Why need *complex* wavefunction?

In summary: I don't know, people who don't already understand complex numbers?In summary, complex numbers are just a convenient way of representing geometric information, and there's no fundamental reason why the wavefunction has to be complex. It could be represented without complex numbers, if someone wanted to do so. The fact that the wavefunction has two components is due to the fact that it needs to incorporate both the magnitude and the phase of the wave.
  • #1
feynmann
156
1
Does anyone know a deeper reason why the quantum mechanical wavefunction has to be complex? Is it to incorperate time dependence?

Or maybe the operator/eigenvector formulation is special and since it includes the scalar product, having complex variables is more general and necessary?

Or maybe the fact that there is some spin with its transformation means that wavefunction should be complex?
 
Physics news on Phys.org
  • #2
Complex numbers merely provide convenient notation for some geometric stuff. It is possible to formulate QM without complex number too, if one, for some mysterious reason, would want to do so. For example, replace a wave function [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{C}[/itex] and the Schrödinger's equation

[tex]
i\hbar\partial_t\Psi(t,x) = \Big(-\frac{\hbar^2}{2m}\nabla^2 + V(x)\Big)\Psi(t,x)
[/tex]

with a wave function [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{R}^2[/itex] and a Schrödinger's equation

[tex]
\hbar\partial_t\left(\begin{array}{c}
\Psi_1(t,x) \\ \Psi_2(t,x)
\end{array}\right)
=\left(\begin{array}{cc}
0 & -\frac{\hbar^2}{2m}\nabla^2 + V(x) \\
\frac{\hbar^2}{2m}\nabla^2 - V(x) & 0 \\
\end{array}\right)
\left(\begin{array}{c}
\Psi_1(t,x) \\ \Psi_2(t,x)
\end{array}\right)
[/tex]

and so on.
 
  • #3
jostpuur said:
Complex numbers merely provide convenient notation for some geometric stuff. It is possible to formulate QM without complex number too, if one, for some mysterious reason, would want to do so. For example, replace a wave function [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{C}[/itex] and the Schrödinger's equation

[tex]
i\hbar\partial_t\Psi(t,x) = \Big(-\frac{\hbar^2}{2m}\nabla^2 + V(x)\Big)\Psi(t,x)
[/tex]

with a wave function [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{R}^2[/itex] and a Schrödinger's equation

[tex]
\hbar\partial_t\left(\begin{array}{c}
\Psi_1(t,x) \\ \Psi_2(t,x)
\end{array}\right)
=\left(\begin{array}{cc}
0 & -\frac{\hbar^2}{2m}\nabla^2 + V(x) \\
\frac{\hbar^2}{2m}\nabla^2 - V(x) & 0 \\
\end{array}\right)
\left(\begin{array}{c}
\Psi_1(t,x) \\ \Psi_2(t,x)
\end{array}\right)
[/tex]

and so on.

It's well known that a complex number is equivalent to a set of two real number.
But why wavefunction has to have two components, not three or four components?
What's so unique about vectors/wavefunctions with two components?
 
Last edited:
  • #4
I think it's 2 components because it needs to incorporate 2 things: magnitude and phase. Experiments reveal the need for both of these, but do not (yet) reveal the need for any additional information.
 
  • #5
Sometimes people attach unnecessarily mysterious interpretations to the complex numbers, so I wanted to debunk that part right away. Merely to make sure that unclarity would not remain hanging around the complex numbers.

feynmann said:
But why wavefunction has to have two components, not three or four components?
What's so unique about vectors with two components?

Ok, this question goes onto different level. Isn't this kind of stuff that first we observe nature, and then find out what mathematical constructs seem to describe it? Sometimes some relations are solved between seemingly independent constructs, but there are always some fundamental questions, which don't get answered. Good luck if you are interested in getting philosophical with this topic. I wouldn't spend too much time in it.

At least non-relativistic electrons should be described with wave function [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{C}^2[/itex] if the spin is relevant. So it has four real components. It follows that in many cases [itex]\Psi:\mathbb{R}\times\mathbb{R}^3\to\mathbb{C}[/itex] is an approximation which works only when spin is not relevant. IMO it is a good philosophy to think that all mathematical constructs we use may later turn out to be approximations of something else. I'm probably not alone with this opinion?
 
  • #6
One way to think about it is that the wavefunction assigns both probability and phase to each eigenstate, and that phase is critical for explaining interference effects. For any complex number x + iy, you can get a real number (the square of its absolute value or 'complex modulus) by multiplying by the complex conjugate x - iy, and in QM this is how you get the probability from the complex amplitude associated with each eigenstate. But note that for any point in the complex plane, if you draw a circle around the origin with the correct radius so that this point lies on the circle, then every point on that circle gives the same number when multiplied by its complex conjugate; so, knowing the complex amplitude associated with an eigenstate as opposed to just the real probability associated with it gives you an additional piece of information about the angle you have to go from the origin to get to that number, and this angle is essentially the phase associated with that eigenstate.

I wonder if it would be possible to actually rewrite the equations of QM so they wouldn't use complex numbers at all, and instead the wavefunction would just assign a real-valued probability and a real-valued phase angle to each eigenstate at every moment...probably it would make the equations less elegant but maybe it'd be helpful from the point of view of teaching QM to students...
 
  • #7
Even in the simplest quantum mechanical system, a spin 1/2 particle whose spin can either be only up or down when measured, we can see the need for complex numbers.

A general rotation in 3 dimensions can be accomplished by acting on vectors (x,y,z) by multiplying them with a 3x3 matrix R. If you described the rotation with 3 angles, or by specifying an axis and a magnitude of rotation about that axis, then the matrix R for that rotation would be full of sines and cosines of these angles. It turns out that rotation matrices are always symmetric and have determinant 1. The set of all 3x3 symmetric matrices with real entries having determinant 1 is called SO(3).

For a 2 state spin system, such as a spin 1/2 particle, part of our task is to find a 2x2 matrix representation of SO(3), which is only possible with complex numbers, and this is because SO(3) is an isomorphic group to SU(2), where SU(2) is the set of all conjugate symmetric 2x2 matrices with complex entries having determinant 1.
 
  • #8
I think there might be a deep mathematical answer in *-algebras, maybe the mathematicians will tell us in a couple of years ;)
 
  • #9
I'm inclined to believe that complex numbers are not required fundamentally to describe spinless non-relativistic wavefunctions. As mentioned before if you have spin and stuff they rotate from frame to frame with complex numbers (similarly in the relativistic case, Dirac particles are complex Grassman variables: however, this isn't fundamental, as Majorana particles are just real Grassman variables). However, with spinless non-relativistic wavefunctions, if you can make the Hamiltonian pure imaginary (which requires that the matrix also be anti-symmetric for Hermicity) with a change of basis, then your wavefunction can be always real, all the time. So certain types of Hamiltonians allow the wavefunction to be real I think.
 
  • #10
Yes, but complex numbers are simpler (in algebra) then real ones, so there is nothing wrong with having some physical variables complex.
 
  • #11
I think one argument is based on the fact that the differential equation governing time evolution needs to be a wave-equation and it needs to be a first order diff. eq. in t.

It needs to be wave-equation because experiments suggest wave-like properties. It needs to be first order in t because the state should be completely determined by the wavefunction. Thus the only information (initial condition) that should be required to find out the wave-function at some time t=t_f should be the wave-function at time t=0.

A real wave equation (with real solutions) would have to be second order and then we would need not only the value of the wave function at t=0 but also it's derivative at t=0. Then the wave-function cannot completely describe the state of the system.
 
  • #12
The Bohmian approach has what is perhaps a better answer to this question. If we write the wave function as [tex]\Psi(x,t)=A(x,t)e^{iS(x,t)/\hbar}[/tex], then the Schrodinger equation separates into two equations in terms of A and S.

One equation, in the limit of vanishing Planck's constant, corresponds to the classical Hamilton-Jacobi equation of a point particle. The other, if we interpret A^2 as the pdf, corresponds to the conservation of probability.

If we had a classical problem expressed in Hamiton-Jacobi form in which the initial position of the particle was uncertain and expressed in terms of a pdf, then we would have a problem expressed in analogous quantities S and A^2.

For quantum systems a non-zero Planck's constant complicates their coupling further.
 
  • #13
RedX said:
I'm inclined to believe that complex numbers are not required fundamentally to describe spinless non-relativistic wavefunctions.
Same here. It seems very similar to the basic wave approximation of light, where only one component (amplitude) is necessary but complex numbers are used instead (effectively doubling the equations and with the second half being thrown away at the end of any calculation) because this lends itself to more concise notation.

jensa said:
A real wave equation (with real solutions) would have to be second order
Why?
 
  • #14
cesiumfrog said:
Why?

I wish I could back up that statement in rigorous way (perhaps it is not even generally true), I am more or less repeating an argument I heard back when I took QM.
As far as I can tell classical wave-equations are usually second order. And the "real valued version" (wick rotated) of the Schrödinger equation is a diffusion equation. Can someone present a first order (in time) wave-equation which is not of the schrödinger form and does not have complex solutions?

My main point was that the naive answer to the OPs question would be that wavefunctions must be complex because the solutions to the Schrödinger equation are complex. So I am suggesting the question should be rephrased as, why does the Schrödinger equation look the way it does? I guess there are a few different ways to argue that it has to have the form it has, but I am not sure they are irrefutable.

EDIT: removed (obvious) false statements
 
Last edited:
  • #15
I realized that my previous post may need further clarification to tie it to the OP's question.

The point is that classically you would need at least two quantities for a quantum-like situation in which we describe the motion of a particle whose position is uncertain: one that governs the dynamical evolution of a particle given its initial value, and another quantity that contains the probability distribution of that initial value. The dynamics will then tell you the probability distribution of the particle's postion at later times.

The same necessity for two quantities carries over to the quantum realm.

It might take a lengthy post to satisfy one why this is this indeed the case but I think this much is sufficient to answer the initial question "Why two quantities?"
 
  • #16
The Schrödinger equation could be used to describe both bound systems, like a particle in a box, and open systems where electrons travels like matter-waves through a structure. It can be shown (no magnetic field) that the current of electron (probability current density) becomes zero if the wavefunction is purely real (or mulriplied by an arbitrary complex phase factor). For bound systems real wave functions are sufficient, but not for open wavelike solutions. Phase-difference (as described by complex number) allows for electrons to "jump" between different places.
 
  • #17
0xDEADBEEF said:
I think there might be a deep mathematical answer in *-algebras, maybe the mathematicians will tell us in a couple of years ;)

*-algebras are used because the wavefunction is complex. So consequently you are working in a vector space with an involution. The wavefunction does not incorporate spin so that's not the reason why it is complex (and it doesn't need to be complex, you can have a real valued wavefunction). I think the reason for them being complex is because a wave is best described in a complex field.
 
  • #19
The wave equation need not be complex. There exists two coupled real functions that will solve the Schrodinger equation. The complex function is simply easier notation.
 
  • #20
DeepThought42 said:
The wave equation need not be complex. There exists two coupled real functions that will solve the Schrodinger equation. The complex function is simply easier notation.

But then the question is why are there *two* coupled real function. The question is why is the equation of this structure.
 
  • #21
Gerenuk said:
But then the question is why are there *two* coupled real function. The question is why is the equation of this structure.

It is a product of a mathematical system that did not originally incorporate all numbers, that is, the square root of negative numbers.

Once they were accepted we had to incorporate them into our current system through the use of the complex numbers a + ib.

The side effect, equations that require two coupled real functions which is the same as a single complex function.
 
  • #22
feynmann said:
Does anyone know a deeper reason why the quantum mechanical wavefunction has to be complex?

There is an approach, called "quantum logic", which tries to deduce the formalism of quantum mechanics (Hilbert space, operators, etc.) from a set of axioms. These axioms, basically, generalize the axioms of classical logic and probability. In 1960's it has been proven that axioms of quantum logic can be satisfied by only three kinds of Hilbert spaces: those based on real numbers (R, 1-dimensional), complex numbers (C, 2-dimensional), and quaternions (Q, 4-dimensional). R-quantum mechanics is not interesting (I think, it contradicts experiment, but I am not sure how exactly). For single particles Q-quantum mechanics looks very similar to C-quantum mechanics. However, there are difficulties in describing multiparticle systems due to the non-commutativity of quaternions. Some people continue working on this approach, but I haven't heard about any interesting results there. So, our best choice is C-quantum mechanics.

If you are interested in references, try to google names C. Piron and J. M. Jauch.
 
  • #23
A real wave equation with real solutions does not have to be second order. The first chapter of G.B. Whitham's book Linear and Nonlinear Waves is devoted to first order wave equations that give rise to shock waves.

Some of you may be interested in Geometric Algebra for Physicists by Doran and Lasenby. Geometric algebra is a Clifford Algebra. In Chapter 8 the authors show how to write the Dirac equation "in the real algebra of spacetime with no need for a unit imaginary." ( p. 283, Their italics).There are a number of papers on the applications of Geometric Algebra by the above authors and their collaborators available at http://www.mrao.cam.ac.uk/~clifford/
Geometric algebra appears to have some nice features. Briefly, very very briefly, it unites the scalar product and the cross product into one operation called the geometric product. This combining of the two operations in a consistent way has, among other things, the interesting consequence that the Pauli matrices form a matrix representation of the geometric algebra of space.

Based on the development in the book I cited above and work I have seen, I would have to say that there no a priori reason why quantum mechanics has to be written in terms of complex wave functions. Historically, part of the reason is that in the 19th century when there was some contention between whether to use Clifford algebras, quaternions, or vector algebras, J. W. Gibbs' vector algebra (vector analysis) won out.
 
Last edited:
  • #24
AEM said:
A real wave equation with real solutions does not have to be second order. The first chapter of G.B. Whitham's book Linear and Nonlinear Waves is devoted to first order wave equations that give rise to shock waves.

I stand corrected, thanks AEM.

Based on the development in the book I cited above and work I have seen, I would have to say that there no a priori reason why quantum mechanics has to be written in terms of complex wave functions.

I suppose you are right. I only wonder if this is not also a type of, what some might call, "cheating", i.e. replacing one complex equation with two real ones. It seems hard to imagine that one could incorporate for example QED if there is no internal structure of the field. Granted one may be able to replace the U(1) gauge group with the isomorphic group SO(2) thereby eliminating the need for complex fields but then instead we have to deal with two-component vectors, which again seems a bit like "cheating".

On the other hand I know gauge theories have a geometric interpretation in terms of fiber-bundles so I guess it is not unreasonable that it can be naturally incorporated in terms of what you describe. Probably I am just not mathematically sophisticated enough for this stuff :)

[Disclaimer: I am quite novice in field theory areas so please correct me if I am making incorrect statements]

EDIT: Probably not a good idea to bring up EM coupling since it assumes the particles have charge ... I guess the issue under discussion is disconnected from the issue of U(1) gauge theory, am I right?
 
Last edited:
  • #25
meopemuk said:
There is an approach, called "quantum logic", which tries to deduce the formalism of quantum mechanics (Hilbert space, operators, etc.) from a set of axioms...
If you are interested in references, try to google names C. Piron and J. M. Jauch.

Great suggestion! That sounds what I'm looking for even though I suppose it will be too theoretical for me to understand (for the moment).
 
  • #26
AEM said:
Geometric algebra appears to have some nice features. Briefly, very very briefly, it unites the scalar product and the cross product into one operation called the geometric product.

When I wrote the above, I was under time pressure and what I wrote is misleading at best. What I should have written was: Briefly, very very briefly, it unites the scalar product and a cross product-like operation called the wedge product into one operation called the geometric product.

There are significant differences between the cross product and geometric algebra's wedge product. For one, the cross product cannot be defined in spaces of more than three dimensions while the wedge product can be. Further, the wedge product allows the definition of geometric entities whose square is -1 thus allowing one to write equations without using the imaginary unit.

I
 
  • #27
AEM said:
Geometric algebra appears to have some nice features. Briefly, very very briefly, it unites the scalar product and the cross product into one operation called the geometric product.

I had the feeling that Quaternions do the same?
 
  • #28
Gerenuk said:
I had the feeling that Quaternions do the same?

Because Geometric Algebra and Quaternions are Clifford Algebras, they have a lot in common. However, there are some subtle differences. See page 34 of the book by Doran and Lasenby for the details. Briefly, Hamilton attempted to identify pure quaternions (no scalar part) with vectors. Doran and Lasenby show that they actually are a left-handed set of bivectors (objects constructed from vectors and the wedge product). Both Geometric Algebra and Quaternions are particularly efficient in expressing rotations. This is why they are often used in computer graphics. One might anticipate that they would be useful in quantum mechanics as well.
 
  • #29
I am afraid I have repeated this several times in other threads, but this question does arise frequently. Not many people know about Shroedinger's very short article (Nature (1952), v.169, p.538), where he shows that quantum mechanics can actually do without complex numbers or, equivalently, two real numbers for the wave function, as, e.g., for any solution of the equations of the Klein-Gordon-Maxwell electrodynamics (a scalar charged field \psi interacting with electromagnetic field) there exists a physically equivalent solution with one real (not complex) field, which can be obtained from the original solution by a gauge transform (see some fine print in thread https://www.physicsforums.com/showthread.php?t=98603). Thus, the entire range of physical phenomena described by the Klein-Gordon-Maxwell electrodynamics may be described using real fields only. Shroedinger's comment: "That the wave function ... can be made real by a change of gauge is but a truism, though it contradicts the widespread belief about 'charged' fields requiring complex representation."
 
  • #30
Wait,wait, *EVERYTHING* you can do with complex numbers you can do using pairs of real numbers. What is a point? Is it about the 'complex numbers can not have any physical meaning because they are complex, and all physical observables are real'? nonsense.

For mathematician complex numbers are much easier then writing pairs of real numbers. Real numbers are mentally impared. They are not alrebraicly closed.
 
Last edited:
  • #31
Dmitry67 said:
Wait,wait, *EVERYTHING* you can do with complex numbers you can do using pairs of real numbers. What is a point? Is it about the 'complex numbers can not have any physical meaning because they are complex, and all physical observables are real'? nonsense.

For mathematician complex numbers are much easier then writing pairs of real numbers. Real numbers are mentally impared. They are not alrebraicly closed.

If you are replying to me, I certainly appreciate that you can always replace a complex number with two real ones. Schroedinger, however, demonstrated that you can replace the complex wavefunction of the Klein-Gordon-Maxwell system with just one real function (moreover, you can easily eliminate the wavefunction altogether, and the resulting equations will describe independent evolution of the electromagnetic field). The price you have to pay for that - you cannot choose the gauge arbitrarily anymore. Whether it's worth it or not (or in which situation that may be useful), is a matter of opinion. My point is just that you can do with one real wavefunction in some situations.
 
  • #32
Yes, but what is a real :) point?
I do believe the underlying (but hidden) subject of this topic is a mysterious hatred of physicists versus the complex numbers, like, 'use see, we can do it without these bad complex numbers, which do not have any physical meaning!'
 
  • #33
jensa said:
I stand corrected, thanks AEM.



I suppose you are right. I only wonder if this is not also a type of, what some might call, "cheating", i.e. replacing one complex equation with two real ones.

No, this formulation of the Dirac equation does not replace one equation with two real ones. The thing that is different about Geometric Algebra is that there are geometric objects that square to minus 1 for each space ( [tex] R^n [/tex] )you are working in. I personally do not know about any application of Geometric Algebra to QED.
 
  • #34
Do not say, that complex numbers can be written with two real functions, because then you misunderstood the question.

Complex numbers have equivalent representations with two real numbers or even 2x2 matrices. That's all the same. The question is why is one single real number for example insufficient.

There were some really good answers and I hope a some point I'll have enough knowledge to understand their principle.
 
  • #35
meopemuk said:
There is an approach, called "quantum logic", which tries to deduce the formalism of quantum mechanics (Hilbert space, operators, etc.) from a set of axioms.

I will look up that theory at some point.

OK, which dimension to use for the numbers can be justified there. But does it also justify why operators on vectors and scalar products for probabilities should be used? Sorry, if this question doesn't make sense. I'm good at QM at a "physicists level" only.
 

Similar threads

  • Quantum Physics
Replies
11
Views
2K
Replies
2
Views
1K
Replies
6
Views
1K
Replies
9
Views
1K
  • Quantum Physics
Replies
15
Views
13K
Replies
18
Views
2K
  • Quantum Physics
Replies
17
Views
1K
Replies
8
Views
1K
Back
Top