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?


14 Answers 14


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


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.


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.

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    $\begingroup$ nice one for a questioning stupid hack, i say if more questioning stupid hacks do this, it is possible the general understanding will be raised by quite an amount $\endgroup$ – Nikos M. Jul 5 '14 at 8:47
  • $\begingroup$ Could you elaborate on the statement about $U(n)$, $Sp(2n, \mathbb{R})$, $O(n)$, $GL(n, \mathbb{C})$? $\endgroup$ – qazwsx Aug 10 '14 at 16:25
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    $\begingroup$ @Problemania: $U(n)=Sp(2n,\mathbb R)\cap O(2n)\cap GL(n,\mathbb C)$; however, the intersection of any 2 of the groups on the RHS is sufficient, and in particular $U(n)=Sp(2n,\mathbb R)\cap O(2n)$; complexity arises naturally when we deal with compatible symplectic and orthogonal structures; of course it's equally valid to say that symplectic structures arise naturally from compatible orthogonal and complex structures or orthogonal ones from compatible symplectic and complex ones; but complex structures are arguably less well motivated from a physical (or perhaps 'philosophical') point of view $\endgroup$ – Christoph Aug 10 '14 at 22:08
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    $\begingroup$ Does the dynamics necessarily have a symplectic structure? What about dynamics that doesn't come from a classical Hamiltonian? $\endgroup$ – Jules Dec 5 '17 at 2:46
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    $\begingroup$ @qazwsx for a precise definition on compatibility, see en.wikipedia.org/wiki/Unitary_group#2-out-of-3_property $\endgroup$ – Falcon Sep 8 '18 at 23:02

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.

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    $\begingroup$ Interesting. However, when you start off the argument with $e^{iHt}$, aren't you assuming complex Hilbert spaces to already? $\endgroup$ – Argyll Jun 6 '15 at 4:38
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    $\begingroup$ @Argyll: i is a 2 by 2 real matrix which squares to -1. Eigenvalues are complex even when the theory is real. $\endgroup$ – Ron Maimon Jun 6 '15 at 11:58
  • $\begingroup$ Then what are the dimensions of $H$ and $t$? Why is it being multiplied by a 2-by-2 matrix on the left? $e^{iHt}$ come's from Stone's theorem on one parameter family of unitary group. What is your version of $e^{iHt}$ based on? $\endgroup$ – Argyll Jun 7 '15 at 14:34
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    $\begingroup$ @Argyll: The real dimension of H is twice the original complex dimension. The Hamiltonian is multiplied by "i" to make the eigenvalues imaginary. This version, which isn't really mine, is just the obvious way to turn complex vector spaces into real vector spaces, by doubling the dimension, and considering a real basis consisting of the original basis vectors e_1 ... e_n and i times these ie_1 .... ie_n, so 2n basis vectors. It is easy to do in both finite dimensions and infinite dimensions, and there are no subtleties. You can check that multiplication by i acts as the 2 by 2 matrix I gave. $\endgroup$ – Ron Maimon Jun 7 '15 at 17:40

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. :-)

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    $\begingroup$ But doesn't just change his question to "QM without shmultiplication"? $\endgroup$ – Alfred Centauri Jul 20 '12 at 14:00
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    $\begingroup$ I do like complex numbers a lot. They are extremely useful and convenient, in connection to the fundamental theorem of algebra, for example, or when working with waves. I'm just trying to understand. $\endgroup$ – Frank Jul 20 '12 at 15:30
  • $\begingroup$ Alfred - yes. That would be the point. I was wondering if there could be, I don't know, a matrix formulation of the same physics that would use another tool (matrices) than complex numbers. Again, I have no problem with complex numbers and I love them. $\endgroup$ – Frank Jul 20 '12 at 15:53
  • $\begingroup$ also note that you can model QM on a space of states on a sphere in $\mathbb{C}^n$ with radius $|x|^2+|y|^2+...=1$. These spheres have dimension $2n$ for the reals. $\endgroup$ – kηives Jul 20 '12 at 16:53
  • $\begingroup$ @kηives: the space of states for a finite-dimensional quantum system is the complex projective space $P_n\mathbb{C}\cong\mathbb{C}^{n+1}\setminus\{0\}/\mathbb{C}^*\cong U(n+1)/U(n)\times U(1)\cong S^{2n+1}/S^1$ $\endgroup$ – Christoph Jul 20 '12 at 17:37

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 not 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: see his Volume 1, Chapter 22 of "The Feynman Lectures in Physics.

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!"

[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.

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    $\begingroup$ Hi Terry: I think I just corrected a typo (changed "disrespecting" to "not disrespecting") (please check carefully) and added a link to support Feynman's fascination with $e^{i\,\pi}+1=0$ (his wonderful lecture on "Algebra"). $\endgroup$ – Selene Routley Jul 9 '15 at 2:41
  • $\begingroup$ Hi Rod, wow, thanks, you are exactly right about the typo! Thanks also for that link. I love that Caltech has made a clean electronic copy with inline formulas! Feynman never, ever lost track of the fact that physics is fundamentally about finding simplicity within seeming complexity, not the other way around, and his Lectures capture that theme beautifully. $\endgroup$ – Terry Bollinger Jul 9 '15 at 13:19
  • $\begingroup$ Indeed Caltech has done a beautiful job. I actually like browsing their online version better than my own paper copies. $\endgroup$ – Selene Routley Jul 9 '15 at 13:41
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    $\begingroup$ Searchability is also a major plus! Sometimes rather extraordinary Feynman tidbits can be hidden away as a asides in an overall text that does not seem relevant to the topic. There are even rare cases where the audio has insights not in the text, though the online version of course does not help there. $\endgroup$ – Terry Bollinger Jul 9 '15 at 17:54

Let the old master Dirac speak:

"One might think one could measure a complex dynamical variable by measuring separately its real and pure imaginary parts. But this would involve two measurements or two observations, which would be alright in classical mechanics, but would not do in quantum mechanics, where two observations in general interfere with one another - it is not in general permissible to consider that two observations can be made exactly simultaneously, and if they are made in quick succession the first will usually disturb the state of the system and introduce an indeterminacy that will affect the second." (P.A.M Dirac, The principles of quantum mechanics, §10, p.35)

So if I interpret Dirac right, the use of complex numbers helps to distinguish between quantities, that can be measured simultaneously and the one which can't. You would loose that feature, if you would formulate QM purely with real numbers.

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    $\begingroup$ Dear asmaier, it is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) In general in such situations, please consider one of the following options: (i) Delete three of your answers. (ii) Flag for duplicate posts and delete three of your answers. (iii) If you think the four posts are not duplicates, then personalize each answer to address the four different specific questions. $\endgroup$ – Qmechanic Nov 2 '13 at 23:13
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    $\begingroup$ Dear Qmechanic, isn't is also frown upon to copy-paste identical comments? ;-) However I admit, that my answers were too similar. So I tried to follow your suggestion (iii) and personalized my answers to address the specific question in a better way. However I still believe the quote from Dirac is very relevant and important, so I will refer to it in every answer. $\endgroup$ – asmaier Nov 3 '13 at 14:48
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    $\begingroup$ @asmaier: I looked at the quote in the book,and I tend to interpret it as follows: in a general case, it is not possible to measure a complex dynamical variable. So I don't quite understand how you make your conclusion: "the use of complex numbers helps to distinguish between quantities, that can be measured simultaneously and the one which can't" $\endgroup$ – akhmeteli Nov 3 '13 at 15:45
  • $\begingroup$ I'm not sure if that is a good example, but think about the wave function described by Schrödingers equation. One could split Schrödingers equation into two coupled equations, one for the real and one for the imaginary part of the wave function. However one cannot measure the phase and the amplitude of the wave function simultaneously, because both measurements interfere with each other. To make this manifest, one uses a single equation with a complex wave function, and generates the observable real quantity by squaring the complex wave function. $\endgroup$ – asmaier Nov 3 '13 at 16:23
  • $\begingroup$ @asmaier: I still don't quite see how this supports your conclusion that I quoted. By the way, as you mentioned the Schrödinger equation, you might wish to see my answer to the question. $\endgroup$ – akhmeteli Nov 3 '13 at 18:47

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.

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    $\begingroup$ I guess I'm wondering if those complex numbers are "intrinsic" or just an arbitrary computing device that happens to be effective. $\endgroup$ – Frank Jul 20 '12 at 2:56
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    $\begingroup$ @Frank: you could ask the same thing about the real numbers. Who ever measured anything to be precisely $\sqrt 2$ meters, anyhow? $\endgroup$ – Niel de Beaudrap Jul 20 '12 at 4:29
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    $\begingroup$ What does it mean though that complex numbers "appear" in AC circuit analysis? The essence of AC is a sinusoidal driving components. You could say the nature of these components come from geometry factors, made electrical from a dot product in generators. Once we have sinusoidal variables interacting in an electrical circuit, we know the utility of complex numbers. That, in turn, comes from the equations. What does that all mean though? $\endgroup$ – Alan Rominger Jul 20 '12 at 13:36
  • $\begingroup$ It means that if the sources in the circuit are all of the form $e^{st}$, the voltages and currents in the circuit will be of that form. This follows from the nature of the differential equations that represent the circuit. The fact that we choose to set $s = j\omega$ for AC analysis and then select only the real part of the solutions as a "reality" constraint doesn't change the mathematical fact that the differential equations describing the circuit have complex exponential solutions. $\endgroup$ – Alfred Centauri Jul 20 '12 at 13:49
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    $\begingroup$ Alan - it probably means nothing. It happens to be a tool that so far works pretty well. $\endgroup$ – Frank Jul 20 '12 at 15:55

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 ).


Complex numbers are used for practical reasons only: QM includes helices and similar functions. The Euler formula

$${e^{i\alpha}} = \sin \alpha + i \cos \alpha$$

is describing three-dimensional helices in a very simple way, but if you want to use it you must replace one real axis by an imaginary axis. This is why QM generally works with an imaginary axis. For the same reason complex numbers are used in engineering: In every case a helix-like process is to be described.


The rotation group, its representations, and their carrier spaces are fundamental parts of quantum mechanics. Every object in the universe is either a spin=0, 1/2, 1, 3/2, 2,… object. For the integer spin objects, the rotation group is O(3), and the rotation matrices contain only real numbers. However, there are half integer spin particles in the world, and matrices with complex numbers in them are needed to rotate them. The group that covers all rotations is SU(2) which has a 2 x 2 array of generators $J$. The 3 rotation angles must be encoded into the 2 x 2 array of lie group parameters $\Theta$. The group element (ie: the rotation matrix) is then $$ R=e^{\Theta^{mn} J^{nm}} $$ The special “S” in SU(2) means $det(R)=1$ which implies $Trace(\Theta)=0$. R is unitary which makes $(\Theta^{nm})^* + \Theta^{mn} = 0$. If the elements in $\Theta$ are real, then $\Theta$ is antisymmetric. So $$ \Theta = \begin{bmatrix} a & b \\ -b & -a \end{bmatrix} $$ Notice there is no way to stuff a third angle c into $\Theta$ without using $ i $. Then using $i$ the 3 rotations can be put into $\Theta$. $$ \Theta=(i/2)\begin{bmatrix} -\theta_z & \theta_x + i\theta_y \\ \theta_x - i\theta_y & \theta_z \end{bmatrix} $$

Therefore, a reason complex numbers are needed in quantum mechanics is because there exists half-integer spin particles.

  • $\begingroup$ This is IMHO essentially the reason Sakurai gives - I like it! $\endgroup$ – Frank Aug 25 '15 at 20:41
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    $\begingroup$ But Schrödingers equation doesn‘t care about spin, and still needs complex numbers. $\endgroup$ – asmaier Mar 17 '18 at 19:16
  • $\begingroup$ @asmaier Correct, others present the argument that Schrödinger's equation requires complex numbers. However, Schrödinger's equation (describing the time translation of objects) is not the only piece of QM. How objects are rotated is another piece of QM. I didn't see any other answers address the need for complex numbers in this part of QM. $\endgroup$ – Gary Godfrey Mar 18 '18 at 3:34

I am not very well versed in the history, but I believe that people doing classical wave physics had long since noted 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 involving 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.

  • $\begingroup$ Can a complex infinite dimensional Hilbert space be written as a real Hilbert space with complex structure ? It seems plausible that it can be done but could there be any problems due to infinite dimensionality ? $\endgroup$ – user10001 Jul 20 '12 at 2:45
  • $\begingroup$ The underlying field you choose, $\mathbb{C}$ or $\mathbb{R}$, for your vector space probably has nothing to do with its dimensionality. $\endgroup$ – Frank Jul 20 '12 at 2:52
  • $\begingroup$ @dushya: There are no problems due to infinite dimensionality, the space is separable and can be approximated by finite dimensional subspaces. $\endgroup$ – Ron Maimon Jul 20 '12 at 18:37

There is in fact a natural way to think of quantum mechanics without using complex numbers. This is closely related to the Hamiltonian-Jacobi (HJ) formulation of classical mechanics and gives an interesting perspective on the link between classical and quantum mechanics!

Classical Mechanics

The HJ formalism is first-order(!) in time, where the velocity is given by $$ \dot {\boldsymbol x}(t) = \frac{\boldsymbol{\nabla} S(\boldsymbol x(t),t)}{m} $$ where $S$ is called Hamilton's principal function, satisfying $$ \boxed{ \partial_t S(\boldsymbol x,t) = - \frac{| \boldsymbol \nabla S(\boldsymbol x,t)|^2}{2m} + V(\boldsymbol x) } \;. $$ Sanity check: if the potential $V=0$, then we can solve the latter equation with $S(\boldsymbol x,t) = \boldsymbol{k \cdot x} - \frac{|\boldsymbol k|^2}{2m}t$, with the equation of motion thus giving us $ \dot {\boldsymbol x}(t) = \frac{\boldsymbol k}{m} $.

Before generalizing this to the quantum case, it is useful to observe that we can rewrite the equation of motion in terms of the density $\rho(\boldsymbol x,t)$ as $$ \boxed{ \partial_t \rho + \boldsymbol{\nabla \cdot}\left( \rho \boldsymbol v \right) = 0 \qquad \textrm{with } \boldsymbol v(\boldsymbol x,t) = \frac{\boldsymbol{\nabla} S(\boldsymbol x,t)}{m} } \; .$$ The special case of $\rho(\boldsymbol x,t) = \delta(\boldsymbol x - \boldsymbol x(t))$ recovers the earlier equation. The above two boxed equations capture classical Newtonian physics.

Quantum Mechanics---without complex numbers

The claim is that quantum mechanics is given by the same continuity equation above (i.e., the second boxed equation), but now we just add a new term to the equation for $S$: $$ \boxed{ \partial_t S(\boldsymbol x,t) = - \frac{| \boldsymbol \nabla S(\boldsymbol x,t)|^2}{2m} + V(\boldsymbol x) + Q(\boldsymbol x) \qquad \textrm{where } Q(\boldsymbol x):= -\frac{\hbar^2}{2m} \frac{ \nabla^2 \left(\sqrt{\rho}\right)}{\sqrt{\rho}} }. $$ This new term is sometimes called the ''quantum potential''. Note that in the limit $\hbar \to 0$, it disappears and we recover classical physics.

The connection to the Schroedinger equation? Simply define $\Psi(\boldsymbol x,t) = \sqrt{\rho} e^{i S/\hbar}$ and you can check that this obeys the Schroedinger equation. This formulation of quantum mechanics is thus also pretty useful for generating semi-classical approximations. In case you are curious what the special case $\rho(\boldsymbol x,t) = \delta(\boldsymbol x - \boldsymbol x(t))$ corresponds to in this set-up: they describe the paths of the ''hidden variables'' of the de Broglie-Bohm / pilot-wave representation of quantum mechanics.


Explicit example of Schrodinger equation "without" complex numbers

Just to give one completely explicit equation of the one particle case in position basis formulated only with real numbers (but two wave functions instead of one $\Psi_{real}$ and $\Psi_{img}$):

$$ \begin{align} - \frac{\partial \Psi_{img}(t, x, y, z)}{\partial t} &= -\frac{\partial^2 \Psi_{real}(t, x, y, z)}{\partial^2 x} -\frac{\partial^2 \Psi_{real}(t, x, y, z)}{\partial^2 y} -\frac{\partial^2 \Psi_{real}(t, x, y, z)}{\partial^2 z} \\ &+ V(t, x, y, z) \Psi_{real}(t, x, y, z) \end{align} $$

$$ \begin{align} \frac{\partial \Psi_{real}(t, x, y, z)}{\partial t} &= -\frac{\partial^2 \Psi_{img}(t, x, y, z)}{\partial^2 x} -\frac{\partial^2 \Psi_{img}(t, x, y, z)}{\partial^2 y} -\frac{\partial^2 \Psi_{img}(t, x, y, z)}{\partial^2 z} \\ &+V(t, x, y, z) \Psi_{img}(t, x, y, z) \\ \end{align} $$

where $\Psi_{real}$ and $\Psi_{img}$ are both real valued functions $\mathbb{R}^4 \to \mathbb{R}$ and represent of course the separate real and imaginary parts of the more standard equivalent Schrodinger equation:

$$ i \frac{\partial \Psi(t, x, y, z)}{\partial t} = -\nabla^2 \Psi(t, x, y, z) + V(t, x, y, z)\Psi(t, x, y, z) $$

Note that the equivalence holds because the potential $V$ must be real valued, otherwise conservation of probability is not observed.

While I don't have any super deep philosophical reasons to why the imaginary number appears (perhaps the high level intuitive "deduction" of the equation will provide the best clues?), the explicit real number form does make the following insight clearer to me:

  • the PDE that we are dealing with is actually a system of PDEs with two equations and two unknown functions

  • although the original equation looks like a heat equation due to the first order time derivative, we know that the Schrodinger equation exhibits wave-like oscilatory behaviour, which is more like the wave equation

    With the explicit real form, this becomes much more believable, because $\partial \Psi_{img}/\partial t$ kind of depends on $\Psi_{real}$, and in turn $\partial \Psi_{real}/\partial t$ kind of depends on $\partial \Psi_{img}$. So by "analogy" with reduction to a first order system in ODEs this looks like a second derivative effectively.

    If we for a moment forget about the Laplacian part and take a constant potential, we have a super classic linear first order system of ODEs which can have sin/cos/exp solutions.

Finally, if you were trying to solve the equation numerically, then you would likely go for the explicit real form, as there aren't really any intrinsically complex-valued operations to that need to be done. In some sense, the complex numbers of the Schrodinger equation can be fully split into two separate real/imaginary equations without problem, as there isn't anything hardcore such as complex differentiation to deal with.

  • $\begingroup$ $\Psi=a+bi$, so $i\left(\frac{\partial{a}}{\partial{t}}+i\frac{\partial{b}}{\partial{t}}\right)=-\left(\frac{\partial^2a}{\partial{x^2}}+i\frac{\partial^2b}{\partial{x^2}}\right)+V(a+bi)$, so $i\left(i\frac{\partial{b}}{\partial{t}}\right)=-\frac{\partial^2a}{\partial{x^2}}+Va$ and $i\frac{\partial{a}}{\partial{t}}=-i\frac{\partial^2b}{\partial{x^2}}+iVb$, so $-\frac{\partial{b}}{\partial{t}}=-\frac{\partial^2a}{\partial{x^2}}+Va$, and $\frac{\partial{a}}{\partial{t}}=-\frac{\partial^2b}{\partial{x^2}}+Vb$ $\endgroup$ – Anders Gustafson Sep 2 '20 at 0:59
  • 1
    $\begingroup$ I used equations for a particle in one dimension of space, but I found that I got a different answer from you, so which one of us is correct? $\endgroup$ – Anders Gustafson Sep 2 '20 at 1:00
  • $\begingroup$ @AndersGustafson thanks for pointing this out, the signs were wrong. Can you check that they match now? $\endgroup$ – Ciro Santilli 新疆再教育营六四事件法轮功郝海东 Sep 2 '20 at 7:38
  • $\begingroup$ Yes, they match. $\endgroup$ – Anders Gustafson Sep 8 '20 at 8:17

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.

  • $\begingroup$ You don't need polar form, just take the real and imaginary parts. $\endgroup$ – Ron Maimon Jul 20 '12 at 10:00
  • $\begingroup$ The most convincing I've heard so far is that since there are "waves" in QM, complex numbers formulation happen to be convenient and efficient. $\endgroup$ – Frank Jul 20 '12 at 15:56

Just to clarify, it sounds like you can.

Today, complex numbers are still taught in universities and still advocated by some. They linger on in physics and engineering where sinusoidal waves or motion are involved, though even here there is (nearly) always a published alternative approach that is free of imaginary numbers.

One famous area of physics where complex methods still have a virtual stranglehold is quantum mechanics. Although vector alternatives exist they are not promoted strongly at present and the dominant approach is to use imaginary numbers. Some even claim this is essential but this cannot be true. Hamilton showed, long ago, that a system of algebra with the same outward behaviour can be defined that lacks references to $𝑖$.


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