# On the single/two-slit diffraction problem and superposition.

1. Nov 27, 2009

### Neo_Anderson

The mechanism by which light diffracts through a two-slit experiment is alledgedly due to the superposition of a single photon at the two-slit interface, causing the photon to either constructively or destructively interfere with itself.
If this is the case, then can you please explain the following phonomena?:

This is called aperture diffraction; and as you can see, there are no two slits or two apertures that the photons are interfering with, yet the diffraction pattern remains!

2. Nov 27, 2009

### Redbelly98

Staff Emeritus
There are not just two apertures that the photon passes through, but many many apertures. These "apertures" are all the points lying within the round opening, and the "waves" produced by all these possible source points interfere to give the pattern shown in your picture.

3. Nov 28, 2009

### alxm

That's a pretty basic optical phenomenon there; this isn't a QM question.

4. Nov 28, 2009

### Staff: Mentor

In order to calculate the interference pattern, you have to integrate the contributions from all parts of the aperture. It's fairly straightforward to set up the integral (see equations 2 and 3 in the PDF linked below), but not trivial to solve it:

Last edited by a moderator: May 4, 2017
5. Nov 28, 2009

### Neo_Anderson

For single-slit diffraction, if the slit width is 600 nm, and if the wavelength of the plane wave is 600 nm, then the first minimum of intensity is roughly equal to pi, or 180 degrees. That is, for a screen that's 10 cm from the slit, the first minimum intensity will be about 10.6 cm from the center of the screen, for a total distance of 21.2 cm for M=0 (first maximum = 10.6+10.6 cm, or 21.2 cm). In other words, we don't see the first minimum of intensity until we look at the slit from the side! Yet with the limit of resolution for aperture diffraction--which is the same basic scenario (aperture diameter = incident wavelength)--, we see the Airy disk with concentric interference fringes radiating outwards. How come? Why doesn't the edge of the Airy disk itself (first minimum intensity for aperture diffraction) cover the same 180 degrees as did the single-slit? After all, the aperture is now the same dimeter as the wavelength of the incident wave...

Last edited by a moderator: May 4, 2017
6. Nov 28, 2009

### Neo_Anderson

Since the current idea is that the photon superposes itself through both slits (or at the edge of the aperture versus the center of the aperture as jtbell's .pdf illustrated (thx, jtbell)); and since superposition of particle/waves is strictly a quantum and non-classical in nature, this thread is very much "QM."

7. Nov 28, 2009

### alxm

That's not what's going on. This is just an ordinary classical diffraction, and those airy disc calculations are done from classical theory (and were first done by Lord Rayleigh well before quantum theory) This is taught in every introductory course in classical physics (e.g. Young and Freedman "University physics", chapter 38 - Diffraction). There's no quantum mechanics in the lecture notes jtbell linked to.

No, diffraction patterns are classical "wave-like" behavior. The point of the double-slit experiment is that this diffraction pattern will arise even if you're dealing with individual quanta of light being emitted. Hence the 'wave-particle duality'.

Your light source, a laser beam, is not so low in intensity that you're recording individual photons, so it's not analagous to the double-slit experiment. It behaves like a classical light wave. But regardless, the quantum mechanical diffraction pattern corresponds to the classical one anyway.

Quantum mechanics isn't needed to explain basic diffraction patterns. If you think it is, then you need to go back and learn classical physics. I've seen single-slit diffraction explained in high-school textbooks, even.

8. Nov 29, 2009

### Staff: Mentor

Why do you think it doesn't?

The angular distance of the first minimum from the center of the diffraction pattern for a circular aperture is given (in the Frauhnofer approximation) by

$$\sin \theta = 1.22 \frac {\lambda} {d}$$

where d is the diameter of the aperture. Setting $\theta = \pi / 2$ gives you $d = 1.22 \lambda$. With this diameter aperture, the central peak of the diffraction pattern completely fills the hemisphere beyond the aperture.

9. Nov 29, 2009

### Neo_Anderson

I think I now see the light, guys.
Current thinking is that there's a probability that the photon will superpose anywhere on the other side of the aperture. The larger the aperture, the lower the probability a photon will be found outside the first maximum on the screen (resultng in a smaller Ariy disk, and smaller diffraction rings). But since the probability that a photon will be found anywhere outside the first maximum approaches unity when the aperture diameter nears the wavelength of the incident light, we should expect to see a greater distribution of photons on that screen. What do we see in experiment? A greater distribution of photons on that screen, of course!

10. Jan 15, 2011

### wangimagine

You don't need to care how a photon behaves at the silt. I think the observation object is the change of state. If you put a screen before the slit, the picture on the screen would only be a single point. Attention! I call this picture a state, which describes the distribution of a large number of photons' behavior at the place where you put a screen.
After the experiment, the result on the screen behind the slit is the changed state (a new state). You only know that the experiment let the state of a large number of photon changed. (Because every single photon's state is changed.) You can't know further about why the state changes. You observe the state, not the cause of it.

11. Jan 17, 2011

### jondoe

But what if individual photons were sent through the aperture? Would we still get the same diffraction pattern building up over time as with the laser beam? In that case, it surely would be analogous to the double-slit experiment and hence require some form of QM explanation?

12. Jan 17, 2011

### Redbelly98

Staff Emeritus
Welcome to PF.

Yes, over time, we would get the same diffraction pattern.
Yes, if observing individual photons is involved, QM is required to explain it. But since this discussion concerns the classical wave-like nature of light, and not the quantized properties, a classical explanation will suffice. The OP showed a classical optical diffraction pattern.

Diffraction is a property of waves. Light is classically described as a wave. It is not necessary to use photons in order to explain diffraction.

13. Jan 17, 2011

### jondoe

Thanks for your reply. I think the OP may have been driving at a QM explanation though, because he initiated the discussion by describing the QM explanation for the double-slit experiment. I'd personally be interested in understanding the QM explanation if it is possible to expand on it.

14. Jan 17, 2011

### Redbelly98

Staff Emeritus
Here is my version of the QM explanation...

It's all about what can be measured or observed; nothing else is relevant.

Light is quantized, and occurs in discrete energy quanta that we call photons. These light quanta still posess the wave properties that we are familiar with from classical optics. This is a statement of duality, that light has both discrete and wave-like properties -- not just one or the other, but both.

A photon traveling through a slit or aperture will diffract and/or interfere with itself, consistent with it's wave properties. The resulting wave function, projected onto an observation screen, has the familiar patterns calculated for waves in classical optics.

Any attempt to measure the photon's location, i.e. observing the screen, will collapse the wavefunction, so that the observation shows the single photon to be located at a small area on the screen. This can be explained quantum mechanically as collapsing the wavefunction by performing a measurement of the photon's position. The probability of observing the photon at some location on the screen is given by the square of it's wavefunction.

Observing many many photons -- in essence, repeating the single-photon experiment many many times -- will reveal how the single-photon location probability (i.e., square of the wavefunction) is distributed over the screen.

Hope that helps.

15. Jan 17, 2011

### wangimagine

Seee the attachments.

#### Attached Files:

• ###### interference of single photon with itself.doc
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16. Jan 18, 2011

### jondoe

Thanks for the replies they were very informative!

17. Jan 19, 2011

### purpledog

I did a bit of research online and could not find any experiments dealing with "single slit" and "one photon at a time".

Also, if this is the case that this setup would lead to individual photons building up the diffraction pattern, why "everybody" talks about the two slits experiment as an example of the way probably waves collapse ? The single slit experiment would demonstrate this with a simpler settings.

Finally, the wiki page says that the three slit experiments doesn't work : "only two slits would produce these results, while three or more slits would not alter the result."

OK, there's not of physics or maths in my post, but here we go: are you guys 100% sure that individual photons build up the diffraction pattern over time ?

18. Jan 20, 2011

### DrChinese

Welcome to PhysicsForums, purpledog!

The reason for the double slit being used as an example is simple: it is easy to demonstrate that the sum of slit A by itself and slit B by itself does not lead to the sum of A and B together, which shows interference. So the example highlights the idea that the photon went through both slits. Actually, as pointed out, the photon takes many paths - not just 2.