# What is the wavefunction of the Young Double Slit experiment?

I have never seen the wavefunction for this experiment and would like to know how to derive it using the Schrodinger equation. I specifically want to see how the electron wave function leaves the source, then goes through the slits, and produces the characteristic diffraction pattern on the other end.

• Welcome to the reality of the double slit experiment: it's usually being fudged, even in the theoretical treatment. Why is it being fudged? Because its details have no meaning. Having said that, if you want to do it right, I would suggest to limit yourself to a two dimensional potential and to calculate with an infinitesimally high, infinitesimally thin potential barrier that is being hit by a plain wave. – CuriousOne Sep 12 '15 at 22:33
• Yeah that's what I'm doing actually, putting a delta potential at the x=0 axis with a step function to split the wall into holes. But even this seems funky cause there is transmission through the wall. So what gives haha, I mean what am I supposed to trust if I can't trust the math? – user92640 Sep 12 '15 at 22:36
• The math tells you the right thing... look up the potential barrier problem. – CuriousOne Sep 12 '15 at 22:38

Well, there are many things you could do. You could:

1. consider two gaussian beams (the linked article is for electrodynamics)
2. apply some paraxial approximation (which would be more appropriate to treat electrons with a high forward momentum)
3. do a cheap/cheesy symmetric point source approximation using Green's functions.

I can do number three for you :)

If you take $\hbar=1$, $m=\frac{1}{2}$, then the equation in question becomes $i \dot{\varphi}+\nabla^2 \varphi=0$, which has a solution:

$$\left(\frac{a}{a+2 i t}\right)^{3/2} \exp \left( {-\frac{x^2+y^2+z^2}{2 a+4 i t}}\right)$$

Then you can add two of these point sources together and translate them:

$$\left(\frac{a}{a+2 i t}\right)^{3/2} \exp \left( {-\frac{x^2+y^2}{2 a+4 i t}}\right)\left(\exp\left( \frac{(z-h)^2}{2 a+4 i t} \right)+\exp\left( \frac{(z+h)^2}{2 a+4 i t} \right) \right)$$

The fact that these wave packets aren't moving is a bit of a cheat, but you can always "boost" to a moving frame by using the answer here: Galilean invariance of the Schrodinger equation (or if you're really on top of your quantum mechanics game you can apply the translation operator $e^{-i\hat{x}\cdot \hat{p}}$)

Voila, an appropriate wavefunction.

Here's an XZ slice of the initial condition $|\psi|^2$:

An XZ slice of $|\psi|^2$ at a later time and offset Y:

And an animation of $|\psi|^2$ (uploaded on imgur)

I used Mathematica to expand psi squared of the previous equation. You can see exactly where the interference comes in (the cosine term)

$$|\psi(x,y,z,t)|^2=\frac{\left(a^2\right)^{3/2}}{\left(a^2+4 t^2\right)^{3/2}} \cdot \left(\exp\left(-\frac{a \left(h^2+2 h z+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)+\exp\left(-\frac{a \left(h^2-2 h z+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)+2 \cos \left(\frac{4 h t z}{a^2+4 t^2}\right) \exp \left(-\frac{a \left(h^2+x^2+y^2+z^2\right)}{a^2+4 t^2}\right)\right)$$

So the oscillatory/important part is $\cos \left(\frac{4 h t z}{a^2+4 t^2}\right)$. This brings up the obvious problem with this approach: it doesn't directly provide the nice result that you usually want, relating the momentum of the particle to the "wavelength" of the interference pattern. The interference pattern reaches its maximum frequency at $t=\frac{a}{2}$, so I'll leave it as an exercise to the reader to see if there's a relation between the momentum ($\hat{p}^2$ maybe?), the de Broglie wavelength, and the usual peak/trough formulas diffraction formulas(this sort of thing)

• This is some pretty spiffy stuff I'll go ahead and accept this as the answer for the time being but I need some time to work it through and think about it! – user92640 Sep 12 '15 at 23:38
• @StevenGrigsby no rush! Accepting an answer discourages other people from answering though! – user12029 Sep 12 '15 at 23:40