How Big is a Photon and What are its Dimensions?

[jnorman]

carl - if i read cohen, et al, correctly, it sounds to me like they are implying that localization effect of photons in energy space may be an artifact of RMT strategy. http://physics.bgu.ac.il/~dcohen/ARCHIVE/wbr_PRL.pdf

still, energy space is hardly a concrete or measureable entity, and to indicate that photons demonstrate "point particle" characteristics similar to that of an electron (which can actually be identified as having a scpecific location, HUP notwithstanding), surely seems a misconception to me. i do not doubt that my understanding of this issue is incomplete, but i am just not following your logic here. i stand by my comment that you cannot say anything meaningful about a photon during its "travels" - it simply cannot be defined as having any location. i do agree, of course, that photons are always detected as particles, but this results from collapse of the probablilty function at the time of absorption, and not from any macro-perspective measureable characteristic of photons. feel free to elucidate further for my edification. danke.

 

[teal4two]

Light scattering.
A visible light photon comes from the sun and enters the atmosphere. An atom in the atmosphere has electrons, whose positions can be described as probabilities (orbitals). The electric field of the photon can cause an electron in the atom to vibrate and in turn the electron radiates another photon.
What determines whether a given photon and a given electron will interact? Is this an all-or-nothing interaction (photon is either scattered or it isn't)? If the photon is on one side of the Earth and the atom on the other, presumably there will be no interaction. How close do they have to be?

 

[lightarrow]

Light scattering.
A visible light photon comes from the sun and enters the atmosphere. An atom in the atmosphere has electrons, whose positions can be described as probabilities (orbitals). The electric field of the photon can cause an electron in the atom to vibrate and in turn the electron radiates another photon.
What determines whether a given photon and a given electron will interact? Is this an all-or-nothing interaction (photon is either scattered or it isn't)? If the photon is on one side of the Earth and the atom on the other, presumably there will be no interaction. How close do they have to be?

The photon is the quantized interaction between light and matter. Just my opinion, of course.

 

[reilly]

No Virginia, there ain't no point particles, except on the pages of physics books and papers. Any particle with charge necessarily has an internal structure due to a cloud of (virtual) photons, e-p pairs, quarks, ... it then follows that any particle, massless or massive cannot be a point particle, but necessarily has an internal structure.

The problem is: we don't know how to compute, even approximately, such structures, form factors, if you will. But the structure of QFT interactions says those structures must exist. Further, we can measure nucleon form factors with electron scattering, for example. But whatever structure electrons or photons have is beyond our capability to measure or compute -- other than basic QED phenomena like the perturbative results for the Lamb Shift or the anomalous magnetic moment of the electron.

Nonetheless, QFT almost always deals with point particles, for mathematical convenience. For the most part, the physics community accepts this approximation as appropriate for comon practice.

If one, for the moment, accepts point particles, then position operators for massive spinning particles, at least, are a bit peculiar. As is noted in the work referenced by CarlB, the Lorentz transformation of a standard position operator, +i d/dp, in 1-D with the momentum=p, gets tangled up with spin, and, more generally, with angular momentum. this happens because, in the Poincare Group, the generators of Lorentz transformations do not commute with the generators of rotation, which in turn causes the product of two Lorentz transformations to result in a rotation followed by a Lorentz xform, or vica versa.

So, the net effect is that under a Lorentz transformation, the transformed position of a particle depends on the spin state -- that is the transformed position of an spin-up electron at (T,X) will not be the same as the position of a spin-down similarly at (T,X). While I realize the dangers of ascribing classical ideas to the arcane phenomena of the quantum world, I think of this as suggesting an extended object -- and there is some classical justification for this, as discussed by Moller in his Relativity text -- a classical object with an internal angular momentum, the Earth or a baseball thrown as a curveball,..., has a radius greater than |L|/(c M0), where L is the angular momentum, M0 is the rest mass, and c is the speed of light.(See section 64 in Moller's Theory of Relativity.)

It seems to be the case that these fine points of position, spin, and Lorentz transformations are, at present, more curiousities that anything else. but certainly, these ideas make the notions of positions of point particles with spin rather a dicey proposition.

Regards,
Reilly Atkinson

 

[Careful]

If one, for the moment, accepts point particles, then position operators for massive spinning particles, at least, are a bit peculiar. As is noted in the work referenced by CarlB, the Lorentz transformation of a standard position operator, +i d/dp, in 1-D with the momentum=p, gets tangled up with spin, and, more generally, with angular momentum. this happens because, in the Poincare Group, the generators of Lorentz transformations do not commute with the generators of rotation, which in turn causes the product of two Lorentz transformations to result in a rotation followed by a Lorentz xform, or vica versa.

So, the net effect is that under a Lorentz transformation, the transformed position of a particle depends on the spin state -- that is the transformed position of an spin-up electron at (T,X) will not be the same as the position of a spin-down similarly at (T,X). While I realize the dangers of ascribing classical ideas to the arcane phenomena of the quantum world, I think of this as suggesting an extended object -- and there is some classical justification for this, as discussed by Moller in his Relativity text -- a classical object with an internal angular momentum, the Earth or a baseball thrown as a curveball,..., has a radius greater than |L|/(c M0), where L is the angular momentum, M0 is the rest mass, and c is the speed of light.(See section 64 in Moller's Theory of Relativity.)

It seems to be the case that these fine points of position, spin, and Lorentz transformations are, at present, more curiousities that anything else. but certainly, these ideas make the notions of positions of point particles with spin rather a dicey proposition.

Regards,
Reilly Atkinson

I agree; let me first add that spin has to enter the position operator for massless particles if you want it to commute with the helicity operator.
On the other hand it wouldn't be unreasonable at all that the spin operator is included even for massive spinorial particles (apart from the obvious reason that you want the limit for m -> 0 to be well defined) : Zitterbewegung (due to spin) makes it hard to ``localize'' -say- an electron, so it is not unreasonable to think that position and spin operators are noncommuting. Indeed, in the book ``Relativistic quantum mechanics and introduction to field theory'', F.J. Ynduarain gives at page 191 a ``Newton - Wigner'' operator for spin 1/2 particles containing the spin operators.

 

[epv]

http://www.pbs.org/wgbh/nova/programs/ht/tm/3501.html?site=10&pl=wmp&rate=hi&ch=10
Talks about slowing down light and length of photon.

 

[ZapperZ]

http://www.pbs.org/wgbh/nova/programs/ht/tm/3501.html?site=10&pl=wmp&rate=hi&ch=10
Talks about slowing down light and length of photon.

I saw this when it first aired. Where exactly did it talk about "length of photon"? Maybe length of a light pulse, but not length of a photon.

You do know that this thread is more than a year old, don't you?

Zz.

 

[CarlB]

I'd forgotten about this thread. Margaret Hawton's position operator was recently discussed briefly on sci.physics.research in the topic of the title
"EM Field of Photon":
http://groups.google.com/group/sci.physics.research/topics?hl=en

Hawton herself chimed in. I put together a blog post showing the relationship between her photon operator and the consistent histories interpretation of quantum mechanics and she linked it into the discussion. The blog post is here:
http://carlbrannen.wordpress.com/2008/01/14/consistent-histories-and-density-operator-formalism/

As you can see from the sci.physics.research discussion, the local expert is not convinced that a photon position operator makes sense, but he also makes it quite clear that he doesn't bother to read the articles that are linked in, LOL. This was from a few days ago, and I suspect that Hawton is busily working on another paper extending these ideas.

 

[jambaugh]

According to Quantum Mechanics, a photon, like any other point particle, has no size at all. It is a point, if treated as a particle. If it is treated as a wave, it has wavelength and amplitude instead, amplitude being the "size".

Not quite. True it "has no size at all" but that means it nether has zero size nor some other value. It isn't a classical object with a size (even zero) while a "point particle" is such an animal. A photon is a process of quantized electro-magnetic energy transmission. This can happen in the very small volume of the nucleus of an atom or equally spread over a planet sized antenna array. "It has no size" in that size of any numerical value isn't a property of the photon.

 

[agkyriak]

Well, a photon IS a point particle...BUT in energy space. A photon is defined as a chunk of energy, nothing more. Ofcourse it is NOT defined as a point particle that has finite spatial boundaries, so asking about the "magnitude" of a photons is useless because this concept is defined (in most cases ) based upon spatial coordinates.

Finally, the only photon quantity we know that is defined using spatial coordinates is the photon's wavelength.

It is absolutely right. From a theoretical point of view the proof was given long years ago in the work of Landau and Peierls and confirmed recently in the works of other scientists Cook, Inagaki and others. Let us consider briefly this proof, using the book of (Akhiezer and Berestetskii, 1969):

The wave function of photon is here introduced as follows. The vectors of the EM field \vec {{\rm E}} and \vec {{\rm H}}, as the solutions of the wave equation of the second order, which follow from the Maxwell equations, are considered as the classical wave functions \vec {\varepsilon }\left({\vec {r},t} \right) and \vec {H}\left( {\vec {r},t} \right).

Representing the wave equation as multiplication of two equations for the advanced and retarded waves, we obtain two linear equations, which correspond to the wave vector \vec {f}_k and is a certain generalization of vectors of EM field. The equation for this function is equivalent to the system of the Maxwell equations. For this reason it is possible to consider the Maxwell's equation as the equation of one photon (Gersten, 2001). The quantization of classical wave function is produced by means of the quantization of energy of this wave by the introduction of the relationship \varepsilon =\hbar \omega. It turned out that in this case the function \vec {f}_k could be interpreted as the quantum wave function of photon in the momentum space.

But with the attempt to introduce the function of photon in the coordinate representation was revealed the insurmountable difficulty According to the analysis of Landau and Peierls (Landau and Peierls, 1930), and later of Cook (Cook, 1982a; 1982b) and Inagaki (Inagaki, 1994), the wave function of photon by its nature is nonlocal (see also the review (Bialynicki-Birula, 1994)). .

Actually, after completing the inverse Fourier transformation of above function \vec {f}_k we obtain:
\frac{1}{\left( {2\pi } \right)^3}\int {\vec {f}_k e^{i\vec {k}\vec {r}}d^3k=\vec {f}\left( {\vec {r},t} \right)}.

It seems that it is possible to determine \vec {f}\left( {\vec {r},t} \right) as the wave function of photon in the coordinate representation. Actually, because of normalization condition for \vec {f}_k the function \vec {f}\left( {\vec {r},t} \right) will be also normalized by the usual method: \int {\left|{\vec {f}\left( {\vec {r},t} \right)} \right|} ^2d^3x=1

However, the value \left| {f(\vec {r},t)} \right|^2 will not have the sense of the probability density distribution to find the photon at the given point of space. Actually, the presence of photon can be established only by its interaction with the charges.

This interaction is determined by the values of the EM field vectors \vec {{\rm E}} and \vec {{\rm H}} at the given point, but these fields are not determined by the value of the wave function \vec {f}\left( {\vec {r},t} \right) at the same point, and they are defined by its values in entire space.

In fact, the component of the Fourier field vectors, expressed byf_k, contain the factor \sqrt k. Formally this can be written down in the form
\[\vec {\varepsilon }\left( {\vec {r},t} \right)=\sqrt[4]{-\Delta }\vec {f}\left( {\vec {r},t} \right)\]
where \Delta is the Laplace operator. But \sqrt[4]{-\Delta } is integral operator, and therefore the relationship between \vec {\varepsilon }\left( {\vec {r},t} \right) and \vec {f}\left( {\vec {r},t} \right) is not local, but integral. In other words, \vec {f}(\vec {r},t) is not determined by field value \vec {{\rm E}}(\vec {r},t) at the same point, but it depends on field distribution IN A CERTAIN REGION, WHOSE SIZE IS THE ORDER OF WAVELENGTH.

This means that localization of photon in the smaller region is impossible and, therefore, the concept of the probability density distribution to find the photon at the fixed point of space does not have a sense.

This conclusion of theory is confirmed by experiment, since all measurements with the use of EM waves or photons (interference, diffraction and so forth) can be carried out to the region, not smaller as wavelength.

Akhiezer, A.I. and Berestetskiy, V.B. (1969). Quantum electrodynamics.
Bialynicki-Birula, Iwo (1994) On the wave function of the photon. Acta physica polonica, 86, 97-116),
Cook, R.J. (1982a). Photon dynamics. A25, 2164
Cook, R.J. (1982b). Lorentz covariance of photon dynamics. A26, 2754
Gersten, A. (2001) Maxwell of equation - the of one-photon of quantum of equation. Found. of Phys., Vol.31, No. 8, August).
Inagaki, T. (1994). Quantum-mechanical of approach to a of free of photon. Phys. Rev. A49, 2839.
Landau, L.D and Peierls, R. (1930). Quantenelekrtodynamik in konfigurationsraum. Zs. F. Phys., 62, 188.

 

[peter0302]

The "amplitudes" of the EM field of a photon at any given point refer not to physical space in which there is a field, but rather to the strength of the EM field at that point, right?

 

[Mentz114]

agkyriak :

Nice post. It's always good to see one's prejudices (only mass can be localised) supported by good maths. I've long argued against ascribing 'particle' properties to photons.

 

[Alfi]

You do know that this thread is more than a year old, don't you?lol - It's brand new to me. I've only just read just now.
In a year from my 'now' this thread will be a year old.
time must be some kind of illusion. The 'now' is very much like a point particle.

but interesting none the less :)

 

[lightarrow]

The 'now' is very much like a point particle.

...or as a non-localized field.
Yes, it's a completely different point of view...

 

[heterotic_]

I'm bringing this conversation back from the dead, because I feel like it has some sort of meaning.

If we talk about size in a non-conventional sense of the word, but as the average minimum amount of space it occupies, doesn't everything have to have a size? If two variables "try to" occupy the same amount of space, at points defined by x, y, z, in our universe, don't they interact (either by repulsion [in which they do not occupy the same space], attraction [bind to form something "new"] or annihilation), instead of occupying the same point in space?

 

[Eynstone]

I wonder what we'd get if we tried to calculate the 'radius' of photon on the lines of the classical radius of electron.
On the other hand, don't we treat particles as points right at the outset of quantum mechanics?

 

[heterotic_]

The actual radius of an electron is calculable, it is just dependent on its environment, so shouldn't this be also true of a photon?

Everything has a center, even if it is so small, its entire volume is its center. It is how far it is spread out that is changeable (or rather, how concentrated the energy is in relation to its center, or a "point") I believe that point must be the minimum space it can occupy, and therefore must have a size, even if we have no way to fathom it.

 

[PhilDSP]

The wave function of photon is here introduced as follows. The vectors of the EM field \vec {{\rm E}} and \vec {{\rm H}}, as the solutions of the wave equation of the second order, which follow from the Maxwell equations, are considered as the classical wave functions \vec {\varepsilon }\left({\vec {r},t} \right) and \vec {H}\left( {\vec {r},t} \right).

That's a wonderful analysis! But isn't it dependent on the simple assumption that the photon IS the wave rather than the alternative that the photon is the structure that generates the wave?

P.S. Sorry, I didn't realize at first that that post was several years old.

 

[Jivesh]

I was also perplexed as to what possible answer could be there when we talk about the size of the photon. Even the concept of measuring the radius of the electron is not justified in quantum theory, because if we consider the electron to be a sphere with the charge smeared on its surface or throughout the volume of the sphere, then we will have to explain the spin of the electron in terms of the an actual spin, as we do for the case of, say, the Earth's spin. But that is completely wrong description of the electron, as it results in absurd values of speed of the surface of the electron.
I think that quantum mechanics starts with the assumption that the elementary particles are point like objects, which do not have any dimensions. Furthermore, if we talk about quantum mechanics, then visualising the electrons, photons, etc. as classical particles would be a mistake. The only thing that we have to guide us in QM is the wavefunction of these supposed "particles", and their evolution according to the Schrodinger's equation.

 

[heterotic_]

That's a wonderful analysis! But isn't it dependent on the simple assumption that the photon IS the wave rather than the alternative that the photon is the structure that generates the wave?

P.S. Sorry, I didn't realize at first that that post was several years old.

Redundancy would be silly, so I brought it back.

I would say that we cannot calculate the "classical" radius of a photon like an electron, because a photon always moves at the speed of light.

But what about calculating the mass of a photon depending on its energy? I know people say the photon has no mass, based on the definition of mass, but it does. It doesn't have a "rest mass", because it is never at rest, but the best definition of mass is that it is "the measure of inertia". Inertia is "the amount of resistance to change in velocity". Change in velocity is "an increase or decrease in Kinetic Energy (and Potential Energy, since Total energy is conserved)".

So, inertia is also "resistance to change in Energy"; the inertia of a body is dependent on its energy content. So mass then, is a measure of resistance to change in energy.

So instead of asking, can we calculate the mass of a photon, we are asking, can we calculate the measure of a photon's resistance to change in energy? YES.

Next, if we talk about the fact that a photon exhibits the characteristics of both a particle and wave, we cannot assume that a photon is only a wave. How it behaves is what it is. So how does it behave?

Now, bear with me, I'm only an undergrad, so if I'm way off base, feel free to correct me. If the photon has a center, and a wavelength, if we look at an electron, which has similar properties of a photon in that its radius is variable and it exhibits the wave-particle duality, we can get an idea of what a photon probably is. In a Hydrogen atom, the wavelength of its electron is equal to the Bohr circumference of the electron's orbit. Can we use this information to describe the duality of a photon?

We know that a photon is polarized and has a spin. So what if, it is a sphere (energy concentrated at a center, and pointing outward in all directions to create a force field) that swells and contracts? Wouldn't that explain the particle-wave duality? If this were true, the radius of a photon would then be variable, but at its swell would be its greatest radius, which we could calculate using its wavelength.

My thoughts:

r(photon-E) \leq \frac{ch}{2E}

and

m(photon-E) = \frac{h^2}{Ec}

I was also perplexed as to what possible answer could be there when we talk about the size of the photon. Even the concept of measuring the radius of the electron is not justified in quantum theory, because if we consider the electron to be a sphere with the charge smeared on its surface or throughout the volume of the sphere, then we will have to explain the spin of the electron in terms of the an actual spin, as we do for the case of, say, the Earth's spin. But that is completely wrong description of the electron, as it results in absurd values of speed of the surface of the electron.
I think that quantum mechanics starts with the assumption that the elementary particles are point like objects, which do not have any dimensions. Furthermore, if we talk about quantum mechanics, then visualising the electrons, photons, etc. as classical particles would be a mistake. The only thing that we have to guide us in QM is the wavefunction of these supposed "particles", and their evolution according to the Schrodinger's equation.

My problem with QM and GR for that matter, is that they both work, but not really together. That just is not a good answer for me.

We talk about the spin of an electron and a photon, but what if it's not a spin, like you said in terms of the Earth? What if it is more like... the electricity shifting in a uniform motion throughout its volume in one direction or the other, as a result of the polarization?

 

[Salman2]

From this book: Lecture Notes in Physics: The Nature of the Elementary Particle, 1978. Malcolm H. MacGregor, it is stated (in quotes):

"The particle properties of the photon emerge most clearly from Compton scattering, in which a high-energy photon makes a billiard-ball type of collision with an electron, and hence delivers its energy and momentum into a volume that has dimensions on the order of 10^-11 cm (the Compton wavelength of the electron)."

The "particle" aspect of the photon from Compton scattering experiments represents a "MeV photon" compared to an "optical-frequency wave photon" that is about 10^6 times larger in dimension.

"In dealing with the photon, we must take into account both its wave aspects and its particle aspects. But since these two aspects differ dimensionally by many orders of magnitude, they are in a practical sense separate entities, so that we can discuss the wave properties of the photon without having to consider its particle properties, and vice versa."

 

[heterotic_]

From this book: Lecture Notes in Physics: The Nature of the Elementary Particle, 1978. Malcolm H. MacGregor, it is stated (in quotes):

"The particle properties of the photon emerge most clearly from Compton scattering, in which a high-energy photon makes a billiard-ball type of collision with an electron, and hence delivers its energy and momentum into a volume that has dimensions on the order of 10^-11 cm (the Compton wavelength of the electron)."

The "particle" aspect of the photon from Compton scattering experiments represents a "MeV photon" compared to an "optical-frequency wave photon" that is about 10^6 times larger in dimension.

"In dealing with the photon, we must take into account both its wave aspects and its particle aspects. But since these two aspects differ dimensionally by many orders of magnitude, they are in a practical sense separate entities, so that we can discuss the wave properties of the photon without having to consider its particle properties, and vice versa."

Realistically, that is only a good conclusion when you discuss the outcomes the two produce independent of each other.

However, we are discussing bridging the gap, so saying not to, is just illogical. If we think about what a wave really is, we envision that oscillating line we produce with technology, or a ripple from a linear perspective so we see the amplitude of a physical wave. The problem with that is that it is a perspective, it isn't the entirety of the actual wave itself. It is only a description from a reference point, and requires ignoring the rest of the "picture", as it were. It is also representing a point · moving across an oscillating line from the "side", but a straight line from the "top". You really cannot picture the wave that way, it is merely a representation to represent wavelength and amplitude, based on what we observe as physical waves from a sideview, like the water oscillating in a ripple.I think a "real view" of the wave would produce a different picture more like:

O>·<O>·<O>·<O>·<O>·<O

With that (granted, think of the swells with more of the same wave length of a particular frequency), we could even find the radii of all the "sizes" that the photon takes on spherical calculations of each moment of the swell.

 

[Pinewater]

Just an fyi, in scientific literature the 'size' of a photon generally refers to its coherence length. (which is definitely NOT to say that this is the only correct way to think about a photon's size!)

You can look up coherence length on the web, but I understand it best in the context of an interferometer. In this setting, the coherence length is the maximum distance that a single photon can travel and still interfere with itself (which means that it produces an interference pattern at the screen).

 

[Pianoasis]

I think the two-slit expirements should be done in a vacuum. The molecules in the air obviously cause some interference with the photons natural behavior. In fact it could be the only reason the photons act like a particle. The photon waves hit the molecules in the air, send them along the photon wave's path, and hit the screen in the appearance of a particle. If these tests were to be done in a vacuum we would have much clearer results.

 

[Vanadium 50]

This thread is two years old.