1 erg = 10^−7 Joules. So, 10^{-14} erg = 10^−21 Joules

If I modify your problem a bit, as the photon is enclosed in cavity of volume V and then its energy density is 10^−21 J, then since energy is equally distributed in E and B fields. This implies, magnetic energy density is 0.5 * 10^−21. Then, B = sqrt(2 * mu * 0.5 * 10^−21) = sqrt (mu* 10^−21) =10^-10.5 * sqrt(mu), where mu is the permeability of the medium. If its vacuum, then  B = 10^-10.5 * sqrt (4 pi 10^-7) = 2 * 10^-14 * sqrt (pi)  = 3.5456 * 10^-14 Tesla

In the illustration above I have assumed Maxwell's classical theory of EMW. In the light of quantum mechanics and Quantum Field theory (QFT), it can get quite tricky. The energy carried by electromagnetic radiation can be calculated from the amplitude of the electric (or magnetic) wave in the classical picture; or from the number of photons in the quantum picture. So, I have not considered the idea of "ONE" photon in your question, but rather treated it as the energy density of EMW in an enclosed volume (cavity).

Thank you for your answer in the framework of classical theory.

However, the concept of a photon occurs in the quantum theory, where the most probable zero-point energy (proportional to the averaged value of B^2) of the photon field has the infinite value (see "Quantum Electrodynamics", Berestetsky,Lifshitz,Pitaevsky) .

It seems in that  there is the origin of ambiguity quantification of the magnetic field.

(If I'm not in mistake)

It is important. For example, in the process of photosynthesis, where a photon creates a field in which a chemical compound of the electron shells of the "right" (selected) atoms occurs. For a quantitative description of this process you need to know this field, which creates a single photon. (It is necessary to know in details, as , for example, this has been obtained for the Landau levels in a magnetic field to the non-relativistic quantum theory.)

Its an old problem.. and it is essentially due to incompatibility between classical mechanics and quantum mechanics.

@ Vikash: I am confused by the claim that the ``energy density’’ is 10^{-21} J. The units are wrong: energy density should be J/m^3. My guess we can proceed as follows: the photon energy divided by hbar is the frequency, the latter yields the wavelength, a photon under ``normal’’ circumstances is confined in a volume of the order of (wavelength)^3, so that the photon energy divided by this volume yields the energy density, which itself yields an order of magnitude estimate for the squared magnetic field.

@ Leyvraz

I implied the given magnitude as the energy density.

The ultimate motive behind the question was to show the inconsistency! As you can see, magnetic field manifests due to a time-varying electric field. Electric field emerges when there is a charged particle; but photon has no charge! So, the question magnetic field of a photon does not makes any sense though of the wave it does. Classically I gave the solution above, quantum mechanically I guess it would be rigorous.

@ Vikash:  I disagree. A photon is a quantised oscillator corresponding to an electromagnetic mode. The two fields E and B are canonically conjugate, and it makes as much sense to ask what is the magnetic field of a photon as it does to ask what is the posiition of an oscillator. In both cases you need some additional information, such as the oscillator frequency, to make the problem determinate. 

I think Lugovoi asked only the numeric value to his question because he needs to confirm a result. Anyway Pandey has given a first number and nevertheless Leyvraz's considerations on energy and energy density are right. Then I have calculated the result searching for considering the energy density in the physico-mathematical model of photon and I have obtained a new number: B=1.75x10-8 Tesla.

@ Leyvraz: I think it is mere due to interplay of words. A photon corresponds to E and B fields, Agreed. But considering charge-less photon, magnetic/electric field created by a photon leads to an ambiguity!

@ Vikash:   Yes, seems, till now, there is not logically consistent quantum theory. This is problem..

@ Leyvraz:  Well, let me some additional information:   the energy of oscillator (photon) is 10^{-14} erg    -->    the cyclic frequency of oscillator (photon) is 10^{+13} seconds^{-1}    -->  the wavelength of photon is 1.9x10^{-2} cm.

 A first QUESTION is:  what is the magnitude of magnetic field which could be (?) created for the ion localized in segment 10^{-6} cm, during the time 10^{-14} seconds ?  These numbers show that, for the period of 10^{-14} seconds, the photon ( with the velocity of 3x10^{+10} cm/sec )   moved on 0.016 part (or on 2 percents) of its wavelength, and the length of segment is in  more than 10 000 times  less than wavelength of photon. 

In frame of these numbers the second QUESTION is: could we tell that discussed here magnetic field during that time of  10^{-14} sec is approximately constant for that mentioned ion, located in that mentioned segment of 10^{-6} cm?  Of course, the magnetic field is changed continuously from some min to max.

The third QUESTION (equivalent to second QUESTION) is:    But, during 10^{-14} sec, the magnetic field takes the approximately  constant any value within the interval between   these mentioned min and max?     

@ Lugovoi,

Sometimes trivial language settings can create unnecessary confusion. So, to be clear, a photon does not "creates" E or B field, but rather charges do. And, a photon is just a nomenclature for quantized aspect of EMW. In essence, your question "magnetic field created by one photon"  is misleading. It should better be rephrased as "magnetic field corresponding to one photon".

Vladimir

The traditional figures for energy density are given with respect to the cubic meter, so it seems to me that since individual photons are most probably localised in volumes much smaller that calculating the magnetic field calculated with respect to the 1 cubic meter volume are likely to be lower than physical reaiity.

Do you have any ball park figure that you tentatively arrived at as to the possible or probable range of magnetic field intensity you are looking for?

Above I have asked three, far not simple, quantitative questions addressed to F. Leyvraz, or to any one, who could give me physically argued answers.

This is the part of calculations for arising the quantum structures (arising in two-dimension structures or in molecular biology) with some of properties which could be found (I hope) in the offered sufficiently simple experiment.

I'm sorry, but non-related answers in common words do not interesting for me absolutely.

Vladimir

I read your 3 questions and also the other answers. I did some calculation and arrived at a much higher intensity value for the magnetic field than Vikash. So high that this makes me wonder and ask if you had yourself calculated some range of possible values to see if my value would be in the range.

Since the photon magnetic field constantly varies from min to max in the de Broglie hypothesis of the double particle photon, as you assume it does in physical reality, the set LC equations detailed in the last par ot this paper may give you a precise set of answers to your 3 questions.

The equations are meant for SI values, but can easily be converted to CGI.

Here is the link to the de Broglie solution:

http://www.omicsonline.com/open-access/on-de-broglies-doubleparticle-photon-hypothesis-2090-0902-1000153.php?aid=70373

Sorry, at the moment for me it's not interesting to discuss any "Photon Hypothesis".

I have addressed to RG to get three answers.

I hope for the time being...

Fine with me. Just thought these equations could help.

I agree with Vikash Pandey. It is better to say:" the magnetic field corresponding to one- photon state". the N-photon states are the eigenstates of Hamiltonian operator(H=E^2+B^2). But the E-filed and also B-field do not commute with H. Thus, the N-photon states are not the eigenstates of B-field. It is possible to show that the expectation values of E and B-fields are  zero for N-photon states. Therefore, the B-field measurement (if it is experimentally possible) on a single-photon state can give different values with different probabilities.  Please see this book: "Greiner, Quantum Mechanics, Special chapters (Page: 16)  " 

@ Vladimir: You are wrong in assuming that there *is* a definite value of the magnetic or electric field for a photon. It is a quantum phenomenon, so that the E and B fields have a distribution of values. The width of this distribution can be obtained approximately as I told you: the energy determines the frequecy, which determines the wavelength, the cube of which gives the volume in whihc the photon can be expected to be confined. The energy divided by this volume gives an energy density, which can be identified with a value of E^2 or B^2, at least in rational units. In the SI system, one needs the appropriate epsilon_0.

Confusion between quantum phenomena and indeterministic phenomena is very popular in postmodern physics that is firm in obsolete standard models. If a infrared quantum with a definite value of 10-21J exists, and it exists like a definite frequency and wavelength exist, it is incomprehensible to think a definite value of amplitude of magnetic field and of electric field don't exist since energy quanta are electromagnetic nanowaves that are described by Maxwell's equations.

Daniele

I fully agree with you here. Maxwell came to the conclusion that the magnetic field and the electric field mutually induce each other based on the idea of displacement cur¬rent, a current that would come into being without the presence of matter in this case, and was the foundation of his theory about light.

Ref: Sears, Zemansky and Young (1984) University Physics, 6th Edition, Addison Wesley, page 625.

This is why when mathematizing localized photons, it is possible to mathematically freeze, so to speak, the oscillation at any point of photon frequency cycle to establish the exact amount of magnetic field and electric field at this instant.

André, I prefer to talk now about "magnetoelectric induction" rather than displacement current, in analogous manner to term "electromagnetic induction".

@ F. Leyvraz :       thanks I got it. But it remains an open question for me about the relationship between the static magnetic field and relativistic quantum physics, in which a theoretical concept of the photon is formulated as the quantum of the electromagnetic field, that is, in fact, as a "piece" of the field. As these "pieces" form a constant magnetic field?

The magnetic field is about 10^{-6}T. It is given by B=E^2/(2ehc^2), where e is the electronic charge, h is the Planck constant and c is the speed of light.

Zero, 0 if you prefer. Photons do not produce a magnetic field. Or at least none has ever been measured.

Yes, "none has ever been measured" this, but it, seems, is very important for micro (sub-nano) processes in Nature ...

Can only one photon create a molecular bond? Take your time to respond. This question is not stupid: if the photon's wave function overlaps two valence electrons and carries energy, all this can stimulate the exchange interaction between these valence electrons, and this is the beginning of the molecular bond!

@Vladimir

I hope I did not treat your question as stupid. I did not mean to offend.

I would say that photons are vital in creating molecular bonds, but magnetism is not the intermediary force that allows this.

Without any real evidence I would suggest that magnetism is a resultant rather than a primary force.

Take reflection from a metallic mirror as an example. The most likely explanation for this is that a photon interacts with the electron cloud fairly randomly, which is then in an unstable energy state. The electron cloud then the cloud emits a photon, again randomly, to achieve a stable energy state.

No magnetic effects have been detected and, logically, there is no mechanism to produce a magnetic field in similar scenarios.

Atomic bonds are again likely to be the result of photonic interaction between electrons at various relative 'positive or negative' states of energy and the sharing of photonic energy quanta to achieve some kind of mutual energy balance or minimum energy state.

Regards

Ian

I did not even think about it.

It seems to me that modern theory greatly simplifies everything, leaving the role of a single photon only as an energy carrier. Perhaps, therefore, there are many insurmountable difficulties, for example, in obtaining an artificial process of photosynthesis, ....

A photon has no charge, so it cannot produce other photons and interact with other photons. Then, why do we need "many" photons to get, for example, a uniform field in an "infinitely" small volume?

I am not certain you are correct when you say photons cannot I react with each other. The Einstien equationt E = mc2 tells us that matter can be converted to energy packets (photons) and back again.

If this is a correct interpretation of facts, then photons in some state or other clearly interact. 'Charge' is not a necessary condition for interaction, it may like magnetism be the result!

Thus the conditions for a field, uniform or otherwise, is that energy must coexist with matter.

This would lead to the conclusion that there is a minimum size for a charge, electrical or magnetic, which is probably exactly the size of an electron or possibly one of the baryons.

Physics is quantitative science. We are talking about a photon with an energy of 10 ^ {- 14} erg = 10 ^ {- 2} eV. In order for a photon to produce an electron-positron pair, its energy must be at least equal to the sum of the masses of the electron and positron = 10 ^ {+ 6} eV.

Maxwell's equations can be written for a "set" of photons, each of which does NOT have an electric charge and therefore does NOT interact with other photons from this "set" of photons.

I must agree with you over the nature of the science. However I will always prefer measurement over theory (theories change so often!).

Unfortunately the equipment available to us for accurately measuring photons does not exist. We are forced to speculate from limited experiments, where the results are vague, on specific results. The energy levels you quote are simply a 'best guess' based on current theories over the nature of a photon at some median value.

Dear Proffitt,

In a recent theory I found that the magnetic field induced by a photon is given by B=E^2/(2ehc^2), where E is the photon energy, e is the electronic charge, h is the Planck constant and c is the speed of light.

I appreciate your comment on the plausibility of the formula.

Thanks in advance.

Arbab

Is the photon motionless? Obviously not; it is moving at light speed. Then where 'its magnetic field' might be observed? One mile behind this photon, two miles? Stationary observer should experience time-varying field (electric and/or magnetic) or the energy flux in other words. If so, then the photon's energy should steadily decrease, thus its frequency should tend to null. Nothing like that is observed, although there are some hypotheses about 'tired light'. In conclusion: the amplitude of a single photon magnetic field is exactly zero.

Or in the other way: photon does not carry any electric charge and therefore produces no (time-varying!) electric field - so why it would be the source of magnetic field, which is uniquely related to electric field?

@ Arbab I. Arbab

Prof Arbab,

I regularly read and enjoy your work, which is both novel and challenging in many ways.

However with respect to your formula above I would challenge it on 3 counts.

1 philosophically. A photon is defined as a quanta of energy, variable with the frequency of the light source. A magnetic field is defined as an accumulation of energy quanta from the Maxwell equations describing it.

2 measurement. A magnetic field has never been measured surrounding even the most intense photonic sources (terawatt lasers) where they are in vacuo. Only when light interacts with matter to produce plasma do we detect a magnetic field.

3 mathematically. Individual photons have been detected at 10^+14eV. (Very High Energy Gamma Rays). If we slot these into your formula then B = (10^14)^2/((2×1.6×10^-19)×(6.626×10^-34)×(3×10^8)^2) This gives B=(10^28/10^39) or 10^64 Teslas for each photon from such a source.

A magnetic field of such strength would pull matter from the sun as it passes and would devastate our modern technology.

Correction

B=(10^28/10^-37) or 10^65 Teslas

Ian

Dear Prof. Proffitt,

Thank you very much for your kind words and your interest in my work. Since a single photon carries a unit of an electromagnetic energy it is legitimate to associate a unit electric and magnetic field with it. I think you forgot to convert the eV into Joule in the above formula, which will make the magnetic field B~10^26 T. In practical units one has B(mT)=1.34 E(eV)^2.

Good question ... been trying to answer this same question via a new definition of the speed of light! Check the following article.

Preprint Amplitude Velocity of a Wave

Dear Marek Wojciech Gutowski ,

From the point of view of classical physics, Maxwell equations, you are absolutely 100% right! Maxwell considers a continuous field, and not an isolated photon, as a quantum of an electromagnetic field. Why does everyone reason only in the framework of classical physics? The most fundamental and therefore interesting processes take place seems to be in the world of quanta. Another thing is that these fundamental processes are repeated many times, and we see in the experiment the "interference" of these repetitions. I really want to find a way to learn the method of calculating this particular “fundamental” act. Everything must be very "simple", from the first principles of quantum mechanics. Maybe I'm wrong, but I believe in it ..

Dear Vladimir,

I think I got your point of view. Maybe it is right, maybe only unnecessarily complicated. In some sense the situation is similar to the description of an ideal gas. We know it consists of single atoms but it is too troublesome to track every such atom to describe the behavior of a system as a whole. It is much easier to 'invent' things like temperature, pressure and entropy. Here we have photons and electromagnetic waves - is this an analogy? Perhaps, but there are problems like existence of time independent electric or magnetic fields.

Anyway, I do not see any clue why the photon should be magnetized.

Dear Marek Wojciech ,

Not exactly an analogy, since one photon is already an electromagnetic wave. Therefore, maybe a time-independent static electric or magnetic field is such a "standing" wave (a set of "standing" waves), that is, the waves do not stand, they propagate "back and forth" in a closed explored space. This is after all the world of quantum mechanics: and, perhaps, in the case of these "standing" waves, the concepts of a quantum mechanical probability wave and the electromagnetic wave (Maxwellian) "locked" (for example, by circular currents) in a limited area are closely mixed. According to the laws of quantum physics, this will inevitably lead to the discreteness of energy, that is, to the appearance of quanta - photons, the wave functions of which can be significantly non-zero in the region between the field source (circular current creating the field) and the test charge, which helps us measure the field.

The electric field does not localize the movement. The magnetic field localizes the motion (Landau levels). Nature herself tells us: a photon falls on a living leaf of a plant, the localization of electrons is obtained (molecular bond).

There are many theories of photosynthesis and not a single artificial repetition of photosynthesis based on these theories, ....

Dear Vladimir,

I think that number ought to be zero. Why that? I remember one of my oral examinations quite a few years back. One of the question was to elaborate on the expectation values of and of one photon states which turn out to be zero. Even though you don't want a book reference, I'll nevertheless point you towards good old Messiah for this. (probably a chapter on semiclassical quantization of the em field)

Dear Kai,

According to the paragraph 5 (in the book "QUANTUM ELECTRODYNAMICS",1965,by authors Berestetsky and

Akhiezer) : "The properties of a quantum system approach the classical ones when there are large quantum

numbers that determine the stationary states of the system. For a free electromagnetic field (in a given volume),

this means that quantum numbers must be large oscillators, that is, the numbers of photons must be large. "

In contrast, we are trying to understand the properties of a single photon. Therefore, the semiclassical

approximation is not something one wants to know about.

The second.

In ["QUANTUM ELECTRODYNAMICS",1965,by authors Berestetsky and Akhiezer]

see formulas (2.2), (2.4), (2.17), (2.18) - (2.20).

As I understand it:

Here (see (2.20)) the electric and magnetic fields for one photon are introduced!

This electric and magnetic field of one photon (see (2.20)) is expressed in terms of the vector potential of one photon (see (2.18)) !

Then the operator of the electric field (and magnetic field) is considered as the sum (see (2.19)) of the electric (and magnetic) fields of photons (with multiplying each of these fields by the creation or destruction operator,

I understand it as, to specify the actually considered number of photons).

My question is:

So why this number can not be equal to 1.

And we are considering one photon?

And, accordingly, for one photon we consider the magnetic field?

Dear Constantin,

Thanks for the answer

Could you give me the amplitude of the magnetic field exactly (in a metric Gaussian system with all coefficients). And let me ask one more question, how to correctly assess the degree of homogeneity of the magnetic field created by one photon (with a given energy) in a short (what?) time interval at a short (what?) distance?

Сould you send copies of these articles, as access to them is limited.

Thank you in advance.

Vladimir

The electric field of a single photon is given by E=(2 pi) U^2/(hc)=15 kV/m .