QM without complex numbers

Question

I am trying to understand how complex numbers made their way into QM. Can we have a theory of the same physics without complex numbers? If so, is the theory using complex numbers easier?

Answer 1

The nature of complex numbers in QM turned up in a recent discussion, and I got called a stupid hack for questioning their relevance. Mainly for therapeutic reasons, I wrote up my take on the issue:

On the Role of Complex Numbers in Quantum Mechanics

Motivation

It has been claimed that one of the defining characteristics that separate the quantum world from the classical one is the use of complex numbers. It's dogma, and there's some truth to it, but it's not the whole story:

While complex numbers necessarily turn up as first-class citizen of the quantum world, I'll argue that our old friend the reals shouldn't be underestimated.

A bird's eye view of quantum mechanics

In the algebraic formulation, we have a set of observables of a quantum system that comes with the structure of a real vector space. The states of our system can be realized as normalized positive (thus necessarily real) linear functionals on that space.

In the wave-function formulation, the Schrödinger equation is manifestly complex and acts on complex-valued functions. However, it is written in terms of ordinary partial derivatives of real variables and separates into two coupled real equations - the continuity equation for the probability amplitude and a Hamilton-Jacobi-type equation for the phase angle.

The manifestly real model of 2-state quantum systems is well known.

Complex and Real Algebraic Formulation

Let's take a look at how we end up with complex numbers in the algebraic formulation:

We complexify the space of observables and make it into a $C^*$-algebra. We then go ahead and represent it by linear operators on a complex Hilbert space (GNS construction).

Pure states end up as complex rays, mixed ones as density operators.

However, that's not the only way to do it:

We can let the real space be real and endow it with the structure of a Lie-Jordan-Algebra. We then go ahead and represent it by linear operators on a real Hilbert space (Hilbert-Schmidt construction).

Both pure and mixed states will end up as real rays. While the pure ones are necessarily unique, the mixed ones in general are not.

The Reason for Complexity

Even in manifestly real formulations, the complex structure is still there, but in disguise:

There's a 2-out-of-3 property connecting the unitary group $U(n)$ with the orthogonal group $O(2n)$, the symplectic group $Sp(2n,\mathbb R)$ and the complex general linear group $GL(n,\mathbb C)$: If two of the last three are present and compatible, you'll get the third one for free.

An example for this is the Lie-bracket and Jordan product: Together with a compatibility condition, these are enough to reconstruct the associative product of the $C^*$-algebra.

Another instance of this is the Kähler structure of the projective complex Hilbert space taken as a real manifold, which is what you end up with when you remove the gauge freedom from your representation of pure states:

It comes with a symplectic product which specifies the dynamics via Hamiltonian vector fields, and a Riemannian metric that gives you probabilities. Make them compatible and you'll get an implicitly-defined almost-complex structure.

Quantum mechanics is unitary, with the symplectic structure being responsible for the dynamics, the orthogonal structure being responsible for probabilities and the complex structure connecting these two. It can be realized on both real and complex spaces in reasonably natural ways, but all structure is necessarily present, even if not manifestly so.

Conclusion

Is the preference for complex spaces just a historical accident? Not really. The complex formulation is a simplification as structure gets pushed down into the scalars of our theory, and there's a certain elegance to unifying two real structures into a single complex one.

On the other hand, one could argue that it doesn't make sense to mix structures responsible for distinct features of our theory (dynamics and probabilities), or that introducing un-observables to our algebra is a design smell as preferably we should only use interior operations.

While we'll probably keep doing quantum mechanics in terms of complex realizations, one should keep in mind that the theory can be made manifestly real. This fact shouldn't really surprise anyone who has taken the bird's eye view instead of just looking throught the blinders of specific formalisms.

Answer 2

Update: This answer has been superseded by my second one. I'll leave it as-is for now as it is more concrete in some places. If a moderator thinks it should be deleted, feel free to do so.

I do not know of any simple answer to your question - any simple answer I have encountered so far wasn't really convincing.

Take the Schrödinger equation, which does contain the imaginary unit explicitly. However, if you write the wave function in polar form, you'll arrive at a (mostly) equivalent system of two real equations: The continuity equation together with another one that looks remarkably like a Hamilton-Jacobi equation.

Then there's the argument that the commutator of two observables is anti-hermitian. However, the observables form a real Lie-algebra with bracket $-i[\cdot,\cdot]$, which Dirac calls the quantum Poisson bracket.

All expectation values are of course real, and any state $\psi$ can be characterized by the real-valued function $$ P_\psi(·) = |\langle \psi,·\rangle|^2 $$

For example, the qubit does have a real description, but I do not know if this can be generalized to other quantum systems.

I used to believe that we need complex Hilbert spaces to get a unique characterization of operators in your observable algebra by their expectation values.

In particular, $$ \langle\psi,A\psi\rangle = \langle\psi,B\psi\rangle \;\;\forall\psi \Rightarrow A=B $$ only holds for complex vector spaces.

Of course, you then impose the additional restriction that expectation values should be real and thus end up with self-adjoint operators.

For real vectors spaces, the latter automatically holds. However, if you impose the former condition, you end up with self-adjoint operators as well; if your conditions are real expectation values and a unique representation of observables, there's no need to prefer complex over real spaces.

The most convincing argument I've heard so far is that linear superposition of quantum states doesn't only depend on the quotient of the absolute values of the coefficients $|α|/|β|$, but also their phase difference $\arg(α) - \arg(β)$.

Update: There's another geometric argument which I came across recently and find reasonably convincing: The description of quantum states as vectors in a Hilbert space is redundant - we need to go to the projective space to get rid of this gauge freedom. The real and imaginary parts of the hermitian product induce a metric and a symplectic structure on the projective space - in fact, projective complex Hilbert spaces are Kähler manifolds. While the metric structure is responsible for probabilities, the symplectic one provides the dynamics via Hamilton's equations. Because of the 2-out-of-3 property, requiring the metric and symplectic structures to be compatible will get us an almost-complex structure for free.

Answer 3

Yes, we can have a theory of the same physics without complex numbers (without using pairs of real functions instead of complex functions), at least in some of the most important general quantum theories. For example, Schrödinger (Nature (London) 169, 538 (1952)) noted that one can make a scalar wavefunction real by a gauge transform. Furthermore, surprisingly, the Dirac equation in electromagnetic field is generally equivalent to a fourth-order partial differential equation for just one complex component, which component can also be made real by a gauge transform (http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (an article published in the Journal of Mathematical Physics) or http://arxiv.org/abs/1008.4828 ).

Answer 4

Frank, I would suggest buying or borrowing a copy of Richard Feynman's QED: The Strange Theory of Light and Matter. Or, you can just go directly to the online New Zealand video version of the lectures that gave rise to the book.

In QED you will see how Feynman dispenses with complex numbers entirely, and instead describes the wave functions of photons (light particles) as nothing more than clock-like dials that rotate as they move through space. In a book-version footnote he mentions in passing "oh by the way, complex numbers are really good for representing the situation of dials that rotate as they move through space," but he intentionally avoids making the exact equivalence that is tacit or at least implied in many textbooks. Feynman is quite clear on one point: It's the rotation-of-phase as you move through space that is the more fundamental physical concept for describing quantum mechanics, not the complex numbers themselves.[1]

I should be quick to point out that Feynman was disrespecting the remarkable usefulness of complex numbers for describing physical phenomena. Far from it! He was fascinating for example by the complex-plane equation known as Euler's Identity, $e^{i\pi} = -1$ (or, equivalently, $e^{i\pi} + 1 = 0$), and considered it one of the most profound equations in all of mathematics.

It's just that Feynman in QED wanted to emphasize the remarkable conceptual simplicity of some of the most fundamental concepts of modern physics. In QED for example, he goes on to use his little clock dials to show how in principle his entire method for predicting the behavior of electrodynamic fields and systems could be done using such moving dials.

That's not practical of course, but that was never Feynman's point in the first place. His message in QED was more akin to this: Hold on tight to simplicity when simplicity is available! Always build up the more complicated things from that simplicity, rather than replacing simplicity with complexity. That way, when you see something horribly and seemingly unsolvable, that little voice can kick in and say "I know that the simple principle I learned still has to be in this mess, somewhere! So all I have to do is find it, and all of this showy snowy blowy razzamatazz will disappear!"

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[1] Ironically, since physical dials have a particularly simple form of circular symmetry in which all dial positions (phases) are absolutely identical in all properties, you could argue that such dials provide a more accurate way to represent quantum phase than complex numbers. That's because as with the dials, a quantum phase in a real system seems to have absolutely nothing at all unique about it -- one "dial position" is as good as any other one, just as long as all of the phases maintain the same positions relative to each other. In contrast, if you use a complex number to represent a quantum phase, there is a subtle structural asymmetry that shows up if you do certain operations such as squaring the number (phase). If you do that do a complex number, then for example the clock position represented by $1$ (call it 3pm) stays at $1$, while in contrast the clock position represented by $-1$ (9pm) turns into a $1$ (3pm). This is no big deal in a properly set up equation, but that curious small asymmetry is definitely not part of the physically detectable quantum phase. So in that sense, representing such a phase by using a complex number adds a small bit of mathematical "noise" that is not in the physical system.

Answer 5

The complex numbers in quantum mechanics are mostly a fake. They can be replaced everywhere by real numbers, but you need to have two wavefunctions to encode the real and imaginary parts. The reason is just because the eigenvalues of the time evolution operator $e^{iHt}$ are complex, so the real and imaginary parts are degenerage pairs which mix by rotation, and you can relabel them using i.

The reason you know i is fake is that not every physical symmetry respects the complex structure. Time reversal changes the sign of "i". The operation of time reversal does this because it is reversing the sense in which the real and imaginary parts of the eigenvectors rotate into each other, but without reversing the sign of energy (since a time reversed state has the same energy, not negative of the energy).

This property means that the "i" you see in quantum mechanics can be thought of as shorthand for the matrix (0,1;-1,0), which is algebraically equivalent, and then you can use real and imaginary part wavefunctions. Then time reversal is simple to understand--- it's an orthogonal transformation that takes i to -i, so it doesn't commute with i.

The proper way to ask "why i" is to ask why the i operator, considered as a matrix, commutes with all physical observables. In other words, why are states doubled in quantum mechanics in indistinguishable pairs. The reason we can use it as a c-number imaginary unit is because it has this property. By construction, i commutes with H, but the question is why it must commute with everything else.

One way to understand this is to consider two finite dimensional systems with isolated Hamiltonians $H_1$ and $H_2$, with an interaction Hamiltonian $f(t)H_i$. These must interact in such a way that if you freeze the interaction at any one time, so that $f(t)$ rises to a constant and stays there, the result is going to be a meaningful quantum system, with nonzero energy. If there is any point where $H_i(t)$ doesn't commute with the i operator, there will be energy states which cannot rotate in time, because they have no partner of the same energy to rotate into. Such states must be necessarily of zero energy. The only zero energy state is the vacuum, so this is not possible.

You conclude that any mixing through an interaction hamiltonian between two quantum systems must respect the i structure, so entangling two systems to do a measurement on one will equally entangle with the two state which together make the complex state.

It is possible to truncate quantum mechanics (at least for sure in a pure bosnic theory with a real Hamiltonian, that is, PT symmetric) so that the ground state (and only the ground state) has exactly zero energy, and doesn't have a partner. For a bosonic system, the ground state wavefunction is real and positive, and if it has energy zero, it will never need the imaginary partner to mix with. Such a truncation happens naturally in the analytic continuation of SUSY QM systems with unbroken SUSY.

Answer 6

Just to put complex numbers in context, A.A. Albert edited "Studies in Modern Algebra" - from the Mathematical Assn of America. C is one of the Normed Division Algebras - of which there are only four: R,C,H and O. One can do a search for "composition algebras" - of which C is one.

Answer 7

If you don't like complex numbers, you can use pairs of real numbers (x,y). You can "add" two pairs by (x,y)+(z,w) = (x+z,y+w), and you can "multiply" two pairs by (x,y) * (z,w) = (xz-yw, xw+yz). (If don't think that multiplication should work that way, you can call this operation "shmultiplication" instead.)

Now you can do anything in quantum mechanics. Wavefunctions are represented by vectors where each entry is a pair of real numbers. (Or you can say that wavefunctions are represented by a pair of real vectors.) Operators are represented by matrices where each entry is a pair of real numbers, or alternatively operators are represented by a pair of real matrices. Shmultiplication is used in many formulas. Etc. Etc.

I'm sure you see that these are exactly the same as complex numbers. (see Lubos's comment: "a contrived machinery that imitates complex numbers") They are "complex numbers for people who have philosophical problems with complex numbers". But it would make more sense to get over those philosophical problems. :-)

Answer 8

Complex numbers "show up" in many areas such as, for example, AC analysis in electrical engineering and Fourier analysis of real functions.

The complex exponential, $e^{st},\ s = \sigma + i\omega$ shows up in differential equations, Laplace transforms etc.

Actually, it just shouldn't be all that surprising that complex numbers are used in QM; they're ubiquitous in other areas of physics and engineering.

And yes, using complex numbers makes many problems far easier to solve and to understand.

I particularly enjoyed this book (written by an EE) which gives many enlightening examples of using complex numbers to greatly simplify problems.

Answer 9

I am not very well versed in the history, but I believe that people doing classical wave physics had long since notes the close correspondence between the many $\sin \theta$s and $\cos \theta$s flying around their equations and the behavior of $e^{i \theta}$. In fact most wave related calculation can be done with less hassle in the exponential form.

Then in the early history of quantum mechanics we find things described in terms of de Broglie's matter waves.

And it works which is really the final word on the matter.

Finally, all the math involing complex numbers can be decomposed into compound operations on real numbers so you can obviously re-formulate the theory in those terms there is no reason to think that you will gain anything in terms of ease or insight.