Proof that complex numbers are required for quantum mechanics? [closed]

A perennial question in quantum mechanics is whether or not complex numbers are truly required for its formulation. I have always intuitively believed that they are required to allow for the time-evolution of quantum states. However, I am no pure mathematician and my attempts at trying to demonstrate this rigorously have typically fallen short of the mark.

Below, I have come up with a fairly simple argument, which I believe holds water. If not, I am hoping that someone can point out the flaws in it (or, even better, fix them!)

The argument

In QM, a physical system may be represented by a unit vector $|\Psi\rangle$ in a separable Hilbert space $\mathcal{H}$. If $\{|\psi_{k}\rangle\}$ is a complete, orthonormal basis set spanning $\mathcal{H}$, then we can represent a time-dependent state vector $|\Psi(t)\rangle$ as

\begin{align} |\Psi(t)\rangle &= \sum_{k} a_{k}(t)|\psi_{k}\rangle, \tag{1} \end{align}

where the $a_{k}(t) = \langle\psi_{k}|\Psi(t)\rangle$ are scalar coefficients. Here, the basis vectors $|\psi_{k}\rangle$ are taken to be time-independent and can be taken to have real elements. The normalisation requirement then implies that

\begin{align} \langle\Psi(t)|\Psi(t)\rangle &= \sum_{k} |a_{k}(t)|^{2} = 1. \tag{2} \end{align}

(Note that at this point the squared modulus notation should not be taken to imply that the $a_{k}(t)$ are complex - it just allows that they might be).

We may now construct a new, unnormalised vector from $|\Psi\rangle$ by multiplying each term in (1) by an arbitrary real number $r_{k}$ to obtain

\begin{align} |\Phi(t)\rangle &= \sum_{k} r_{k}a_{k}(t)|\psi_{k}\rangle. \tag{3} \end{align}

We then have

\begin{align} \langle\Phi(t)|\Phi(t)\rangle &= \sum_{k} r_{k}^{2}|a_{k}(t)|^{2} = C^{2}, \tag{4} \end{align}

where $C$ is some constant real number. Taking the time derivative of this, we have

\begin{align} \frac{\partial}{\partial t}\langle\Phi(t)|\Phi(t)\rangle &= \sum_{k} r_{k}^{2}\left[\frac{\partial a_{k}^{*}}{\partial t}a_{k} + a_{k}^{*}\frac{\partial a_{k}}{\partial t}\right] = 0. \tag{5} \end{align}

Since the $r_{k}$ have been chosen arbitrarily, in order for this to hold in all cases, each term in the square brackets must be zero. That is

\begin{align} \frac{\partial a_{k}^{*}}{\partial t}a_{k} &= -a_{k}^{*}\frac{\partial a_{k}}{\partial t} = \left(a_{k}^{*}\frac{\partial a_{k}}{\partial t}\right)^{*}. \tag{6} \end{align}

This can only be satisfied for real $a_{k}$ if either $a_{k}$ or its time derivative equals zero, meaning that either it does not appear in the summation (3) or it is a constant and does not contribute to the time dependence of the vector.

In order to satisfy (6) and contribute to the time dependence, the factor $a_{k}^{*}\partial a_{k}/\partial t$ must be pure imaginary. Hence, the time evolution of vectors in a Hilbert space requires (at least) complex numbers (since we might also satisfy (6) using quaternions).

Comments

Now, in anticipation of certain general objections, I am aware that many authors claim to construct quantum mechanics using real Hilbert spaces. However, whilst these Hilbert spaces have only real elements, closer inspection of the operators acting on them shows that the complex elements of the equivalent complex Hilbert space have actually been replaced by their $2\times2$ matrix equivalents

\begin{align} a + b\mathrm{i} \to \left[\begin{array}{cc} a & -b \\ b & a\end{array}\right]. \end{align}

Since these $2\times2$ matrices are isomorphic to their scalar counterparts they are still complex numbers. The nominally real Hilbert spaces constructed in this way therefore retain an intrinsic complex structure. A more detailed discussion is given in this arxiv article: https://arxiv.org/abs/2604.17086.

The real vs complex Hilbert space issue merits a completely separate discussion. What I am really asking about is whether the particular `proof' given above has any mathematical or conceptual flaws in it?