Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an explanation that I need some help with: that the de Broglie wavelength and the wavelength of an elastic wave do not show similar properties under a Galilean transformation. He basically says that both are equivalent under a gauge transform and also, separately by Lorentz transforms. This, accompanied with the observation that $\psi$ is not observable, so there is no "reason for it being real". Can someone give me an intuitive prelude by what is a gauge transform and why does it give the same result as a Lorentz tranformation in a non-relativistic setting? And eventually how in this "grand scheme" the complex nature of the wave function becomes evident.. in a way that a dummy like me can understand.


A wavefunction can be thought of as a scalar field (has a scalar value in every point ($r,t$) given by $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{C}$ and also as a ray in Hilbert space (a vector). How are these two perspectives the same (this is possibly something elementary that I am missing out, or getting confused by definitions and terminology, if that is the case I am desperate for help ;)


One way I have thought about the above question is that the wave function can be equivalently written in $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{R}^2 $ i.e, Since a wave function is complex, the Schroedinger equation could in principle be written equivalently as coupled differential equations in two real functions which staisfy the Cauchy-Riemann conditions. ie, if $$\psi(x,t) = u(x,t) + i v(x,t)$$ and $u_x=v_t$ ; $u_t = -v_x$ and we get $$\hbar \partial_t u = -\frac{\hbar^2}{2m} \partial_x^2v + V v$$ $$\hbar \partial_t v = \frac{\hbar^2}{2m} \partial_x^2u - V u$$ (..in 1-D) If this is correct what are the interpretations of the $u,v$.. and why isn't it useful. (I am assuming that physical problems always have an analytic $\psi(r,t)$).

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    $\begingroup$ Hi Yayu. I've always found interesting a paper by Leon Cohen, "Rules of Probability in Quantum Mechanics", Foundations of Physics 18, 983(1988), which approaches this question somewhat sideways, through characteristic functions. Cohen comes from a signal processing background, where Fourier transforms are very often a natural thing to do. Fourier transforms and complex numbers are of course pretty much joined at the hip. $\endgroup$ Commented Apr 5, 2011 at 18:08
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    $\begingroup$ Here are a few straightforward observations that might be helpful. (1) You can describe standing waves with real-valued wavefunctions, e.g., one can almost always get away with this in low-energy nuclear structure physics. (2) The w.f. of a photon is simply the electric and magnetic fields. These are observable and real-valued. (3) If the electron w.f. was real and observable, the wavelength would have to be invariant under a Galilean boost, which would violate the de Broglie relation. (4) Even for real-valued waves, operators are complex, e.g., momentum in the classically forbidden region. $\endgroup$
    – user4552
    Commented May 30, 2013 at 23:08
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    $\begingroup$ @yayu A complex analytic function is a function from complex numbers to complex numbers. And the Cauchy-Riemann equations are about such functions. To pick on x and t as if the t axis were an imaginary axis and the x axis were a real axis and y and z didn't exist is very confusing. $\endgroup$
    – Timaeus
    Commented Jan 2, 2015 at 4:33
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    $\begingroup$ Related: physics.stackexchange.com/q/32422 $\endgroup$
    – valerio
    Commented Oct 22, 2016 at 8:01
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    $\begingroup$ I am a bit concerned about the fact that many physicists think that the wave-function has to be complex because of arguments which inherently already assume that we are entering complex realms (as seen in the answers). This might be a huge obstacle on the way to an intuitive interpretation of the fundamental laws of nature. Of course the wave function is not inherently complex. Complex numbers (as many constructs in math) are just an elegant way to write down things. Complex numbers are a notational tool to wrap polar coordinate systems into "numbers" which we are more familiar with. $\endgroup$
    – M. Winter
    Commented Dec 5, 2017 at 9:12

15 Answers 15


More physically than a lot of the other answers here (a lot of which amount to "the formalism of quantum mechanics has complex numbers, so quantum mechanics should have complex numbers), you can account for the complex nature of the wave function by writing it as $\Psi (x) = |\Psi (x)|e^{i \phi (x)}$, where $i\phi$ is a complex phase factor. It turns out that this phase factor is not directly measurable, but has many measurable consequences, such as the double slit experiment and the Aharonov-Bohm effect.

Why are complex numbers essential for explaining these things? Because you need a representation that both doesn't induce nonphysical time and space dependencies in the magnitude of $|\Psi (x)|^{2}$ (like multiplying by real phases would), AND that DOES allow for interference effects like those cited above. The most natural way of doing this is to multiply the wave amplitude by a complex phase.

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    $\begingroup$ But what are the differences between the sound waves and the wavefunction? Why the second must be complex, while the first also may interfere? And we may write the our wavefunction through sines and cosines, so the value $\psi^{T}\psi$ also refers to the invariant in this case. $\endgroup$ Commented Mar 27, 2014 at 5:08
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    $\begingroup$ @AndrewMcAddams: the difference is that the amplitude of a sound wave is an observable, while only the amplitude of the modulus squared is an observable in quantum mechanics. I can see the phase of a water wave, but I can only see the phase of an electron wave through interference effects. $\endgroup$ Commented Mar 27, 2014 at 13:15
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    $\begingroup$ This is the most concise and easily understood explanation I've ever read concerning 'why'. Thank you. So many textbooks on quantum mechanics fail to communicate this fact. $\endgroup$
    – docscience
    Commented Nov 9, 2014 at 15:03
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    $\begingroup$ But @Jerry Schirmer, you say using complex phase is the most natural way to model quantum behavior, is it the ONLY way? $\endgroup$
    – docscience
    Commented Nov 9, 2014 at 15:07
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    $\begingroup$ @docscience: of course not -- you don't even need complex numbers to do the math of complex numbers, after all. It's just a nice, easy way to do them. And people have tried to reformulate quantum mechanics using quarternions, but I don't know how far they've really gotten, that's outside of my field of expertise. $\endgroup$ Commented Nov 9, 2014 at 23:14

This year-old question popped up unexpectedly when I signed in, and it's an interesting one. So I guess it's OK just to add an intuition-level "addendum answer" to the excellent and far more complete responses provided long ago.

Your kernel question seems to be this: "Why is the wave function complex?"

My intentionally informal answer is this:

Because by experimental observation, the quantum behavior of a particle far more closely resembles that of a rotating rope (e.g. a skip rope) than it does a rope that only moves up and down.

If each point in a rope marks out a circle as it moves, then a very natural and economical way to represent each point along the length of the rope is as a complex magnitude. You certainly don't have to do it that way, of course. In fact, using polar coordinates would probably be a bit more straightforward.

However, the nifty thing about complex numbers is that they provide a simple and computationally efficient way to represent just such a polar coordinate system. You can get into the gory details mathematical details of why, but suffice it to say that when early physicists started using complex numbers for just that purpose, their benefits continued even as the problems became far more complex. In quantum mechanics, their benefits became so overwhelming that complex numbers started being accepted pretty much as the "reality" of how to represent such mathematics.

That conceptual merging of complex quantities with actual physics can throw off your intuitions a bit. For example, if you look at moving skip rope there is no distinction between the "real" and "imaginary" axes in the actual rotations of each point in the rope. The same is true for quantum representations: It's the phase and amplitude that counts, with other distinctions between the axes of the phase plane being a result of how you use those phases within more complicated mathematical constructions.

So, if quantum wave functions behaved only like ropes moving up and down along a single axis, we'd use real functions to represent them. But they don't. Since they instead are more like those skip ropes, it's a lot easier to represent each point along the rope with two values, one "real" and one "imaginary" (and neither in real XYZ space) for its value.

Finally, why do I claim that a single quantum particle has a wave function that resembles that of a skip rope in motion? The classic example is the particle-in-a-box problem, where a single particle bounces back-and-forth between the two X axis ends of the box. Such a particle forms one, two, three, or more regions (or anti-nodes) in which the particle is more likely to be found.

If you borrow Y and Z (perpendicular to the length of the box) to represent the real and imaginary amplitudes of the particle wave function at each point along X, it's interesting to see what you get. It looks exactly like a skip-rope in action, one in which the regions where the electron is most likely to be found correspond one-for-one to the one, two, three, or more loops of the moving skip rope. (Fancy skip-ropers know all about higher numbers of loops.)

The analogy doesn't stop there. The volume enclosed by all the loops, normalized to 1, tells you exactly what the odds are on finding the electron along any one section along the box in the X axis. Tunneling is represented by the electron appearing on both sides of the unmoving nodes of the rope, those nodes being regions where there is no chance of finding the electron. The continuity of the rope from point to point captures a rough approximation of the differential equations that assign high energy costs to sharp bends in the rope. The absolute rotation speed of the rope represents the total mass-energy of the electron, or at least can be used that way.

Finally, and a bit more complicated, you can break those simple loops down into other wave components by using the Fourier transform. Any simple look can also be viewed as two helical waves (like whipping a hose around to free it) going in opposite directions. These two components represent the idea that a single-loop wave function actually includes helical representations of the same electron going in opposite directions, at the same time. "At the same time" is highly characteristic of quantum function in general, since such functions always contain multiple "versions" of the location and motions of the single particle that they represent. That is really what a wave function is, in fact: A summation of the simple waves that represent every likely location and momentum situation that the particle could be in.

Full quantum mechanics is far more complex than that, of course. You must work in three spatial dimensions, for one thing, and you have to deal with composite probabilities of many particles interacting. That drives you into the use of more abstract concepts such as Hilbert spaces.

But with regards to the question of "why complex instead of real?", the simple example of the similarity of quantum functions to rotating ropes still holds: All of these more complicated cases are complex because, at their heart, every point within them behaves as though it is rotating in an abstract space, in a way that keeps it synchronized with points in immediately neighboring points in space.

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    $\begingroup$ I'm not sure whether the OP is aware of this, but it emphasises your comment "it doesn't have to be this way". Real matrices of the form $\left(\begin{array}{cc}a&-b\\b&a\end{array}\right) = I a + i b$ where now $I$ is the $2\times2$ identity and $i= \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$ form a field wholly isomophic to $\mathbb{C}$. In particular, a phase delay corresponds to multiplication by the rotation matrix $\exp\left(-i\,\omega\,t\right)=\left(\begin{array}{cc}\cos\omega t&-\sin \omega t\\ \sin\omega t&\cos\omega t\end{array}\right) = I \cos\omega t + i\sin\omega t$. $\endgroup$ Commented Jul 29, 2013 at 0:47
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    $\begingroup$ Rod, yes. A similar trick can be done for quaternions. I'm actually a quaternion bigot: I like to think of many of the complex numbers used in physics as really being overly generalized quaternions, ones in which our built-in 3D bias keeps us from noticing that the imaginary axis of a complex number is actually just a quaternion unit pointer in XYZ space. You lose a lot of representation richness by doing that, since for example you inadvertently abandon the intriguing option of treating changes in the quaternion-view i orientation as a local symmetry of XYZ space. $\endgroup$ Commented Jul 31, 2013 at 22:36
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    $\begingroup$ Although I guess from the OPs point of view, it would be wrong to call it a trick - there are many ways to encode the kinds of properties complex numbers do and this one IS complex numbers (an isomorphic field). As for quaternions, yes, it's a shame that Hamilton, Clifford and Maxwell never held sway over Heaviside. $\endgroup$ Commented Aug 2, 2013 at 0:05
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    $\begingroup$ @Terry Bollinger Feynman would be proud of your answer $\endgroup$ Commented May 18, 2019 at 19:15
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    $\begingroup$ I hope I'm not violating Stack Exchange etiquette by saying thank you for your kind remark. I always regret that I never had a chance to meet Richard Feynman in person. He was unique. $\endgroup$ Commented May 18, 2019 at 19:31

Alternative discussion by Scott Aaronson: http://www.scottaaronson.com/democritus/lec9.html

  1. From the probability interpretation postulate, we conclude that the time evolution operator $\hat{U}(t)$ must be unitary in order to keep the total probability to be 1 all the time. Note that the wavefunction is not necessarily complex yet.

  2. From the website: "Why did God go with the complex numbers and not the real numbers? Answer: Well, if you want every unitary operation to have a square root, then you have to go to the complex numbers... " $\hat{U}(t)$ must be complex if we still want a continuous transformation. This implies a complex wavefunction.

Hence the operator should be: $\hat{U}(t) = e^{i\hat{K}t}$ for hermitian $\hat{K}$ in order to preserve the norm of the wavefunction.

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    $\begingroup$ Personally I prefer Jerry Schirmer's answer because it requires less postulate and instead uses experimental fact directly. =) $\endgroup$
    – pcr
    Commented Apr 8, 2011 at 4:56
  • $\begingroup$ I very much like your answer, as much as Jerry's. But I would add two things: firstly, the square root thing is a bit obtuse: I would put it as follows for those like me who are a bit slow on the uptake: ....(ctd)... $\endgroup$ Commented Aug 5, 2013 at 4:44
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    $\begingroup$ "All eigenvalues of unitary operators have unit magnitude. So the only nontrivial unitary operator with all real eigenvalues is one with a mixture of +1s and -1s as eigenvalues- say $M$ -otherwise it is the identity operator $I$. Since $U(t)$ and its eigenvalues vary continuously, $U(t)$ cannot reach $M$ from its beginning value $U(0)=I$ unless at least one eigenvalue goes through all values on the unit semicircle to reach the value -1". ...(ctd)... $\endgroup$ Commented Aug 5, 2013 at 4:44
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    $\begingroup$ Secondly, the argument won't quite fly as is: there are nontrivial, real matrix valued unitary groups $\mathbf{SO}(N)$ (whose members have complex eigenvalues but nonetheless are real matrices) that will realise the $U(t)=\exp(i\,K\,t)$ in your argument, so quantum states can still be all real wavefunctions if they are real at $t=0$. I don't quite have a fix for this, maybe you could appeal to an experiment. It is a pretty argument, though, so I'll keep thinking. $\endgroup$ Commented Aug 5, 2013 at 4:46

Among other things, the OP reprinted a page of a textbook, asking what "it is all about". I think it is impossible to answer this kind of questions because what is the OP's problem all about is totally undetermined, and the people who offer their answers could be writing their own textbooks, with no results.

The wave function in quantum mechanics has to be complex because the operators satisfy things like $$ [x,p] = xp-px = i\hbar.$$ It's the commutator defining the uncertainty principle. Because the left hand side is anti-Hermitian, $$ (xp-px)^\dagger = p^\dagger x^\dagger - x^\dagger p^\dagger = (px-xp) = -(xp-px),$$ it follows that if it is a $c$-number, its eigenvalues have to be pure imaginary. It follows that either $x$ or $p$ or both have to have some non-real matrix elements.

Also, Schrödinger's equation $$i\hbar\,\, {\rm d/d}t |\psi\rangle = H |\psi\rangle$$ has a factor of $i$ in it. The equivalent $i$ appears in Heisenberg's equations for the operators and in the $\exp(iS/\hbar)$ integrand of Feynman's path integral. So the amplitudes inevitably have to come out as complex numbers. That's also related to the fact that eigenstates of energy and momenta etc. have the dependence on space or time etc. $$\exp(Et/i\hbar)$$ which is complex. A cosine wouldn't be enough because a cosine is an even function (and the sine is an odd function) so it couldn't distringuish the sign of the energy. Of course, the appearance of $i$ in the phase is related to the commutator at the beginning of this answer. See also

Why complex numbers are fundamental in physics

Concerning the second question, in physics jargon, we choose to emphasize that a wave function is not a scalar field. A wave function is not an observable at all while a field is. Classically, the fields evolve deterministically and can be measured by one measurement - but the wave function cannot be measured. Quantum fields are operators - but the wave function is not. Moreover, the mathematical similarity of a wave function to a scalar field in 3+1 dimensions only holds for the description of one spinless particle, not for more complicated systems.

Concerning the last question, it is not useful to decompose complex numbers into real and imaginary parts exactly because "a complex number" is one number and not two numbers. In particular, if we multiply a wave function by a complex phase $\exp(i\phi)$, which is only possible if we allow the wave functions to be complex and we use the multiplication of complex numbers, physics doesn't change at all. It's the whole point of complex numbers that we deal with them as with a single entity.

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    $\begingroup$ thanks for answering. I have one question, not knowing about Feynman path integrals yet, I take it that what you are saying is the same thing as: if we make the transformation $\psi(r,t) = e^{i\frac{S(r,t)}{\hbar}}$ then the Schrodinger equation reduces to the classical hamilton Jacobi equations (if terms containing $i$ and $\hbar$ were negligible)? $\endgroup$
    – yayu
    Commented Apr 5, 2011 at 5:35
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    $\begingroup$ Dear yayu, thanks for your question. First, the appearance of $\exp(iS/\hbar)$ in Feynman's approach is not a transformation of variables: the exponential is an integrand that appears in an integral used to calculate any transition amplitude. Second, $\psi$ is complex and $S$ is real, so $\psi=\exp(iS/\hbar)$ cannot be a "change of variables". You may write $\psi=\sqrt{\rho}\exp(i S/\hbar)$, in which case Schrödinger's equation may be (unnaturally) rewritten as two real equations, a continuity equation for $\rho$ and the Hamilton-Jacobi equation for $S$ with some extra quantum corrections. $\endgroup$ Commented Apr 5, 2011 at 5:39
  • $\begingroup$ I edited my question removing the reprints and trying to state my problem without them.. it will take some time to think about some points you made in the answer already, though. $\endgroup$
    – yayu
    Commented Apr 5, 2011 at 6:00
  • $\begingroup$ I think a better explanation would not use the idea of operator formalism, since when Schrödinger came up with his equation, the formalism wasn't developed yet. $\endgroup$
    – Sidd
    Commented Oct 24, 2015 at 23:57
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    $\begingroup$ Sorry but Schrödinger only came with his "wave mechanics" almost a year after quantum mechanics was discovered by Heisenberg and pals in the form of the "matrix mechanics". Despite popular misconceptions, Schrödinger isn't even one of the founders of quantum mechanics and he never correctly understood the meaning of the theory. $\endgroup$ Commented Oct 25, 2015 at 7:09

If the wave function were real, performing a Fourier transform in time will lead to pairs of positive-negative energy eigenstates. Negative energies with no lower bounds is incompatible with stability. So, complex wave functions are needed for stability.

No, the wave function is not a field. It only looks like it for a single particle, but for N particles, it is a function in 3N+1 dimensional configuration space.


This question has been asked since Dirac

In fact Dirac's answer is available for $ 100 from JSTOR in a paper by Dirac from I think 1935 ?

A recent answer from James Wheeler - is that the zero-signature Killing metric of a new, real-valued, 8-dimensional gauging of the conformal group accounts for the complex character of quantum mechanics

Reference is Why Quantum Mechanics is Complex , James T. Wheeler ArXiv:hep-th9708088

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    $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$
    – Gonenc
    Commented Sep 30, 2015 at 16:33

EDIT add:
My Answer is GA centric and after the comments I felt the need to say some words about the beauty of Geometric Algebra:
On 2nd page of Oersted Medal Lecture (link bellow):

(3) GA Reduces “grad, div, curl and all that” to a single vector derivative that, among other things, combines the standard set of four Maxwell equations into a single equation and provides new methods to solve it.

Geometry Algebra (GA) encompasses in a single framework for all this:
Synthetic Geometry, Coordinate Geometry, Complex Variables, Quaternions, Vector Analysis, Matrix Algebra, Spinors, Tensors, Differential forms. It is one language for all physics.
Probably Schrödinger, Dirac, Pauli, etc ... would have used GA if it existed at the time.
To the Question: WHY is the wave function complex? This Answer is not helpful: because the wave function is complex (or has a i on it). We have to try something different, not written in your book.
In the abstracts I bolded the evidence that the papers are about the WHYs. If someone begs a fish I'll try to give a fishing rod.
I'm an old IT analyst who would be unemployed if I had not evolved. Physics is evolving too.
end EDIT

Recently I've found the Geometric Algebra, Grassman, Clifford, and David Hestenes.

I will not detail here the subject of the OP because each one of us need to follow paths, find new ideas and take time to read. I will only provide some paths with part of the abstracts:

Overview of Geometric Algebra in Physics

Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics (a good start)

In this lecture Hestenes is arguing for a reform of the way in which mathematics is taught to physicists. He asserts that using Geometric Algebra will make it easier to understand the fundamentals of physics, because the mathematical language will be clearer and more uniform.

Hunting for Snarks in Quantum Mechanics

Abstract. A long-standing debate over the interpretation of quantum mechanics has centered on the meaning of Schroedinger’s wave function ψ for an electron. Broadly speaking, there are two major opposing schools. On the one side, the Copenhagen school (led by Bohr, Heisenberg and Pauli) holds that ψ provides a complete description of a single electron state; hence the probability interpretation of ψψ* expresses an irreducible uncertainty in electron behavior that is intrinsic in nature. On the other side, the realist school (led by Einstein, de Broglie, Bohm and Jaynes) holds that ψ represents a statistical ensemble of possible electron states; hence it is an incomplete description of a single electron state. I contend that the debaters have overlooked crucial facts about the electron revealed by Dirac theory. In particular, analysis of electron zitterbewegung (first noticed by Schroedinger) opens a window to particle substructure in quantum mechanics that explains the physical significance of the complex phase factor in ψ. This led to a testable model for particle substructure with surprising support by recent experimental evidence. If the explanation is upheld by further research, it will resolve the debate in favor of the realist school. I give details. The perils of research on the foundations of quantum mechanics have been foreseen by Lewis Carroll in The Hunting of the Snark!


Abstract. A reformulation of the Dirac theory reveals that i¯h has a geometric meaning relating it to electron spin. This provides the basis for a coherent physical interpretation of the Dirac and Sch¨odinger theories wherein the complex phase factor exp(−iϕ/¯h) in the wave function describes electron zitterbewegung, a localized, circular motion generating the electron spin and magnetic moment. Zitterbewegung interactions also generate resonances which may explain quantization, diffraction, and the Pauli principle.

Universal Geometric Calculus a course, and follow:
III. Implications for Quantum Mechanics

The Kinematic Origin of Complex Wave Functions
Clifford Algebra and the Interpretation of Quantum Mechanics
The Zitterbewegung Interpretation of Quantum Mechanics
Quantum Mechanics from Self-Interaction
Zitterbewegung in Radiative Processes
On Decoupling Probability from Kinematics in Quantum Mechanics
Zitterbewegung Modeling
Space-Time Structure of Weak and Electromagnetic Interactions

to keep more references together:
Geometric Algebra and its Application to Mathematical Physics (Chris Thesis)

(what lead me to this amazing path was a paper by Joy Christian 'Disproof of Bell Theorem')
'Bon voyage', 'good journey', 'boa viagem'

  • $\begingroup$ Why the Down votes? $\endgroup$ Commented Apr 5, 2011 at 15:03
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    $\begingroup$ @yayu; No, I shall upvote it because I've read the papers linked and know they are exactly appropriate for the question. I can give a short description of the argument: As soon as you use spin-1/2 you have that $\sigma_x\sigma_y\sigma_z = i$ is an imaginary unit in that it squares to -1 and commutes with the other elements. This is also inherent to the Dirac equation. What Hestenes goes on to show is that the "i" in Schroedinger's equation arises from looking at the Pauli equation with a fixed spin. $\endgroup$ Commented Apr 6, 2011 at 5:01
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    $\begingroup$ @yayu; Answers on Stack Exchange are read by more than just the person who asks. Spin-1/2 (and the Pauli spin matrices) will be covered in any introduction to QM; it's the simplest non trivial Hilbert space possible. It doesn't get much simpler than that. But in general, even if it were suitable only for Albert Einstein it has to be posted here. On SE, duplicate questions are closed. This is the only opportunity to answer the question for all readers. $\endgroup$ Commented Apr 6, 2011 at 5:31
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    $\begingroup$ @yayu spinless particle? It seems that there is no need to be complex. KINEMATIC link above or the comment by Carl previous to your 'spinless' comment. $\endgroup$ Commented Apr 6, 2011 at 15:41
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    $\begingroup$ Removed the downvote by adding a +1. I found this post informative and detailed and I don't really care if it precisely answered the original question or not. $\endgroup$ Commented Dec 6, 2012 at 1:15

From the Heisenberg Uncertainty Principle, if we know a great deal about the momentum of a particle we can know very little about its position. This suggests that our mathematics should have a quantum state that corresponds to a plane wave $\psi(x)$ with a precisely known momentum but entirely unknown position.

A natural definition for the probability of finding the particle at the position $x$ is $|\psi(x)|^2$. This definition makes sense for both a real wave function and an imaginary wave function.

For a plane wave to have no position information is to imply that $|\psi(x)|$ does not depend on position and so is constant. Therefore we must have $\psi$ complex; otherwise there would be no way to store the information "what is the momentum of the particle".

So in my view, the complex nature of wave functions arises from the interaction between the necessity for (1) a probability interpretation, (2) the Heisenberg uncertainty principle, and (3) plane waves.

  • $\begingroup$ Please clear some doubts for me. 1. The probability interpretation: I think it followed since the wavefunction was complex and physical meaning could only attributed to a real value. If we make a construction $\psi^*\psi$ then we arrive at the continuity equation from the schrodinger equation and the interpretation can now be made that the quantity $\rho=\psi^*\psi$ is the probability density. Starting from an interpretation like $\rho=\psi^*\psi$, I do not see any way to work backwards and convincingly argue that the amplitude $\psi$ must be complex. $\endgroup$
    – yayu
    Commented Apr 6, 2011 at 18:09
  • $\begingroup$ the uncertainty relations follow from the identification of the free particle as a plane wave. I am guessing your answer points in the right direction, I am working on (2) as suggested in Lubos' answer as well and trying to get why $\psi$ is complex valued as a consequence, however I fail to see how anything except (2) is relevant for showing it conclusively. $\endgroup$
    – yayu
    Commented Apr 6, 2011 at 18:16
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    $\begingroup$ @yayu: see my post--there are two essential experimental facts: 1) phase is not directly measurable; 2) interference effects happen in a broad range of quantum materials. It's hard to reconcile these things without using complex numbers. $\endgroup$ Commented Apr 7, 2011 at 3:39
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    $\begingroup$ While I agree with the core thoughts in this answer, I don't agree with the conclusion that this requires complex numbers. There is nothing lost, for example, expressing the (complex) Fourier transform as two real sine/cosine Fourier transforms. This does not require complex numbers, although they may be convenient. $\endgroup$
    – anon01
    Commented Feb 16, 2017 at 2:17
  • $\begingroup$ @ConfusinglyCuriousTheThird Hi! Does my modest contribution (currently the 3rd one below this one), expanding a little on Carl's answer, provide a reply you can accept? Rgds – iSeeker $\endgroup$
    – iSeeker
    Commented Jan 17, 2018 at 8:36

THIS LATE ANSWER (Jan 2018) expands a little on Carl Brannen’s straightforward, and (IMO) underappreciated, answer (showing a little way above, at time of posting), which reminded me of another simple and convincing argument as to why the wave-function should be complex, set out many years ago in Dicke & Wittke's “Introduction to Quantum Mechanics” (1960; pp. 23-24).

Given their review in Ch 1 of why a quantum mechanical wave is subject to wave-particle duality/the De Broglie relation, they proceed as follows:

For a wave-particle of sharply-defined momentum:

λ = h/p (and thus, by Δxp>= h/4π, completely uncertain – essentially uniform – position)

…the probability distribution |ψ|^2 of a plane wave should be uniform in position, which cannot be satisfied by a real-valued plane wave

ψ = A sin(kx - ωt + α)

…but is satisfied (generalising to an arbitrary position) by

ψ = A exp [i (k.x- ωt)], where the propagation vector k = p/(h/2π).

Dicke & Wittke then discuss how the complex-valued wave function accounts for interference effects (Out of copyright and safely available via https://archive.org/details/IntroductionToQuantumMechanics).

[NB Care googling the book title/pdf - many online sources, unlike the one above, are unsafe]


The question is a good one and has been asked also by Ehrenfest (1932): "Einige die Quantenmechanik betreffende Erkundigungsfragen". The answer was given by Pauli (1933): "Einige die Quantenmechanik betreffenden Erkundigungsfragen". Unfortunately I'm not aware of a english translation of these two publications. However one can find a slightly different form of the answer also in Pauli's book "General Principles of Quantum Mechanics" p.16. In that book Pauli writes

a single real function is not sufficient in order to construct from wavefunctions of the form (3.1) a non-negative probability function that is constant in time when integrated over the whole space.

I will try to summarize his arguments here:

A wave packet to describe a single particle (basically deBroglie's idea) can be written generally like $$ u(x,t) = \int U(k) e^{i(kx-\omega t)} dk = \int U(k) e^{ikx} dk \, e^{-i\omega t} $$ where $U(k)$ is the Fourier transform of $u(x,0)$. The complex conjugate of this wave packet is $$ u^*(x,t) = \int U^*(k) e^{-i(kx-\omega t)} dk = \int U^*(k) e^{-ikx} dk \, e^{i\omega t} $$ One can also define such wave packets in electrodynamics. But in quantum mechanics we have an additional condition, namely that the probability $P(x,t)$ to find a particle must always be positive and the total probability to find a single particle somewhere must be one, so $$ P(x,t) \geq 0 \\ \int P(x,t)\, dx = 1 $$ Pauli argues that the simplest ansatz to construct such a function $P(x,t)$ from $u(x,t)$ is a definite quadratic form from the functions $u$ and $u^*$, that means $$ P(x,t) = a u^2 + b {u^*}^2 + c u u^* $$ Now from the form of $u(x,t)$ and $u^*(x,t)$ we see that $$ u^2 \sim e^{-2i\omega t}\ \text{and}\ {u^*}^2 \sim e^{2i\omega t} $$ and an integral over space over these two functions can never be time independent. So the constants $a$ and $b$ must be zero in the ansatz for $P(x,t)$. Only the product of a wave packet and it's complex conjugate will yield a time independent total probability: $$ 1 = \int P(x,t)\, dx = \int uu^*\, dx = \iiint U(k)U^*(k') e^{i(kx-k'x)} \, e^{-i\omega t} e^{i\omega t} dk dk' dx \\ = \iint U(k)U^*(k') \delta(k-k') dk dk' = \int \left|{U(k)}\right|^2 dk = \int P(k)\, dk $$ Since the product $uu^* = Re[u]^2 + Im[u]^2$ it follows that - as Pauli said - in order to compute a meaningful probability from wave packets of the form $u(x,t)$ one needs the real and the imaginary part of $u(x,t)$ and the wave function in quantum mechanics must be complex.


Already great answers to this often asked question. Very simply put tho, quantum eigenstates have associated relative phase values, and the in-phase and quadrature plane (also strangely but conventionally referred to as the Real/Imaginary plane) provides for specifying these phases when plotting or otherwise specifying a wave function.

As pointed out in other answers, there are other mathematical ways to do this, but the “complex” way is mathematically very convenient.


Since the physical point of view, the wave function needs to be complex in order to explain the double-slit experiment, as well mentionated in the book of The Feynman Lectures on Physics-III, I suggest you that review chapters 1&3, where it is explained how $\psi$ has to be considered of probabilistic nature, according to the pattern of interference, because "something" has to behave like a wave at the time of crossing through "each one" of the slits. Furthermore, Bohm proclaims that path of the particle (electron,photon, etc.) can be considered classic, so as a consequence you may watch this one, as it follows the rules already known at the macro... in that sense, you can see next reference or this one to consider the covariance of the laws of mechanics.


The wavefunction $\psi(x)$ is the projection of the physical system's state vector $|\psi\rangle$ onto the $\hat{x}$ eigenket $|x\rangle$ of eigenvalue $x$, viz. $|\psi\rangle=\int dx\psi(x)|x\rangle$. You mustn't confuse the scalar-value $\psi(x)=\langle x|\psi\rangle$ with the vector $|\psi\rangle$ that lives in a Hilbert space.

The first sentence of your first question is, in technical terms: why is this Hilbert space over the field $\mathbb{C}$ rather than, say, $\mathbb{R}$? If you Ctrl+F to "Real vs. Complex Numbers" here, you'll get a detailed discussion of several motivations for why quantum mechanics ought to look like that. One advantage of a complex wavefunction is it has both an amplitude and a phase, but only the former affects the probability density $|\psi|^2$, and the latter gives us quantum interference because of trigonometric identities such as $|1+\exp i\theta|=2|\cos\frac{\theta}{2}|$. However (to continue your Q1), a Galilean transformation needs to include a phase shift so the Schrödinger equation will be invariant; see here and here for more information. A gauge transformation such as the Galilean one is simply a way of transforming coordinates, or fields (which come to the same thing in a Lagrangian field theory), which leave the action and its equations of motion invariant. (By the way, you need to be careful not to confuse the words transform and transformation.)

Your Q2 also hinges on not confusing $\psi$ with $|\psi\rangle$. The ray is the set of values of $\exp i\theta|\psi\rangle$ with $\theta\in\mathbb{R}$, but switching from one value of $\theta$ to another leaves $\psi$ invariant because $\langle x|$ gets multiplied by $(\exp i\theta)^\ast=\exp -i\theta$.

As for Q3, it's more useful to work with the modulus and phase of the complex $\psi$ rather than its real and imaginary part, because under unitary transformations the former is invariant and so are differences in the latter.


I like to think about the complex nature of the wave function by comparing traveling and standing waves in both the classical/quantum settings. Suppose we try to characterize the quantum state of a free particle moving in one dimension by a real wave function, such that $|\psi(x,t)|^2$ gives the probability density of finding the particle in position $x$ at time $t$. So far, no problem, as we can describe each monochromatic component of the wave packet that describes the particle as both real and complex, and total integral $\int_{-\infty}^{\infty}|\psi(x,t)|^2dx$ will be conserved and equal to $1$.

However, we will encounter problems using real-valued wave functions in the quantum setting of a particle in a box (which can also be thought as a superposition of two monochromatic waves traveling in opposite directions). Suppose we write the real-valued wave function of the particle at moment $t=0$ in the same form of a standing wave in a string:

$$\psi(x,t) = A\mathbb{sin}(kx)\mathbb{cos}(\omega t)$$

One can easily see that except at times $t = \frac{n\pi}{\omega}$, the value of the integral $\int_{0}^{L}|\psi(x,t)|^2dx$ will not be conserved and equal to $1$, which means that real-valued wave functions can't represent a probability density function for the particle.

In the classical setting of a vibrating string one would say that the total energy of the string (potential+kinetic) is conserved, despite that kinetic energy is converted to potential energy and vice versa during the oscillation. In the quantum setting of a particle in a box we speak of probabilities (not total energy), and the only way to go around this obstacle of non-conserved total probability is to use complex-valued wave functions.


Since both the amplitude and the wavelength cannot be known with precision simultaneously, I think of this as meaning that there is some missing information that must still be dealt with continuously. That information is conveniently stored in the imaginary part of a complex number.

  • 4
    $\begingroup$ This is not nearly substantiated enough to be an answer, and besides, I'm quite sure that's not a good way to think about that. $\endgroup$
    – ACuriousMind
    Commented Sep 10, 2014 at 0:17

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