Can I differentiate a wavefunction with respect to a quantum number?

So I was reading some papers, mainly in the Green's functions theory of the time-independent Schroedinger equation, and came across an equation that had a term similar to:

$$\frac{\partial \Psi_n^*(x)\Psi_n(x')}{\partial E_n}$$

Basicially asking to evaluate the derivative of an energy eigenstate with respect to the energy of that eigenstate. This did not make a lot of sense to me, even if I could be able to do it mathematically. But then I started to think about a system with discrete spectrum - let's say a harmonic oscillator.

In the case of the harmonic oscillator, instead of the derivative with respect to the energy, I might just as well take the derivative with respect to the quantum number. Then, apart from the constant factor, I'd have to evaulate the following expression:

$$\frac{\partial \frac{1}{2^n} \exp(-x^2/2) H_n(x)}{\partial n}$$

Now, again, mathematically this might be doable, since I can find integral representations for the Hermite polynomials. However, I am really wondering, can I do this physically? If I would use the mathematical definitions for derivative, wouldn't I implicitly assume that there is an energy eigenstate with $$n' = n+ \epsilon$$ quantum number?

• Whether or not this derivative makes any sense depends on the physical context, so could you provide a bit more? Commented Mar 27, 2020 at 22:02
• Yes, if the expression is analytic in the quantum number. Commented Mar 27, 2020 at 22:12

Well, for continuous spectra, say the free particle, $$E=\frac{\hbar^2k^2}{2m}$$, you do it al the time, in transforming integral measures, $$dE=2\frac{\hbar^2}{2m} kdk$$.

You are basically asking: "when can I analytically continue discrete spectra in-between discrete (~integral) indices"? The answer would, of course, depend on the context, but most experienced practitioners (in resolvent theory) would reassure you "most of the time". Where there is a will, there is a way.

Since you brought up the oscillator, of course Hermite polynomials are analytic in n. But this does not mean that there are extra bona-fide normalizable ("physical") solutions of the oscillator TISE that were somehow missed in textbooks.

By far the most tasteful formal definition is the differential one $$H_n(x)= e^{-\partial_x^2/4} ~ (2x)^n ,$$ of enormous utility in umbral calculus. Note this is not a polynomial for non integer n.

$$H_n(x)$$ is the inverse Weierstrass transform of a power, $$x^n$$, which underlies their constituting an Appell sequence, replicating properties of monomials, $$\partial_x H_n(x) = 2n H_{n-1}(x).$$

From their generating function, $$\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!} = e^{2xt-t^2},$$ you may see how the discrete index n may be traded for a continuous conjugate one (t), much easier to handle in sundry applications.

To sum up, continuous and discrete spectra are not as irreconcilably different as you might think at first.

NB. Fine print : The above Hermite discussion is heuristic, consciously damning the torpedoes (Feynman's favorite phrase). So lots of fussing care must be put in defining a meaningful inverse Weierstrass transform. There exists a serious paper, Bilodeau 1962, to which, however, I have no access.

• That's an interesting perspective. Another approach to defining $H_n$ for $n\in\mathbb R$ is via the Rodrigues's formula $H_n(x)= (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$, where the fractional differentiation operator is defined via the Fourier transform (transform forward, multiply by $(ik)^n$, transform backward). The result appears to be the same. This approach easily generalizes to other families of orthogonal polynomials, for which there's a Rodrigues's formula. Commented Mar 28, 2020 at 14:43
• @Ruslan Sure, absolutely. I avoided that Rodrigues formula here, because analyticity in $\partial_x$ would spook the faint of heart who might have no problem with monomials in x. Commented Mar 28, 2020 at 14:49
• Hmm, monomials? I wouldn't call $(2x)^{1/2}$, for example (for $H_{1/2}$), a monomial. Actually I've spent quite a bit of time to make sure (empirically) that the definition via the Rodrigues's formula gives the same result for non-integer $n$ as the definition via inverse Weierstrass transform. The series for the exponential of differentiation operator doesn't actually converge for non-integer $n$. It seems to be an asymptotic series for $x\to+\infty$ in this case. So I think there's more to be spooked by the inverse Weierstrass transform than by analyticity :) Commented Mar 28, 2020 at 15:04
• @Ruslan. I added the fussing Bilodeau paper in the footnote addressed to the cognoscenti. This outranges the scope of the trail map answer here, and would best be picked dry by mathematicians at the math SE. A bit of a race with the devil, there, as they often strive to prevent answers rather than finding them... Commented Mar 28, 2020 at 15:13
• Could you maybe provide a general reference (eg. a textbook or a review article) about the practices used in resolvent theory/Greens functions? It is exactly the field that I want to get into, but I do not really have an excellent maths background.
– user112876
Commented Mar 29, 2020 at 13:54