Quantization as an eigenvalue problem

2. E < 0.

In this case the condition (15) is not eo ipso excluded, but for the moment we hold onto its exclusion, as arranged. Then, according to (14′′) and (17), U1 grows unlimited for r = ∞, whereas U2 vanishes exponentially. Our entire transcendent U (and the same holds true for χ) will then remain finite if and only if U is identical to U2, a part from a numerical factor. However, this is not the case. One realises this so: choosing in (12) for the integration contour L a closed circuit about both the points c1 and c2 (due to the integerness of the sum α1 + α2 such contour is then really closed on the Riemann surface of the integrand), upon fulfillment of the very condition (13), one can easily show that the integral (12) represents then our entire transcendent U . Namely, it can be developed in a series of positive powers of r, which always converges for sufficiently small values of r, hence it satisfies the differential equation (7′), and thus the power series must coincide with that for U . So: U is represented by (12) when L is a closed contour about both the points c1 and c2. However, this closed contour can be distorted so that it appears as the combination of the two previously considered integration paths, those related to U1 and U2, and indeed without vanishing factors, say 1 and e2πiα1 . QED17

Then, our entire transcendent U , which is the only possible solution of the variation problem among the solutions of (7′), does not stay finite for large r, under the prescriptions made. — Under reserve of the investigation of the completeness, that is of the proof that our procedure allows to find all the linearly independent solutions, we can then claim:

For those negative values of E which do not satisfy condition (15), our variation problem has no solution.

We have now to investigate only that discrete set of negative values of E which do satisfy condition (15). α1 and α2 are then both integer. Of the two integration paths which have earlier delivered to us the fundamental system U1, U2, the first must certainly be modified in order to yield something non-vanishing. Since α1 − 1 is certainly positive, the point c1 is then now neither a branch point nor a pole of the integrand, but an ordinary zero. The point c2 can also become regular, if indeed also α2 − 1 is non-negative. In any case, however, two suitable integration paths can be easily provided and the integration along them can be even carried out in a closed form, through known functions, so that the behaviour of the solutions is completely controlled.

Let namely (15′) me2 K√−2mE = l ; l = 1, 2, 3, 4 . . . . Then, according to (14′′) (14′′′) α1 − 1 = l + n , α2 − 1 = −l + n .

One has now to distinguish the two cases l ≤ n and l > n. To begin with, let a) l ≤ n. Then c2 and c1 lose any singular character, but gain the suitability to serve as initial or endpoints of the integration path, in order to satisfy condition (13).

A third point suitable for this is the the negative real infinite. Any path between two of these three points delivers a solution and of these three solutions two by two are linearly independent, as one easily verifies by calculating the integral in closed form. In particular, the entire transcendent solution is delivered through the integration path from c1 to c2. Then, one recognises, without calculating it, that this integral stays regular for r = 0. I stress this because the actual calculation is rather likely to conceal this fact. On the other hand, it shows that the integral grows beyond any limit for positive infinitely large r. For large r, one of the other two integrals remains finite, but in turn it becomes infinite for r = 0. So, we get in the case l ≤ n no solution to the problem. b) l > n. Then, according to (14′′), c1 is a zero and c2 is a pole of at least first order of the integrand. Two independent integrals are then delivered: one via the path that from z = −∞ leads to the zero, taking care of avoiding the pole; the other by the residual of the pole. The latter is the entire transcendent. We want to give its calculated value, but at once multiplied by rn, so that we obtain,

17Translator’s note: in German, w.z.b.w.: Was zu beweisen war, i.e. “which was to demon- strate”.

according to (9) and (10) the solution of the originally presented equation (7). (The irrelevant multiplicative constant is arbitrarily adjusted.) One finds:

One recognises that this is really a usable solution, since it remains finite for all real, non-negative r. Furthermore, the surface condition (6) is guaranteed by its exponential vanishing. We summarise the results for negative E: For negative E, our variation problem has solutions in the case, and only in the case, in which E satisfies condition (15). To the integer number n, which gives the order of the spherical harmonic appearing in the solution, can be then always assigned only values smaller than l (of which there is at least one always available). The part of the solution dependent on r is given by (18).18 By counting out the constants in the spherical harmonics (2n + 1, as is well known) one finds further: The solution found contains, for an admissible combination (n, l), exactly 2n+1 arbitrary constants; for a given value of l then l2 constants.19 We have thus proved in the main traits the claims established at the beginning, about the spectrum of the eigenvalues of our variation problem, although there are still gaps.

In the first place, the completeness of the whole established system of eigenfunctions. I do not want to deal with that in this note. According to other experiences one can suspect that no eigenvalues have escaped to us.

In the second place, it is to be remembered that the eigenfunctions established for positive E do not readily solve the variation problem in the form in which it was posed at the beginning, because at infinity they go to zero only as 1/r, thus ∂ψ/∂r on a large sphere goes only as 1/r2 to zero. The surface integral (6) remains then just of the order of δψ at infinity. If one really wishes to obtain also the continuous spectrum, one must then add one extra condition to the problem: for example that δψ vanishes at infinity, or at least that it should tend to a constant value independent from the direction along which one goes to spatial infinity; in the latter case the spherical harmonics make the surface integral to vanish.