1
$\begingroup$

We know that by performing the Double Slit Experiment we get an interference pattern on the detector screen. To explain this most sources talk about the state of the particle: calling $|\psi _l \rangle $ the state in which the particle crossed the left slit, and $|\psi _r\rangle$ the state in which the particle crossed the right slit, we have the following total state for the single unmeasured particle ($|\psi\rangle$):

$$|\psi \rangle=\frac{1}{\sqrt{2}}\Big(|\psi _l\rangle+|\psi _r\rangle \Big)$$

Then usually books on the topic talk about the particle's probability of landing left versus the probability of landing right, calculate those, and show that there are interference terms in the expression of the probability, job done for them usually.

However I am not satisfied: experimentally seeing a continuous interference pattern on the screen (with a certain shape!) implies that the single particle crossing the slits must have, at the time of impact, a wave function with the same shape.

My question, calling the time of impact $t_i$, is: what is the simplest (easiest to understand) way to derive the monodimensional wave function $\psi(x,t_i)$ of the single particle crossing the slits? (Better if the explanation starts from the postulates of QM)

I feel that theoretically finding the explicit form of this wave function is the only proper way of explaining the double slit experiment fully. In fact this is the only way to properly predict the shape of the figure of interference that will appear on the screen. Based on experimental results the wave function we find should of course look something like this:

enter image description here

$\endgroup$
5
  • 1
    $\begingroup$ My sense is you would not be asking this question if you had carefully gone through the classic relevant masterful discussion in A Tomonaga's Quantum Mechanics, Vol. 2: New Quantum Theory ... $\endgroup$ Jun 3, 2022 at 15:13
  • 1
    $\begingroup$ "Then usually books on the topic talk about the particle's probability of landing left versus the probability of landing right" The computation of the interference pattern is probably often not that detailed in QM texts because it's isomorphic to the classical derivation of the interference pattern for e.g. an electromagnetic wave, but I have no idea what this "landing right" and "landing left" is supposed to denote. This question would likely prompt better answers if you detailed a bit more what sort of derivation you're talking about here and what you find dissatisfying about it. $\endgroup$
    – ACuriousMind
    Jun 6, 2022 at 21:23
  • $\begingroup$ In particular, the stress on "continuous" in your question is strange - your state $\lvert \psi\rangle$ should have a corresponding wave function $\psi(x) = \langle x \vert \psi\rangle$ that gives you exactly the "continuous" probability amplitude you're looking for, what's wrong with that? $\endgroup$
    – ACuriousMind
    Jun 6, 2022 at 21:24
  • $\begingroup$ @ACuriousMind I now get that is not exactly clear what I wanted to say in some parts of my question, but the central point should be clear: I am wondering what is the simplest way to prove that the distribution of hits on the detector screen must be shapen like the picture in my question (i.e. the classic figure of interference of the double slit experiment that we all know and love). $\endgroup$
    – Noumeno
    Jun 8, 2022 at 16:07
  • $\begingroup$ I'd just like to point out that in a real experiment, $t$ would not be fixed, rather the screen would just wait for particles to hit at any time. For this reason it would not be so simple as to solve the Schrödinger equation for $\psi(x,t)$ and plug in a fixed value of $t$. $\endgroup$ Jun 9, 2022 at 11:09

1 Answer 1

4
+25
$\begingroup$

I think the most straightforward way to get the pattern is to first consider an isotropic point source of particles with momentum $k$. Here I consider the $d=2$ case. The wave function for the particles coming out of the source can be found by using the relation for probability current density: $$\vec{J}=\frac{i\hbar}{2m}(\psi \nabla \bar{\psi}-\bar{\psi}\nabla \psi) \tag{1}$$ and the fact that to have particle number conservation we should have (again for $d=2$): $$\vec{J}\propto \frac{\hat{r}}{r} \tag{2}$$ where the source is assumed to be at the origin. It is easy to check that $\psi$ has the following functional form: $$\psi(r)=A\frac{e^{ikr}}{\sqrt{r}} \tag{3}$$ Where A depends on the intensity $A\propto \sqrt{I}$.

Now consider two sources of with amplitudes $A_1$ and $A_2$ at points $r_1$ and $r_2$ with initial phase difference $\Delta \varphi$, then the total wave function is the superposition of these two and we have: $$\psi(\vec{r})=A_1 \frac{e^{ik|\vec{r}-\vec{r}_1|}}{\sqrt{|\vec{r}-\vec{r}_1|}}+A_2 \frac{e^{ik|\vec{r}-\vec{r}_2|+i\Delta \varphi}}{\sqrt{|\vec{r}-\vec{r}_2|}} \tag{4}$$

For the double slit experiment, we can put $A_1=A_2=A$ and $\Delta \varphi=0$. We also assume the slits to be at $(\pm a,0)$. Then, the wave function on a wall at $y=L$ is given by (working in units where $A=1$): $$\psi(x,L)=\frac{e^{ik\sqrt{L^2+(x-a)^2}}}{(L^2+(x-a)^2)^{1/4}}+\frac{e^{ik\sqrt{L^2+(x+a)^2}}}{(L^2+(x+a)^2)^{1/4}} \tag{5}$$

If you plot $|\psi|^2$ you'll see the interference pattern with a decay.

The more complicated way of doing this is to solve the time independent Schrodinger equation between two walls in the presence of two holes, which would give qualitatively the same picture for $L\gg a$ while for smaller values of $L$ it would also take the effect of successive reflections of particles of the walls into account.

$\endgroup$
2
  • $\begingroup$ Wonderful answer (I took the liberty to number your equations), but could you expand (by editing your answer) on how to get (3) from (1) and (2), and also on why particle number conservation implies (2)? I feel this would improve the general understandability of your answer a lot $\endgroup$
    – Noumeno
    Jun 9, 2022 at 11:10
  • $\begingroup$ Why in this answer do we use a solution to the time-independent Schrödinger equation? Surely in each run of the experiment we have a time-dependent moving particle which finally hits the screen. This solution to the TISE is for a stationary, spacially-extended particle whose position probability never changes in time. $\endgroup$ Apr 9, 2023 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.