# I The complex numbers in QM

#### fresh_42

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2018 Award
Summary
Which properties of $\mathbb{C}$ are actually necessary?
The following is speculative as well as a honestly meant question about the way QM is modelled. I don't want to create a new theory, just understand the necessities of the old one.

Physicists use complex numbers for QM. But why are they necessary? I understand that the existence of eigenvalues must be granted, but any algebraically closed field does it. I understand that characteristic two is problematic, but what if the characteristic is a high prime, and I assume something like $99989$ or maybe even $89$ already counts as high. I understand that a topology is necessary to perform differentiation, but does it have to contain $\mathbb{R}$? What's used from the symmetry groups looks as if finite groups could do, i.e. mainly discrete calculations. Why isn't discreteness an intrinsic demand from the start? The Lie algebra representations wouldn't look much different on the algebraic closure of $\mathbb{F}_{89}$.

And finally, if complex numbers are necessary, what could be achieved by transcendental extensions like $\mathbb{C}(t)\,?$ This way the entire spacetime could simply be part of the scalar field. Isn't this more natural to do than to look for a way to attach it afterwards?

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#### PeterDonis

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For clarity: it seems like you're not asking why we have to use $\mathbb{C}$ instead of $\mathbb{R}$ in QM, but why we have to use the continuous algebraically closed field $\mathbb{C}$ instead of a discrete one?

#### PeterDonis

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Assuming my understanding in post #2 is correct, one obvious response to me is that there are observables that have a continuous spectrum.

#### fresh_42

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2018 Award
For clarity: it seems like you're not asking why we have to use $\mathbb{C}$ instead of $\mathbb{R}$ in QM, but why we have to use the continuous algebraically closed field $\mathbb{C}$ instead of a discrete one?
Yes, since a discrete one which is large enough shouldn't be very much of a difference. Algebraic closure or even transcendental extensions could be used. My question is, why characteristic zero, and not characteristic with a large prime.

#### PeterDonis

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I understand that a topology is necessary to perform differentiation, but does it have to contain $\mathbb{R}$?
It does if spacetime is continuous, since that is what gives a continous spectrum to spacetime observables (position, momentum, energy).

#### fresh_42

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It does if spacetime is continuous, since that is what gives a continuous spectrum to spacetime observables (position, momentum, energy).
And in the end quantize it for a couple of eigenspaces from Lie algebra representations? That's where my question arose. Those representation aren't very different from prime characteristic anymore if the prime is large enough. We won't find spin 500 particles.

Anyway, there is yet the question, why we do not use $\overline{\mathbb{C}(t,x,y,z)}$ as field and build the theory with spacetime involved as part of the scalar field. I admit that I have no idea how physically absurd eigenvalues in such a field would be, or whether they could still be interpreted. But at least EFE would be there from the start possibly leading to a factor ring of $\overline{\mathbb{C}(t,x,y,z)}$ as scalar domain. Has this been tried before? It sounds more natural to me than the super-world does.

I mean, we consider e.g. a Clifford algebra or a Graßmann algebra as a quotient of the tensor algebra. So why not do the same for the scalar domain.

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#### PeterDonis

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We won't find spin 500 particles.
Spin is a discrete observable, so it's not the kind I was talking about. Doing quantum mechanics on a discrete field means modeling space and time as discrete. Without any experimental evidence of such discreteness, treating spacetime observables like position, momentum, and energy as continuous seems like the simplest hypothesis.

why we do not use $\overline{\mathbb{C}(t,x,y,z)}$ as field
I don't understand what this means.

#### fresh_42

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2018 Award
I don't understand what this means.
The algebraic closure of the complex numbers extended by time, and space coordinates. This way we could define all necessary equations, build the ideal $I$ those equations define in $\mathbb{C}[t,x,y,z]$ and consider $\mathbb{C}[t,x,y,z]/I$ as our scalar domain. Something like what we do with $SO(3,1)$ or with Graßmann algebras as $V^{\otimes}/\langle x\otimes x \rangle$. This makes equations automatically true: $x \wedge y = - y \wedge x$ because $(x+y) \otimes (x+y) =x \otimes y + y \otimes x$ is made zero.

The scalar field is a possible screw that can be adjusted. I understood the argument with the continuous spectra. And I probably should have restricted my question to QFT. But there are still more fields or just rings or algebras of characteristic zero available. EFE are just a system of equations, so they generate an ideal in some tensor algebra. Now take this tensor algebra and factor out this ideal. This way we have a domain which automatically respects EFE. And from there on build the entire theory on this algebra instead of adjusting things afterwards. I'm not sure whether this can be done on the level of field extensions or at least division algebras. However, this was the second thought. My first thought was just $U(1,\mathbb{F}_p[ i ]) \times SU(2,\mathbb{F}_p[ i ]) \times SU(3,\mathbb{F}_p[ i ])$ with, say $p>5$, instead of searching for $E_8$ or $SU(5)$ extensions. The physical results should be the same, if only $p$ is big enough. However, this is a hypothesis, and I would like to find an argument why the results might not be the same. The continuous observables are such an argument in QM. But what about QCD?

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#### PeterDonis

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the complex numbers extended by time, and space coordinates
I don't understand what this means either.

#### fresh_42

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2018 Award
It means the complex function field of Laurent polynomials, quotients of complex series in four variables and their inverses. $at^2+bx+c(yz)^{-1}$ would be an element, a scalar.

#### PeterDonis

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the complex function field of Laurent polynomials, quotients of complex series in four variables and their inverses
Ok. But it looks like what you're proposing is a theory of quantum gravity, since you're expecting spacetime itself to pop out of the equations instead of having to be put in. So a simple response to your original question in this thread would be that even if all that complexity (pun intended ) turns out to be necessary for quantum gravity, that doesn't mean it's necessary for ordinary QM.

#### PeterDonis

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the complex function field of Laurent polynomials, quotients of complex series in four variables and their inverses.
Is this field not continuous?

#### fresh_42

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2018 Award
Is this field not continuous?
It is. We often proceed by the principle: eliminate what disturbs. So putting in EFE from the start instead of at the end seems logical. Spacetime is a (pseudo?) Riemannian manifold. Somebody must have tried using it as the basis for QFT just like in algebraic geometry where geometric objects (solutions of polynomial equations, zeros) are turned into ring theory. This results in additional algebraic tools.

#### A. Neumaier

One cannot dispense with the notion of order, e.g., causality requires the concept of ''later'', and dissipation means energy ''decreases'', being is lost to the environment. Thus the Hermitian part of the field of scalars must be a totally ordered field. This makes it a subfield of the reals.

Mathematical convenience requires completeness, e.g., to have the intermediate value theorem. Thus it must be the field of all reals.

The Schrödinger equation involves a purely imaginary number, forcing the full field of scalars to be precisely the field of complex numbers.

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#### PeterDonis

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putting in EFE from the start instead of at the end
Isn't that what standard GR does?

#### meopemuk

Within quantum logic, there is a theorem stating that the Hilbert space of quantum mechanics can be built only over real, complex or quaternionic scalars:

C. Piron, "Foundations of Quantum Physics", (W. A. Benjamin, Reading, 1976).

M. P. Soler, "Characterization of Hilbert spaces by orthomodular spaces", Comm. Alg. 23 (1995), 219.

So, if you contemplate any other fields of scalars, be prepared to have some funny logical relationships between experimental propositions.

Eugene.

#### fresh_42

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2018 Award
Isn't that what standard GR does?
I don't know. Is there a version which approaches GR by ring theory, where the solutions to EFE are translated into ringed spaces, schemes and sheaves, i.e. instead of doing analysis on a manifold, doing ring theory on local rings. Somebody must have tried, but I don't know the literature.

With your and @A. Neumaier's arguments, prime characteristics seems to be off the table, although I still wonder why we need $\mathbb{C}$ in representations of the Gauge groups. However, it would be unnatural to change the fields at this point.

That leaves us with the question, why extensions of $\mathbb{C}$ aren't considered. E.g. according to Wikipedia (sorry for the source), the position operator uses $L^2(\mathbb{R}^3,\mathbb{C})$ as Hilbert space. Shouldn't we use $L^2(\mathbb{F})$ instead and work with functionals which already carry the coordinates not as variables but as scalars, since the existence of a stress-energy tensors affects the behavior of the coordinates? $\mathbb{F}$ would then be $\mathbb{C}(t,x,y,z)$, leading to a space where $L^2(\mathbb{R}^3,\mathbb{C})$ is only a subspace. An all-in-one environment, so to say.

#### PeterDonis

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Is there a version which approaches GR by ring theory, where the solutions to EFE are translated into ringed spaces, schemes and sheaves, i.e. instead of doing analysis on a manifold, doing ring theory on local rings
I'm not aware of any work along these lines, but there's a lot of literature I haven't read, particularly in the realm of math as opposed to physics, which is where I would think it more likely to find something like this.

However, I'm still confused by your terminology. What do you mean by "putting in the EFE at the start instead of at the end"? You say you don't know whether GR does this or not, so now I'm not sure what you mean by that phrase.

#### PeterDonis

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the position operator uses $L^2(\mathbb{R}^3,\mathbb{C})$ as Hilbert space
More precisely, the position operator for a single particle uses this as Hilbert space. But for more than one particle, the configuration space is no longer $\mathbb{R}^3$, but $\mathbb{R}^{3N}$, where $N$ is the number of particles. And that's leaving out the spin degrees of freedom, which introduce further complications.

Shouldn't we use $L^2(\mathbb{F})$ instead
Do you mean, for example, $L^2(\mathbb{R}^3,\mathbb{F})$ for a single particle?

and work with functionals which already carry the coordinates not as variables but as scalars
I'm not sure why you would want this. The coordinates are not observables or amplitudes. They are labels for points in spacetime. You can change the labels (transform the coordinates) without changing any physics. But changing a scalar--an observable or an amplitude--would change the physics.

the existence of a stress-energy tensors affects the behavior of the coordinates
I'm not sure what you mean by this.

#### fresh_42

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2018 Award
I meant the following construction: We consider a tensor algebra $T$, e.g. $T(\mathbb{R}^4)$. If I understood it correctly, a metric tensor $g$ can be considered as an element of $\mathbb{R}^4 \otimes (\mathbb{R}^4 \otimes \mathbb{R}^4)^*$, and similar for the Ricci-tensor. So if we consider the ideal of $I\subseteq T$ generated by the EFE, then $A:= T/I$ is a domain in which EFE hold automatically true. Hence every calculation done in this vector space, and I'm sure it can be turned into a Hilbert space, automatically respects GR. I just think this is a natural construction which is used elsewhere a lot of times, that naturally builds a framework to base QFT on. Its solutions would automatically be relativistic.

#### PeterDonis

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If I understood it correctly, a metric tensor $g$ can be considered as an element of $\mathbb{R}^4 \otimes (\mathbb{R}^4 \otimes \mathbb{R}^4)$∗
I don't understand why there are three copies of $\mathbb{R}^4$ there.

if we consider the ideal of $I\subseteq T$ generated by the EFE, then $A:= T/I$ is a domain in which EFE hold automatically true
I don't understand this. $T / I$ is a quotient space, correct? Why should the EFE be true, or even well-defined, on the quotient space? To me this is like saying that, since the even integers are an ideal $I$ of the integers that satisfy the equation $n = 2 k$ for some integer $k$, the quotient space of the integers over the even integers, which is just $Z_2$, also satisfies that equation, which doesn't even make sense.

#### fresh_42

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2018 Award
Do you mean, for example, $L^2(\mathbb{R}^3,\mathbb{F})$ for a single particle?
I meant $L^2(\mathbb{F},\mathbb{F})$. But I see that such a generalization at the beginning probably won't make anything easier, and all calculations would simply start with a condition that restricts the situation again to the actually necessary subspace.

The original thought of this thread had been: If we only consider a few representations of the Lie algebra of $U(1)\times SU(2)\times SU(3)$ and of rather low dimensions, what do we need the complex numbers for? Any sufficiently large algebraic closed field of characteristic not two would lead to the same results. And if so, why not vary the scalar field instead of the groups ($E_8,SU(5)$) or the Lie algebras ($\mathbb{Z}_2$ grading) to extend the theory.

#### fresh_42

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2018 Award
I don't understand why there are three copies of $\mathbb{R}^4$ there.
The first for the location, the second and third for the quadratic form.
I don't understand this. $T / I$ is a quotient space, correct? Why should the EFE be true, or even well-defined, on the quotient space? To me this is like saying that, since the even integers are an ideal $I$ of the integers that satisfy the equation $n = 2 k$ for some integer $k$, the quotient space of the integers over the even integers, which is just $Z_2$, also satisfies that equation, which doesn't even make sense.
It does. If we have a system of equations, say of the form $2k=x^2$ for all even $x,k$, and $R:=2\mathbb{Z}[x,k]/\langle 2k-x^2 \rangle$, then all equations in $R$ satisfy $2k=x^2$. The complex numbers are nothing else: $\mathbb{C}=\mathbb{R}[x]/\langle x^2+1\rangle$.

#### PeterDonis

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The original thought of this thread had been: If we only consider a few representations of the Lie algebra of $U(1)\times SU(2)\times SU(3)$ and of rather low dimensions, what do we need the complex numbers for? Any sufficiently large algebraic closed field of characteristic not two would lead to the same results. And if so, why not vary the scalar field instead of the groups ($E_8$,$SU(5)$) or the Lie algebras ($\mathbb{Z}_2$ grading) to extend the theory.
This looks like a question more suited to the Beyond the Standard Model forum; if any work has been done in this area, regulars there would probably be more likely to know about it.

#### PeterDonis

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Huh? $Z_2$ consists of two elements, and only one of them is even. So not all elements of $Z_2$ satisfy the equation that generates the ideal you used to form the quotient space $Z_2$ from the ring of integers.

"The complex numbers in QM"

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