Photon Position: Is It Less Classical Than Massive Particles?

[zonde]

There's no other quantum which is less adequate described as a classical particle than the photon. It doesn't even admit to define a position observable etc. etc. We have discussed this indeed endlessly in this forum.

I am puzzled by that statement. I do not intend to question the statement that "photon doesn't admit to define a position observable". My confusion is with the conclusion that photon is less "classical" because it does not have position observable as I can reach only weaker conclusion that photons are different than massive particles. And that includes possibility that photons are more "classical" than massive particles.

 

[Orodruin]

A typical property of a classical particle is that you can observe its position. For massive particles, you can define a position observable and so this is at least in the same spirit as the classical particle. For the photon, you cannot define the position observable.

 

[zonde]

But absence of position observable does not mean that position can't be observed. When photon detector clicks we can infer (given we speculate that photon has position) that photon was there at the time of the click.

 

[Orodruin]

But absence of position observable does not mean that position can't be observed.

Yes it does, by definition.

 

[hilbert2]

How does the absence actually appear? If I write the Klein-Gordon position operator as an integral that contains the creation and annihilation operators for different modes, is the integral not convergent in the massless limit? This isn't really in my field of study but it seems interesting.

EDIT: not to mean that photons would be described by the KGE, just thought that the same absence of position observable would also hold in the case of any massless field

 

[vanhees71]

But absence of position observable does not mean that position can't be observed. When photon detector clicks we can infer (given we speculate that photon has position) that photon was there at the time of the click.

What you measure here is that an electromagnetic field (maybe a single photon, if prepared as such) lead to the emission of a photoelectron which was amplified to make a click (supposed, the detector works via the photoeffect, but the detailed microscopic mechanism is not so important here).

 

[zonde]

Yes it does, by definition.

Can you explain what you mean? Surely clicks in detectors are experimental facts. It does not make sense to deny experimental facts.

 

[zonde]

What you measure here is that an electromagnetic field (maybe a single photon, if prepared as such) lead to the emission of a photoelectron which was amplified to make a click (supposed, the detector works via the photoeffect, but the detailed microscopic mechanism is not so important here).

Yes, we are on the same page.
So my take on this is that we can speak about position of some particle like excitation of electromagnetic field. Do you mean something else?

 

[Orodruin]

Can you explain what you mean? Surely clicks in detectors are experimental facts. It does not make sense to deny experimental facts.

Did you even read post #6?

 

[Orodruin]

How does the absence actually appear? If I write the Klein-Gordon position operator as an integral that contains the creation and annihilation operators for different modes, is the integral not convergent in the massless limit? This isn't really in my field of study but it seems interesting.

EDIT: not to mean that photons would be described by the KGE, just thought that the same absence of position observable would also hold in the case of any massless field

I recommend this thread and links therein.

 

[zonde]

Did you even read post #6?

Yes, several times.

 

[Orodruin]

Yes, several times.

So then you have understood that the detector click is not "measuring the photon position"?

 

[zonde]

So then you have understood that the detector click is not "measuring the photon position"?

No, I don't understand why detector is not measuring photon position.
Say if I look at the ghost imaging using downconverted photons then the image is reconstructed taking into account detector positions in detector array. So it matters in which detector the "click" happens.

 

[vanhees71]

Yes, we are on the same page.
So my take on this is that we can speak about position of some particle like excitation of electromagnetic field. Do you mean something else?

Again: You cannot define "position" for an electromagnetic field or of photons, which are certain states of this field (namely single-photon Fock states). Nobody thinking about classical electromagnetic waves would ever come to the idea to ask, what the position of a field might be. The very definition of a field is that of a space-time dependent quantity, e.g., the temperature in my office, is a scalar field, which I can measure with a thermometer at different points and at different times. It doesn't make sense to say "the temperature has a certain position", but I can measure the temperature with a thermometer placed at some position.

For an electromagnetic field you measure intensities, i.e., the energy density of the field at given places. For a photon (one-photon Fock state) it gives the probability to register a photon at a given place determined by the position of the photodetector. The photon itself has no position, because you cannot define what position might be. For an overview, see

http://arnold-neumaier.at/physfaq/topics/position.html

 

[zonde]

Again: You cannot define "position" for an electromagnetic field or of photons, which are certain states of this field (namely single-photon Fock states).

Yes, it is meaningless to speak about position of electromagnetic field.
No, it is perfectly meaningful to speak about position of certain configuration of electromagnetic field.
If for you "electromagnetic field"="photon" then I am at loss.

Nobody thinking about classical electromagnetic waves would ever come to the idea to ask, what the position of a field might be. The very definition of a field is that of a space-time dependent quantity, e.g., the temperature in my office, is a scalar field, which I can measure with a thermometer at different points and at different times. It doesn't make sense to say "the temperature has a certain position", but I can measure the temperature with a thermometer placed at some position.

Yes, it doesn't make sense to say "the temperature has a certain position".
But it makes sense to say "the highest temperature is at certain position".

And would you say it does not make sense to speak about position of soliton?

For an electromagnetic field you measure intensities, i.e., the energy density of the field at given places. For a photon (one-photon Fock state) it gives the probability to register a photon at a given place determined by the position of the photodetector.

Never heard about direct measurement of energy density of the field. There are different mechanisms how electromagnetic radiation of different wavelength can be absorbed. So any measurement would be specific to certain wavelength range.
So basically it's the other way around. You measure probabilities to register photons of different wavelength and from that you can calculate energy density of the field.

The photon itself has no position, because you cannot define what position might be. For an overview, see

http://arnold-neumaier.at/physfaq/topics/position.html

The only related sentence that I found was this:
"As a consequence of our discussion, photons (m=0, s=1) and gravitons (m=0, s=2) cannot be given natural probabilities for being in any given bounded region of space."
But then there is no problem to give operational definition for position of photon - it's where detector "clicks".

 

[vanhees71]

Yes, it is meaningless to speak about position of electromagnetic field.
No, it is perfectly meaningful to speak about position of certain configuration of electromagnetic field.
If for you "electromagnetic field"="photon" then I am at loss.

No, for me a photon is only a certain class of states of the electromagnetic quantum field, namely single-photon Fock states, and again, I stick to my opinion that it doesn't make sense to talk about the position of a photon or an electromagnetic field.

Even in classical electrodynamics, the electromagnetic field is defined everywhere in spacetime. At each point of space time the em. field has a specific value. That's the big breakthrough in our picture of nature since Faraday.

Yes, it doesn't make sense to say "the temperature has a certain position".
But it makes sense to say "the highest temperature is at certain position".

Sure, you can look for the spacetime point, where the temperature reaches a maximum. So what?

And would you say it does not make sense to speak about position of soliton?

Of course not, it's a special state of the field under consideration, and maybe it makes sense to talk about the center of the wave (usually defined by energy-density weighted mean "positions").

Never heard about direct measurement of energy density of the field. There are different mechanisms how electromagnetic radiation of different wavelength can be absorbed. So any measurement would be specific to certain wavelength range.
So basically it's the other way around. You measure probabilities to register photons of different wavelength and from that you can calculate energy density of the field.

Well, the photon-detection probabilities are given by the energy density. How else, do you want to define it in a gauge-invariant and Poincare-covariant way. I think, I finally have to sit down and write my photon FAQ article...

The only related sentence that I found was this:
"As a consequence of our discussion, photons (m=0, s=1) and gravitons (m=0, s=2) cannot be given natural probabilities for being in any given bounded region of space."
But then there is no problem to give operational definition for position of photon - it's where detector "clicks".

Yes, you can give "detector-click probabilities", but not position of photons. That's what I said in the very beginning...

 

[Vanadium 50]

Hows can this be answered at B level if what a photon is - and certainly what a position operator is - is I level?

 

[Orodruin]

Hows can this be answered at B level if what a photon is - and certainly what a position operator is - is I level?

If not A-level ... The EM field is one of the more complicated things there is to quantise. There is a reason QFT texts usually start with scalar fields and then treat Dirac fermions only to treat the quantisation of gauge fields a few chapters later.

 

[DrDu]

But absence of position observable does not mean that position can't be observed. When photon detector clicks we can infer (given we speculate that photon has position) that photon was there at the time of the click.

I think the problem here is that when we talk of the position operator of e.g. an electron, this operator does not destroy the electron. While when we try to measure the position of a photon, e.g. with a detector based on the photoeffect, the photon gets destroyed. So we can measure the position of a photon, but not in a non-destructive way.

 

[vanhees71]

I think that even classical electrodynamics is on the more complicated side of the theoretical-physics standard curriculum. In my opinion one should change the usual order of the BSc-theory curriculum which usually is mechanics, classical electrodynamics, quantum theory to mechanics, quantum mechanics, classical electrodynamics, because you need a lot of the math that's easier introduced within QM 1 in classical electrodynamics. In QM 1 you usually deal with scalar non-relativistic particles and the non-relativistic Schrödinger equation. E.g. the treatment of angular momentum in QM1 introduces in the most simple way spherical harmonics and the corresponding multipole expansion of fields. The em. field is, in comparison, a pretty complicated structure with its 6 components. Switching to QED makes the issue even more complicated, because you need to introduce necessarily the potentials (or better four-vector potential since it's better to use the manifestly covariant formalism from the very beginning) and has to deal with gauge symmetry and a rather complicated construction of the Hilbert space already for the free field.

 

[PeterDonis]

Hows can this be answered at B level if what a photon is - and certainly what a position operator is - is I level?

It can't. This topic should be at least "I" level, if not "A" level as Orodruin suggests.

This thread is now closed.