# Frequency or wavelength of a photon

1. Jan 4, 2007

### touqra

I don't understand what it means by the frequency or wavelength of a photon.
I know that troughs and crests of a wave is associated with intensity, which translates to the number of photons that can be detected at that particular time and location, and hence frequency and wavelength. And that's how I understand why the Doppler effect.
But what does it mean by the frequency or wavelength of a photon.

2. Jan 5, 2007

### James R

If you look at light as a wave, it has frequency and wavelength.

Frequency is the number of wave crests passing a fixed point in space every second. Wavelength is the spatial distance between two wave crests.

In the photon picture of light, each photon has an energy related directly to the frequency the light would have in the wave picture. The frequency (or wavelength) of light determines its colour. So, if you take, say, yellow light, and consider it as a wave, that wave may have a wavelength of 500 nm. If you consider a beam of yellow light to be composed of photons, each photon will have an energy of E=hf, where f is the frequency value corresponding to 500 nm.

Frequency and wavelength of light are related by $c = f\lambda$, where c is the speed of light, $\lambda$ is the wavelength and f is the frequency.

3. Jan 5, 2007

### Jimmy Snyder

A photon is a quantum (minimum quantity) of light. The frequency of a photon is the frequency of the associated light.

4. Jan 6, 2007

### touqra

A clarification...
Is it that although the wavelength of the light is related to one photon's energy,
the wavelength of light has a physical meaning only when we see it as a wave, and that troughs and crests corresponds to the amount of photon on that spot, and hence a wave due to the sinusoidal pattern in the photon distribution?
Cause I don't think we can directly measure a photon's wavelength just by detecting one and only one photon. Well, unless that photon happens to excite some electron in an atom etc, that we can calculate the lambda through E = hc/(lambda)...

Confusing...

5. Jan 6, 2007

### Jimmy Snyder

As I understand the wave-particle duality, the only time that light behaves as a particle is during its interaction with other particles. If that is the case, then as you say, one way to measure the wavelength of a photon would be to have it interact with an atom.

6. Jan 6, 2007

### marlon

If that were true, then why do we observe an interference pattern in the doubble slit experiment ? Then wy can a photon even self interfere ?

Nono, the particle wave duality is an inherent property of the QM formalism and is ALWAYS VALID.

marlon

7. Jan 6, 2007

Perhaps, during the time any single QM photon begins process of travel through the two slits || and ||, it splits, then comes together on the other side to "self-interfere", and at the point where the two QM waves meet (see attached figure), via superposition principle, the two QM waves express their joint QM particle nature ?

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8. Jan 6, 2007

### ZapperZ

Staff Emeritus
If this is true, then you would detect a signal at both slits. So where is it? And how does it know how many it needs to split into if I have multiple slits?

Would you also argue for the same thing to occur with electrons? How does an electron splits into two (while still observing conservation laws)? How many does it have to split into to produce all those patterns in, let's say, a LEED measurement?

Zz.

Last edited: Jan 6, 2007
9. Jan 6, 2007

### marlon

Actually, the (self)-interference happens between the initial starting point and the final observation point. That is all we can say about it. One can certainly NOT be talking about self interference AFTER the photon has passed through the slits because the interference leads to the superposition of "passing through slit 1 " + "passing through slit 2". What you say here implies that we can observe the photon going through the slits, which is ofcourse impossible in this context of the doubble slit experiment. The interference happens between all possible paths that the photon may follow between initial and final point.

marlon

10. Jan 8, 2007

Is there not a logical contradiction to this statement As seen in this diagram of the Young double slit experiment:http://images.google.com/imgres?img...slit+experiment+image&svnum=10&hl=en&lr=&sa=N the starting point is well before the slits and the observation point is a screen some distance (L) away from the slit. So, if what you say above is true, and the self interference of a single photon is not after it passes through the slit, where does it occur ? -- before the slits ? But this cannot be, for we read here at this link:http://images.google.com/imgres?img...slit+experiment+image&svnum=10&hl=en&lr=&sa=N

We see that the electron must in some sense pass through both slits at once and then interfere with itself as it travels towards the detector. Young's double slit experiment has been performed many times in many different ways with electrons (and other particles). The inescapable conclusion is that each electron must be delocalised in both time and space over the apparatus.

So, it is clear that the self-interference of a photon (as with the electron) occurs after the single photon travels through both slits "at once" then self-interfers "as it travels toward the detector" (that is, after the slits). And this is all I was trying to show in my initial post.

Now if both of these web sites provide false information I would appreciate an explanation why.

Last edited by a moderator: Apr 22, 2017
11. Jan 8, 2007

I am not aware of any double slit experiments where the observation point screen is "at both slits", the only ones I know about have the screen some distance (L) from the slits. In QM, it is possible for a wave function to be at two or more places at the some time--does this not answer your second question

12. Jan 8, 2007

### ZapperZ

Staff Emeritus
No, it does not. Remember, it was YOU who claimed this:

This contradicts an explanation where a photon "interferes" with itself, since you are claiming that a photon splits into two to pass through each slit, and then "come together" afterwards. This is nothing more than a 2-photon interference, something that occurs VERY SELDOM. Single-photon interference is not the same as 2-photon interference.

So based on what you claim above, I asked you if there is any detection of such a thing. You don't need a screen. Just put a detector at both slits. Now shoot photon one at a time. This has been done many times in many different experiments. If your claim is true, there should be a detection of a photon that came out of that split and passing through each of the slit. Where is this evidence?

Zz.

13. Jan 9, 2007

### Schrodinger's Dog

Ok here's how I think it works:-

If you let a single photon hit the back of the screen, and do this repeatedly, eventually an interference pattern will build up, logical conclusion? The photon must be interfering with itself(since there is only one photon present)
However, if we place a detector at both slits, we detect a photon at one or the other slits but not both? So how does it interfere with itself then if it's not detectablly in a superposition, surely it must have some superposition in order for it to affect itself?

The answer is fairly obvious if you think about it, the very act of observing the photon decoheres it's superposition, giving only one position for the photon and that is at one or the other of the slits. Since the photons no longer posesses a superposition they strike the plate at the back without causing an interference pattern, this is a proof of superpostion, although infered and is a fundemental experiment in understanding the ideas behind the Copenhagen Interpritation.

The act of observation/detection, means we can not see the exact nature of matter, because by detecting it we change its nature.

Last edited: Jan 9, 2007
14. Jan 9, 2007

### lightarrow

And this is the same as to say that "the exact nature of matter" doesn't exist without the act of measuring it.

15. Jan 9, 2007

### marlon

How exactly can you see from these figures that the self interference happens BEFORE the photon passes through the slits ? Does this imply there is no self interference after the slits ? How ?

ps : i will answer to the second question myself in the next paragraf.

How do you see that ?

Well, anyhow, what i said is that the self interference happens in between the starting point and the final detection point. Where in between ? WE DO NOT KNOW, but we KNOW FOR SURE it happens because of the observed profile at the detector.

If you would measure at one of the slits, the interference cannot be detected and does not happens because the superposition of the photon's wavefunction consisting out of the path trhough slit 1 + path through slit 2 is broken. This is called breakdown due to measurement. Again, the self interference happens between all the POSSIBLE prajectories of the photon. If you measure at one slit, you are looking at only one such trajectory and thus all the knowledge of the other trajectories is gone. So no superposition anymore, you see ?

marlon

16. Jan 9, 2007

### touqra

To resolve this issue, can we not use detectors placed at all slits, but these detectors are entangled, just like a two-particle entangled system? Have a photon pass through the slit, and see what happens to the click result of the detectors?

17. Jan 9, 2007

### marlon

No because measuring at one slit brakes the superposition over all possible photon paths. The latter is the ESSENTIAL ingredient of the observed (self) interference.

marlon

18. Jan 9, 2007

OK, thank you, I have no evidence of what I posted, it was just an idea. And since the evidence does not exist, then clearly my idea was in error--that's why I'm here at this forum--to learn.

19. Jan 9, 2007

Yes, thank you--you (and ZapperZ) make it clear that the "single" photon cannot go through both slits at the same time in the double slit experiment--never been observed--ever. But now I find this to be strange, since QM predicts that photons "in theory" can be at two different places at the same "time"--so now another question results. Are photons outside rules of QM and that is why they do not go though both slits at same time ? Or, is QM theory incorrect and it is NOT true that QM entities such as photons can be at two places (say two slits || and ||) at same time ? One of the above statements must be true, but it is not clear to me which one. Thanks for your time.

20. Jan 9, 2007

### ZapperZ

Staff Emeritus
Whaaaaa????!!!

I never said that. In fact, the whole point of me pointing out to you that single-photon interference is different than two-photon interference means what to you?

What I said that not happen is a single photon splitting into 2 and then having each one passing through both slits. This is what you were claiming. How from that to having me claim that the "single photon cannot go through both slits at the same time", I have no idea!

Zz.