Maxwell’s Equations for Photons

In the quantum theory of electromagnetic fields Maxwell’s equations play two roles.

One is to describe the behaviour of the actual field observables which measure the combined effects of all possible numbers of photons, and the other is that they are satisfied by something that is as close as possible to being the “wave function” for a single photon.

I say “as close as possible” and put “wave function” in scare quotes because it does not satisfy all the properties of a non-relativistic wave function. In relativistic theories, the concept of strict localization does not exist. It can only be approximated for massive particles in frames where they have low momentum, and cannot be done at all for photons. But nonetheless, (as discussed in this survey article (.pdf) by Iwo Bialynicki-Birula) with appropriate normalization, a function satisfying the complex form of Maxwell’s equations can be used to generate probabilities for detection of a single photon in various experimental contexts.

See also the answer by ‘Chiral Anomaly’ to this question at physics stack-exchange.

Source: (1000) Alan Cooper’s answer to If particle nature of light is involved then what are Maxwell equations? – Quora

Schrödinger’s Equation for Light

A Quora Question Asks: Is it true that there is no Schrodinger equation for light because the Schrodingers’ equation is only for massive particles, and that only Maxwell’s equations can be used for light?

My answer is that it depends on what you mean by the Schrödinger Equation.

The first time many of us see that name is in reference to the position space wave equation for a single massive particle in non-relativistic quantum mechanics (and that is indeed historically appropriate, but is not the only equation that is correctly attributed to Schrödinger). In terms of this usage, the Schrödinger Equation clearly does not apply to light because the behaviour of light is not Galilean-covariant (and, unlike particles, does not even have a useful Galilean-covariant approximation).

But the name “Schrödinger Equation” is also used for the equation that governs the time evolution of any quantum system (which is basically just a statement of the fact that any unitary representation of the one dimensional translation group has a self adjoint generator – which, in the case of time translations, we identify as corresponding to the total energy of the corresponding spacelike time slice). And in that sense, as noted by Mark John Fernee, (with some further elaboration by Diógenes Figueroa) it certainly does apply to the quantized electromagnetic field and so to light.

There is also a Schrödinger Equation in this second sense for the Quantum Field Theory of electrons and positrons; however the relativistic wave equation for a single free electron is not called a Schrödinger Equation but rather is known as the Dirac Equation instead.

For the case of a single relativistic spinless particle, there is no wave equation that is first order in time, and the best we can do is the second order Klein-Gordon Equation; but there is nonetheless a Schrödinger Equation in the second sense for the corresponding scalar quantum field theory.

There is a sense in which Maxwell’s Equations play a similar role to the Dirac and Klein-Gordon equations (and the first interpretation of Schrödinger’s) as the wave equation for a single photon, but the use of Electric and Magnetic fields to extract a probability density is complicated by having to take account of gauge invariance and polarization state.

Source: (1000) Alan Cooper’s answer to Is it true that there is no Schrodinger equation for light because the Schrodingers’ equation is only for massive particles, and that only Maxwell’s equations can be used for light? – Quora


Infinite Wavelength of Stationary Particle

A particle does not have a wave function with respect to itself; but for any observer, the uncertainty principle tells us that if a particle could be known to have any exact velocity (and in particular if it is known to be in the observer’s rest frame with a velocity of zero) then its position would be completely unknown – and in the case of zero velocity this would make the wave function constant. (Of course, in practice we never know either the position or the momentum exactly, and this corresponds to the mathematical fact that the constant amplitude wave-function is not normalizable.)

A typical realistic position space wave function is in the form of a wave packet which has an amplitude representing the probability density multiplied by a complex phase factor which oscillates (or more precisely rotates around the unit circle in the complex plane) at a frequency corresponding to the average observed velocity. As the velocity goes to zero, the wavelength of those phase oscillations goes to infinity and the wavefunction just looks like a bump of almost constant phase. But this infinite wavelength does not mean that the wavefunction is constant, and the shape of its amplitude envelope means that its fourier transform includes contributions from frequencies other than that corresponding to the average observed velocity (and so the momentum space wave function is also a bump with width related to that in position space by the uncertainty principle).

Source: (1000) Alan Cooper’s answer to Debroglie says wavelength is inversely proportional to velocity. But velocity is relative, and it is 0 in that particle’s resting frame of reference. So what does it mean for a particle to have infinite wavelength? – Quora

“Basis Vector” Confusion

A Quora question asks: “The wave function is contained in a Hilbert space while its basis vectors aren’t because plane waves are not square-integrable functions. Is this true for all Hilbert spaces or only for the square-integrable sub-space?”

My response: There are a number of ways you are misusing the language here, and I thought at first that your main misunderstanding may be to think that an element of a Hilbert space has its own set of basis vectors – while in quantum theories the choice of a relevant basis is more often related to an observable than to a state. But perhaps you are not associating the “basis” with a particular wave function, and rather thinking of it as associated with the position space representation as a whole. That makes more sense so let’s go back and try to describe the situation more clearly.

A wave function is a representation of an element of the Hilbert space of quantum states by a square integrable function (but the space of such functions is actually isomorphic to the entire state space as a whole and not just a proper sub-space). There are many such representations, sometimes with associated ways of identifying bases and sometimes not. In particular for any observable with purely discrete spectrum (such as the Hamiltonian of a harmonic oscillator) there is a basis of eigenvectors, and every state is represented by a square summable sequence corresponding to its spectral decomposition. Unfortunately the position and momentum observables have pure continuous spectrum and no eigenvectors, so the corresponding representations involve elements from some larger space. The usual “position space” wave function corresponds to the spectral decomposition of the position operator, and the analogue of basis vectors are actually not functions at all but rather distributions (in particular delta functions). The plane waves on the other hand are (position-space representations of) eigenfunctions of the momentum operator (but again not eigenvectors since not in the Hilbert space).

So in the end I might answer your question by saying that there is only one Hilbert space of states but that any complete set of observables can be used to represent it in terms of square integrable functions (or sequences); and that it is only in the case of pure point spectrum that the resulting spectral decomposition can be described in terms of an actual basis, while observables with continuous spectrum generally require some kind of generalized basis involving functions or distributions that do not actually correspond to normalizable states.

Source: (1000) Alan Cooper’s answer to The wave function is contained in a Hilbert space while its basis vectors aren’t because plane waves are not square-integrable functions. Is this true for all Hilbert spaces or only for the square-integrable sub-space? – Quora