Are there any "natural" physical observables which have non-empty point spectrum which consists of numbers which are not algebraic numbers?

  • $\begingroup$ Weird, but good question. I'm not sure whether it's well-posed, though; surely you have to fix a system of units? And then you have the problem of constants: is the electron charge $e$ transcendental or not? $\endgroup$ – Greg Graviton Jul 7 '11 at 12:16
  • 3
    $\begingroup$ @Greg is right. You always have scaling and shifting freedom in units. Which means first two eigenvalues (in the case the operator is bounded from below and has a "mass gap") can always be taken to be 0 and 1. In any case, I am not sure about the physical value of this question... $\endgroup$ – Marek Jul 7 '11 at 12:38
  • 1
    $\begingroup$ I assume that you are considering theories that take only rational or algebraic numbers as input. (Otherwise just observe the fine structure constant). And only dimensionless ratios of observables. One would have to imagine that "almost" all observables have transcendental spectra, but proving this for particular cases is hard, because we can't calculate the spectra and its hard to prove transcendentality. $\endgroup$ – BebopButUnsteady Jul 7 '11 at 13:38

I don't have a complete answer, but maybe the following is useful for your purposes:

Consider the Laplacian $\Delta$ on a circular drum of unit radius. As explained on the wikipedia page, the axially symmetric eigenvectors $\Delta u(r) = -\lambda^2 u(r) $ are Bessel functions $u(r)=J_0(\lambda r)$. Obviously, the boundary condition requires $J_0(\lambda)=0$. In other words, the eigenvalues of this Laplacian correspond to zeroes of Bessel functions, and I would be very surprised if these numbers are not transcendental. In fact, Mathworld mentions that the first zero has been proven to be transcendental by Le Lionnais.

The corresponding quantum mechanical situation would be the Hamilton operator $H=-\hbar^2/2m\cdot \Delta$ of a free 2D electron confined to the unit disk. The boundary conditions are the same, $\psi|_{ \partial\Omega}=0$.

Of course, the electron has the problem that it's not clear whether $\hbar= h/2\pi$ should be counted as transcendental or algebraic, physicists frequently set $\hbar = 1$.

For instance, consider a 1D electron inside a box of length $L$, i.e. $\psi(0)=\psi(L)=0$. The eigenstates are simply standing waves, and the eigenvalues are

$$ E_n = \frac{\hbar^2}{2m_e} \left(\frac{\pi n}{L}\right)^2.$$

If you count $h$ as algebraic, then this is algebraic. But if you count $\hbar$ as algebraic, then this is transcendental. Your way out is probably to turn the question into a relative one: is there a physical observable whose eigenvalues are algebraically independent of the $E_n$? Clearly, you only need to consider the eigenvalues of the Laplace operator now.

  • $\begingroup$ Is this example worked out somewhere in the literature? $\endgroup$ – Łukasz Grabowski Jul 7 '11 at 16:35
  • $\begingroup$ @Łukasz Grabowski. The wikipedia page already solves the problem of the circular drum completely, but it doesn't answer the question of whether the zeroes of the Bessel functions are algebraic or transcendental. I don't know any reference for the latter problem. $\endgroup$ – Greg Graviton Jul 7 '11 at 17:28
  • 1
    $\begingroup$ @Greg Graviton At least the smallest zero of $J_0(x)$ is transcendental. $\endgroup$ – mmc Jul 7 '11 at 22:58
  • $\begingroup$ @Greg: I like this answer because it's very explicit, but is the quantum mechanical part worked out somewhere? (For example the boundary conditions you mention) Also, I know nothing about Bessel functions but it could be that tose roots which correspond to eigenvalues are actuall algebraic. BTW: what would happen for a free 2D electron confined to a sphere? (so that one doesn't have to care about the boundary) $\endgroup$ – Łukasz Grabowski Jul 8 '11 at 14:40
  • $\begingroup$ @Łukasz Grabowski: Unfortunately, I haven't seen this electron worked out somewhere. Fortunately, I was wrong about the boundary condition, the usual $\psi=0$ is correct. The free 2D electron on a sphere seems to correspond to the eigenvalues of the Laplacian on a sphere, but this is not quite correct, you have to include tunneling effects, i.e. information about the embedding of the sphere in $\mathbb R^3$. See the following answer, in particular the article by Costa. $\endgroup$ – Greg Graviton Jul 9 '11 at 9:11

Transcendental numbers often pop out in higher-loop Feynman diagram calculations. One example is the spectrum of local operators in $\mathcal{N}=4$ supersymmetric Yang-Mills theory, where the anomalous dimensions at higher loops in general contain $\zeta$-functions. As an example, the dimension of the operator $\mathop{Tr} \phi^a\phi^a$ is given perturbatively by $$\gamma = 4 + 12g^2 - 48g^4 + 336g^6 + (-2496+576\zeta(3)-1440\zeta(5))g^8 + \mathcal{O}(g^{10})$$ (I don't think there is a rigorous proof showing that $\zeta(n)$ is non-algebraic for positive and odd values of $n$, but it seems likely to me...)

  • 2
    $\begingroup$ "Perturbatively"! How do you know that the real answer isn't actually algebraic? $\endgroup$ – Siyuan Ren Jul 7 '11 at 13:01
  • $\begingroup$ @Karsus: I guess that's true. I just wanted to given an example where non-algebraic numbers appear naturally. By the way, can an algebraic function have a convergent series expansion around 0 with non-algebraic coefficients? $\endgroup$ – Olof Jul 7 '11 at 13:44
  • 1
    $\begingroup$ $ 1=\sum_{n=0}^{\infty}\frac{e^{-1}}{n!} $. And I remember reading somewhere that a lot of series obtained by perturbation is not convergent at all; they are asymptotic. $\endgroup$ – Siyuan Ren Jul 7 '11 at 14:33
  • $\begingroup$ @Karus: That's a nice example, thanks! A perturbative series may in general asymptotic, but the perturbative expansion of $\mathcal{N}=4$ SYM is supposed have a finite radius of convergence. $\endgroup$ – Olof Jul 7 '11 at 14:40
  • 1
    $\begingroup$ @Karus: Sorry, I'm not sure it's such a good example after all. While it is true that your series have non-algebraic coefficients but sum up to an algebraic number, I don't think of that series as an expansion of the constant function 1. And something like $\sum_{n=0}^{\infty} \frac{e^{-1} x^n}{n!} = e^{x-1}$ is clearly not an algebraic function. Anyway, I'm afraid we are getting a bit off-topic... $\endgroup$ – Olof Jul 7 '11 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.