# I Is there a difference between EM waves and photon wavefunctions?

1. Feb 5, 2016

### sciencejournalist00

Matter has a wavefunction associated to it. But what about light? Does it have both a electromagnetic wave described by Maxwell's equation and a wavefunction described by Schroedinger's equation?

Or is the electromagnetic wave considered to be the wavefunction of the photon?

I read somewhere that in interference patterns, the highest intensity peaks of a EM wave correspond to greatest probability of observing a photon. Elsewhere that photons have no wavefunction associated to them because massless particles have no position eigenstate.

Which one is true?

Intensity of EM waves = probability distribution for photon location

No wavefunction for masless particles

or

Both EM wave and wavefunction for photons

2. Feb 5, 2016

### sciencejournalist00

From Wikipedia

Both (photons and material) particles such as electrons create analogous interference patterns when passing through a double-slit experiment. For photons, this corresponds to the interference of a Maxwell light wave whereas, for material particles, this corresponds to the interference of the Schrödinger wave equation. Although this similarity might suggest that Maxwell's equations are simply Schrödinger's equation for photons, most physicists do not agree. For one thing, they are mathematically different; most obviously, Schrödinger's one equation solves for a complex field, whereas Maxwell's four equations solve for real fields. More generally, the normal concept of a Schrödinger probability wave function cannot be applied to photons. Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate , and, thus, the normal Heisenberg uncertainty principle does not pertain to photons.

3. Feb 6, 2016

### blue_leaf77

No, it is not. At least, from the fact that a wavefunction is not an observable quantity on the other hand EM field is, you should see that they are two different things. In the case of photons having certain wavefunction, the observed electric and magnetic fields are the average values of the field operators.
By "wavefunction", I believe that source was referring to the wavefunction in position representation, photons do not have this kind of wavefunction. But they do have states, as any quantum objects do.

4. Feb 6, 2016

### A. Neumaier

This is a very superficial difference.

One of the nice representations of the electromagnetic field is in terms of Silberstein vectors, which are 3-dimensional complex vectors, for which field equations can be derived. With an appropriate inner product, these are precisely the functions making up the 1-photon Hilbert space, and hence define the pure states of a single photon.

This is not correct. it is well-known that photon location cannot be described by a probability distribution.

See the slides of my lectures Classical and quantum aspects of light and Optical models for quantum mechanics for a more appropriate correspondence.

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5. Feb 6, 2016

The probability of finding a photon, i.e. the photon density, is proportional to the second power of the electric field amplitude E. The total energy is the number of photons multiplied by the energy of each photon, Ep=h*f. (The letter Ep is used for photon energy, but E for electric field is not energy.) The intensity "I" and energy density U are both proportional to the square of the electric field amplitude E. The "I", U, and N*Ep where N is number of photons are all proportional to the second power of the electric field amplitude E. The electric field is proportional to the the quantum mechanical wave function "psi". (Greek letter). The psi*psi also gives a probability density for photons.

6. Feb 6, 2016

### A. Neumaier

Can you substantiate your claims with a reference? The first few sentences flatly contradict well-known results about the nonexistence of a position operator and therefore of a Schroedinger-type probability interpretation. The remaining part is also incorrect - A wave function is necessarily complex while the electric field is real.

7. Feb 6, 2016

### blue_leaf77

This is what I would call technical understanding of photons. Theoretical and quantum physicists have their own, which is the correct, description of photons. For this last one, I would call theoretical understanding of photons. Many scientists like to use the term "photon density" to actually refer to EM wave density - in any textbooks which uses these words in particular instances within the book, I am almost sure that they are actually referring to the equation $\rho = 1/2\epsilon_0|E|^2+1/2|B|^2/\mu_0$. The reason why many scientists mix up the use of the word "photon" with "EM radiation" is most likely because a matter of convenience. Since the introduction of photons as light particles by Max Planck (if I am not mistaken), the easy way of thinking is that "light are made up of photons". This has made the hand-waving, but blunt association between EM radiation and photons came into existence among scientists. The only thing in your comment which is unheard of to me is
I have never heard of this claim though.
As I have a little bit elaborated above, there are actually physicists who indeed make use of the word "photons" in a broader sense than it should have been. While such description is not correct, that's just the way some of us understand EM radiation as being made up of a stream of photons. If you are lucky, you can find technical physics textbooks (some of them must be those in the interdisciplinary category) which abuse the literal use of "photon density", which should not exist in the correct theoretical view. Some online examples are and http://www.medilexicon.com/medicaldictionary.php?t=23602. The last one even establishes it as one of their vocabulary collection.

8. Feb 6, 2016

My post is for the most part simply Classical Electrodynamics results, (see for example J.D. Jackson's Classical Electrodynamics textbook), but the electric field can be written as a complex quantity. The Q.M. wavefunction for the photons is of a many-particle type boson form. One obstacle that is encountered is if there are multiple photons in the same photon mode (e.g. at the same wavelength) as in a laser, how can the individual electric fields add in phase?, and the answer to this is that apparently the phases are random for multiple photons in the same mode. (The phasor diagram is a 2-D random walk.) The complete quantum mechanical description of the electromagnetic field with second quantized field operators (see e.g. textbooks on relativistic quantum field theory by Bjorken and Drell) is rather difficult to follow-(I struggled with the concepts as a graduate student), but the items contained in my previous reply are considerably simpler, and I believe them to be correct.

9. Feb 6, 2016

Additional item: A photodiode measures photons giving typically one electron of photocurrent for each photon with typically about an 85%-90% quantum efficiency. In working in the visible and IR parts of the spectrum, it is photons (=photocurrent from a photodiode) that are measured rather than a time-varying electric field (e.g. as with radio waves). The wave nature of the light shows up when using instrumentation such as diffraction gratings (diffraction grating spectrometers), and interferometers (e.g. the Michelson interferometer.) Incident visible light of irradiance S(watts/cm^2)=1 milliwatt/cm^2 (using the nomenclature of J.D. Jackson textbook where S=Poynting vector for electromagnetic wave ), contains billions of photons per second per square centimeter. S=c*E^2/(4*pi) (c.g.s. units) where E is electric field amplitude, c=speed of light.) Localizing the energy and/or the photons seems to be straightforward. The derivation of the Planck blackbody radiation formula is one very good example of the use of both the wave nature of the light along with the particle nature of the photon. In this derivation, the photon modes are first counted inside a cavity, the numbers of photons is then computed using the mode mean occupancy number for each mode for bosons (as a function of the temperature), and the rate of photons that effuse out of an aperture is computed along with their energy to arrive at the spectral formula for blackbody radiation. (see e.g. Statistical Physics by F. Reif)

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10. Feb 7, 2016

### A. Neumaier

Einstein reintroduced in 1905 the particle picture of light already used by Newton, in a modified quantum form. The name photon appeared much later (1936), and even then it took a very long time to be generally accepted as a valid description of light.

11. Feb 7, 2016

### A. Neumaier

No, there is no notion of photons in classical ED.
How you relate the notion of photons to CED is completely nonstandard and mistaken.

The correct quantum mechanical description can be found in the quantum optics book by Mandel and Wolf (Optical Coherence and Quantum optics), which is easy to follow, though it requires some serious study, including background material. It draws a picture substantially different from yours.

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12. Feb 7, 2016

### blue_leaf77

That's right, that's what some of us have been talking about. However, the problem with post #5 is that you associated EM wave energy density with photon's spatial probability density which is supposed to be non-existent. From this and
I have got the impression that you associate fields with photon wavefunction. They are fundamentally different, photon wavefunction (or more properly photon state) is mathematical object introduced as a part of quantum description of light and it can be complex vector in the Fock state basis, but a state is not an observable physical quantity. Fields, on the other hand, can be physically measured. Despite sometimes being written complex, it is usually done so to simplify the calculation; every time a field is written as a complex number, the complex conjugate of this number with the right sign must also be written (or at least implied) to make the entire field expression real. Most importantly, in QM description of light, fields are assigned with operators, as any other physical quantities in quantum mechanics are, so they are by no means the state of the photon itself. Although in some instances, the time dependency of the fields do carry signature of which kind of photon state these observed fields are subject to. For example, photons in Fock state, coherent state, and squeezed state have their own forms of the time-dependent fields.

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13. Feb 7, 2016

### vanhees71

Take any textbook on quantum optics. Obviously is the detection probability for photons taken as proportional to the energy density of the em. (quantum) field a very succuessful model. It's to a certain extent taken from classical electromagnetics, where the intensity of light is also taken as the (time-averaged) em. field-energy density.

Also more complicated observables (like HBT correlations) are defined via electromagnetic field-strength-tensor correlators. See, e.g.,

M. O. Scully, M. S. Zubairy, Quantum Optics, Cambridge University Press 1997

It's of course true that photons have no position operator but still you have intensity measures and detection probabilities for single photons at the place where your detector is located, and that's usually defined via the expectation value of the field-energy density or correlation functions of the electromagnetic-field operator.

14. Feb 7, 2016

### A. Neumaier

It is not the detection probability but the photodetection rate that is proportional to the energy density of the electromagnetic field. Since it doesn't matter whether the latter is a classical or a quantum field, the photodetection rate it is not a property of photons. Certainly not a photon density as specified in post #5.

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15. Feb 7, 2016

### vanhees71

Ok, it's the photodetection rate (although I'd rather say that's the modulus of the (expectation value of) the Poynting vector, i.e., the energy-flow density, but that's semantics). A photon-number density is indeed problematic to define, because there's no conserved photon number, i.e., no photon-number-density four-current that obeys the continuity equation, and thus photon numbers, defined in a naive way with the "photon number operators" are not a Lorentz-invariant (scalar) quantity, while for total field energy and momentum that's the case.

16. Feb 7, 2016

Most of the time, when measuring light sources, you can simply count photons using a photodiode-i.e. there are billions of them each second. Normally there is no concern about photon creation or photon annihilation processes, where the photon number changes. For a semi-classical approach the photons are proportional to the energy and the photon number can be considered to be conserved. In the case of a Doppler shift, I think the photon number can be considered to be conserved. It appears you are taking a more advanced and detailed approach.

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17. Feb 7, 2016

### A. Neumaier

In a semiclassical treatment the electromagnetic field is classical and there are no photons. What is proportional to the energy density is the number of detector clicks, which can be counted, not the number of photons, which cannot exist in a model only containing quantum electrons and a classical external electromagnetic field.

18. Feb 7, 2016

Undoubtedly, to get a completely accurate treatment of the electromagnetic field, a detailed quantum mechanical approach is necessary. I believe Bjorken and Drell's textbooks, Gordon Baym's Quantum Mechanics, and J.J. Sakurai's Advanced Q.M. all treat this subject, but I found this topic very difficult to get a good handle on.

19. Feb 7, 2016

### A. Neumaier

The book by Mandel and Wolf cited in #11 is much better, and very comprehensive. it explains both the semiclassical and the full quantum situation, including squeezed states and parametic downconversion, which are true quantum phenomena. But even then, the energy density cannot be equated with a photon density.

20. Feb 7, 2016