Quantum theory over the algebraic closure $\overline{\mathbb{Q}}$ of the rational numbers $\mathbb{Q}$ Consider classical quantum theory described through the København interpretation over the complex numbers.
Is there any need for transcendental numbers at all to describe state vectors ? I think I never saw papers which use transcendental numbers as coefficients, and I doubt if they are needed at all.
I understand that generally one wants an algebraically closed field such as $\mathbb{C}$ for having a well-behaved eigenvalue theory at hand, but why not work over the algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ ?
This field is contained in any algebraically closed field in characteristic $0$, so is more universal than $\mathbb{C}$, while on the other hand it is also countable (whereas $\mathbb{C}$ is not).
 A: Quantum mechanics is not purely algebraic, it involves topology and calculus as well. It makes sense to use $\mathbb C$, as it is the (topological) completion of $\bar{\mathbb Q}$.

I think I never saw papers which use transcendental numbers as coefficients, and I doubt if they are needed at all.

Two examples :

*

*Consider a two level system with Hamiltonian (in matrix form)  :
$$\hat H = \hbar \omega\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$$
In this case, the evolution operator is :
$$e^{-i t \hat H/\hbar } = \begin{pmatrix}
\cos(\omega t) & i\sin(\omega t) \\
i\sin(\omega t)& \cos(\omega t)
\end{pmatrix}$$
Unless you want to restrict $t$ to a weird subset of $\mathbb R$, you will need so transcendental coefficients. Actually, even for a one-dimensional space (or a subspace generated by an eigenstate of the Hamiltonian), the phase $e^{iE_0t/\hbar}$ is not algebraic for all $t\in\mathbb R$.


*Consider a free particle living in $\mathbb R$ (or more generally a Riemanian manifold). The space of states is usually $L^2(\mathbb R,\mathbb C)$. If you restrict to the subspace of functions taking values in $\bar{\mathbb Q}$ (almost everywhere), then you don't have a Hilbert space anymore and the spectral theorem, which is central in QM, does not work anymore. On top of that, the hermitian product of two $\bar{\mathbb Q}$-valued wavefunctions is not necessarily algebraic. When the Hamiltonian has a discrete spectrum, you could in principle diagonalize it and then consider the vector space over $\bar{\mathbb Q}$ generated by the eigenvectors. But then you would run into the same problem as in 1. : time evolution becomes ill-defined.
Conclusion : the topological properties of $\mathbb C$ are very important to the calculus and functional analysis on which the usual formulation of quantum mechanics relies. That being said, you are free to try to develop the theory using only algebraic numbers. I personally think that there is not much hope of finding something that is not horribly cumbersome, far from the elegance of ordinary QM.
A: You'd have to be careful with what you mean by quantum mechanics over $\overline{\mathbb Q}$.
From an intuitive perspective, the wavefunction of a particle is a square-integrable function $\psi:\mathbb R\rightarrow \mathbb C$, which has norm $\Vert\psi\Vert := \sqrt{\int \mathrm dx |\psi(x)|^2}$.   If the vector space is over the field $\overline{\mathbb Q}$, then the norm takes its values in $\overline{\mathbb Q}\cap\mathbb R$.  Are you saying that only functions with $\overline{\mathbb Q}\cap \mathbb R$-valued norm should be allowed?  How would you enforce this very restrictive constraint? Are wavefunctions to be avoided entirely?
Avoiding the question of wavefunctions and working instead with abstract vectors, how does convergence work in this theory? Let's say that $\{\hat e_i\}$ is an orthonormal basis for your space.  Then the sequence of partial sums $\psi_N = \sum_{k=1}^N \frac{1}{k^2} \hat e_k$ is Cauchy, but fails to converge to a function with $\overline{\mathbb Q}\cap \mathbb R$-valued norm, so presumably is not allowed in the space.  But that means that not all Cauchy sequences converge, so we can't be working with a Hilbert space (which is explicitly a complete inner product space).  Many of the results which are crucial to quantum theory (e.g. the spectral theorem) rely on this Hilbert space structure, so they would have to be re-worked as well.
Even if it were possible to repair these and all the other issues which would arise from your proposal - and I am not particularly inclined to think it is possible - what are we gaining from this? Who cares if the field underlying our vector space is countable? Why should transcendental numbers be avoided in this way?
