Which linear operator has a spectrum that corresponds to the physical light spectrum?

Question

If there is only one spectrum for light, which is a continuous spectrum, and Wikipedia says the spectrum that physicists use is the same as the spectrum from math, Which linear operator corresponds to the spectrum for physical light?

Answer 1

The relevant quote (taken from here) is as follows:

The name spectral theory was introduced by David Hilbert in his original formulation of Hilbert space theory, which was cast in terms of quadratic forms in infinitely many variables. The original spectral theorem was therefore conceived as a version of the theorem on principal axes of an ellipsoid, in an infinite-dimensional setting. The later discovery in quantum mechanics that spectral theory could explain features of atomic spectra was therefore fortuitous. Hilbert himself was surprised by the unexpected application of this theory, noting that "I developed my theory of infinitely many variables from purely mathematical interests, and even called it 'spectral analysis' without any presentiment that it would later find application to the actual spectrum of physics."

The question that we're interested in is this: what exactly is the connection between the "spectrum" as it used in mathematics and as it is used in physics?

First of all, let's define "spectrum" as it's used in this context. The "spectrum" (in either case) is a collection of discrete (real number) values. The fact that such a collection is called the spectrum "of" the object to be studied gives us an idea as to its use: ultimately, such a spectrum allows us to identify the object in question. We can clearly see how this is done with (atomic) emission spectra: by looking at the spectrum of frequencies that are emitted by an atom, we are given enough information about the nature of the atom to be able to figure out exactly which kind of atom (i.e. which element) we are looking at.

The situation regarding the study of compact self-adjoint operators is similar. The spectral theorem tells us that any such operator has a "spectrum" of real eigenvalues, and that if two such operators have the same spectrum, then those operators must be "the same" (i.e. unitarily similar). In fact, this spectral property of compact self-adjoint operators is ultimately what allows us to identify atoms: associated to any particular atom is a certain Hamiltonian operator governing electron behavior. Two distinct atoms will have distinct Hamiltonians, which is to say that the two Hamiltonians will have different eigenvalue spectra.

The connection between the emission spectrum of an atom and the eigenvalue spectrum of its Hamiltonian is as follows. Any electron of an atom (in a stable configuration) must satisfy the time-independent Schrödinger equation, which is to say that it must occupy a (eigen-)state corresponding to a certain eigenvalue of the Hamiltonian; this eigenvalue is the electron's energy level. When an electron descends from a high-energy state corresponding to eigenvalue $\lambda_1$ to a low-energy state corresponding to eigenvalue $\lambda_2$, the electron emits a photon whose frequency is proportional to this change of energy, namely $\lambda_1 - \lambda_2$. The set of all possible emission frequencies (or more typically, the set of such frequencies that are visible) is the "emission spectrum" of the atom.

To summarize: for any two (necessarily real) eigenvalues $\lambda_1>\lambda_2$ of the Hamiltonian operator governing the electron behavior of an atom, the difference $\lambda_1-\lambda_2$ gives us one of the frequencies of the atom's emission spectrum. Conversely, every frequency of the emission spectrum corresponds to such a difference of eigenvalues.

Answer 2

In natural units, $\hbar =1$, the free relativistic hamiltonian, so, then, the 3-momentum operator in the direction of motion, $$ \hat H = \hat p ~. $$ The eigenvalues of this operator are continuous and semipositive, as it is defined, and, naturally, amount to the frequency of the traveling photons.

(There is lots of fine print about meaningfully taking the square-root of $\hat p^2$, but I gather this is not at the root of your question.)

Note the origin of this photon and its energy/momentum is not discussed, or asked about: it could be emission, Bremsstrahlung, etc. The spectrum of light is the energies $\propto$ frequencies of the individual photons comprising it, traveling about.