Is position not an observable of a photon?

[blenx]

Is position not an observable of a photon?

 

[Fredrik]

It isn't. For massive particles in relativistic quantum field theories, you can construct a position operator called the Newton-Wigner position operator, but this doesn't work for massless particles. (I'm sure there's a good reason why you can't define a position operator for massless particles, but I don't know it).

 

[unusualname]

Is position not an observable of a photon?

No, but you can detect the position of its interaction with another particle, the interaction destroys the photon.

A photon has no rest frame wrt which you could specify its position.

Theoretically it is not proven that photons move at exactly "the speed of light", so the possibility remains that a very small mass photon exists which does have a rest frame. Then we'd just need to rename the term "speed of light" to be the the speed invariant under a lorentz transformation.

But in current models we talk about photon number and energy density of the em field rather than photon position.

 

[blenx]

No, but you can detect the position of its interaction with another particle, the interaction destroys the photon.

Here I have a question:

Suppose I emit a light pulse whose length is very short, so we can consider the pulse contains only a few phtons, for example 2 photons.

If this 2-photon pulse interact with an atom at point(x,t) and one phton is destroyed, can we say that the position of another photon at time t is x?

 

[unusualname]

Here I have a question:

Suppose I emit a light pulse whose length is very short, so we can consider the pulse contains only a few phtons, for example 2 photons.

If this 2-photon pulse interact with an atom at point(x,t) and one phton is destroyed, can we say that the position of another photon at time t is x?

No, you can only say that one photon interacted with an atom at position x at time t

 

[blenx]

No, you can only say that one photon interacted with an atom at position x at time t

Thanks for your answer. But can you tell me why I can't talk about the position of another photon?

 

[blenx]

A photon has no rest frame wrt which you could specify its position.

Neutrino also move at speed c. Can we specify its position?

 

[Andrey]

Thanks for your answer. But can you tell me why I can't talk about the position of another photon?

Well you can talk about the position of another proton; actually you can talk about the position of all protons. Except the fact that you would have to be outside the space-time continuum to do that. It's like a 2D and a 3D plane. The 2D plane doesn't know that 3D plane exist and the 2D plane cannot see all the interactions of 2D plane because it is within the 2D plane itself. It is interacting with 2D plane, it within 2D plane so it's ability to see mupliple positioning of proton becomes limited, in fact imposible. In 3D plane, however, we can see the interactions of protons in a 2D plane. We can see them all. Why because we are no longer percieving reality from a 2D plane. However, not that we steped into 3D plane, we cannot see all the posibilities of 3D plane until we move into 4D plane and so on. think of it like a Russian doll!

:)

 

[Demystifier]

It isn't. For massive particles in relativistic quantum field theories, you can construct a position operator called the Newton-Wigner position operator, but this doesn't work for massless particles. (I'm sure there's a good reason why you can't define a position operator for massless particles, but I don't know it).

The Newton-Wigner position operator is not the only way to construct a position operator in relativistic quantum physics.

 

[DaTario]

I am not pretty sure about it, but

Suppose you have a one slit experiment. The picture bellow shows the situation. If the green dot in the ecran is a detected photon spot, at time t2, then it seems that ona can infer, using simple c velocity kinematics, that at a given time t1 < t2 the photon was inside the slit and therefore, has a certain position, with a certain reality content.

View attachment quantum fig1.pdf


Best Regards

DaTario

 

[blenx]

Well you can talk about the position of another proton; actually you can talk about the position of all protons. Except the fact that you would have to be outside the space-time continuum to do that. It's like a 2D and a 3D plane. The 2D plane doesn't know that 3D plane exist and the 2D plane cannot see all the interactions of 2D plane because it is within the 2D plane itself. It is interacting with 2D plane, it within 2D plane so it's ability to see mupliple positioning of proton becomes limited, in fact imposible. In 3D plane, however, we can see the interactions of protons in a 2D plane. We can see them all. Why because we are no longer percieving reality from a 2D plane. However, not that we steped into 3D plane, we cannot see all the posibilities of 3D plane until we move into 4D plane and so on. think of it like a Russian doll!

:)

We are talking about the position of the photon, not proton.

 

[blenx]

The Newton-Wigner position operator is not the only way to construct a position operator in relativistic quantum physics.

Can you show us some examples of other position operators?

 

[sweet springs]

Hi All.
We can say something about position of light, e.g. trace of beams, focusing light by rens in geometric optics.
In these cases threshold or limit of position observation is h'/p =λ　wavelength.
Regards.

 

[Demystifier]

Can you show us some examples of other position operators?

For example, the spatial part of the spacetime position operator in
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595]

 

[kexue]

It isn't. For massive particles in relativistic quantum field theories, you can construct a position operator called the Newton-Wigner position operator, but this doesn't work for massless particles. (I'm sure there's a good reason why you can't define a position operator for massless particles, but I don't know it).

One good reason is that photons are masseless and move at the speed of light and have no rest frame! Then also they are bosons, so you can't tell which are which.

Also, the position operator is a very fishy concept in relativistic quantum physics, since at Compton wave-length there are no single particles anymore, but particles pop in and out of existence. Relativistic quantum physics is a multiparticle theory, that's why the position operator is completely abandoned and instead becomes a parameter to a quantum field. (At least thta's the story in all the standard textbooks and lecture notes.)

 

[Fwiffo]

The position operator for a photon is a very big problem. A good review article on the subject is
http://adsabs.harvard.edu/abs/2005PhR...411...1K
by Ole Keller

Another person who has done a lot of work on the subject is Margaret Hawton.

It's hard to count the problems with this idea. In general a non relativistic (schrodinger) approach is not good for photons (they travel too fast), and in relativistic quantum (field) theory there is no concept of single photons. In general position is a problem in relativistic quantum theory, but it's much more of a problem for photons. It turns out that even during the emission process the photon is smeared over a volume larger then the Bohr radius of the atom.

Other problems are that attempts at a wigner like position operator give a position operator with non commuting parts (i.e X does not commute with Y). Another approach gives a well localized electric field if the magnetic field is non localized and vice versa. Remember that a photon is an electromagnetic wave.

Hope this helps

 

[Demystifier]

One good reason is that photons are masseless and move at the speed of light and have no rest frame! Then also they are bosons, so you can't tell which are which.

Also, the position operator is a very fishy concept in relativistic quantum physics, since at Compton wave-length there are no single particles anymore, but particles pop in and out of existence. Relativistic quantum physics is a multiparticle theory, that's why the position operator is completely abandoned and instead becomes a parameter to a quantum field. (At least thta's the story in all the standard textbooks and lecture notes.)

I think neither of these arguments is a good argument against position operator of a photon. Let me explain.

A classical massless particle also moves with the velocity of light. Does it mean that its position is not well defined? Of course not.

In nonrelativistic QM, the position operator is well defined even for symmetric (bosonic) wave functions.

In relativistic quantum physics, particles pop in and out only if there are interactions. This does not happen for free particles (Klein-Gordon or Dirac equation). Even with interactions, the vacuum and the lowest-mass one particle states are stable.

Relativistic quantum mechanics (Bjorken and Drell I) can be well interpreted as a single particle theory, or a theory with a fixed number of particles. It is INTERACTIONS, not relativity, which causes physical particle creation and destruction.

(Yes, you are right that this is explained so in many introductory textbooks, but these explanations are wrong.)

 

[Fwiffo]

Relativistic quantum mechanics (Bjorken and Drell I) can be well interpreted as a single particle theory, or a theory with a fixed number of particles. It is INTERACTIONS, not relativity, which causes physical particle creation and destruction.

Regardless of interactions there is no position operator for photons. True localization is less of a problem for free photons, but the operator ( d/dp for example) does not work. Other choises can be constructed for spinless (massless) relativistic particles but a photon position operator is still a problem. The problem is however much reduced in the case of no interactions where some kinds of position operators can be defined (such as the one Hawton uses) . There is as far as I know no cannonical photon position operator.

 

[Demystifier]

and in relativistic quantum (field) theory there is no concept of single photons.

This is simply wrong. A one-photon state is a well defined state in the Hilbert space of relativistic quantum field theory. Even if the number of photons change due to interactions, that only means that the state is a superposition of states with different numbers of particles. But each term in the superposition is well defined (otherwise the superposition itself would not make sense), which means that the states with definite number of particle are well defined.

Other problems are that attempts at a wigner like position operator ...

As I already mentioned in a previous post, there are also other definitions of the position operator.

Another approach gives a well localized electric field if the magnetic field is non localized and vice versa. Remember that a photon is an electromagnetic wave.

A 1-photon quantum state is NOT an eigenstate of the operator of electromagnetic field. That means that electromagnetic field is not well defined for a 1-photon state, but it does not mean that the position of such photon is not well defined. Indeed, there exists a 1-photon wave function which is well localized in space at a given time.

 

[Fwiffo]

Indeed, there exists a 1-photon wave function which is well localized in space at a given time.

Where can I find this?

 

[Demystifier]

Regardless of interactions there is no position operator for photons. True localization is less of a problem for free photons, but the operator ( d/dp for example) does not work. Other choises can be constructed for spinless (massless) relativistic particles but a photon position operator is still a problem. The problem is however much reduced in the case of no interactions where some kinds of position operators can be defined (such as the one Hawton uses) . There is as far as I know no cannonical photon position operator.

Maybe it is not canonical (whatever that means), but there is a way to introduce a position operator that does not depend on interactions. After all, in the Schrodinger picture of quantum theory the operator is something that can be defined BEFORE specifying interactions.

 

[Fwiffo]

Maybe it is not canonical (whatever that means), but there is a way to introduce a position operator that does not depend on interactions. After all, in the Schrodinger picture of quantum theory the operator is something that can be defined BEFORE specifying interactions.

By cannonical I mean one which is used in most cases. You are right there are ways to define a position operator , but it usually has some drawbacks such as non commuting components, or being non relativistic, or not giving the position. One that seems to overcome a lot of problems is the one of Hawton http://pra.aps.org/abstract/PRA/v59/i2/p954_1 , but it has not been used very often (based on citations).

In the Schrodinger picture following maxwell's equations it is (as far as i know) not possible to get well localized photons.

 

[Demystifier]

Where can I find this?

The Appendix of
http://xxx.lanl.gov/abs/0804.4564
explains how to calculate the wave function for any relativistic 1-particle state. By choosing an appropriate function c(q), you can obtain any function of space at a fixed time you want, including a delta function. Equations are written for a mass m, but there is no problem to put m=0 in the equations. Equations are written for a spinless particle, but it is not a problem to include the spin as well.

 

[Fwiffo]

The Appendix of
http://xxx.lanl.gov/abs/0804.4564
explains how to calculate the wave function for any relativistic 1-particle state. By choosing an appropriate function c(q), you can obtain any function of space at a fixed time you want, including a delta function. Equations are written for a mass m, but there is no problem to put m=0 in the equations. Equations are written for a spinless particle, but it is not a problem to include the spin as well.

Like you said this works for spinless particles. I don't think including spin in this wavefunction is as simple as you make it out to be.

 

[Demystifier]

Like you said this works for spinless particles. I don't think including spin in this wavefunction is as simple as you make it out to be.

But it is simple. Essentially, all you have to do is to add an additional index to quantities such as a, c and psi. This means that the wave function has a few components, but it can still be localized in the sense that each independent component can be localized. In fact, it should not be surprising, because one usually says that it is relativity or masslessness that causes problems, not the spin.

The TRUE problem with relativity and masslessness is that you cannot have a STATIONARY wave function localized in space. But if you don't require the wave function to be stationary, then it is not a problem at all to make it localized at a given time. Of course, since it is not stationary, it will not be localized at another time. In fact, at infinitesimally later time the wave function may not be even approximately localized. Some refer to this fact as Hegerfeldt paradox, but this is misleading too, because there is nothing paradoxical about it.

 

[Fwiffo]

But it is simple. Essentially, all you have to do is to add an additional index to quantities such as a, c and psi. This means that the wave function has a few components, but it can still be localized in the sense that each independent component can be localized. In fact, it should not be surprising, because one usually says that it is relativity or masslessness that causes problems, not the spin.

Adding an index is not enough. This is not a Schroedinger equation but a second quantized approach, you have to use a whole different method. Normalization is not trivial, eqs 49 and 53 cannot be used with just an index added.

 

[Demystifier]

First, the problem of normalization is not directly related to the problem of localization. Second, the problem of normalization is also easily solved: just sum over repeated indices.

But this is all nothing but a red herring. The problem of localization in relativistic QM is actually the following. You can easily construct a wave function localized in space at a fixed time, but such a wave function (as any wave function) is not localized in time. This, of course, is true also in nonrelativistic QM. However, in relativistic QM, it makes sense to look at what happens in ANOTHER Lorentz frame. And it is easy to see that, in another Lorentz frame, this same wave function is not localized in space. Therefore, the claim that a wave function is localized in space IS NOT LORENTZ INVARIANT. But it does not mean that you cannot localize a particle at a given time. You can do it with respect to a given Lorentz frame, e.g., the one with respect to which your measuring apparatus is at rest.

 

[unusualname]

Is there any way to measure a photon's position without destroying it?

 

[Fwiffo]

This thread has diverged into looking at 3 related topics. Photon position operators; Photon detection; and photon localization. All three are hard problems. I'll start with the easiest one which is detection/emission of (single) photons. Here the really big problem is that relativistic field theory (the laws of physics as we know them) does not handle single photons in a very good way. Although in theory single photons can exist in free space (as Demystifier wrote above), detection necessarily involves interaction, an at this point photon numbers are not conserved. Single photon detection is in fact not possible with perfect fidelity due to dark counts which are related to the uncertainty in the number of photons (even in vacuum). There may be ways of detecting photons without destroying them but they are very specialized methods which only work for certain kinds of setups where the photons are in a certain configuration (say entangled with other photons), again such methods can never work with perfect fidelity due to dark counts. The same goes for creating (emitting) single photons.

The second problem is that of a photon position operator. I don't remember/know all the details of why this is such a hard problem, but over the years people have been trying to work out this problem with varying degrees of success. There is no accepted way to write down this operator, and the main problems with finding it are the spin of the photons and relativity.

The third problem is localization, again people have been working on this for years. The problem is threefold 1. photons are inherently relativistic, which means that any localization should be covarient. 2. When interacting photons can be created which means that single photons are a problem (see detection). and 3. Photons have spin, this means that the wavefunction has more then one component,but these are not independent (they obey Maxwell's equations) , there is enough freedom to localize one component but what happens is that the others are not localized at this point

 

[Demystifier]

Is there any way to measure a photon's position without destroying it?

In principle yes, but efficiency of such a detector would be ridiculously small. Namely, for such a detector you need an interaction Lagrangian quadratic in photon field, which you can obtain in effective action associated with higher-order Feynman diagrams in QED.

 

[unusualname]

Yes, I guess by "observable" people usually mean a property that can be measured (without destroying the particle), but thanks for the links and clarifications on what is a very subtle and hard question Fwiffo and Demystifier.

btw, http://xxx.lanl.gov/abs/quant-ph/0609163 is an excellent review article on QM in general (I believe Demystifier is the author)

 

[Demystifier]

btw, http://xxx.lanl.gov/abs/quant-ph/0609163 is an excellent review article on QM in general (I believe Demystifier is the author)

Let me just say that a similarity between my name and the title of this article is not a coincidence.

 

[blenx]

…… Remember that a photon is an electromagnetic wave.

When talking about what a photon is, I always hear about different versions: Some says it is a short pulse, some says it is a particle, and some says it is a plane wave with determined momentum and polarization…… Different viewpoints make me very confused.

If a photon is just a wave, can we say that its position is where the wave amplitude has a maximum? Otherwise how can we know a photon is also moving at speed c?

If a photon is a particle, how large does it occupies in space? Can we specify its position from its scattered by other particles?

 

[Fwiffo]

This is your basic wave/particle duality. The photon has wave like properties, these properties obey the maxwell equations (i.e the equations for electromagnetic waves). However it is quantized (a particle property) so two photons cannot interfere with each other in the same way two electromagnatic waves can, this is (i'm not being precise here) the photoelectric effect.

This is true for any particle but the equations are different, electrons follow the schrodinger (or dirac) equations etc...

Confusing? I know!

 

[blenx]

This is your basic wave/particle duality. The photon has wave like properties, these properties obey the maxwell equations (i.e the equations for electromagnetic waves). However it is quantized (a particle property) so two photons cannot interfere with each other in the same way two electromagnatic waves can, this is (i'm not being precise here) the photoelectric effect.

This is true for any particle but the equations are different, electrons follow the schrodinger (or dirac) equations etc...

Confusing? I know!

I know any particle have the wave-particle duality. Contrast to photon, it's easy to understand that an electron locates at some point in space. However, wave-particle duality has not told us that whether a photon does the same and what the term "photon" is indeed referred to ---- a small energy packet locates at some point and moves at speed c or just an EM wave whose length is whether long or short, will disappear entirly when part of it interact with things such as an atom in the ground state. It seems that the term "quantum excitation of the EM field" has not made things clearer.

One more question, you said a photon is a wave, then can you write down the mathmatical expression of this wave to us?