How many photons are reflecting?

[Alvydas]

Hello,
Lets we have some small diode laser pointing to a mirror.
Lets say 10mW red laser something like this maybe
http://www.aliexpress.com/wholesale?SearchText=10mw+laser+red&catId=0&manual=y

The question: how many photons are reflecting now?

 

[Simon Bridge]

Depends on the mirror, and the intensity of the laser.

 

[Dale]

The energy of a 650 nm photon is E = hc/λ = 3E-19 J. So a 10 mW output would correspond to 3.3E16 photons/s from the laser. Then you would need to know what the reflectivity of the mirror is to determine how many are absorbed vs. reflected.

 

[Alvydas]

The energy of a 650 nm photon is E = hc/λ = 3E-19 J. So a 10 mW output would correspond to 3.3E16 photons/s from the laser. Then you would need to know what the reflectivity of the mirror is to determine how many are absorbed vs. reflected.

You have calculated how many photons was emitted per second.
But I am asking how many of them are touching mirror at current moment.

I think we need length of photon also.

Lest say reflectivity is 100%.

 

[Drakkith]

You have calculated how many photons was emitted per second.
But I am asking how many of them are touching mirror at current moment.

I think we need length of photon also.

Lest say reflectivity is 100%.

We don't know. We can't know. It is statistical in nature. And this ignores the difficulties of saying a photon "touches" the mirror in the first place. At the quantum level things are not like they are at our scale.

 

[Dale]

You have calculated how many photons was emitted per second.
But I am asking how many of them are touching mirror at current moment.

I think we need length of photon also.

I don't think that photons have a length. I cannot even think of a way that the length of a photon could be measured in principle.

As far as the expectation for how many are touching "now", the answer would be zero. This should be obvious from the well known properties of statistical distributions. If you want to know the expectation for how many are touching over some finite time period of duration Δt then simply take 10 mW times Δt to get the energy and divide that by the energy per photon above.

 

[Drakkith]

I don't think that photons have a length. I cannot even think of a way that the length of a photon could be measured in principle.

Well, I think many people think of it as the wavelength of the EM wave. However this really depends on what you view photons as. Personally I only use photons to describe the quantized interaction of the wave, in which case the wavelength of a photon doesn't make any sense. However that is just my own way of thinking of it, mostly as a result of multiple discussions here on PF about what a photon is.

 

[Dale]

Personally I only use photons to describe the quantized interaction of the wave, in which case the wavelength of a photon doesn't make any sense.

Yes, I agree. Roughly speaking I think of a photon as a quantized excitation of the EM wavefunction. I guess you could define the photon's length by taking its wavefunction and doing some sort of 95% confidence interval, but that seems awfully artificial.

 

[sophiecentaur]

I don't think that photons have a length. I cannot even think of a way that the length of a photon could be measured in principle.

As far as the expectation for how many are touching "now", the answer would be zero. This should be obvious from the well known properties of statistical distributions. If you want to know the expectation for how many are touching over some finite time period of duration Δt then simply take 10 mW times Δt to get the energy and divide that by the energy per photon above.

Yes: if they had "a length" and, of course, all photons of the same energy would have to be identical, then they would all have to have the same "length", whatever mechanism happened to produce them. The only possible value of length that would satisfy this requirement is Zero. This agrees with the accepted 'extent' of a photon.

Consider this, also. The wavelength of visible light is conveniently small to allow you to consider the photons as having a nice short length (in an arm-waving kind of a way and something related to the wavelength) but how long would the photons of an lf radio wave need to be - with a wavelength of a couple of km? Best not try to allocate a length, I think.

 

[Dale]

Yes: if they had "a length" and, of course, all photons of the same energy would have to be identical, then they would all have to have the same "length", whatever mechanism happened to produce them. The only possible value of length that would satisfy this requirement is Zero. This agrees with the accepted 'extent' of a photon.

I never like asserting that the length or size of a particle is zero since that always unfortunately conjures up the idea of classical point particles zipping around, and even non-classically the wavefunction is not a dirac delta in general. So I would rather say that it doesn't have a length than to say that it has a length and that length is zero.

What is your opinion of saying that it doesn't have a length rather than saying that its length is 0?

 

[ZapperZ]

You people should not argue with each other on this matter and let the OP get away with asking a rather meaningless question. I mean, "... Touching mirror at current moment..."? No one questioned that?

Zz.

 

[Alvydas]

But to emit instantaneously electron could move much quicker then c ?

 

[Dale]

But to emit instantaneously electron could move much quicker then c ?

Huh? No, an electron has mass, so it cannot move quicker than c. That said, I have no idea how you think that fits in with the rest of everything.

Btw, coherent states, like lasers, don't have a definite number of photons.

 

[Alvydas]

Huh? No, an electron has mass, so it cannot move quicker than c. That said, I have no idea how you think that fits in with the rest of everything.

Btw, coherent states, like lasers, don't have a definite number of photons.

So very short time after emission we have the same atom like before,
because any parts of it can not change their dislocation due c limit so quickly
+ just emitted new photon?

 

[Dale]

So very short time after emission we have the same atom like before,
because any parts of it can not change their dislocation due c limit so quickly
+ just emitted new photon?

What does "dislocation due c limit" mean or have to do with emitting a photon. Are you under some confusion that a massive particle must move at c in order to emit a photon? If so, that is not correct, I don't know why you would think that.

 

[Bobbywhy]

Jan27-12 08:23 AM

Alvydas, Have you read the Rules of this forum, as I suggested to you recently? If you had, or if you do, and you have any understanding of the English language, then you will NOT post this speculation here. Remember what happened 10 days ago to this same ficticious whacko off-the-wall stuff. Your thread was locked until you reference some peer-reviesed journal/source. You have only rewittten the same old broken-down false ideas from you last attempt at posting here. Not to mention your ridiculous comparisons of theoretical subatomiic particles such as "fruits" and "apples"! You are either really stupid, or unbalanced, or drunk, or so emotionally attached to these uproven phony wastebasket garbage theories that you cannot control your own actions. Bye bye.

 

[Alvydas]

What does "dislocation due c limit" mean or have to do with emitting a photon. Are you under some confusion that a massive particle must move at c in order to emit a photon? If so, that is not correct, I don't know why you would think that.

I do not think that electron must have some proper velocity to emit in classical way.
But instantaneous emission makes some confusion.
Like I mentioned after short time after emission we have the same atom + a new photon.
I am right?
If so (this atom is still is the same) why it can not emit 10 the same photons more?

 

[Alvydas]

Jan27-12 08:23 AM

Alvydas, Have you read the Rules of this forum, as I suggested to you recently? If you had, or if you do, and you have any understanding of the English language, then you will NOT post this speculation here. Remember what happened 10 days ago to this same ficticious whacko off-the-wall stuff. Your thread was locked until you reference some peer-reviesed journal/source. You have only rewittten the same old broken-down false ideas from you last attempt at posting here. Not to mention your ridiculous comparisons of theoretical subatomiic particles such as "fruits" and "apples"! You are either really stupid, or unbalanced, or drunk, or so emotionally attached to these uproven phony wastebasket garbage theories that you cannot control your own actions. Bye bye.

I nothing had prosed at this tread. Do you ban even questions?

 

[sophiecentaur]

I never like asserting that the length or size of a particle is zero since that always unfortunately conjures up the idea of classical point particles zipping around, and even non-classically the wavefunction is not a dirac delta in general. So I would rather say that it doesn't have a length than to say that it has a length and that length is zero.

What is your opinion of saying that it doesn't have a length rather than saying that its length is 0?

Either would be fine with me. No value or a value of zero would both avoid a specific length. Excluding the possibility of a value could be hard for some people to accept, though. But the whole idea of a wave function 'collapsing' when a photon is detected implies instantaneous transfer of information etc. etc. which is also hard to accept. It's all a bit too much for a normal brain, I think.

 

[Drakkith]

I do not think that electron must have some proper velocity to emit in classical way.
But instantaneous emission makes some confusion.
Like I mentioned after short time after emission we have the same atom + a new photon.
I am right?
If so (this atom is still is the same) why it can not emit 10 the same photons more?

Hold on. If the emission of a photon is by an electron dropping energy levels in an atom or molecule, the electron does NOT move anywhere. to my knowledge the change in energy levels can be instant because there is no motion here. It can't emit 10 photons because the change in energy levels gives up energy in the form of the photon and the electron is now in a new state. It can't go back and drop energy levels again, we would have to excite it first, which requires energy.

 

[Dale]

I do not think that electron must have some proper velocity to emit in classical way.
But instantaneous emission makes some confusion.
Like I mentioned after short time after emission we have the same atom + a new photon.
I am right?
If so (this atom is still is the same) why it can not emit 10 the same photons more?

In order to emit a photon (whether stimulated or spontaneous) the atom has to be in an excited state. After the emission it is in a non excited state. The energy difference between the two states is equal to the energy in the emitted photon. It cannot emit more photons unless it is re excited.

 

[sophiecentaur]

In order to emit a photon (whether stimulated or spontaneous) the atom has to be in an excited state. After the emission it is in a non excited state. The energy difference between the two states is equal to the energy in the emitted photon. It cannot emit more photons unless it is re excited.

It can move to its ground state in more than one step, though, emitting two or more photons of the appropriate energy as it passes through intermediate states.

 

[Drakkith]

It can move to its ground state in more than one step, though, emitting two or more photons of the appropriate energy as it passes through intermediate states.

True, bu they wouldn't all be the same energy, as the poster was asking. At least, I think that's what he was asking.

 

[GarageDweller]

Jan27-12 08:23 AM

Alvydas, Have you read the Rules of this forum, as I suggested to you recently? If you had, or if you do, and you have any understanding of the English language, then you will NOT post this speculation here. Remember what happened 10 days ago to this same ficticious whacko off-the-wall stuff. Your thread was locked until you reference some peer-reviesed journal/source. You have only rewittten the same old broken-down false ideas from you last attempt at posting here. Not to mention your ridiculous comparisons of theoretical subatomiic particles such as "fruits" and "apples"! You are either really stupid, or unbalanced, or drunk, or so emotionally attached to these uproven phony wastebasket garbage theories that you cannot control your own actions. Bye bye.

Bit harsh, he has probably not received any tuition in this field but has interest.
I suggest learnin the real quantitative physics, grab a book.

 

[Dale]

It can move to its ground state in more than one step, though, emitting two or more photons of the appropriate energy as it passes through intermediate states.

Yes, that is why I said "non excited" rather than "ground." Although, I guess "less excited" would have been even more accurate.

 

[Alvydas]

Btw, coherent states, like lasers, don't have a definite number of photons.

Why? You may spread light onto very big photosensitive equipment and count them individually. Or decrease number of photons by semitransparent mirrors and count them individually also.

 

[Alvydas]

If the emission of a photon is by an electron dropping energy levels in an atom or molecule, the electron does NOT move anywhere. to my knowledge the change in energy levels can be instant because there is no motion here.

Yes, you understood what I am talking about.
Do atom's size really stays the same after emission?

If you think photon is a point than maybe question can be transformed to:
what size is space where we may find it at current moment?

 

[Alvydas]

There is one more interesting sequence from idea about point like photon.
First experiments for photoelectric effect maybe was not right understood.
They have varied intensity of red light and no photoelectric effect was obtained,
but even weak blue light made this effect.

But if photon is a point it can not overlap each other and therefore always acts alone on an atom. Even if experimenters arise intensity of red photons the atom can not catch two of them simultaneously because of their small/zero size.

But maybe here is possible some big enough intensity to produce photoelectric effect with weak photons? This maybe may suggest that they start overlap each other?

 

[Drakkith]

Why? You may spread light onto very big photosensitive equipment and count them individually. Or decrease number of photons by semitransparent mirrors and count them individually also.

True, but if you take a slice of time, let's say 1 second, and count all the photons that impact the detector it will NOT match the next 1 second slice, or the next slice, or the one after. Each will vary in counts. If we have on average 100 counts per second, there is a 64% chance that each 1 second slice will have 90-110 counts, a 86% chance it will have 80-120 counts, etc. I didn't do the math so the numbers may not be exactly right, but that's the general idea, that the photons are generated randomly.

Yes, you understood what I am talking about.
Do atom's size really stays the same after emission?

If you think photon is a point than maybe question can be transformed to:
what size is space where we may find it at current moment?

I hesitate to comment on their size, as it's not really that simple. At the quantum level we'd need to determine where we wanted to say the atom is since all its component particles exist as wavefunctions which don't have a defined position. As for the position of the photon, that is found by the probability amplitude of its wavefunction which depends on its energy and what it is interacting with. But the specific location is still random.

There is one more interesting sequence from idea about point like photon.
First experiments for photoelectric effect maybe was not right understood.
They have varied intensity of red light and no photoelectric effect was obtained,
but even weak blue light made this effect.

But if photon is a point it can not overlap each other and therefore always acts alone on an atom. Even if experimenters arise intensity of red photons the atom can not catch two of them simultaneously because of their small/zero size.

But maybe here is possible some big enough intensity to produce photoelectric effect with weak photons? This maybe may suggest that they start overlap each other?

No, what would happen is that the atom would absorb multiple photons at once and transfer their energy to an electron to cause it to jump out of the material. This has nothing to do with photons overlapping, but with the sheer number of photons simply making it likely that an atom will absorb many at one time. (I'm not actually sure exactly what would happen, but I think an atom can absorb multiple photons at once)

 

[Simon Bridge]

@Alvydas ... It would be helpful if you would address the problems that have come up with your previous statements before throwing in some more - it makes you look somewhat "trollish".

eg. What did you mean by "length of a photon"?
(BTW: @ sophiecentaur etc, is is common for people to define "the wavelength of a particle" through the deBroglie relation as you well know. Isn't it somewhat pointless, however, to argue the toss before OP has explained what was meant?)

eg. what do you mean by "touching the mirror"? You know - considering that, on the scales that talking about "photons" makes sense, there is no such thing as a "surface" for the mirror.

For me, I don't understand the difficulty with the your question: what is it about reflection re photons that you don't understand?

Usually I'd take issue with the vagueness of your wording too - but I'm guessing that English is not your first language. Is that the case?

For instance:

So very short time after emission we have the same atom like before,
because any parts of it can not change their dislocation due c limit so quickly
+ just emitted new photon?

... you seem to be saying that you think the atom is unchanged by interactions with photons. However, this is not correct. When an electron absorbs a photon, for example, it is not re-emmitted right away.

When you are considering the physics of reflection, you have a choice of models depending on what you are using to look at it. For the photo-electron level, you will do well with QED - which is pretty much the go-to model at this level.

... that's lecture 1, see all 8. He deals with reflection from a "surface" as one of the examples.

 

[Alvydas]

but I'm guessing that English is not your first language. Is that the case?

Yes third only, sorry for language mistakes.

Maybe someone may estimate what density of light is requared so that notable amount of cases atoms would absorb two photons simultaneously (instead of 1).

Lets say 20% of cases we can consider like notable.

This also would give me some info I was looking by asking indirect question like one at the beginning of this tread.

So how strong light could be to make photoeffect with photons a bit weaker (longer wave lenght) then usually needed.

 

[ZapperZ]

Please do a search on multiphoton photoemission. This is a very well-known phenomenon that we have already been using. The original photoelectric effect is a single-photon photoemission phenomenon!

This thread has meandered all over the place.

Zz.

 

[Dale]

Why? You may spread light onto very big photosensitive equipment and count them individually. Or decrease number of photons by semitransparent mirrors and count them individually also.

I don't know if you are aware of the uncertainty principle in quantum mechanics. For coherent states, like a laser, there is an uncertainty relationship between the phase and the number of photons.

http://en.wikipedia.org/wiki/Coherent_states

 

[Dale]

If you think photon is a point than maybe question can be transformed to:
what size is space where we may find it at current moment?

I do not think that a photon is a point, but I briefly discussed the approach you are suggesting here back in post 8.

 

[Simon Bridge]

Yes third only, sorry for language mistakes.

No worries: we are all very skilled communicators so we can compensate as I'm sure you do when someone tries to talk to you in your first language when they are not used to it... well most of us anyway. Your skills are better than a lot of native English speakers I've met ;)

This also would give me some info I was looking by asking indirect question like one at the beginning of this thread.

It is better to be direct - especially when you are talking to scientists. You will get much better help a lot sooner. Don't worry about appearing silly or anything - we've all been there: we understand.

ZapperZ has given you a good lead to follow. Enjoy :)

 

[Dale]

But if photon is a point it can not overlap each other and therefore always acts alone on an atom.

First, as I said above, it is bad to think of a photon as a classical point particle, which appears to be what you are describing. However, because an atom does have size even if photons do not you could still get multiple photons interacting with the same atom at a given time.

 

[ZapperZ]

Furthermore, the multiphoton effect has nothing to do with the "size" of the photon, but rather the time scale of the interactions. In multiphoton photoemission, it is the time for an excited state to decay back to its ground state. If a photon can excite a particular electron while it is in an excited (but bound state) of the solid before it decays back to the valence band, then it has a chance to be excited to an even higher state. The factor that affects this includes the photon density per unit area, and the decay rate of the excited state. It has nothing to do with the size of the photon.

Zz.

 

[Alvydas]

So I can send strong infrared beam to a mirror and get reflected visible light ?
But not due surface heating, just because of reemitting photons with double energy by the mirror.
Am I right?
If so what (approximately) power of incoming laser light I need to have at 1 square mm of mirror's surface? (to be effect visible)

 

[sophiecentaur]

You are referring to Laser Up-converting, which is a process that can be used successfully. Google it - there are a lot of links. Most lasers work with a single 'pumping' wavelength applied and a longer wavelength laser light is produced as atoms return to a lower state. However, it is possible to pump some lasers with more than one wavelength and the resulting EM will have a higher energy than either of the pumping sources because the atoms are pumped to a level equal to the sum of the energies.

This will only happen under rare circumstances where the probability of an atom that has already been 'pumped' once, will decay is less than the probability that another incoming photon will pump it further. In most circumstances, the once-excited atom will decay before this can happen.

 

[ZapperZ]

So I can send strong infrared beam to a mirror and get reflected visible light ?

That has not been shows to can occur. You are applying different physics to different phenomenology. The multiphoton effect is on a photoemission phenomenon, NOT a "reflection".

There's just something not quite right with what you are thinking and with your question.

Zz.

 

[sophiecentaur]

That has not been shows to can occur. You are applying different physics to different phenomenology. The multiphoton effect is on a photoemission phenomenon, NOT a "reflection".

There's just something not quite right with what you are thinking and with your question.

Zz.

Yes, I agree it's a bit tenuous but, there again, when light is reflected in a mirror, the interaction isn't just one photo - one atom, in any case, but with the bulk material. Perhaps that should have been emphasised more earlier on.
I thought this up-conversion laser effect occurred in the solid state so the effect would be not 'specular' reflection but would still involve light coming off a surface when illuminated with two other, longer, wavelengths.

 

[Dale]

So I can send strong infrared beam to a mirror and get reflected visible light ?
But not due surface heating, just because of reemitting photons with double energy by the mirror.
Am I right?

Not to my knowledge, no. Reflection has little, if anything, to do with atomic emission.

 

[ZapperZ]

Yes, I agree it's a bit tenuous but, there again, when light is reflected in a mirror, the interaction isn't just one photo - one atom, in any case, but with the bulk material. Perhaps that should have been emphasised more earlier on.
I thought this up-conversion laser effect occurred in the solid state so the effect would be not 'specular' reflection but would still involve light coming off a surface when illuminated with two other, longer, wavelengths.

I didn't want to get into that because the OP has a lot more issues with other basic understanding.

It is my opinion that this thread is nothing but a jumble of disconnected information that the OP has no ability to process correctly. This is one of those times where I often wish that people would ask just simple, basic question first before attempting to find something beyond what they are capable of understanding.

Zz.

 

[sophiecentaur]

Yep.
I agree entirely. I think people are reluctant to ask the most basic questions for fear of appearing ignorant.

 

[Alvydas]

Thank you all.
I have got some useful info.
I will try to adjust my style of asking to be more clear.