# Frequency of a single photon

1. Apr 2, 2005

### locke

I'm having trouble understanding how you can associate a frequency with a single photon. What exactly does it mean when one talks about a photon having a particular freq? Does it refer to the individual fronts in it's wave-packet? or to something else entirley?

2. Apr 2, 2005

### Billy T

You obviously can't measure it like one meaures the frequency of a continuous electromagnetic wave (for example an unmodulated TV carrier signal. Perhaps the best way to measure the frequency of a single photon would be to arrange for it to be absorbed by a single atom (really a ion), one in a low intensity beam that has been passed thru electromagnetic "velocity filter" something like a mass spectrograph. Then you would know the ion's velocity. If the photo is absorbed into a long lived excited state, and intersected the ion beam transversely, then its energy and momentum would be added to the ion. The transverse momentum gain could be measured by letting the ion intersect a spatially resolving detector (fine grained photograph film would probably work fine - but I am not sure) Both the energy and the momentum are directly porportional to the frequency. You would need an ion that has energy level appropriately above the ground state to absorb the photon and this would be a problem if you did not basically already know the frequency rather well. Thus one might need to use a very small aggregate of matter in the beam (both mass and velocity know as it has passed thru both mass spectrograh and a velocity filter. Then again the transvers momentum given to some particles in the beam are the measure of your frequency. Not an easy problem - measuring the frequency of a single photon, but I think it probbly can be done somewhat along the lines I have outlined, but I have not done any analysis to see what accuracy could be achieved (low I expect)

3. Apr 2, 2005

### SpaceTiger

Staff Emeritus
It depends on the energy of the photon. Approximate energies of photons are routinely measured in X-ray detectors, since each one can excite many electrons in a CCD (the uncertainty in the derived energy would be dependent upon the square root of the number excited). It certainly is a harder problem for lower energies, however.

4. Apr 3, 2005

### Mariko

A photon can be either a packet of light or a particle(It can be either)the reason it has a frequency(wavelength) is that its source is a hot - (a photon described in my college astronomy Book states;Photon:a discreet UNIT of electromagnetic radiation)- ok and heres the definition of electromagnetic radiation:Radiation consisting of WAVES propigated through regularly varying electric and magneticfields including radio infrared,visible light,ultravoilet,xrays and gammarays- everything is a wavelegth in some form or another

5. Apr 3, 2005

### locke

So...If you say a photon has frequency x, it just means that if you took a whole pile of these photons and put them together you would get monochromatic light of frequency x?

Does this mean then, that the concept of frequency exists only in the abstract for a photon by itself?

6. Apr 4, 2005

### Billy T

Correct - I have in fact used several large NaI crystals (with attached photo -multiplier tubes) to measure the energy released simultaneously in two gama ray from neutron excited Chlorine. I can't tell why as is classified.

Later,via edit: It is also worth adding to SpaceTiger's comments that with X-rays, and a good scattering crystal (Bragg refraction) and a piece of pre sensitized film (It takes two light photon absortions for photo senstive film crystal to be developed, but this may not be necessary for X-rays. I don't know. One X-ray photon may do. If so, don't presensitize by weak exposure to light.) the energy might be reasonably well defined.

I once processed a "lauie diagram" (badly spelled) I made as graduate student to get the crystal type and atomic spacing of the "unit cube", but if the unit cube is known, then the frequency is determined. In fact, to make good (really sharp) "lauie diagram" one should double scatter off two crystals - the first scattering serves as a spectrograph, selecting one frequency thru some lead stops to be incident on the second crystal.

Last edited: Apr 4, 2005
7. Apr 4, 2005

### Staff: Mentor

1=1=1=1=1=1=1

If they're all the same, they're all the same. Each individual photon has a frequency and it is measurable on an individual basis under certain conditions.
No. How does that follow?

8. Apr 5, 2005

### locke

If someone says "frequency" to me, I understand it be the number of periodic events happening in a unit time. Perhaps my question should be: "Does a photon exhibit any kind of periodic behaviour?"

9. Apr 5, 2005

### dextercioby

The photon is a particle.It has 4 momentum.Period.

That "frequency" is the frequencly of the classical electromagnetic radiation/wave whose quanta of energy is one photon.

So we use "frequency" when discribing a photon for 2 main reasons
a)History (see Planck's work in the autumn of 1900 and Einstein's article [1] of 1905).
b)Natural (Heaviside-Lorentz) units,in which $\hbar=1$,so the energy of a photon is equal to its frequency.

Daniel.

-------------------------------------------------------------------------------
[1]A.Einstein,"Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt"
("On a Heuristic Viewpoint Concerning the Production and Transformation of Light")
,Ann.der Physik,June 9-th,1905.

10. Apr 5, 2005

### Billy T

I hope you will acknowledge the following facts about that "particle" and reconsider your view.

With a low pressure gas discharge light source of "particles" and some simple colored glass filters, it is possible to make a beam of light in which all of the "particles" have essentially the same energy. If this mono-energetic beam is then made to pass thru a highly reflecting (but very slightly transmitting) "mirror" set at 45 degrees to the beam direction, then the few "particles" that do pass thru and continue on in the original beam direction can be so infrequent that 99+ percent of them are the only "particle" existing during their flight time. That is, less than 1% of the time would there be two particles existing at the same time. Even in the rare event that two were "alive" at the same time, they are not likely to near each other and interact with any matter they come near completely independently. (To keep the room very dark, the reflected beam, should enter the open end of a long cone with very black interior and the light source is completely covered except for the path of the beam.)

Now if in this weak beam of "particles", two fine parallel slits, quite close together are placed and behind the mask in which these slits were cut in, there is a piece of photographic film (Also covered except for the path from the slits so it is in a very dark region and long exposures are possible). (The film is not in contact with the slit mask, but perhaps 40 times the slit separation behind it, or more if coarse grained film is used.) Then the distribution of exposed areas on the file will have the same pattern as if a short exposure were made with the slightly transparent "mirror" removed.

Each of these "particles" (in some sense difficult for humans to understand) goes thru both slits. After a large number of them have passed thru the slits, one at a time, the "interference pattern" characteristic of "waves" is exposed on the photographic film.

Summary: In some experimental arrangements, these "particles", observed individually, one at at time, exhibit all of the characteristics of "waves" and none of the characteristic of "particles." For this reason it is not correct to make the statement you emphatically do, quoted above.

Technical note for thoses who like to be careful with the facts: The long exposure pattern will be exactly the same, (size and shape) but the intensity distribution will not be. The weakly exposed areas will be even weaker in the long exposure than they were in the short exposure.

The failure of slowly delivered photons to produce the same exposure is called the "reciprocity failure." It is has to due with fact that a second photon must strike the same photographic grain, before the effects of the first (Production of a "color center" like lattice defect.) has relaxed back to the undisturbed state. - In typical use of a film camera, with sub-second exposures, it makes essentially no difference if the lens is open wide and the exposure is brief or if the same number of photons enter the camera in a longer exposure with the lens stopped down. If the exposure takes hours, "reciprocity failure" becomes important, if the film is not keep cold, but that has its problems also.

Last edited: Apr 5, 2005
11. Apr 5, 2005

### Billy T

Absolutely! It is sometimes only an electric field and at other times only a magnetic field. The collapse of the magnetic field is what creates the electric field (that is easy to understand) and conversely the growing electric field makes a "current" which has an magnetic field associated. (This is a little harder to understand and the timing is not easy to express in words)
Maxwell's contribution was to unify all the already existing equations of magnetism and electricity and add one new term to the set. That new term is called the "displacement current" because it plays almost the same role as a real moving charge.

Summary: photons are energy alternating between pure magnetic form and pure electrical form with frequency directly related to the energy of the photon. They also have another characteristic, usually called the "coherence length" which is the extent that their energy is spread out along their path of travel. If you try to demonstrate interference with a interferometer that is of the type using beam splitter to make two different paths that later recombine and you make the difference in the two paths more than the "coherence length" you will get no interference. The "coherence length" from most common mono-energetic sources (see other adjacent post I just made) is a meter or less. If you use a atomic transition that is almost forbidden (long life time for the excited state) "coherence length" can get to be much longer (very sharp spectral line).

12. Apr 5, 2005

### locke

*penny drops* Thanks, thats exactly what I wanted to know.

13. Apr 5, 2005

### Staff: Mentor

...just not a classical one. I think its confusing if you say it that way to someone who hasn't dealt with the issue before.

Though it may not be physically true, at a basic level it may be helpful to visualize a photon as being a single crest-to-crest wave of a particular frequency. Just for a start...

14. Apr 5, 2005

### ExecNight

Well in my opinion every photon seems to have an individual frequency because of its vectoral movement's difference from our vectoral movement in space-time..

Just my opinion :shy:

15. Apr 5, 2005

### locke

No need to worry, I'm familar with basic quantum phenomena. If such a thing exists...

16. Apr 5, 2005

### Almeisan

I would like to add a question. I am not sure if it is a relevant one.

I understand a photon is a flip flopping field of magnetism or electricity, the one creates the other.

Now I do not know the math of alof of physics and I am always told you cannot understand the elegane of physics without knowing the math.

I would assume there is a formula which shows this elegance, the interchange of both kinds of energy, the field of one type changing into the other, am I right? Could someone point me into the right direction?

17. Apr 6, 2005

### Billy T

Yes there is. It is Maxwell's equations of EM waves, but as you say you don't know much physics, the following may help:

The energy stored in either a magnetic or electric filed is proportional to the square of the field amplitude. You can think of the photon's electric field as a function of time as given by a sin function and its magnetic field as described by a cos function. When the sin function is at its max amplitude, the cos has zero amplitude and conversely. The photon does not lose any energy as it switches back and forth between magnetic and electrical storage because:

SIN^2 + COS^2 = 1

from all times (all phases of the alternation between magnetic and electric energy storage).