{1}{\pi\sqrt{2a}}\int_{-\infty}^{\infty}\frac{\sin(ka)}{k}e^{i\left(kx - \frac{\hbar k^{2}}{2m}t\right)}dk$$

<Apparently there is no way to do this integral except on a computer… (T_T)>

OK now we have to come back to that velocity thing!!!

Normalization looks resolved, but why!!! does the velocity from classical mechanics not match the velocity you read off the wave?!

Honestly, I’d recommend reading the “phase velocity vs. group velocity” part of a modern physics book before coming back to the quantum mechanics text on this.

http://gdpresent.blog.me/220461554515

Modern Physics I Studied #11. Wave Properties of Particles

Up until just before, we learned that waves have particle-like properties. About light, known as electromagnetic waves, particles…

blog.naver.com

OK, one last little thing.

$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{i\left(kx - \frac{\hbar k^{2}}{2m}t\right)}dk$$

Let’s rename a variable so this visibly looks like a wave:

$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\Phi(k)e^{i(kx - \omega t)}dk$$$$\omega = \frac{\hbar k^{2}}{2m}$$

This is called the dispersion relation — the relationship between $\omega$ and $k$.

Electromagnetic waves have their own dispersion relation. Water waves have their own. Every kind of wave has a characteristic dispersion relation — and this is the one for matter waves.

And! In a wave, $\omega$ is equivalent to energy, and $k$ is equivalent to momentum.

(Properly speaking, the dispersion relation is really the relationship between momentum and energy, not wave number and angular frequency.)

$$\begin{aligned} E &= h\nu \\ 2\pi E &= h\omega \\ E &= \hbar \omega \end{aligned}$$$$\begin{aligned} k &= \frac{2\pi}{\lambda} \\ hk &= \frac{2\pi}{\lambda}h = 2\pi\frac{h}{\lambda} = 2\pi \cdot p \\ k &= \frac{2\pi}{h}p = \frac{p}{\hbar} \end{aligned}$$

Honestly — the free particle section in Griffiths feels pretty thin.

(Not just this part. The whole book is written to be easy-going, so… that’s kinda how it goes?)

I think I’ll need to actually study this properly again later. Sakurai? People say that book is good — I wonder what it’s like. heh heh, sigh.

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