What is the actual shape of a photon and how can it be measured directly?

[oldphysicist]

I am unable to find a practical answer to this in any of the conventional texts. Suppose an excited atom drops back down to the ground state and emits a single photon of energy h-nu. The conventional description of this photon is that it is a single excitation of the EM field contained in an arbitrary but specific normalizing volume V, and that it has a definite wavenumber k and polarization vector e. This "wave" can then travel half way across the universe (e.g.) and interact with an identical atom, exciting it back up to the original state. What is the actual shape of the wave that propagates forward? The normalizing box V does not really exist, so what is the precise value of E and H in this wave in its free space flight? Does it really extend to +/- infinity in the perpendicular plane, which is what the definition of a plane wave implies. Also, since this wave must have some of the characteristics of a pulse - after all, it is a single photon - what is the shape in time of the pulse? It is hard to establish the form of a Poynting vector for this field, from which the total energy (h-nu) could be determined. This transition between quantum and classical theory for EM fields seems to fall between the cracks of conventional discourse.

 

[phyzguy]

Perhaps the point that you've missed (I certainly missed it for a long time) is the following. We really use the word photon to describe two things. On the one hand, we refer to a photon as being an elementary excitation of the EM field, which has a definite energy, wavelength, and frequency. These photons are the eigenstates of the EM field, and extend to +/- infinity. On the other hand, we also use the word photon to refer to the wave packet emitted by an atom when it drops from one energy level to another. But this latter photon is not an energy eigenstate. It does not have a definite frequency, because the atom is not in the upper and lower energy states for an infinite time, so there is a time uncertainty which leads to an energy uncertainty. So the wave packet contains a range of frequencies, and multiple measurements of its energy would lead to a distribution of probable values. Because there is a range of frequencies and wavelengths, the wave packet is also bounded in space and time. The "size" of the wave packet (let's say the duration in time) depends on how long the atom is undisturbed. If the atom is in a very undisturbed environment, with a long time between collisions, then the wave packet is very sharp, with a long duration (Delta-t), and a small distribution of energies (Delta-E). If the atom is in an environment where collisions are frequent, then the wave packet is more spread out, with a short duration (Delta-t), and a broad distribution of energies (Delta-E). Does this help?

 

[Naty1]

yes there seem to remaining cracks in a lot of places:

..

experiments confirm that the photon is not a short pulse of electromagnetic radiation; it does not spread out as it propagates, nor does it divide when it encounters a beam splitter".[47] Rather, 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; that hypothesis was also refuted by the photon-correlation experiments cited above. 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

and see also the uncertainty discussion which follows ..
http://en.wikipedia.org/wiki/Photon#Wave.E2.80.93particle_duality_and_uncertainty_principles

 

[sheaf]

The normalizing box V does not really exist, so what is the precise value of E and H in this wave in its free space flight?

For the expectation value and variance of the E field in n-particle states, see equations III-3a and III-4b in http://people.seas.harvard.edu/~jones/ap216/lectures/ls_3/ls3_u3/ls3_unit_3.html"

 

[A. Neumaier]

Suppose an excited atom drops back down to the ground state and emits a single photon of energy h-nu. What is the actual shape of the wave that propagates forward?

A radial wave, i.e., a vector version of the wave you get when you emit a small stone into a calm lake.

since this wave must have some of the characteristics of a pulse - after all, it is a single photon - what is the shape in time of the pulse?

Theoretically, a single photon can have any shape described by a solution of the Maxwell
equations, not just a pulse. In the above case, you get not a pulse but a spherical wave if the atom is isolated and decays spontaneously.

For the nature of photons, see
http://arnold-neumaier.at/ms/lightslides.pdf
http://arnold-neumaier.at/ms/optslides.pdf
and the section ''What is a photon?'' from Chapter B2 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html

 

[Delta2]

Theoretically, a single photon can have any shape described by a solution of the Maxwell
equations, not just a pulse. In the above case, you get not a pulse but a spherical wave if the atom is isolated and decays spontaneously.

Yes but we say that a single photon has a unique frequency, while any solution of Maxwell equations is a sum of waves with different frequencies.

Also a single photon has energy E=hv, while the energy carried by a wave that satisfies Maxwell's equations and has the same frequency isn't necessarily hv. In this case a photon is any portion of the wave that carries the same energy hv?

In any case, what is considered more fundamental? The photon or the EM field?. EM field is like a theoretical construction that seems to exist because photon exists? Or the other way around, photon is a theoretical construction out of EM field which we consider real?

 

[A. Neumaier]

Yes but we say that a single photon has a unique frequency, while any solution of Maxwell equations is a sum of waves with different frequencies.

Only monochromatic photons have a single frequency. The general photon is not monochromatic, but a superposition of monochromatic ones.

In any case, what is considered more fundamental? The photon or the EM field?. EM field is like a theoretical construction that seems to exist because photon exists? Or the other way around, photon is a theoretical construction out of EM field which we consider real?

It depends on how one defines ''fundamental''. Conventionally, the field is taken as fundamental, since it appears in the action defining QED, and the whole discipline is called quantum field theory because of this fundamental role of fields.

The field is measurable in the large, hence independent of the notion of a photon
(though it may be construed as being composed of virtual photons with weird properties).
The (real, nonvirtual) photon is defined as an asymptotic, irreducible excitation of the field (appearing as input or output of scattering processes).

 

[lightarrow]

Perhaps the point that you've missed (I certainly missed it for a long time) is the following. We really use the word photon to describe two things. On the one hand, we refer to a photon as being an elementary excitation of the EM field, which has a definite energy, wavelength, and frequency. These photons are the eigenstates of the EM field, and extend to +/- infinity. On the other hand, we also use the word photon to refer to the wave packet emitted by an atom when it drops from one energy level to another. But this latter photon is not an energy eigenstate. It does not have a definite frequency, because the atom is not in the upper and lower energy states for an infinite time, so there is a time uncertainty which leads to an energy uncertainty. So the wave packet contains a range of frequencies, and multiple measurements of its energy would lead to a distribution of probable values. Because there is a range of frequencies and wavelengths, the wave packet is also bounded in space and time.

Ok up to here.

The "size" of the wave packet (let's say the duration in time) depends on how long the atom is undisturbed. If the atom is in a very undisturbed environment, with a long time between collisions, then the wave packet is very sharp, with a long duration (Delta-t), and a small distribution of energies (Delta-E). If the atom is in an environment where collisions are frequent, then the wave packet is more spread out, with a short duration (Delta-t), and a broad distribution of energies (Delta-E).

There is something I don't understand here. The "size"/duration of the wavepacket, related to the degree of its monochromaticity, shouldn't depend on *the kind* of atomic transition which determines its half-life (see, e.g. metastable levels)?

 

[phyzguy]

Ok up to here.

There is something I don't understand here. The "size"/duration of the wavepacket, related to the degree of its monochromaticity, shouldn't depend on *the kind* of atomic transition which determines its half-life (see, e.g. metastable levels)?

Perhaps I'm wrong, but I think it does depend on the kind of transition. The longer the lifetime of the state, the narrower the energy peak, and the longer the duration of the wave packet. Any other comments?

 

[lightarrow]

Perhaps I'm wrong, but I think it does depend on the kind of transition. The longer the lifetime of the state, the narrower the energy peak, and the longer the duration of the wave packet. Any other comments?

If we agree on this, I don't have other comments

 

[Delta2]

It depends on how one defines ''fundamental''. Conventionally, the field is taken as fundamental, since it appears in the action defining QED, and the whole discipline is called quantum field theory because of this fundamental role of fields.

The field is measurable in the large, hence independent of the notion of a photon
(though it may be construed as being composed of virtual photons with weird properties).

Theoretically the field seems to be the fundamental since the theory starts from quantizing the field to produce the notion of the photon. BUT

What do you mean when you say the field is meazurable in the large? As i understand there is no way to measure the field directly, all we measure is some quantity that comes from or varies due to the force that the photon carries to a charged particle usually the electron.

 

[A. Neumaier]

What do you mean when you say the field is measurable in the large? As i understand there is no way to measure the field directly, all we measure is some quantity that comes from or varies due to the force that the photon carries to a charged particle usually the electron.

Engineers routinely measure electromagnetic fields; they are the expectation values of the corresponding field operators. Electron microscopes measure a bit more indirectly the electron density, expectation values of the operator psi^*(x)psi(x) made from electron fields.

 

[sheaf]

Perhaps the point that you've missed (I certainly missed it for a long time) is the following. We really use the word photon to describe two things. On the one hand, we refer to a photon as being an elementary excitation of the EM field, which has a definite energy, wavelength, and frequency. These photons are the eigenstates of the EM field, and extend to +/- infinity. On the other hand, we also use the word photon to refer to the wave packet emitted by an atom when it drops from one energy level to another. But this latter photon is not an energy eigenstate. It does not have a definite frequency, because the atom is not in the upper and lower energy states for an infinite time, so there is a time uncertainty which leads to an energy uncertainty.

On thinking about this some more, I'm now a little confused by this. If the electron transitions from a high energy level E_n to a low energy level E_m, isn't the energy of the photon (and hence its frequency) completely determined, i.e. E_n-E_m regardless of how long it had been in the higher energy state for ?. (I'm neglecting the effects of atomic motion here, which would introduce an uncertainty into the total energy, I'm just questioning the direct application of the energy/time uncertainty relation in this case).

 

[Delta2]

Engineers routinely measure electromagnetic fields; they are the expectation values of the corresponding field operators. Electron microscopes measure a bit more indirectly the electron density, expectation values of the operator psi^*(x)psi(x) made from electron fields.

Are there devices that measure the em field directly without involving its interaction with electron (or another charged particle)? I don't think so.

 

[Cthugha]

Are there devices that measure the em field directly without involving its interaction with electron (or another charged particle)?

I do not think so, but I do not see how this poses a problem. There are no direct measurements of anything without detectors and electrons are good detectors for em fields. At least Science considered this as a direct measurement and Science has a rather strict peer review system. "Direct measurement of light waves" by E. Goulielmakis et al. (Science 305, 1267 (2004)) is one of the best references on this topic, I think.