I The probability density function for the double-slit experiment

[Ad VanderVen]

TL;DR Summary

I want to know the formula for the probability density function for the interference pattern obtained in the double slit experiment with both slits open.

I am desperate. I've scoured the web for the formula for the probability density function for the interference pattern obtained in the double slit experiment with both slits open. So I want to know the probability density function and not the intensity function. I prefer not to have references to sites on the internet, but want the formula itself. I would also like it if all parameters, such as for example the width of the slits, the distance between the slits, the distance of the slits to the wall, etc., are clearly described. Who can help me?

 

[Nugatory]

By probability density you mean the probability of a particle detection at a given moment as a function of position on the screen? That’s just the intensity scaled appropriately.

 

[PeroK]

I think this question has been asked before and there was a reference to some research paper. It might be this one:

https://arxiv.org/abs/1710.09758

What you need is a purely QM solution to the problem that does not employ a heuristic such as "the uncertainty principle" applied at the slits. That gets complicated!

 

[Ad VanderVen]

By probability density you mean the probability of a particle detection at a given moment as a function of position on the screen? That’s just the intensity scaled appropriately.

How exactly is the formula written?

 

[Ad VanderVen]

I think this question has been asked before and there was a reference to some research paper. It might be this one:

https://arxiv.org/abs/1710.09758

What you need is a purely QM solution to the problem that does not employ a heuristic such as "the uncertainty principle" applied at the slits. That gets complicated!

Please, no references. I am asking for the formula of the probability density function.

 

[Nugatory]

How exactly is the formula written?

Assume for the sake of simplicity that we're working with photons. The intensity at any given point divided by the photon energy will be the expectation value of the arrival rate per unit time at that point and the probability follows from that.

It might be helpful if you could tell us what you're expecting to find from the probability density that you won't get from the intensity. The reason you're so often encountering the formula for the intensity instead of the probability density is that the intensity is more often the most useful way of describing the physics.

 

[Ad VanderVen]

Assume for the sake of simplicity that we're working with photons. The intensity at any given point divided by the photon energy will be the expectation value of the arrival rate per unit time at that point and the probability follows from that.

It might be helpful if you could tell us what you're expecting to find from the probability density that you won't get from the intensity. The reason you're so often encountering the formula for the intensity instead of the probability density is that the intensity is more often the most useful way of describing the physics.

You write:

"The intensity at any given point divided by the photon energy will be the expectation value of the arrival rate per unit time at that point and the probability follows from that."

If it's all so simple, why don't you give the formula? No one seems willing or able to give the actual formula.

The reason I need the probability density function is that I have an alternative formula and want to compare it to the existing formula. I don't have an alternative formula for the intensity.

 

[Ad VanderVen]

Could it be that under certain conditions the function below
$$\displaystyle f \left(x \right) = \frac{2}{\pi} \frac{~\sin \left(x \right)^{2}~\cos \left(b ~x \right)^ {2}}{ ~x ^{2}}$$
could be a function describing the probability density in the double-slit experiment? And what would be those conditions exactly?

 

[DrChinese]

1. I think this question has been asked before and there was a reference to some research paper. It might be this one:

https://arxiv.org/abs/1710.09758

What you need is a purely QM solution to the problem that does not employ a heuristic such as "the uncertainty principle" applied at the slits. That gets complicated!

2. Could it be that under certain conditions the function below
$$\displaystyle f \left(x \right) = \frac{2}{\pi} \frac{~\sin \left(x \right)^{2}~\cos \left(b ~x \right)^ {2}}{ ~x ^{2}}$$
could be a function describing the probability density in the double-slit experiment? And what would be those conditions exactly?

3. If it's all so simple, why don't you give the formula? No one seems willing or able to give the actual formula.

1. Notice that the referenced paper has 67 formulae, and covers the case of only a single slit.

2. Notice that your formula has only a single variable, and therefore can't answer your original question.

3. Notice that no one said anything about this being simple, and in fact the opposite was said.

Entire books (and even careers) are written to give proper treatment to understanding and describing quantum elements of light (although it is not clear from your OP whether you are talking about photons, electrons, or other particles that can be run through a double slit setup). Even "close approximations" get complicated. So it's not like someone can simply "give you the answer".

If you share more with us, I'm sure we could drill in closer to where you are headed.

 

[Ad VanderVen]

I now understand why no one has been able to give me the formula for the probability density of the impact position of a particle on the screen when both slits are open. What is the case. On Wikipedia, in the article Double-slit experiment in the chapter Classic wave-optics formulation, the formula for the intensity is given. It reads as follows
$$\begin{align}
I(\theta)
&\propto \cos^2 \left [{\frac {\pi d \sin \theta}{\lambda}}\right]~\mathrm{sinc}^2 \left [ \frac {\pi b \sin \theta}{\lambda} \right]
\end{align}$$
where b is the width of the slits, d is the distance between the two slits and $\theta$ is the magnitude of the angle between the line from the midpoint between the slits and the position of impact on the screen and the line from the midpoint between the slits perpendicular on the screen. Since $tan(\theta)=x/L$, where $x$ is the position of impact of the particle on the screen and $L$ is the distance between the midpoint between the two slits and the screen, we also have $\theta = arctan(x/L)$ and in this way one can derive an expression for the intensity of $x$, $I(x)$ by replacing in the formula for $I(\theta)$ $\theta$ with $arctan(x/L)$. To determine the final probability density function $f(x)$, it must be possible to solve the integral of the formulas for $I(\theta)$ and $I(x)$, which is not the case. As long as these integrals cannot be determined, the probability density function cannot be determined either. Therefore, no one is able to give the formula for the probability density function. But why does no one say that in this case it is not possible to give the formula for the probability density function?

 

[Lord Jestocost]

To my mind, the intensity at a point on the screen is proportional to the square of the wave function at that point, $P(x)=|\psi (x)|^2$, and thus represents a 'probability density' for 'impact positions'. The probability of detecting an electron within a narrow region of width $\delta x$ at position $x$ is directly proportional to the probability density:
$$Prob(in~\delta x~at~x) = P(x)δx$$

 

[Ad VanderVen]

I know all that.

 

[vanhees71]

The above quoted formula shows you that indeed in the Fraunhofer approximation the diffraction interference pattern is the Fourier transform of the slit openings, and thus closely related to the momentum-space wave function. It's a manifestation of Huygens's principle.

 

[Ad VanderVen]

I don't understand what all this has to do with my question:

But why does no one say that in this case it is not possible to give the formula for the probability density function?

 

[vanhees71]

But you quoted yourself the probability distribution in #10, which is the Fourier transform of the openings making up the double slit. It follows from finding the Green's function for the Schrödinger (or rather the Helmholtz equation derived from it), leading to Huygens's principle and Kirchhoff's approximations in the theory of diffraction. You find this topic in great detail rather in textbooks on optics, e.g., A. Sommerfeld, Lectures on Theoretical Physics, vol. 4.

 

[PeroK]

I don't understand what all this has to do with my question:

But why does no one say that in this case it is not possible to give the formula for the probability density function?

Is the problem that the maximum intensity needed to normalise the pdf does not appear to have a closed form solution in terms of $b,d$ and $\lambda$. But, would have to be calculated numerically from the integral?

 

[Lord Jestocost]

"We could obtain this same double slit pattern by either using photons with wavelength 550 nm (green light) and a slit separation of 55 μm or electrons accelerated through a potential of 10 KV and a slit separation of 1.23 nm. The precise expression for the double slit interference curve is

where $d$ is the slit separation, $a$ is the slit width, $\theta$ is the angle in Figure 3, $\lambda = h/p$, $h$ is Planck’s constant, and $p$ is the momentum of the particle."

See: “The Double Slit Experiment and Quantum Mechanics” by Richard Rolleigh (2010)

The Double Slit Experiment and Quantum Mechanics∗​

 

[Ad VanderVen]

Is the problem that the maximum intensity needed to normalise the pdf does not appear to have a closed form solution in terms of $b,d$ and $\lambda$. But, would have to be calculated numerically from the integral?

Yes!

 

[PeroK]

Yes!

Not everything has a closed form solution. That doesn't affect existence, mathematically or physically.

 

[Ad VanderVen]

Not everything has a closed form solution. That doesn't affect existence, mathematically or physically.

Thank you very much for your wise and reassuring answer.