# How do you assign an observable to spectral lines in Heisenberg's resolution of Rydberg-Ritz?

Ron's comment essentially answers the question below:

Here's what I really want to know:

Suppose I have an experiment that yields 6 spectral lines corresponding to (one-way) transitions between 4 basic frequencies in an "idealized" atom with finitely many energy levels. According to Connes (see below) the observable algebra of this system is a 4x4 complex matrix algebra. According to Quantum Mechanics, one should be able to assign an observable (self-adjoint element) of the observable algebra such that the observations of an experiment (in this case the 6 spectral lines) index the orthogonal eigenspaces of the observable. This, oddly, seems to say that there is a 4 by 4 matrix with 6 eigenvalues. This is clearly nonsense. Where is the problem with my hypothetical experiment?

Original question:

The following is an embarrassingly basic question.

Edit: The following picture is from Connes's book: Noncommutative Geometry. The question came out of trying to connect what is written there with the usual postulates of quantum mechanics.

In classical mechanics, every observable quantity is represented by a real-valued function on the phase space X of a system. In the case of an electromagnetically radiating atom, according to Maxwell's theory each function admits a Fourier expansion which, via the Fourier transform, yields an isomorphism of the algebra of functions on X with the convolution algebra of an additive subgroup of the integers.

The Rydberg-Ritz combination principle, as experimentally verified, shows that the above prediction fails and instead there is a set "I" of frequencies such that only differences of frequencies appearing in I are emitted or absorbed by the atom. The spectral lines are thus labelled by differences in frequencies in I. The convolution algebra that Heisenberg prescribes that we substitute in for the one predicted above is isomorphic to a matrix algebra.

I am trying to understand the basic relation between how observables, or self-adjoint elements, of the matrix algebra above are precisely associated to the observed spectral lines, as postulated in the mathematical formulation of quantum mechanics. For example, (trying to ape this in finite dimensions...just to have a toy model) if "I" has 4 frequencies whose respective differences correspond to allowed transitions, the resulting matrix algebra will consist of 4x4 matrices. But then, since each transition labels a spectral line there ought to be 16 eigenspaces making up the self-adjoint element associated to the observed spectral lines. These are too many subspaces to be associated with any 4x4 matrix acting on $\mathbb{C}^4$.

Am I right in assuming that for every distinct observed quantity (assumed discrete), we are to assign an eigenspace of a hermitian operator in the algebra describing the system? If so, how do we assign such an observable to a set of spectral lines? (Perhaps my finite-dimensionality is simply not doable.)

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I see you read Connes book. Read the Wikipedia page on Matrix Mechanics to get the explanation. –  Ron Maimon Nov 11 '12 at 3:39
I guess the question is mathematical in nature, though. If I observe an atom with an instrument where I see only 4 spectral lines, this indicates 16 energy transitions and in Connes's picture the observable algebra has 4 dimensions. There isn't room for 16 orthogonal subspaces! –  Jon Bannon Nov 11 '12 at 12:44
As Ron points out below, there would only be 6 visible spectral lines...since these index transitions and not the frequencies in "I". Still, though, if the algebra is a 4x4 matrix algebra, no single observable can have 6 distinct eigenspaces if the algebra is viewed as acting on $\mathbb{C}^{4}$. –  Jon Bannon Nov 11 '12 at 16:28

Spectral line frequencies are positive differences between eigenvalues of the hamiltonian. If the Hamiltonian is 4 by 4, there are 4 eigenvalues and therefore 6 positive differences (unless there are degenerate eigenvalues and the number is less).

Thus there are 6 spectral lines. Their frequencies are not independent, though, as they must have the pattern $\omega_1,\omega_2,\omega_3,\omega_1+\omega_2,\omega_2+\omega_3,\omega_1+\omega_2+\omega_3$, as one easily checks. If this pattern is realized then a possible Hamiltonian is $H=Diag(0,\omega_1,\omega_1+\omega_2,\omega_1+\omega_2+\omega_3)$.

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Fabulous, Arnold!!! Thanks for the answer. –  Jon Bannon Nov 12 '12 at 18:21

The answer is on the Wikipedia page on matrix mechanics, but I'll repeat some of it here. The intuition Heisenberg had was from classical mechanics, which is only valid in the correspondence limit, which is large values of the energy levels, large n orbits.

There is a simple classical relation between the frequencies of the classical motion and the outgoing classical radiation (in the dipole approximation): the outgoing radiation intensity is in a periodic function, which produces outgoing periodic waves as the period of d(T), the dipole moment of the atom as a function of time. If you think the nucleus is stationary, this dipole moment is just the position of the electron, and the period is the orbital period.

The outgoing waves are just produced from the oscillating dipole moment, and they are long-wavelength compared to the atom, and they resolve into separated radiation at frequencies: 1/T, 2/T, 3/T, as any periodic function with period T does.

The classical function x(t) representing the position of a classical electron has a fourier decomposition:

$$x(t) = \sum_{k=-\infty}^\infty e^{ik\omega t} X_n$$

Where $X_n = X_{-n}^*$ (because X is real), and $\omega = {2\pi\over T}$. I am neglecting back reaction in this. The dipole moment is just the same thing times -e, and the field produced by this charge gives outgoing waves in a classical superposition of outgoing modes of the same frequencies $k\omega$, with amplitude which goes as $X_n$.

If you have another classical function, say the momentum, you write it as a similar Fourier series:

$$p(t) = \sum_k e^{ik\omega t} P_n$$

and the product $x(t) p(t)$ is, in Fourier space, the convolution of $X_n$ and $P_n$. This is what the "convolution algebra over the additive subgroup" means. The frequencies are integer spaced, with spacing equal to the inverse classical period, and multiplying functions of time together is convolving their Fourier coefficients on this frequency set.

But in an atom, at large n,m, you know that the emitted photon energies are according to the formula:

$$E_n - E_m = {C\over n^2 } - {C\over m^2}$$

Where C is some small multiple of the Rydberg constant. The frequency of the ougoing photon is just the energy (up to $\hbar$, which I take to be 1). These energies are not integer spaced, so the frequencies, even at large n and m where you expect the motion to be nearly classical, are not exactly evenly spaced multiple of an inverse orbital period. This means that the classical orbital picture doesn't work, the frequencies are not classical.

If you assume n and m are big, but their difference is small compared to the absolute size of either:

$$n = N$$

$$m = N + k$$

Then the frequencies are approximately evenly spaced, since

$${1\over n^2} - {1\over m^2} \approx {-2k\over N^2}$$

And if you work out all the constants, this approximate even spacing is exactly the reciprocal period of the classical Bohr orbit corresponding to orbit N (or M, the difference is higher order). This was already known to Bohr, it is called the correspondence principle, and it was the major tool he used for finding the quantization rule. The correspondence principle says that the level spacing is the inverse of the classical period at large quantum numbers, and this allows you to derive the old quantum condition, that it is the action that should be quantized to leading semiclassical order.

Heisenberg is trying to give a dynamical theory which is quantum and complete, which contains a description like X(t) of the electron motion. He chooses to find the quantum analog of the time Fourier coefficients $X_k$. He knows that $|X_k|^2$ is the intensity of classical radiation emitted at frequency $k\omega$, and he also knows that the quantum mechanics thing is emitting with frequency, not $k\omega$, but $E_N - E_{N+k} \approx k\omega $$. So he considers replacing the quantities X_k with X_{nm}, according to the correspondence rule:$$ X^N_k = X_{N (N+k}$$which means, the classical Fourier series of the N-th Bohr orbit is the near-diagonal part of a matrix of quantities X_{mn}. The number of steps you are away from the diagonal is the order k. The time evolution of Fourier coefficients (ignoring radiation) is simple:$$ X_n(t) = e^{in\omega t} X_n(0)$$From this, it is trivial to conclude that$$ X_{nm}(t) = e^{i(E_n - E_m)t} X_n(0)$$in quantum mechanics. This is the Heisenberg equation of motion. The notion of product is convolution classically, so quantum mechanically, if you have X_{nm} and P_{rs}, you want to produce a product of these which reduces to the classical convolution near the diagonal of the matrix. The products$$ X_{nm} P_{rs} $$have a frequency under time evolution which is$$ E_n - E_m + E_r - E_s $$, so if m=r, you get cancellation, and something which can be interpreted as a matrix element of a new quantity. Heisenberg immediately can guess:$$ \sum_n X_{mn} P_{ns} = (XP)_{ms}$\$

because this is what reduces to classical convolution in the correspondence limit. This is matrix multiplication, as Jordan and Born tell him. Next Heisenberg decides to find the form of the matrices from the known semiclassical Harmonic oscillator. This is very easy to guess, the classical motion has only one frequency, so the matrices are only 1-step off diagonal.

Then Heisenberg translates the old quantum condition to matrix mechanics, and finds the on-diagonal canonical commutation relation. With Born and Jordan, they find the full canonical commutation relation in the next paper. All this is from semiclassical reasoning, which is reproduced on Wikipedia in the way I found most clear.

When discussing quantum systems, you construct the operator algebra from the known state-space using some operators which you already know what their state space is supposed to be. For a single particle, you start with the X operator basis where you have a continuous basis of x states. Then diagonalizing the Hamiltonian reproduces Heisenberg's picture.

If you are starting with 4 discrete states, you can make up any 4 by 4 matrices, and these will be some quantum observables. The correspondence between the index of the matrix and the spectral lines (differences of energy levels) is assuming that you are in a basis where the particle Hamiltonian is diagnolized--- that the states are energy states.

It is best if you read physics sources on this to learn it, not mathematics sources.

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Thanks for the answer, Ron. I appreciate your nice treatment of the background, even if Wikipedia could have partially helped. –  Jon Bannon Nov 11 '12 at 12:42
I guess the question is mathematical in nature, though. If I observe an atom with an instrument where I see only 4 spectral lines, this indicates 16 energy transitions and in Connes's picture the observable algebra has 4 dimensions. There isn't room for 16 orthogonal subspaces! –  Jon Bannon Nov 11 '12 at 12:44
@JonBannon: The dimension of the space of operators is not the same as the dimension of the states--- the states are vectors, the operators are matrices. The matrices are dimension 16, the vector is dimension 4. The notion of "orthogonal" is for vectors, the matrices are an operator algebra acting on this 4-complex dimensional Hilbert space. –  Ron Maimon Nov 11 '12 at 14:44
@JonBannon: You are missing out on the fact that each spectral line corresponds to a difference of two energy levels, and it is the levels that are eigenvalues of the energy observable, not the lines themselves. By the way, there are not 16 lines for 4 levels, only 6, it's n(n-1)/2 transitions. –  Ron Maimon Nov 11 '12 at 15:42
@JonBannon: The spectral lines are not the eigenstates, they are the transitions between two eigenstates. The Bohr orbits, the energy levels, these are the eigenstates. –  Ron Maimon Nov 11 '12 at 16:43