Photon Size: Uncertainty Principle & QM Explained

[Dale]

What is the size operator? If there is no such operator then the question is poorly defined. If there is such an operator then it is just a matter of plugging it into a single-photon Fock state to see what the result is.

 

[sophiecentaur]

What is the size operator? If there is no such operator then the question is poorly defined. If there is such an operator then it is just a matter of plugging it into a single-photon Fock state to see what the result is.

We're into (piece of) string, perhaps?

 

[JK423]

It has been shown that a photon can be arbitrarily "localized" in spacetime, i.e. there is not fundamental limit from the theory that forbids the complete localization of a one-photon state.

http://prl.aps.org/abstract/PRL/v79/i9/p1585_1

Edit: Localization is meant in the sense of arbitrarily small extent of its mode function as it travels with 'c', not that you localize it in a position eigenstate.

 

[Naty1]

Some comments on comments:

My understanding is that when we make a measurement we are really poking at the wavefunction, which holds all the measurable information about a particle within it. When the measurement is made it causes any probabilities to collapse and take on a definite value.

That is one view of QM...one which appeals because of its analogy to classical descriptions but fails in a number of situations. One failure is that to explain entanglement it must collapse faster than light speed...

yes:

photons certainly have different propties depending on the source.

I think it's worth pointing out that 'the photon' cannot be pictured as some sort of 'burst of oscillations' ... as a little bullet.

Size isn't something with a precise physical meaning in this context. The size of a photon depends on how you measure it.

[This is analogous to what Susskind says with his 'propeller' description.

the photon has size that interacts with the detector with pointlike properties, but while propagating before detection it has size that is described with wavelike properties distributed over the meter distance.

[yes to the first part, no to the second]

'size' is just not a relevant property for a photon.

[see the following for some further clarification of this valid point.]

///////////////////////

Other descriptions:

Albert Messiah, Quantum Mechanics, pg 66:
The wave packet...

has a center that travels analogous to a classical point particle.

From another discussion,
from Vanhees:

This position of a registration of a photon is a well defined physical quantity that can be measured, the position of a photon in the strict sense of an observable cannot even be defined in principle! For details, see

http://arnold-neumaier.at/ph.../position.html

unsure of this source...

If a photon were truly free, not interacting with any other particles, its PLANE wave would extend across the universe to the cosmological horizon.

{} my clarification:
From Wikipedia:

To account for the particle interpretation that phenomenon [double slit experiment] is called probability distribution but behaves according to Maxwell's equations. However, experiments confirm that the photon is not a short pulse of electromagnetic radiation; it does {MAY} not spread out as it propagates, nor does it divide when it encounters a beam splitter.
The photon seems to be a point-like particle since it is absorbed or emitted as a whole by arbitrarily small systems, systems much smaller than its wavelength, such as an atomic nucleus (≈10−15 m across) or even the point-like electron. Nevertheless, the photon is not a point-like particle whose trajectory is shaped probabilistically by the electromagnetic field, as conceived by Einstein and others. According to our present understanding, the electromagnetic field itself is produced by photons, which in turn result from a local gauge symmetry and the laws of quantum field theory

http://en.wikipedia.org/wiki/Quantum_field_theory#Unification_of_fields_and_particles

The "second quantization" procedure that we have outlined in the previous section takes a set of single-particle quantum states as a starting point. Sometimes, it is impossible to define such single-particle states, and one must proceed directly to quantum field theory. For example, a quantum theory of the electromagnetic field must be a quantum field theory, because it is impossible (for various reasons) to define a wavefunction for a single photon.

[Zapper had posted an identical statement to the boldface portion and I found the above confirmation.]

Also, recall that VIRTUAL photons, complex numbered particles rather than the usual real numbers, are responsible for all electromagnetic interactions as described using quantum field theory. It's QFT that is used in the Standard Model of particle physics. So not only are photon characteristics 'fuzzy' in classical terms, so too they seem to be in mathematical terms.

 

[Naty1]

Anybody have any comments on the paper JK423 posted...The description seems to conflict with what has been discussed in these forums...and it is not all that new...1997...

It has been shown that a photon can be arbitrarily "localized" in spacetime, i.e. there is not fundamental limit from the theory that forbids the complete localization of a one-photon state.

http://prl.aps.org/abstract/PRL/v79/i9/p1585_1

Edit: Localization is meant in the sense of arbitrarily small extent of its mode function as it travels with 'c', not that you localize it in a position eigenstate.

 

[ZapperZ]

I always find it interesting that this question on the "size" of a photon (or the size of anything, really) has come up repeatedly. So let's try to narrow down what we DO know from straight-forward quantum mechanics and see if people do agree on those:

1. There is a position operator in QM, and that gives you the value of the position of the particle.

2. The spread in position of the particle is not the same as the size of the particle.

3. There is no "size operator" yet well-defined in QM.

Do we all not agree on those?

Zz.

 

[Bill_K]

1. There is a position operator in QM, and that gives you the value of the position of the particle.

This right here is the hot button issue. The situation is not so straightforward in relativistic QM, partly because a single particle cannot be localized in a region smaller than a Compton wavelength.

For particles with mass, the Newton-Wigner position operator is the accepted answer. However it is argued that for photons no position operator exists.

 

[ZapperZ]

This right here is the hot button issue. The situation is not so straightforward in relativistic QM, partly because a single particle cannot be localized in a region smaller than a Compton wavelength.

For particles with mass, the Newton-Wigner position operator is the accepted answer. However it is argued that for photons no position operator exists.

That's fine. My argument and intention here are that, when there IS a well-defined position operator, and that it gives a value or an average value, that is the value of a POSITION, and not the value of a "size". I'm seeing people confusing, say, the uncertainty in position as being the "size" of the object.

Zz.

 

[Dale]

3. There is no "size operator" yet well-defined in QM.

This is the key problem that makes the question itself ill-posed. The only way to answer questions about the size of something in QM is to have a size operator and calculate its expectation.

Since there is no generally accepted size operator, I think that the OP is free to define one. Until they do, however, the question is incomplete and cannot be answered.

 

[Bill_K]

Since there is no generally accepted size operator, I think that the OP is free to define one. Until they do, however, the question is incomplete and cannot be answered.

Of course there is. But to talk about the size of an object you have to talk about an object that has a size to begin with! Which leaves out all the elementary particles, including photons, since they are pointlike.

Composite objects such as protons and neutrons have a size. For a proton it's the charge radius. A neutron's size can be defined in terms of its form factors. Nonspherical nuclei have rotational degrees of freedom, and for them the size can be defined in terms of the moment of inertia.

 

[ZapperZ]

Of course there is. But to talk about the size of an object you have to talk about an object that has a size to begin with! Which leaves out all the elementary particles, including photons, since they are pointlike.

Composite objects such as protons and neutrons have a size. For a proton it's the charge radius. A neutron's size can be defined in terms of its form factors. Nonspherical nuclei have rotational degrees of freedom, and for them the size can be defined in terms of the moment of inertia.

But those are not "size operator" the way we have Hermitian operators in QM that give you an observable value. I can also point out similar things such as scattering or absorption cross-section, which one can easily imagine to correspond to some energy-dependent "size".

The determination of sizes of composite particles is not as simple as having an operator that spits out some average value.

Zz.

 

[Naty1]

I have a semantics question:

Bill_k posts...

The situation is not so straightforward in relativistic QM, partly because a single particle cannot be localized in a region smaller than a Compton wavelength.

Is 'relativistic QM' the same 'relativistic QFT' of the Standard Model??...or should I be aware of some distinctions...

////////////

Zapper:

I'm seeing people confusing, say, the uncertainty in position as being the "size" of the object.

When one does not know all the math and subtending assumptions, like me, and apparently for many who think they do understand, that is an easy step to take. I can attribute that to at least three things that immediately come to mind:

[1] When the Standard Model defines particle measurements, that is interactions, say a point of a detection screen, and leaves unspecified what the particle 'is' between interactions, that just begs for personal interpretations galore.

[2]When interacting point particles of the Standard Model have different position uncertainities between,say, free plane wave particles [cosmological horizon to horizon]
versus confinement as, say, an orbital electron, versus a different confinement characteristic in a metal lattice, all of which we have discussed elsewhere, no wonder people like me sometimes think Sounds like it takes on different SIZES instead of interaction characteristics.

[3] Classical analogies and classical thinking never fits completely. When you quantize the classical physical electromagnetic wave, a 'physical' interaction field, stuff changes in QM in subtle ways. Nobody even calls the [Schrödinger] wave equation a 'field'...how it becomes a 'probability distribution' seems a mystery...where did my 'physical field' go?? [LOL]

 

[Mandragonia]

To talk about the size of an object you have to talk about an object that has a size to begin with! Which leaves out all the elementary particles, including photons, since they are pointlike.

Even for objects that are generally considered to be pointlike, one may postulate that they do have a very small but still finite size. Putting all theoretical and experimental evidence together, one can set an upper limit for this size. Clever experiment can be devised to reduce the upper limit further. Isn't that the way physicists approaches such subjects, for example regarding the existence of neutrino mass or the non-constancy of physical parameters such as the fine-structure constant?

 

[Bill_K]

Is 'relativistic QM' the same 'relativistic QFT' of the Standard Model??...or should I be aware of some distinctions...

Have you seen the two-volume set by Bjorken and Drell? The first book is called "Relativistic Quantum Mechanics", while the second is "Relativistic Quantum Field Theory". That's the kind of distinction I was thinking of. In the first case (aka First Quantization) you treat the Klein-Gordon and Dirac Equation as direct relativistic generalizations of the Schrödinger equation - wave equations for the probability amplitude of a single particle. Thus "position operator" unambiguously refers to position of that single particle. Relativistic QFT (aka Second Quantization) is about multiparticle states from the word go, and negative energy states are replaced by antiparticles.

Relativistic QM is an incomplete theory - one must avoid situations where pair production and negative energy states come into play, and these restrictions lead to the unusual properties of the position operator.

 

[Dale]

Of course there is. But to talk about the size of an object you have to talk about an object that has a size to begin with!

Then please write down the size operator for a particle that has a size.

 

[Bill_K]

Then please write down the size operator for a particle that has a size.

Sure, it's just r². The size of an object is its RMS radius, which is <r²>^1/2.

But not for a particle, I said "composite object". Do this for an atomic electron and you will get the size of the atom, not the size of the electron. The size of a composite object such as a proton or nucleus is determined by the uncertainty in the position of its components.

 

[ZapperZ]

Sure, it's just r². The size of an object is its RMS radius, which is <r²>^1/2.

But not for a particle, I said "composite object". Do this for an atomic electron and you will get the size of the atom, not the size of the electron. The size of a composite object such as a proton or nucleus is determined by the uncertainty in the position of its components.

But that is disingenuous, and could mislead people who don't know what you did!

That is still the position operator. It tells you the location, or average location, of an electron in a particular state. We then turn around, and use that to approximate the size of an atom. It is still not a "size operator" the way I described earlier. You do not have to have such redefinition or context to define the position, momentum, etc. operator.

Let's try to get at least some consensus here before we try to apply this to more complicated or specific situation. I listed those three statements, and at the very least, we should agree on something at the fundamental level. Or else, we'll be talking about apples and oranges, and this discussion will never go anywhere productive.

Zz.

 

[Dale]

Sure, it's just r². The size of an object is its RMS radius, which is <r²>^1/2.

I could accept that iff the object is located at the origin. However, since we cannot generally localize the object to the origin what this actually gives us is not just the size of the object but also the probability of finding it a certain distance from the origin. So this isn't a good candidate for a size operator.

But not for a particle, I said "composite object". Do this for an atomic electron and you will get the size of the atom, not the size of the electron. The size of a composite object such as a proton or nucleus is determined by the uncertainty in the position of its components.

If you had a valid size operator then you should simply be able to apply it to any particle's wavefunction to get the size of the particle. Whether it is a fundamental particle or not shouldn't matter.

If a valid size operator existed and had the property that it gave undefined results for fundamental particles, then it would be natural to say that fundamental particles don't have a size, but to say that such an operator exists but we are forbidden from applying it to a class of particles simply because you assert that they have no size doesn't make sense to me.

 

[Maui]

In some practical sense, can it be said that the 'classical' size of photons is the following - consider why ordinary light microscopes fail below 200 nm resolution - light travels as a wave and photons wavelength is too big and diffracts below this threshold and cannot penetrate further. On the other hand, photons are always detected at a point much smaller than 200 nm -- is the above an accurate description of why light can't penetrate further than 200 nm at all? If it is(it seems so according to Olympus' website discussing light diffraction) is that the 'classical' size of photons?

 

[Dale]

the 'classical' size of photons

There are no photons classically.

 

[Maui]

There are no photons classically.

You could state the same of all elementary particles, that's why 'classically' was in quotes denoting that size seems to play a practical role(e.g. the example of light diffraction in optics).

 

[Dale]

You could state the same of all elementary particles, that's why 'classically' was in quotes denoting that size seems to play a practical role(e.g. the example of light diffraction in optics).

If a photon were a classical particle with some size then it wouldn't diffract at all.

 

[Maui]

If a photon were a classical particle with some size then it wouldn't diffract at all.

I agree but I didn't say a photon was a classical particle. Only that in some practical respects photon's wavelength attributes a physical size that appears to be a limit in photography and optics in general(microscopes, etc.).

 

[vanhees71]

The question is, how to define the radius of an (elementary or composite) particle, atom, or molecule. In physics such quantities have to be defined as a measurable quantity. Usually one measures charge radii of the (sub)atomic object, usually by elastic scattering of electrons. Measuring the differential cross section as a function of the scattering angle gives the charge radius. Such experiments measure, the root-mean-square radius, which is defined as
$$r_{\text{ch}}^2=\langle r^2 \rangle_{\rho}=\frac{\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \vec{x}^2 \rho(\vec{x})}{\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \rho(\vec{x})},$$
where $\rho$ is the charge distribution of the object.

For photons I'm not aware of any proper definition of its radius. It's anyway pretty wrong to assume that photons can be thought of as little classical particles at all! There's not even a well-defined position operator for a photon!

 

[Bill_K]

There's not even a well-defined position operator for a photon!

Yes, for sure that's the majority conclusion, but there seems to be some controversy remaining about it, and maybe the difficulty has been overstated. There's a candidate position operator that was found in 1948 by Pryce,

x ≡ i ∇_p + p x s/p²

where s is the photon's spin vector. The shocking thing about this x is that its three components don't commute,

[x_i, x_j] = -i ε_ijk p_k p·s/p³

Nevertheless, there's an argument by Mourad that this is not so bad. He says,

The three components of the Pryce operator cannot be simultaneously diagonalized, so that one cannot find states localized exactly at a given point in space. Newton and Wigner argued that this makes the photon non-localizable. However, we will show that although the commutator of the three components is non-vanishing, one can find states which are ”localized" within an arbitrarily small region.

 

[sophiecentaur]

I agree but I didn't say a photon was a classical particle. Only that in some practical respects photon's wavelength attributes a physical size that appears to be a limit in photography and optics in general(microscopes, etc.).

Wavelength and size are not related. Diffraction calculations are based on an infinitely long wave train. The effect of one photon is just an individual occurrence. How can you say there is some sort of correspondence between the wavelength and any idea of 'size'? Any 'entity' that can have a 'single' wavelength needs to have infinite extent. An entity with very limited extent needs a very large range of associated wavelengths. That's the (Fourier) relationship between temporal and frequency domains (or the equivalent spatial thing), surely.

 

[Maui]

Wavelength and size are not related.Diffraction calculations are based on an infinitely long wave train. The effect of one photon is just an individual occurrence. How can you say there is some sort of correspondence between the wavelength and any idea of 'size'? Any 'entity' that can have a 'single' wavelength needs to have infinite extent. An entity with very limited extent needs a very large range of associated wavelengths. That's the (Fourier) relationship between temporal and frequency domains (or the equivalent spatial thing), surely.

Are you saying that a photon's wavelength has no associated width? How do you explain diffraction patterns below the 200 nm? It sounds like a contradiction to measure the practically measured wavelength of photons in nm and at the same consider the photon as not spatially extended.

After all the wavelength of photons is a real measureable phenomenon determining properties of the EM field(colors, visible light, etc.).

I was listening to a physics professor lecturing on QM and he raised the question "How big is a photon?" and indicated it had arisen during his PhD defense.

He than began to discuss the accurately known frequency and wavelength of a laser emitted photon (and thus accurately known momentum) in the context of the uncertainty principle.

It seems most physicists think of photons as part of the EM field which is of infinite extent and 'photons' with definite and detectable properties as excitations of that field upon measurement.

 

[sophiecentaur]

Are you saying that a photon's wavelength has no associated width? How do you explain diffraction patterns below the 200 nm? It sounds like a contradiction to measure the practically measured wavelength of photons in nm and at the same consider the photon as not spatially extended.

After all the wavelength of photons is a real measureable phenomenon determining properties of the EM field(colors, visible light, etc.).

It seems most physicists think of photons as part of the EM field which is of infinite extent and 'photons' with definite and detectable properties as excitations of that field upon measurement.

You need to go back to elementary diffraction theory and see what it is really saying. There is nothing to do with 'width' of the waves in the calculation of the interference between two slits. The pattern is related to the spacing of the slits and the wavelength (longitudinal variation). There is nothing 'special' about the wavelength of 200nm; the same theory applies to 200km or 2nm. The 'lateral extent' of a photon, as a concept, is meaningless because there is a probability that it can be measured over the whole area of a 100m sphere after it has 'travelled' for just 300ns from a point source. This is true for whatever wavelength of EM we are considering.

When talking about the interaction with matter, it is more precise to talk in terms of frequency. The energy of a photon is normally given by E =hf, with good reason. The wavelength, locally, is not actually going to be the same as the free space wavelength because, by definition, the photon is not interacting in free space but in the presence of matter. The frequency is going to be unchanged but what actual value of wavelength can you give your photon? (certainly not c/f)

 

[Maui]

There is nothing 'special' about the wavelength of 200nm; the same theory applies to 200km or 2nm. The 'lateral extent' of a photon, as a concept, is meaningless because there is a probability that it can be measured over the whole area of a 100m sphere after it has 'travelled' for just 300ns from a point source. This is true for whatever wavelength of EM we are considering.

But in practice there is something special in that photons at 200 nm are already in the UV portion of the EM field. Ordinary light microscopes hit the diffraction limit of visible light just above the 200 nm limit or at 1 PHz. AFAIK 200-300 nm is exactly the diffraction limit of visible light.

What you seem to be talking about is the probability amplitude of finding an electron which is different from photons wavelength. Or are you saying they are related?

When talking about the interaction with matter, it is more precise to talk in terms of frequency. The energy of a photon is normally given by E =hf, with good reason. The wavelength, locally, is not actually going to be the same as the free space wavelength because, by definition, the photon is not interacting in free space but in the presence of matter. The frequency is going to be unchanged but what actual value of wavelength can you give your photon? (certainly not c/f)

This seems to imply that my impression was correct in stating that -

It seems most physicists think of photons as part of the EM field which is of infinite extent and 'photons' with definite and detectable properties as excitations of that field upon measurement.

 

[Cthugha]

In fact already Maxwell's equations allow for highly localized solutions in free space. Phys. Rev. Lett. 88, 100402 (2002) cites some of them and states that "Remarkably, in contrast to the conclusions reached in previous studies, localization need not be related to photon wavelength nor necessarily associated with a wavelength-scale spatial range."

Getting away from the "size" concept, there are for sure several characteristic length scales of light fields. We have at least the wavelength, the coherence length, a decay length of the electric field/intensity, a decay length of the energy density and a decay length of the detection probability. While in simple cases several of these characteristic length scales will coincide, already the not that complicated case of non-monochromatic single photons leads to strange results like the maximum of the energy density being somewhere else than the highest detection probability and both scaling very differently. The Mandel/Wolf has a chapter on that topic (12.11 in my edition).

So is one of these length scales "better" as a size than the others? I do not think so. These different concepts of length scales exist separately for a good reason.