In Yale's paper, automorphisms on complex numbers $\mathcal{C}$ are shown to be wild, allegedly with cardinality $2^{\mathfrak{C}^{\mathfrak{C}}}$. There are two well-known ("trivial") automorphisms: identity $id$ and complex conjugation $\bar{ }$.

If we think of complex vector spaces as modules with action coming from the automorphisms of $\mathcal{C}$, then given a module $V$ we can identify $V$ as the domain of $id$ and $\bar{V}$ as the domain of $\bar{ }$. If we extend this idea to Hilbert spaces, we can write $H$ and $\bar{H}$ for the corresponding Hilbert spaces of these two vector spaces.

In quantum computing one tacitly assumes the existence of these two Hilbert spaces and their duals, that is, four spaces in total: $H, \bar{H}, H^*, \bar{H^*} = H^{\dagger}$.

If we now consider a non-trivial automorphism of $\mathcal{C}$ instead (i.e. different from $id$ and $\bar{ }$), say $a$, we can identify space $H^a$. Suppose we use this $H^a$ instead of $\bar{H}$. I think (I can't think why not) all calculations of quantum computing should remain the same, e.g.$\langle v | v \rangle = 1$.

So, my question is :

Does the choice of the $\mathcal{C}$-automorphism have any significance for quantum computing, or quantum mechanics in general?

If all operations in QM remain the same with this change of automorphism, then I think it is fair to say the answer is no. But I wanted to post this question to hear of anyone else's thoughts on it.

Moreover, could there be a physical or even philosophical significance to the choice of the automorphism? Could it amount to a choice of the measuring apparatus? This of course sounds too far-fetched, but since I mentioned the word "philosophy" I'm going to leave it.

Many thanks in advance.

  • $\begingroup$ I have no idea what "domain of action" means, but if your question comes down to whether twisting by an arbitrary automorphism of ${\mathbb C}$ preserves all the mathematical structure we care about in quantum mechanics, the answer is clearly no, because (among other things) a typical automorphism is not continuous and (among other other things) does not preserve the real numbers. $\endgroup$
    – WillO
    Oct 19, 2017 at 14:03

2 Answers 2


You cannot "choose" an automorphism other than the identity and complex conjugation and get a Hilbert space $H^a$ as you seem to imply.

The difference between $H$ and $\bar{H}$ is the choice of complex structure on the underlying real vector space, but no other automorphism of the complex numbers preserves the reals (this is theorem 3 in the paper by Yale you refer to), so no other automorphism induces such a choice of complex structure, and these automorphisms have no obvious relevance at all to the formalism of quantum mechanics on complex Hilbert spaces.


thank you for let me know about this very important question. I do not agree with these two answers. Automorphism need not to preserve real numbers, why? Automorphism group of complex numbers is very important for QM. We jointly proposed to consider Hilbert Q[i]-spaces, instead of R[i]-spaces, because a group Aut{Q[i]} consists of exactly only two elements. Instead, a group Aut{R[i]} is rather not known yet. One should consider always ALL automorphisms of a chosen ring of scalars, and not restrict to some choices. There are so many different modules (vector spaces) over this chosen ring of scalar as exactly as many elements in the group of automorphism if this ring of scalars. Therefore your important question is equivalent to question: Why Quantum Mechanics needs necessarily the complex numbers as a ring of scalars, and not other choices, like non-commutative ring of quaternions [Stephen Adler, Quaternionic Quantum Mechanics, Oxford 1995], or other commutative rings, like Q[i]. The first step is a choice-postulate of a ring of scalars. After vector spaces (or modules or bi-modules), and finally iso-morphisms analogous to known Hilbert structures, that should be considered as variables, and not fixed. I am so happy that you still are thinking about our problem we have had 10 or more years in the past. I think that your fundamental question is important for quantum computing.

  • $\begingroup$ This isn't what the question was about. $\endgroup$
    – user4552
    May 23, 2018 at 19:59

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