Delta-Function Potential: Scattering States and Quantum Tunneling

Picking up right where we left off.

Same potential as before — but this time, $E>0$.

If I’d crammed this case into the last post too, the scroll bar would’ve been microscopic, so I cut it here and moved it over.

This is exactly~~~ the kind of situation we’re dealing with~~~

And it’s suuuper similar to the $E<0$ case from before,,, hmm~

Anyway, let’s go!!

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

OK, split the regions!!!!

Let’s look at $x<0$ first.

$$\frac {d^{2}\psi}{dx^{2}}\quad =\quad -\frac {2mE}{\hbar^{2}}\psi \quad =\quad -k^{2}\psi$$

The difference from last time: before, we had to sneak in a minus sign to keep things positive. Here, since $E>0$, it’s already $-k^{2}\psi$ straight up!!!

And a second-order ODE like this? Piece of cake!!!

$$( D+ik)(D-ik)\psi \quad =\quad 0 \\ \psi (x)\quad =\quad Ae^{ikx}\quad +\quad Be^{-ikx}$$

Oh crap, we’re cooked…. Last time these were real exponentials, so $\psi(-\infty)=0$ killed one constant for us.

But now we’ve got complex exponentials — they just keep spinning in circles, magnitude 1 forever…

OMG. We can’t kill anything.

And for $x>0$, same story —

$$\psi (x)\quad =\quad Fe^{ikx}\quad +\quad Ge^{-ikx}$$

Two terms survive on this side too.

Even when we slap on the “$\psi$ is continuous” condition~!!~~!!!!!

All we get is $A+B=F+G$………..ha…………sad.

OK let’s hit the boundary $x=0$!!!!

Just like in the bound state case, integrate both sides of the time-independent Schrödinger equation from $-\epsilon$ to $+\epsilon$.

(The cleanest way to kill a $\delta$? Integrate it, right?!!)

$$-\frac {\hbar^{2}}{2m}\frac {d^{2}\psi}{dx^{2}}\quad -\alpha \delta (x)\psi \quad =\quad E\psi \\ \int _{-\epsilon}^{\epsilon}{\quad}\frac {d^{2}\psi}{dx^{2}}dx\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\int _{-\epsilon}^{\epsilon}{\quad}\delta (x)\psi dx\quad -\frac {2mE}{\hbar^{2}}\left( \int _{-\epsilon}^{\epsilon}{\quad}\psi dx \right) \\ \text{the red one}\quad :\quad \psi \text{ is continuous, so} \to 0$$$$\int _{-\epsilon}^{\epsilon}{\quad}\frac {d^{2}\psi}{dx^{2}}dx\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\int _{-\epsilon}^{\epsilon}{\quad}\delta (x)\psi dx\\ \Delta \left( \frac {d\psi}{dx} \right) \quad =\quad -\frac {2m\alpha}{\hbar^{2}}\psi (0)$$$$\text{left side,}\\ \Delta \left( \frac {d\psi}{dx} \right) \quad =\quad \left| \frac {d\psi}{dx} \right|_{+\epsilon}\quad -\quad \left| \frac {d\psi}{dx} \right|_{-\epsilon}\quad =\quad ik(F-G-A+B)$$$$\text{right side,}\\ -\frac {2m\alpha}{\hbar^{2}}\psi (0)\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\left( A+B \right) \quad =\quad \left( \text{or}\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\left( F+G \right) \right)$$$$ik(F-G-A+B)\quad =\quad -\frac {2m\alpha}{\hbar^{2}}\left( A+B \right) \\ \text{let's line things up.} \\ ik(-A+B)\quad +\quad \frac {2m\alpha}{\hbar^{2}}(A+B)\quad =\quad -ik(F-G)\\ (A-B)\quad +i\frac {2m\alpha}{k\hbar^{2}}(A+B)\quad =\quad (F-G)\\ A\left( 1+i\frac {2m\alpha}{k\hbar^{2}} \right) \quad -B\left( 1-i\frac {2m\alpha}{k\hbar^{2}} \right) \quad =\quad F-G \\ \text{ugh!><}\quad \text{too}\quad \text{messy!!!!><} \\ A\left( 1+i2\beta \right) \quad -\quad B\left( 1-i2\beta \right) \quad =\quad F-G\quad \left( \beta \equiv \frac {2m\alpha}{k\hbar^{2}} \right)$$

Two equations. That’s all we could squeeze out… T_T T_T T_T T_T T_T

$$\begin{cases}{A+B\quad =\quad F+G}\\{A\left( 1+i2\beta \right) \quad -\quad B\left( 1-i2\beta \right) \quad =\quad F-G}\end{cases}$$

Ha…… counting $\beta$, we’ve got 5 unknowns,,,, and only two equations… T_T T_T T_T T_T T_T T_T T_T T_T

T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T T_T ha……… how do we even solve this……..?

But physics students, other than me, are literally the smartest people on earth, right? The future leaders of humanity???~?

“Hey hey hey hey hey hey, you do NOT have to solve for everything!!!?!??^^ Think about what it means and only keep the stuff that’s actually meaningful!”

$$\psi (x)\quad =\quad Ae^{ikx}\quad +\quad Be^{-ikx}\quad \text{right?}\\ \text{oh ho, so}\\ \psi (x,t)\quad =\quad Ae^{i\left( kx-\frac {E}{\hbar}t \right)}\quad +\quad Be^{-i\left( kx+\frac {E}{\hbar}t \right)}\quad \text{and also also also}\quad k\text{ was taken positive, remember}\approx ~~? \\ \text{I'm getting a feel for this.} \\ \text{For }x<0,\quad Ae^{i\left( kx-\frac {E}{\hbar}t \right)}\quad +\quad Be^{-i\left( kx+\frac {E}{\hbar}t \right)}\text{ is}\quad \\ \text{a right-moving wave}\quad Ae^{i\left( kx-\frac {E}{\hbar}t \right)}\quad \text{plus}\quad \text{a left-moving wave}\quad Be^{-i\left( kx+\frac {E}{\hbar}t \right)}\text{, ohhh}?!!!! \\ \text{Then for}\quad x>0\text{, same deal:}\quad \\ \text{right-moving wave}\quad Fe^{i\left( kx-\frac {E}{\hbar}t \right)},\quad \text{left-moving wave}\quad Ge^{-i\left( kx+\frac {E}{\hbar}t \right)}\quad \text{superposed together!}\\ \text{And the amplitudes of each of those waves are what we named }A,B,F,G\text{, ohhh}?!?!!!!!!>3<$$

NOW we can actually read this thing.

We can finally see what’s going on.

For simplicity, picture a wave coming in from the left, heading right. That means we can set $G=0$, right??

Imagine wave $A$ rolled in from waaaaay out at $-\infty$~~~~, cruising along, and somewhere on the ground there’s a big~~deep pit buried in its path.

OK, so what about $B$???~!?!?!?!

※ Warning: mental breakdown incoming ※

$B$ is NOT zero lol lol lol lol lol lol lol lol lol lol lol lol

lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol

lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol

(Feel free to step outside, scream into the void for a solid minute, come back heh heh lol lol lol)

Like, what even is this, for real lol lol lol lol lol lol

$G$ we can set to zero. $B$, we cannot. I literally argued with my professor about this in class, ..

I mean — why? If $G\ne 0$ “doesn’t make sense,” then fine, set $G=0$.

But then $B\ne 0$ also doesn’t make sense, and yet we’re told we have to keep it?? lol

lol lol lol lol lol lol lol lol lol lol lol lol lol lol lol

It genuinely does not make sense lol lol lol lol

OK but — there’s been a “situation change,” so hear me out: don’t set $B$ to zero yet.

$G\ne 0$…. if we allowed that…. it would mess with $A$, which is coming from $-\infty$ toward $x=0$.

So $G=0$!!!! That’s how I got talked into it.

Fine, I accepted it… heh heh.

Keeping $B\ne 0$:

In any macroscopic setting this is obviously wrong. But if we absolutely, really absolutely had to

draw the incomprehensible behavior of the microscopic world by analogy with something macroscopic —

In the micro world: a pitcher throws a ball to a batter, and there’s a (ridiculously tiny, ridiculously deep) pit in between………

With some absurdly tiny probability, even though there’s nothing in front of the ball, (because of the pit) the ball can bounce back and fly toward the pitcher…

I want to die for real lol lol lol lol lol lol lol lol lol lol lol

Why does it reflect when there is literally NOTHING there……………………. (this was the exact moment I decided to switch majors.)

Seriously, what kind of field is this lol lol lol lol lol — it’s a religion. A religion. Quantum physics???? Fun stuff.

Quantum religion. Quantum religion lol lol lol lol — bring out the cult leader, who is it…huh….

OK back to the actual math.

$$\psi (x)\quad =\quad \begin{cases}{Ae^{ikx}+Be^{-ikx}\quad (x<0)}\\{Fe^{ikx}\quad (x>0)}\end{cases}$$

Now… putting the physics student hat back on….

“$A$, $B$, $F$ individually aren’t what matters!!!!!”

How much of $A$ gets reflected:

$$\frac {\left| B \right|}{\left| A \right|}$$

How much of $A$ gets transmitted:

$$\frac {\left| F \right|}{\left| A \right|}$$

Those are the ratios with actual physical meaning, right??!?!?

And with just the two equations we already have, we can get at least this much!

$$\begin{cases}{A\quad +\quad B\quad =\quad F}\\{\quad A(1+i2\beta )\quad -B(1-i2\beta )\quad =\quad F}\end{cases}$$

Plan:

i) Use equation 1 to write $B=F-A$, plug into equation 2, get $F$ in terms of $A$.

ii) Use equation 1 the other way, $F=A+B$, plug into equation 2, get $B$ in terms of $A$.

$$i) \\ A(1+i2\beta )\quad -B(1-i2\beta )\quad =\quad F\\ A(1+i2\beta )\quad -(F-A)(1-i2\beta )\quad =\quad F\\ A(1+i2\beta )\quad +A(1-i2\beta )\quad -F(1-i2\beta )\quad =\quad F\\ 2A\quad =\quad F(2-i2\beta ) \\ \therefore \quad \frac {F}{A}\quad =\quad \frac {2}{2-i2\beta}\quad =\quad \frac {1}{1-i\beta}$$$$ii)\quad \\ A(1+i2\beta )\quad -B(1-i2\beta )\quad =\quad F\\ A(1+i2\beta )\quad -B(1-i2\beta )\quad =\quad A+B\\ A(i2\beta )\quad =\quad B(2-2i\beta ) \\ \therefore \quad \frac {B}{A}\quad =\quad \frac {2i\beta}{2-2i\beta}\quad =\quad \frac {i\beta}{1-i\beta}$$

And that is the $A$-to-$B$ and $A$-to-$F$ relationship — the thing we actually cared about!

“How much transmits~~~~ & how much reflects~~~~~” relative to $A$ — that’s the meaning.

Technically we should use absolute values!!!

But squaring them doesn’t hurt the meaning one bit!!

And here’s how we read them.

Call this the transmission coefficient ($T$) — it tells you how much gets through.

(Squaring isn’t just “square it and hope” — it’s $|\cdot|^2$, so we multiply by the complex conjugate.)

$$T\quad =\quad \frac {\left| F \right|^{2}}{\left| A \right|^{2}}\quad =\quad \frac {1}{1+\beta^{2}}$$

Same deal, call this the reflection coefficient ($R$) — how much bounces back.

$$R\quad =\quad \frac {\left| B \right|^{2}}{\left| A \right|^{2}}\quad =\quad \frac {\beta^{2}}{1+\beta^{2}}$$$$\beta =\frac {m\alpha}{k\hbar^{2}} \\ k=\frac {\sqrt {2mE}}{\hbar} \\ \\ \therefore \beta \quad =\quad \frac {m\alpha}{\hbar \sqrt {2mE}}, \\ \beta^{2}=\quad \frac {m\alpha^{2}}{2E\hbar^{2}}$$

So in terms of energy $E$ and the material property $\alpha$, $\beta$ looks like that.

Therefore,

$$T\quad =\quad \frac {1}{1+\beta^{2}}\quad =\quad \frac {1}{1+m\alpha^{2}/2\hbar^{2}E}\quad$$

&

$$R\quad =\quad \frac {\beta^{2}}{1+\beta^{2}}\quad =\quad \frac {1}{1+2\hbar^{2}E/m\alpha^{2}}\quad$$

wheeeeeeee

And with that, the $\delta$-function potential well case is fully done!!!!!!

But — lucky us — all this work also hands us the $\delta$-function potential barrier for free.

From

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

to

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

That’s the only change.

So in everything we did, we just flip the sign of $\alpha$. Done.

Except —

$T$ and $R$ depend on $\alpha^2$…

So flipping $\alpha \to -\alpha$? The transmission and reflection coefficients don’t change AT ALL…!!!!

OMG!!!!

(I can’t anymore.) First it was “it reflects even though there’s nothing there.”

Now it’s “it passes through a wall” (the probability of getting through is NOT zero (exasperated face)) lol lol lol lol lol lol

This phenomenon is called quantum tunneling,

and it’s thanks to this deeply weird effect that the sun burns bright and gives us light, and that computers and smartphones could pull off all that rapid technological progress,

and apparently a microscope called the STM gave us a whole new way of seeing,,,, or so I’m told.

So the upshot is:

“The fact that the probability of passing through a wall isn’t zero is a technology we’re using right now, heavily.” That’s the takeaway.

(Honestly at this point it’s not even new tech anymore lol lol)

Quantum religion is totally not a cult lol lol lol lol lol lol lol

Time to switch majors, byeee

---

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