Is the $i$ in QM a time component in disguise?

Question

In SR, it is possible to replace the Minkowski metric $\eta_{\mu\nu}$ with a (pseudo) euclidean metric $\delta_{\mu\nu}$ provided that time is measured in imaginary units.

I was wondering if the same trick can be used to get rid of complex numbers in QM. The answer I gave myself is "No, without changing the physics". In fact, the usual commutator of a particle: $$ [q^i, p_j] = i \delta^i_j $$ becomes: $$ [q^\mu,t_{\nu\rho}] = C_\rho \delta^\mu_\nu $$ where $C_\rho$ is real and $t_{i0}=p_i$ in the rest frame: $$ [q^i,t_{j0}] = \delta^i_j $$

Now, my questions are:

• Is my reasoning correct or does this lead to any inconsistencies?
• If it does not, is experimental accuracy enough to distinguish between the two "theories"?
• If this is not ruled out neither by experimental nor experimental evidence, I guess I'm not the first one to have this idea: could you please provide any pointers?

Answer 1

You may redefine various quantities according to $p\to ip_{\rm yours}$ and things like that but that clearly doesn't change physics, just conventions. For example, the mixed-signature spacetime may indeed be emulated by having the ${+}{+}{+}{+}$ signature but with one (or three) components pure imaginary, and Einstein actually favored this convention at some moment.

However, no QM theory can ever get rid of $i$. It appears in $$ [x,p]=i\hbar $$ The commutator of two Hermitian operators is unavoidably anti-Hermitian, i.e. $i$ times a Hermitian operator. It follows from $(xp)^\dagger=p^\dagger x^\dagger$ i.e. $[x,p]^\dagger = -[x^\dagger,p^\dagger]$.

Also, in Schrödinger's or Heisenberg's equations, there has to be an $i$ because the probability-preserving operators are unitary, and unitary operators are $\exp(iH)$ where $H$ is Hermitian.

Equivalently, $i$ must appear in the exponent $\exp(iS/\hbar)$ of Feynman's path integral.

Every big modification of this kind either produces a totally inconsistent theory, or a totally equivalent theory to quantum mechanics.

Answer 2

L. Motl offered quite a few arguments suggesting that complex numbers are required for quantum theory. Surprisingly, this is not quite correct, at least in some general and important cases, and I do not have in mind replacing complex numbers with pairs of real numbers. Schroedinger noted that, in the case of a scalar field interacting with electromagnetic field (the klein-Gordon-Maxwell electrodynamics, or scalar electrodynamics), you can use the so-called unitary gauge, where the scalar field is real. I showed in http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.) (see also http://arxiv.org/abs/1502.02351) that three out of four complex components of the Dirac spinor in the Dirac equation can be algebraically eliminated in a general case. The remaining component can be made real by a gauge transform.

EDIT (05/28/2016): @Luboš Motl criticized my answer as follows: "...the reality of the scalar field has nothing to do with the complexity of the wave function. A scalar field is an observable so its most elementary part is indeed a Hermitian operator. But the wave function is something entirely different than the fields or other observables and it has to be complex." If a well-known physicist apparently did not quite understand my answer, probably my answer was not clear enough, and I regret that. However, other people could misunderstand me as well, and this is why I am writing this EDIT.

It is true that both Schroedinger's paper in Nature and my humble article in Journ. Math. Physics mostly deal with (non-second-quantized) fields. However, I wrote in my article: "To illustrate the parallels with beautiful but little-known Schroedinger’s work..., the Dirac equation is considered as part of spinor electrodynamics, although most derivations are valid without any changes for the Dirac equation in electromagnetic field independently of the Maxwell equations." Let me further explain that.

Let us consider the standard Klein-Gordon equation for a spinless particle in electromagnetic field:

$(\partial^\mu+ieA^\mu)(\partial_\mu+ieA_\mu)\psi+m^2\psi=0$

(I use a system of units where $\hbar=c=1$). Can we regard $\psi$ as a wave function? I don't think anybody would doubt that, otherwise why would we consider $\psi$ in the standard Schroedinger equation a wave function?

L. Motl states that this wave function "has to be complex", but offers no argumentation. However, it is pretty obvious, as Schroedinger noted, that an arbitrary complex function $\psi$ can be made real by a gauge transform (at least locally). A gauge transform gives a physically equivalent theory. Therefore, the wave function does not have to be complex, and if L. Motl disagrees with my answer, he should offer more argumentation.

L. Motl also criticizes me for promoting my own "otherwise unknown papers" and believes that this "violates some rules of this Stack Exchange". I readily agree that my papers are "otherwise unknown" (by the way, the Schroedinger's paper is unfortunately all but forgotten), but until L. Motl explicitly indicates the rule that forbids mentioning my clearly relevant and properly published article, I consider this critique unfounded.