Frequency or wavelength of a photon

[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.

 

[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.

 

[Jimmy Snyder]

But what does it mean by the frequency or wavelength of a photon.

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

 

[touqra]

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.

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

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...

 

[Jimmy Snyder]

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).

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.

 

[marlon]

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 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

 

[Rade]

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

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 ?

 

[ZapperZ]

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 ?

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.

 

[marlon]

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",

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

 

[Rade]

...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...

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.

 

[Rade]

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?

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

 

[ZapperZ]

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

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

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 ?

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.

 

[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 fundamental 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.

 

[lightarrow]

Ok here's how I think it works:-

...

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

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

 

[marlon]

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 ?

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.

-- 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

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

 

[touqra]

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

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?

 

[marlon]

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?

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

 

[Rade]

...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...

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.

 

[Rade]

... So no superposition anymore, you see ?

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.

 

[ZapperZ]

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.

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.

 

[marlon]

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.

That is NOT what i have said. Reread what i wrote in my previous post on the connection between measurement and breakdown of superposition.

marlon

 

[Rade]

To Marlon and/or ZapperZ: Correction in order--my confusion. Now you both make it very clear that a single photon DOES go through both slits || and || at the same time in a double slit experiment (I do not see that we have any other logical possibilities left in this discussion). But, if the single photon goes through both slits at the same time, then why no experimental observation of this phenomenon when detectors are put at the ends of both slits, for as stated by ZapperZ: If this is true [eg, that a single photon goes through both slits at the same time], then you would detect a signal at both slits. So where is it? And this comment makes it clear that no experiment has ever observed a single photon leaving both slits at the same time. So, some clarification would be helpful--that is, if the "single" photon DOES go through both slits at the same time, why is there no experiment evidence of this when you measure for signals at both slits at the exact same time the "single" photon leaves each slit ? Are you saying we can measure signal from only one slit at any time and not at both slits simultaneously ?

 

[ZapperZ]

To Marlon and/or ZapperZ: Correction in order--my confusion. Now you both make it very clear that a single photon DOES go through both slits || and || at the same time in a double slit experiment (I do not see that we have any other logical possibilities left in this discussion). But, if the single photon goes through both slits at the same time, then why no experimental observation of this phenomenon when detectors are put at the ends of both slits, for as stated by ZapperZ: If this is true [eg, that a single photon goes through both slits at the same time], then you would detect a signal at both slits. So where is it? And this comment makes it clear that no experiment has ever observed a single photon leaving both slits at the same time. So, some clarification would be helpful--that is, if the "single" photon DOES go through both slits at the same time, why is there no experiment evidence of this when you measure for signals at both slits at the exact same time the "single" photon leaves each slit ? Are you saying we can measure signal from only one slit at any time and not at both slits simultaneously ?

The interference pattern IS the evidence!

The whole principle of superpostion, as Schrodinger tried to illusrate, is that the moment you measure an observable, the wavefunction for collapse for that observable, producing a definite outcome. The same with the photon. If you tried to measure which slit, you'll only get one or the other, but at the same time, the pattern you see on the screen will also differ.

But when you measure what is known as a "non-commuting observable", then the superpostion due to that first observable is maintained and you can now detect that effect. In the 2-slit experiment, the non-commuting observable to which-slit-observable is the interference pattern (which is really a momentum or frequency observable). The interference pattern, that gives a value for the photon momentum/frequency does not "interfere" with the superposition of position/path that the photon made when it went through the slit.

Again, I've mentioned about the Schrodinger-Cat type experiments a lot, especially lately. The Delft/Stony Brook experiment illustrate such an effect.

Remember, in QM, if two observables A and B do not commute, i.e. [A,B]!=0, then A and B do not share the same eigenfunction. So if you measure A, you still have the wavefunction in a mixed state for B. Example, if you measure the z-comp of the angular momentum Lz, Lx and Ly are still undetermined, because Lx and Ly do not commute with Lz. The is one of the foundation of QM. In fact, the commutation relations between operators have also been called as the First Quantization.

Zz.

 

[Anonym]

Remember, in QM, if two observables A and B do not commute, i.e. [A,B]!=0, then A and B do not share the same eigenfunction. So if you measure A, you still have the wavefunction in a mixed state for B.

The theoretical reality (formalism) is much more sophisticated. [A,B]!=0 is generally correct operator eq., which physically means nothing unless you specify it particular realization and measurable value (see R.Jackiw,JMP,9,339,(1968)).

 

[ZapperZ]

The theoretical reality (formalism) is much more sophisticated. [A,B]!=0 is generally correct operator eq., which physically means nothing unless you specify it particular realization and measurable value (see R.Jackiw,JMP,9,339,(1968)).

.. and you think I need to do that here within the context of this discussion? I'm surprised you didn't just point to Sakurai's text.

Zz.

 

[Anonym]

.. and you think I need to do that here within the context of this discussion? I'm surprised you didn't just point to Sakurai's text.

No. My intention was only to contribute something that maintain “duality” or “complimentarity”. You mentioned important experimental material and I added the important theoretical material for the theoretically oriented kids.By the way, I am not sure that W.Heisenberg understood that point then.

 

[gil7]

How do you describe the photon with the Strng theory?

 

[Schrodinger's Dog]

How do you describe the photon with the Strng theory?

I personally wouldn't bother but that's entirely a matter of choice

Rade, go back and look at my description and try and draw what I've described, when you picture the experiment on paper it's a lot easy to see what is going on, if your a little lazy though this site shows what is happening pretty well.

http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html

And this rather neatly explains what is going on too.

http://thisquantumworld.com/ht/content/view/17/27/

What this means.

http://groups.msn.com/Tyrannicidae/twoslitexperiment.msnw

 

[Rade]

...Rade, go back and look at my description and try and draw what I've described, when you picture the experiment on paper it's a lot easy to see what is going on, if your a little lazy though this site shows what is happening pretty well...

Thank you, the links were very good. I have another question, based on this statement below that you made in your first post:

...if we place a detector at both slits, we detect a photon at one or the other slits but not both

Now I'm going to be difficult and suggest a "null hypothesis" that it is physically impossible to design an experiment to observe what you claim above--that is, it is not possible to detect (AT BOTH SLITS SIMULTANEOUSLY) any single photon going through two slits ||+|| at EXACTLY THE SAME TIME AT BOTH SLITS. {edit} That is, of course we only see what happens at one slit or another, because it is not physically possible to detect at both slits simultaneously, EVEN IF THE PHOTON WAS IN REALITY AT BOTH SLITS SIMULTANEOUSLY--do you understand what I am asking here ? Now, forget about "the pattern that is observed at the screen some distance (L) away from the slits"--I want to know "WHAT IS HAPPENING AT THE SLITS, NOT THE SCREEN--this is what I cannot grasp !

Now, obviously there must be experiments that falsify my null hypothesis, since it would appear that your explanation and the explanations provided in the links you provided "assume" that my "null hypothesis" has been experimentally falsified --sooooo, could you please give me the prime literature citation that has been conducted that has falsified my null hypothesis (that I place in bold text above) so that I can read it. Thank you for your time.

 

[Schrodinger's Dog]

You don't grasp how waves behave when they hit slits is the problem here. So I'll try and introduce some analogies with the real world to show the behaviour of waves in real world terms and in quantum mechanics.

If you look at the object in question as a wave, which has a superposition, like when you wiggle a spring both up and down and side to side, and someone wiggles the spring slightly slower at the other end, this shows within the spring all of these waves mixed to gether in a sort of real world view of what we're talking about.

http://www.phys.hawaii.edu/~teb/java/ntnujava/waveSuperposition/waveSuperposition.html

http://www.answers.com/topic/quantum-superposition

More specifically, in quantum mechanics, any observable quantity corresponds to an eigenstate of a Hermitian linear operator. The linear combination of two or more eigenstates results in quantum superposition of two or more values of the quantity. If the quantity is measured, the projection postulate states that the state will be randomly collapsed onto one of the values in the superposition (with a probability proportional to the square of the amplitude of that eigenstate in the linear combination).

In English they're talking about decoherence here.

Basically two states in a superposition collapse to form one single state randomly, same with any number of states in a superposition.

The reason you can't detect two photons is simply because,their aren't two photons in Feynmans two slit just one,you need to grasp the concept of superposition and that it is decohered by striking a slit all the other "circles" disappear.

Think of it like a water wave hitting the slits, the wave itself behaves much the same in that it spreads out in a pattern which interferes with itself, if you did the same experiment with dyed water you'd find the same pattern emerging on a white screen - make that anaologous to your view of a wave - the difference is that once we try and measure the wave it decoheres so there is no superposition any more, thus no way for the photon's wave/superposition to pass through both slits thus the decoherence is what you are looking at, when we don't try to detect the photon then the wave continuous unimpeded and undecohered and we see the nature of light which isn't decohered.

If there is no detector there is no decoherence, and the photons super position passes through both slits, interfering with itself.

What this suggests is that the very act of measuring causes something to be quantifiable, but at the same time disturbs what we are really trying to measure. It shows amply that light is both a wave and a particle regardless of how we chose to measure it, it shares a duality, which is demonstrated neatly by the experiment, but more importantly it demonstrates by inference that superposition and the Copenhagen intepritation are viable.

http://plato.stanford.edu/entries/qm-copenhagen/

Look at the wave pattern in the first links I gave, can you visualise the super position going through both slits in an open two slit experiment where one isn't closed?

http://www.freewebs.com/mypicturesandsht/water4.gif"

Right...

What happens when when we detect a photon? It's a bizarre thing to think about, but if you feel it's unusual or weird you are not alone

 

[Rade]

...The reason you can't detect two photons is simply because,their aren't two photons in Feynmans two slit just one,you need to grasp the concept of superposition and that it is decohered by striking a slit all the other "circles" disappear...

Thanks very much for your explanation and time. One last question based on the above statement (maybe). Suppose a situation where there were "two photons" (eg, simultaneously at both sites) and not "just one"--do you know of any experimental design at could detect "two photons simultaneouly" at the slits--can you provide a reference ?

 

[marlon]

that is, it is not possible to detect (AT BOTH SLITS SIMULTANEOUSLY) any single photon going through two slits ||+|| at EXACTLY THE SAME TIME AT BOTH SLITS. {edit} That is, of course we only see what happens at one slit or another, because it is not physically possible to detect at both slits simultaneously, EVEN IF THE PHOTON WAS IN REALITY AT BOTH SLITS SIMULTANEOUSLY--

The fact that "the photon passes through both slits" is a direct consequence of the QM formalism, ie the superposition of both slits in the photon's wavefunction. I think we all agree on that. Now, that same formalism clearly proves that if you measure at the slits, the superposition in the photon's wavefunction is broken. So you no longer have this sum of "passing through slit 1" and "passing through slit 2". In this case, the photon will pass either through slit 1 OR slit 2, as we will indeed observe if we measure at the slits. THAT IS ALL THERE IS TO IT.

To be clear, if we do not measure at the slits, the formalism only states that we do NOT know what trajectory the photon is following. We only know that there is wavelike behaviour as we observe at the detector. So, actually talking about "a photon that passes through both slits" is meaningless and useless.


marlon

 

[Rade]

...actually talking about "a [single] photon that passes through both slits" is meaningless and useless.

Thank you Marlon, I now understand the essence of the concept "constraint" in quantum mechanics.

 

[marlon]

Thank you Marlon, I now understand the essence of the concept "constraint" in quantum mechanics.

I am not quite sure what you mean here. Just to be sure, a single photon does give us interference patterns and one could indeed say "that is passes through both slits" but using this language is confusing if one is not familiar with the underlying concept. That is what i wanted to say.

marlon

 

[Haelfix]

Feynman lectures, volume 3, chapter 1.

No clearer account has ever been written about the fundamental problem of QM.