The Delta-Function Potential [Quantum Mechanics I Studied #11]

We’re working through the time-independent Schrödinger equation case by case, sorted by what V looks like.

And now — drumroll — potential #4: a potential shaped like a delta function.

(The finite square well is exactly~~!! the same vibe as the delta function potential discussion!!!!)

Before we dive in for real, juuust a tiny bit of vocab on bound states and scattering states.

1. Bound state

Equation from the original notes

A bound state is when

$E<\left| V(\pm \infty ) \right|$

something like that, or it could be $E<0$,

and honestly most(?) of what we’ve done so far has been bound states —

infinite square well, harmonic oscillator, all bound states.

Why “bound”? Because the particle’s $E$ is less than $V(x)$,

so it absolutely cannot escape the walls of $V(x)$. It’s stuck. Trapped. Bound. …heh

That’s the picture, right!?!? (Is it really trapped though? hehehe~)

2. Scattering state (scattering… what now?)

Equation from the original notes

Scattering state is just the opposite of bound.

$E>\left| V(\pm \infty ) \right|$

that’s the deal,

and obviously when $V=0$, the case $E>0$ is the same thing,

and out of stuff we’ve covered, the free particle is a scattering state, heh

But does $E$ have to be greater than $V$ everywhere for it to be a scattering state??

Equation from the original notes

What about something like this..

Apparently this counts as a scattering state too~!!!

OK, on to the delta function.

The definition of the delta function is

$\delta (x)\quad =\quad \begin{cases}{\infty \quad (x=0)}\{0\quad (x\neq 0)}\end{cases}$

$x$ spikes ridiculously hard!!! at one specific spot, and everywhere else it’s just nothing!!!!

(Like — picture a person who reacts to nothing~~ in life but goes absolutely wild for money!!! If you wrote that person as a delta function:)

$\delta (x-\text{money})\quad =\quad \begin{cases}{\infty \quad (x=\text{money})}\{0\quad (x\neq \text{money})}\end{case$

So basically it’s a function people invented when they wanted to capture that kind of behavior~!!

But then this dude named Dirac picked it up and made it actually mean something.

“Hey, c’mon — where in nature does anything respond to one specific thing with an infinite response (-_-)? It wouldn’t be infinite, it’d just be a ‘really high value’~~~”

$\int _{-\infty}^{\infty}{\delta (x)}dx\quad =\quad 1$

A delta function that satisfies this,

$\delta (x)$

is what we call the Dirac delta function.

What it really means is

$\int _{-\infty}^{\infty}{f(x)\delta (x-a)}dx\quad =\quad f(a)$

OK so before we dive in for real, let me quickly say why~~~ we even bother with the case where the potential is

$\delta (x)$

Two pictures, that’s all.

Equation from the original notes

If you take an electron moving inside some material

and just sketch the energy as a graph,

Equation from the original notes

ta-daaa~~~

OK, now for real.

In the Schrödinger equation we need to plug in $V(x)$ as the function — but instead of just plugging in

$\delta (x)$

let’s be a tiny bit more general and write

$V(x)\quad =\quad -\alpha \delta (x)$

You can think of $\alpha$ as something like the “properties of the material.”

Like — the depth of the potential in uranium is different from the depth in hydrogen, that kind of vibe???

The minus sign just means the delta function bulges downward~~

So the time-independent Schrödinger equation becomes

$-\frac {\hbar^{2}}{2m}\frac {d^{2}\psi}{dx^{2}}\quad -\alpha \delta (x)\psi \quad =\quad E\psi$

We’re going to look at $E>0$ and $E<0$ separately.

Let’s start with $E<0$.

What does a bound electron look like?

Equation from the original notes

Something like this, right~!?!?

Then, just like with the infinite square well, let’s split by region and gogogo

$X<0\quad ,\quad V=0\ -\frac {\hbar^{2}}{2m}\frac {d^{2}\psi}{dx^{2}}\quad =\quad E\psi \ \frac {d^{2}\psi}{dx^{2}}\qua$

Now for $x>0$!!!!!

$\therefore \quad \psi (x)\quad =\quad Fe^{-Kx}$

Same logic as above — to keep it from blowing up at $x=\infty$, only that term survives.!?

OK, now the boundary! !!!!

$\psi(x)$ has to be continuous at $x=0$!!!!!

$\psi (0_{-})\quad =\quad \psi (0_{+})\ A\quad =\quad F$

And we land at

$\therefore \psi (x)\quad =\quad \begin{cases}{Ae^{Kx}\quad (x<0)}\{Ae^{-Kx}\quad (x>0)}\end{cases}$

If you sketch this~~~~

Equation from the original notes

It looks like that — and how sharp the pointy bit at the middle is should depend on the potential, right!? huh~~~?

The sharpness at that boundary probably depends on $\alpha$….. (T_T)

But how do we actually express that?~?~?~~!!

So~

$-\frac {\hbar^{2}}{2m}\frac {d^{2}\psi}{dx^{2}}\quad -\alpha \delta (x)\psi \quad =\quad E\psi$

We’re going to integrate both sides — from $-0$ to $+0$…

Buuut writing it as “$-0$ to $+0$” looks too high-school-ish….

(Not a knock on high schoolers! High schoolers can totally follow this!!! Shoutout to high schoolers everywhere!!!)

$\int _{-\epsilon}^{\epsilon}{\quad \cdot \cdot \cdot}$

Writing it like this is fine, right!?!?

Just imagine $\epsilon$ as $0.000000000000000000000001$.

Then GoGo, integrate both sides GoGo

$-\int _{-\epsilon}^{\epsilon}{\quad}\frac {d^{2}\psi}{dx^{2}}dx\quad -\alpha \frac {2m}{\hbar^{2}}\int _{-\epsilon}^{\ep$

$-\int _{-\epsilon}^{\epsilon}{\quad}\frac {d^{2}\psi}{dx^{2}}dx\quad -\frac {2m\alpha}{\hbar^{2}}\int _{-\epsilon}^{\eps$

Those colored bits up there — we can totally compute them!!!!!

$\left| \frac {d\psi}{dx} \right|_{+\epsilon}\quad =\quad -KAe^{K0}\quad =\quad -KA$

$\left| \frac {d\psi}{dx} \right|_{-\epsilon}\quad =\quad KAe^{-K0}\quad =\quad KA$

$-2KA\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\psi (0)\quad =\quad -\frac {2m\alpha}{\hbar^{2}}A \ \therefore \quad K\qu$

$\text{Originally the constant }K\text{ was defined as }K\quad =\quad \frac {\sqrt {-2mE}}{\hbar}, \ \text{but now }K\te$

Finding $K$ also means we’ve found $E$ — like you saw above —

and it also means we’ve found the wave function $\psi(x)$. OK, last touch.

Normalize so the square actually means a probability!!!

$\therefore \psi (x)\quad =\quad \begin{cases}{Ae^{\frac {m\alpha}{\hbar^{2}}x}\quad (x<0)}\{Ae^{-\frac {m\alpha}{\hbar$

$1=\quad \int _{-\infty}^{\infty}{\psi^{*}(x)\psi (x)}dx\quad =\quad \int _{-\infty}^{0}{A^{2}Ae^{\frac {2m\alpha}{\hbar^$

The fully normalized $\psi(x)$:

$\psi (x)\quad =\quad \frac {\sqrt {m\alpha}}{\hbar}e^{-\frac {m\alpha}{\hbar^{2}}\left| x \right|} \ E\quad =\quad -\fr$

Notice — unlike the infinite square well — there’s no $n$ in the expression!?!?!

Which means: only one state exists. One $E$. That’s it.

And the probability of finding the electron is (obviously) highest right at the atom, right???!!

(Does more of it kinda stick to the atom?) hahahaha I have no idea, don’t ask me.

Equation from the original notes

1. Bound state

2. Scattering state (scattering… what now?)

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