Photon Size: Direction & Probability

[alvaros]

When a photon is emitted it goes in all directions ? or just in one ? N, S, E..
Or there is a probability you can find it in any direction ?

 

[olgranpappy]

emitted by what?

 

[alvaros]

By my body, or by a dipole, or by an excited atom.
Is there any difference ?

 

[Schrodinger's Dog]

A photon does not have a size in any real terms, it has no mass AFAIK or anyone else does, so if you're talking about how "big" is the wavelength or what is its direction that's not really indicative of size of a photon just it's path after emission, it's an unanswerable question. The question is phrased poorly I think...

 

[meopemuk]

A photon does not have a size in any real terms, it has no mass AFAIK or anyone else does, so if you're talking about how "big" is the wavelength or what is its direction that's not really indicative of size of a photon just it's path after emission, it's an unanswerable question. The question is phrased poorly I think...

This is true. In order to have an answerable question one needs to specify how photons are prepared, where are measuring devices, and what are they measuring. Then quantum mechanics will exactly predict the probabilities of measurement results. Still, QM won't be able to tell what each individual photon will be doing.

Eugene.

 

[alvaros]

Schrodinger's Dog wrote:
"The question is phrased poorly I think...", shure it is.

The cuestion arises from:

A dipole ( an antenna= aerial ) is just made up by two wires that can be very thin. A dipole receives photons from a surface much bigger than their surface ( length * whidth of the wires ). So, I infer, the photons whose path is not exactly through the wires of the dipole can be "captured". This will give us a size of the photons.

But, if you put another wire in front of your dipole, it will "capture" more photons. "In front" means in the line between the emitter antenna and the receiver. So, whith this device ( a yagui antenna, used to receive TV signals ) you receive photons from a bigger surface.

Where is the limit ?

 

[meopemuk]

Schrodinger's Dog wrote:
"The question is phrased poorly I think...", shure it is.

The cuestion arises from:

A dipole ( an antenna= aerial ) is just made up by two wires that can be very thin. A dipole receives photons from a surface much bigger than their surface ( length * whidth of the wires ). So, I infer, the photons whose path is not exactly through the wires of the dipole can be "captured". This will give us a size of the photons.

But, if you put another wire in front of your dipole, it will "capture" more photons. "In front" means in the line between the emitter antenna and the receiver. So, whith this device ( a yagui antenna, used to receive TV signals ) you receive photons from a bigger surface.

Where is the limit ?

I guess you are confusing the size of the wavelength of photon's wave function with the particle size. The wavelength can be made as large as you wish. Still the photon should be treated as a pointlike particle.

Eugene.

 

[alvaros]

I guess you are confusing the size of the wavelength of photon's wave function with the particle size. The wavelength can be made as large as you wish. Still the photon should be treated as a pointlike particle.

Eugene.

No, I am not confusing the size of the wavelength of photon's wave function with the particle size.

The photon can't be a "pointlike particle". Remember the experiments about interference: a single photon goes through two holes.

 

[Schrodinger's Dog]

No, I am not confusing the size of the wavelength of photon's wave function with the particle size.

The photon can't be a "pointlike particle". Remember the experiments about interference: a single photon goes through two holes.

This is explained by wavelike behaviour: superposition of a wave, not by particle behaviour? I think you need to look at the two slit experiments carefully, they do not suggest a size of a photon at all or particle behaviour as going through both slits.

The photon itself only ever travels through one slit in terms of a single photon experiment and it does so in a random fashion and with a 50/50 certainty as shown by Feynman's two slit experiment.

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

 

[Gza]

Why does this question come up every other post in this forum?

 

[Schrodinger's Dog]

I don't know perhaps the sources people learn from are rubbish or it's hard to get a grip on the implications of the two slit experiment. Whatever the cause I think this could do with a FAQ, but then what would the question? Although the answers are invariably the same, the questions vary widely.

There's already a "does a photon have mass?" FAQ, but it doesn't cover it completely.

https://www.physicsforums.com/showpost.php?p=1285138&postcount=6

 

[lightarrow]

Why does this question come up in every other post in this forum?

Because it's related to the essence of QM weirdness and to dualism. If physicists talk about particles, and they never say these particles are in the detector, people will, rightly, think that these particles have to fly from source to detector. Then, the conclusion that they go in both slits is straightforward, as are the strange conclusions about their size.

I would like to tell the OP that, even if not original, his question is not a stupid question; it's the result of QM books/QM lesson's schizophrenic assumptions about particles.

 

[alvaros]

To Schrodinger's Dog: Thank you for the link "http://www.upscale.utoronto.ca/Gener...oubleSlit.html" but I am not discussing about duality or Heisenberg Uncertainty Principle.

Define size of photon = surface ( or angle, I don't know ) where it can be detected.

Detector = antenna. The photon is detected if it contributes to the received signal ( if its electromagnetic energy is converted to electric current ).

I think I didnt explain well the cuestion.

A dipole detects photons whose path does not cross the wires. It detects photons whose path is a little up or down the dipole . A parameter of an antenna is the Equivalent Surface.

But if you put another wire in front of the dipole the Equivalent Surface gets bigger.

If you put two wires ... bigger.

The real antennas have a lot of wires in front of the dipole as you can see.

So it seems that if you put enough wires you can detect the photons whose path is 10 m up or down the dipole.

How big is a photon ? ( Where a single photon can be detected ? )

Note that I started the thread from the very beginning: Have the photons a path ?

 

[Proof.Beh]

How big is a photon ? ( Where a single photon can be detected ? )

It is obvious that your second question has "nowhere" answer. Since the calm mass of a photon is zero and therefor you don't detect a free photon in space (In fact it is a point because of that). Your question like that ask: where a single electron that has observed in space-time? Of course both of them (electron and photon) don't equal together really but indeed, we don't answer to these questions to when don't observe them in nature to form of single. But I think thus.

Thanks.
Mr Beh

 

[marlon]

When a photon is emitted it goes in all directions ? or just in one ? N, S, E..
Or there is a probability you can find it in any direction ?

How does light propagate ? Indeed, isotropically...

No, I am not confusing the size of the wavelength of photon's wave function with the particle size.

The photon size is a concept that does not exist because size is defined in terms of spatial coordinates. A photon however is defined as a chunk of energy. Energy is a concept that is NOT defined in terms of spatial coordinates. Do you see the contradictio in terminis ?

The photon can't be a "pointlike particle". Remember the experiments about interference: a single photon goes through two holes.

Yes it can be because the particle (position and momentum) and wave (wavelength, frequency) nature are dual. They are like two different languages to say the same thing : ie a photon is a chunk of energy.

Besides, in an energy base (the coordinates are now values of energy), the photon is a nice point particle.

marlon

 

[Proof.Beh]

The photon size is a concept that does not exist because size is defined in terms of spatial coordinates. A photon however is defined as a chunk of energy. Energy is a concept that is NOT defined in terms of spatial coordinates. Do you see the contradictio in terminis ?

There is no sense for your answer. If a light ray emits from a lighting source, we allocate to that length, diagonal of light ray etc and know that a light ray made of the number of bounded photons. Then the light has spatial coordinates and thus a photon also is this form.

Thanks.
Mr Beh

 

[marlon]

There is no sense for your answer. If a light ray emits from a lighting source, we allocate to that length, diagonal of light ray etc and know that a light ray made of the number of bounded photons.

Yes but that does not imply the photon is a point particle in coordinate space, ie a particle with finite sized soatial boundaries. THAT is the point i was trying to make.

besides, what are "bounded" photons. I always thought that photons do NOT mutually interact (at least up the the first orders of EM interaction).

Then the light has spatial coordinates and thus a photon also is this form.

Thanks.
Mr Beh

How on Earth can you make this conclusion. Not only that, you are also talking about a photon's form. What is that ?

When you talk about form in this context, you are talking about a shape defined by finite spatial boundaries. A photon is NOT defined in this way. If you do not agree with me, i politely ask you to provide me with such a definition. Realise that you cannot use the wavelike photon concept to answer that question because of the reasons in gave in my previous post.


marlon

 

[Anonym]

The photon size is a concept that does not exist because size is defined in terms of spatial coordinates. A photon however is defined as a chunk of energy. Energy is a concept that is NOT defined in terms of spatial coordinates.

You repeat the same statements that you made about year ago. It reminds me the attitude of the average engineer. Energy is eigenvalue (diagonal matrix element) of the Hamiltonian operator when the QM state is the function of space-time coordinates. Moreover, it is the most fundamental conserved quantity. I suggest starting study of quantum theory.

To begin with it is useful to read posts in “size of photon particle” and M&W after that.

Regards, Dany.

 

[Proof.Beh]

Yes but that does not imply the photon is a point particle in coordinate space, ie a particle with finite sized soatial boundaries. THAT is the point i was trying to make.

besides, what are "bounded" photons. I always thought that photons do NOT mutually interact (at least up the the first orders of EM interaction).


How on Earth can you make this conclusion. Not only that, you are also talking about a photon's form. What is that ?

When you talk about form in this context, you are talking about a shape defined by finite spatial boundaries. A photon is NOT defined in this way. If you do not agree with me, i politely ask you to provide me with such a definition. Realise that you cannot use the wavelike photon concept to answer that question because of the reasons in gave in my previous post.


marlon

"Bounded" imply to number of the photons in the length of a light ray that can reffer to finite. Like that in a bounded interval exist finite number of integer numbers. So if we consider the Photo-electric experiment, in a arbitrary time invertal, number of photons that osculate with the surface are bounded or finite. Thus there are concepts that emphasize if we wonna constrict the diagonal of location that photons are crossing from there, then the number of photons that will arrive to the surface reduse. It means that the accumulation of photons occupy a spatial region in coordinates of spase. Therefor can you say me that what is your supposal for relating this subject to energy of a photon? Of course I must go and I couldn't answer fully to your questions!

Thanks.
Mr Beh

 

[marlon]

"Bounded" imply to number of the photons in the length of a light ray that can reffer to finite.

Sorry, but i don't get it :

1) What is the length of a light ray ?
2) What do you mean by "imply to the number of photons in the length of..." ?
3) How can you even be talking about the number of photons ? To what purpose ?

So if we consider the Photo-electric experiment, in a arbitrary time invertal, number of photons that osculate with the surface are bounded or finite.

I am sorry but according to me, there are no photons "osculating" with the surface. I

It means that the accumulation of photons occupy a spatial region in coordinates of spase.
Therefor can you say me that what is your supposal for relating this subject to energy of a photon?

Ok, i get your point here.

First of all, the double slit experiment learns us that we CANNOT make any claim onto the wherabouts of any atomic particle in the region in between the emittor and detector !. If you want to know its position, you need to measure and this causes the wavefunction collapse, remember ? But i guess you know this stuff so i will not get into that.

Secondly, in wave-lingo, let's indeed assume that the EM intensisty is lowered when going from source to detector. This indeed means that, if we now switch to the particle-lingo, some photons have left the bundle (i.e. that have scattered off atoms that constitute the medium through which the EM ray passes). But, that is about all you CAN say. Nothing more. You cannot talk about their position (x,y,z) coordinates because you did not measure. But suppose you WOULD measure, you would indeed observe that there is a photon present through its interaction with the detector (emitted EM radiation for example : the light signal from the detector). The point is however that you are not observing the actual photon, you are not measuring the photon's (x,y,z) coordinates !

You observed a packet of energy (as it is defined) through its interaction with the detector.

The clue is to understand what people mean by "measuring a photon" !

Just ask yourself this :
Suppose that a photon indeed has finite spatial boundaries, how would you measure them ?

greets
marlon

 

[Anonym]

Just ask yourself this: Suppose that a photon indeed has finite spatial boundaries, how would you measure them?

For example, I take the rigid ball with the internal surface being mirror and small inlet. I hope that you know that pi is irrational number. Therefore, I will take rigid rod and will measure it diameter.

But suppose you WOULD measure, you would indeed observe that there is a photon present through its interaction with the detector (emitted EM radiation for example : the light signal from the detector). The point is however that you are not observing the actual photon, you are not measuring the photon's (x,y,z) coordinates !

W.Heisenberg, “The Physical Content of Quantum Kinematics and Mechanics”, Zeitschrift fur Physik, 43, 172 (1927):

“We turn now to the concept of “path of the electron” By path we understand a series of points in space (in a given reference system) which the electron takes as “positions” one after the other. As we already know what is to be understood by “position at a definite time”, no new difficulties occur here. Nevertheless, it is easy to recognize that, for example, the often used expression, the “1s orbit of the electron in the hydrogen atom”, from our point of view has no sense.

There is no difference between the electron and photon with this respect. In addition I desperately tried to explain to you in the past that the position (x,y,z) and the size (delta(x),delta(y),delta(z)) are different self-adjoint operators.

Regards, Dany.

P.S. In addition, the detector absorbs the photons and does not emit them.

 

[Shahin]

I think we can answer the question by antoher method. Has the photon a wavefunction? it it has, we cab use |psi|^2 as a magnitude that can give us an idea of te size of the phton, considering, for example, that the photon size is "the volume" where, again for example, |psi|^2>1/10maximun(|psi|^2). Ii is like whe you want to know the size of a nucleo, you can use several magnitudes to determine this, the mass, the charge... and you say: when de density is less than... i consider the nucleo has ended.

 

[Anonym]

I think we can answer the question by another method. Has the photon a wavefunction? If it has, we can use |psi|^2 as a magnitude that can give us an idea of the size of the photon, considering, for example, that the photon size is "the volume" where, again for example, |psi|^2>1/10maximum (|psi|^2). It is like when you want to know the size of a nucleo, you can use several magnitudes to determine this, the mass, the charge... and you say: when the density is less than... i consider the nucleo has ended.

It is E.Schrödinger conjecture. I am close to prove that. |psi(x)|^2 is what we observe as a photon. However, the photon size is a parameter (number) that defines it (among others). You don’t need guess; you may calculate it and compare with the experimental data.

Regards, Dany.

 

[Proof.Beh]

Suppose that a photon indeed has finite spatial boundaries, how would you measure them?

Why you connect the structure of a photon (such as spatial coordinates) to the "measuring" concept? You said that "they are the packets of energy only that we are observing them", so have these packets the spatial coordinates or not? You did not answer yet my essential question. But first of all, I answer your elementary questions about the PERCEPTION of some parts of my last posting:

1- length of a light ray means that if we consider a laser ray that emmited from its source, then we can measure its duration of arriving for example to point B ("A" is location of sourse), then by name its duration "T" and use the formula c.T=L ("c" is speed of light) we shall define L or length of light ray.

2- imply sth means PREDICATE to sth.

3- Because I meant that show if we guess in the double slit experiment will constrict the diagonal of each slit, then the your energy packets because of their spatial coordinates and volume will cross for the first situation less.

Thanks.
Mr Beh

 

[alvaros]

I think it would be helpful if you speak about operational definitions instead of "chunk" "size" "bounded"...

If a photon has no size nor path, how can you get a photo ?

If a photon makes a spot on the photografic film we can infer that it collided in that point. How big can be this point ?

 

[Schrodinger's Dog]

Prof Beh, volume? What do you mean by this term? As I understand volume it would involve mass yes?

To be frank the question has been answered very well, there is no size, only a mathematical probability of where a photon will progress and this has nothing at all to do with an actual "size of the photon" I think you're confusing the issue.

Wavelength or direction or phase or whatever is the "length" in terms of this, but it does not denote an implication of size on the photon, at least as I understand it, all we're talking about is a spatial vector and that is not a matter of size per se just motion.

I could be wrong here, hell I often am, but size here is a matter of semantics, and is not really important when discussing the OP. I think we're talking about concepts that are not part of a "size" issue, merely vector progression.

I think it would be helpful if you speak about operational definitions instead of "chunk" "size" "bounded"...

If a photon has no size nor path, how can you get a photo ?

If a photon makes a spot on the photografic film we can infer that it collided in that point. How big can be this point ?

It's irrelevant the point is energy not a point of size, of course energy will cause the measurement medium to produce a point ie where it hit, but that does not infer a size only a packet of energy or quanta no?

 

[jtbell]

we can use |psi|^2 as a magnitude that can give us an idea of te size of the photon, considering, for example, that the photon size is "the volume" where, again for example, |psi|^2>1/10maximun(|psi|^2).

How about an electron? Consider a hydrogen atom in its ground state. Would you say that the electron in that atom has a radius that is roughly equal to the Bohr radius (or whatever result your prescription gives), which most physicists would say measures the size of the electron's ground state orbital?

 

[Shahin]

How about an electron? Consider a hydrogen atom in its ground state. Would you say that the electron in that atom has a radius that is roughly equal to the Bohr radius (or whatever result your prescription gives), which most physicists would say measures the size of the electron's ground state orbital?

No no, i didnt mean that thing. You are talking about the bohr radius, that is the radius of certain orbit in a certain system of reference, and has nothing to do with what i wrote. In your example, |psi|^2 is not "big" in all the sphere that has the bohr radius centred in the nucleo, so the condition |psi|^2>1/10maximun(|psi|^2) does not hold. The correct analysis would be, for example, represent |psi|^2 as a function of r, and yo will see that |psi|^2 is only appreciable in a little region around the bohr radius.

For example; you have a photon, alone in the universe for simplicity. You can describe it in some particular sense with his wavefunction. Now, it is natural to supose that this wavefunction, has a maximun and a particular region in which it is very probable to find the photon, and outside this region we can neglect the probability of finding te photon. So we can conclude that, if the photon has size, a volume, it could be the volume of this region.

 

[Proof.Beh]

Prof Beh, volume? What do you mean by this term? As I understand volume it would involve mass yes?

To be frank the question has been answered very well, there is no size, only a mathematical probability of where a photon will progress and this has nothing at all to do with an actual "size of the photon" I think you're confusing the issue.

Wavelength or direction or phase or whatever is the "length" in terms of this, but it does not denote an implication of size on the photon, at least as I understand it, all we're talking about is a spatial vector and that is not a matter of size per se just motion.

I could be wrong here, hell I often am, but size here is a matter of semantics, and is not really important when discussing the OP. I think we're talking about concepts that are not part of a "size" issue, merely vector progression.



It's irrelevant the point is energy not a point of size, of course energy will cause the measurement medium to produce a point ie where it hit, but that does not infer a size only a packet of energy or quanta no?

You answer this: What is reason of reduction the number of photons by constrict the diagonal of a slit in double slit experiment?

Thanks.
Mr Beh

 

[marlon]

Why you connect the structure of a photon (such as spatial coordinates) to the "measuring" concept?

Because if we cannot measure the spatial concepts you are referring to then how can we prove they even exist ? Again i ask you, how do you measure a photon's size ?

You said that "they are the packets of energy only that we are observing them", so have these packets the spatial coordinates or not?

Indeed photons propagate through the x,y,z plane. But that is about ALL you can say.
But that is not the same as saying that a photon is a point particle with x,y and z coordinates (what would be the equation for the size or its trajectory ?), for example like a car which has x,y,z coordinates as a function of time t and t². Let alone that we would be able to observe the trajectory/position of one single photon with absolute accuracy over and over again (i mean the same measurement executed consecutively) as we would be able to do with the car in classical physics! That would violate the HUP !

You did not answer yet my essential question.

Which was ?

1- length of a light ray means that if we consider a laser ray that emmited from its source, then we can measure its duration of arriving for example to point B ("A" is location of sourse), then by name its duration "T" and use the formula c.T=L ("c" is speed of light) we shall define L or length of light ray.

That's what we call distance, not length.

2- imply sth means PREDICATE to sth.

I don't get this.

3- Because I meant that show if we guess in the double slit experiment will constrict the diagonal of each slit, then the your energy packets because of their spatial coordinates and volume will cross for the first situation less.

I am sorry but i really don't understand your English here. Besides, are you talking about volume of rays here ? How does one measure that volume and how does one measure the size of a photon experimentally ? Those questions, i ask YOU.

I am lost, sorry.

marlon

edit : This discussion is analogous to why you cannot measure a photon's path. If you want to deal with concepts like paths, you need to know the particle's position at different times. This means you need to measure that position. Basic QM teaches us (double slit experiment) that in between measurements, the concept of path is not even defined. You can't associate a path to a wavefunction because once a measurement is made, the wavefunction changes. This also applies to "measuring the spatial boundaries of a photon". The problem is that you are trying to apply classical conepts into the QM world which operates in a different way, as shown in this paragraf.

 

[marlon]

this wavefunction, has a maximun and a particular region in which it is very probable to find the photon, and outside this region we can neglect the probability of finding te photon. So we can conclude that, if the photon has size, a volume, it could be the volume of this region.

You just wrote : a region in which there is a certain probability of finding the photon. That's correct. Then you say, the volume of that region is the size of the photon, right ? But then i can say, well you will always have probability 1 (ie absolute certainty) to find that photon in this region because the photon's volume is just as big as that region. Once you are in the region, you will observe the photon because it "fills up" they entire region. Clearly, this is incorrect.

No, the ONLY thing you can say is : there is a certain probability of finding the photon in that region which has a certain volume. THAT IS ALL ! No information is given about the actual shape of the photon ! Just like the electrons in orbitals.

marlon

 

[jtbell]

No no, i didnt mean that thing. You are talking about the bohr radius, that is the radius of certain orbit in a certain system of reference, and has nothing to do with what i wrote. In your example, |psi|^2 is not "big" in all the sphere that has the bohr radius centred in the nucleo, so the condition |psi|^2>1/10maximun(|psi|^2) does not hold. The correct analysis would be, for example, represent |psi|^2 as a function of r, and yo will see that |psi|^2 is only appreciable in a little region around the bohr radius.

The wave function for the ground state of the hydrogen atom is

$$\psi = \frac{1}{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0}$$

with the corresponding probability density

$$|\psi|^2 = \frac{1}{\pi a_0^3} e^{-2r/a_0}$$

where

[tex]a_0 = \frac{\hbar^2}{mq_e^2} = 0.0529 nm[/itex]

is the (first) Bohr radius.

Clearly the probability density is maximum at $r = 0$, with the value

$$|\psi|^2_{max} = \frac{1}{\pi a_0^3}$$

Setting $|\psi|^2 = 0.1 |\psi|^2_{max}$ and solving for $r$, I get

$$r = - \frac{a_0}{2} \ln 0.1 = 1.15 a_0$$.

Using your criterion, then, the size of an electron in a ground-state hydrogen atom is (1.15)(0.0529 nm) = 0.0608 nm.

 

[Proof.Beh]

To be frank the question has been answered very well, there is no size, only a mathematical probability of where a photon will progress and this has nothing at all to do with an actual "size of the photon" I think you're confusing the issue.

Mathematical probability? The mathematical concepts are only in mathematic. There is no reason for that agree with existence of a photon's size is a mathematical probability. Since if it has been true, I'll assume that because I don't know the electron approximately size, according to "mathematical probability" can select a value that differs with reality. But I agree with that because we have not any information about a photon's size, we can connect it to "mathematical probability".

Thanks.
Mr Beh

 

[Shahin]

The wave function for the ground state of the hydrogen atom is

$$\psi = \frac{1}{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0}$$

with the corresponding probability density

$$|\psi|^2 = \frac{1}{\pi a_0^3} e^{-2r/a_0}$$

where

[tex]a_0 = \frac{\hbar^2}{mq_e^2} = 0.0529 nm[/itex]

is the (first) Bohr radius.

Clearly the probability density is maximum at $r = 0$, with the value

$$|\psi|^2_{max} = \frac{1}{\pi a_0^3}$$

Setting $|\psi|^2 = 0.1 |\psi|^2_{max}$ and solving for $r$, I get

$$r = - \frac{a_0}{2} \ln 0.1 = 1.15 a_0$$.

Using your criterion, then, the size of an electron in a ground-state hydrogen atom is (1.15)(0.0529 nm) = 0.0608 nm.

NO NO NO NO. I think it is my fault because i didnt explain myself propely. Obviously, what i was meaning writting |psi|^2 was the density of probability, different in each case. So for the hydrogen atom you have to take r^2|psi|^2 and then you can do the analysis correctly. Anyway, I HAVE NEVER SAID that we can size the electron using this method, because, of course the electron has no size. I have just mentioned it as an idea about how to know the size of a photon.

 

[marlon]

I have just mentioned it as an idea about how to know the size of a photon.

So the electron has no size but the photon has ? I really don't get why you think this is the case. How does this size determining method work ? I tried to reread your previous posts but i did not find a valid explanation. I guess this method does not work according to the stuff you wrote in the volume of the region in which you can find a photon.

marlon