Why need *complex* wavefunction?

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  • #26
AEM
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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
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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
AEM
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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
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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
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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.
 
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  • #31
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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
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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
AEM
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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
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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
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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.
 
  • #36
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Thank you AEM for your reply, I appreciate it. And I will definitely check out the book you referred to sometime in the future when I have more time.
 
  • #37
The question is why is one single real number for example insufficient.
The real numbers are only a small portion of all numbers that exist. There is one single complex number which is a single point on the complex plane.

The real issue is a fault in the multiplication system.
 
  • #38
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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.
Yes, quantum logic provides a foundation for the entire formalism of quantum mechanics, including operators, vectors, scalar products etc. If you already know QM, you may find it very entertaining to see how all this machinery follows from a set of axioms, which are not much different from axioms of classical logic and probability. Actually, only one axiom of the standard Boolean logic needs to be modified - the distributive law.

Unfortunately, I don't know any good and comprehensive introductory level book or review on quantum logic. I got my knowledge from various places:

The original idea is more than 70 years old. This paper is well-written and highly recommended

G. Birkhoff, J. von Neumann, "The logic of quantum mechanics", Ann. Math., 37, (1936), 823

Possibly, the best elementary-level introduction can be found in Section 2-2 of

Mackey, G. W., The Mathematical foundations of Quantum Mechanics, W. A. Benjamin, Inc.
(1963), New York

Piron's book contains (the substantial part of) the proof of the theorem that I've mentioned. However, this book is a bit too heavy on the mathematical side.

Piron, C. Foundations of Quantum Physics, 1976, W. A. Benjamin Inc.
Reading, Massachusetts
 
  • #39
turin
Homework Helper
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Can someone present a first order (in time) wave-equation which is not of the schrödinger form and does not have complex solutions?
The Majorana equation. The catch is that the wave function is a 4-component object.
 
  • #40
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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!'
I don't want to speculate what other physicists love or hate. For me, the point is that this may be important for interpretation of quantum mechanics. For example, in the case of the de Broglie-Bohm interpretation, the electromagnetic field can play the role of the guiding field in the Klein-Gordon-Maxwell system (after you get rid of the complex wavefunction).
 
  • #41
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The Majorana equation. The catch is that the wave function is a 4-component object.
Hi turin, I appreciate your answer. I probably should have pointed out that it should also have to be a scalar wave-function. Anyway, it seems AEM has already called my bluff...

However, since you replied to memaybe I can ask you a question regarding the majorana equation? I am familiar with what we in the condensed matter community call majorana fermions, and I suppose that there should be a clear relationship.

In condensed matter community we can construct two (what we call) majorana (or real) fermions from a single complex one:

[tex]\chi_1=\psi+\psi^\dag, \quad \chi_2=i(\psi-\psi^\dag)[/tex]

with the properties [itex]\chi_i^\dag=\chi_i, \ i=1,2[/itex] (which together with anti-commutation relations of fermions yields interesting physics if one can separate spatially two majorana fermions from each other - has application to topological quantum computing)

The equation of motion (usually a complex one) for [itex]\psi[/itex] becomes a set of real coupled equations for [itex]\chi_i[/itex]. The analogy to what has been presented in this thread in terms of the Schrödinger equation should be obvious. This only replaces one complex equation to two real ones. In this sense majorana does not add anything useful to the discussion.


Going out on a limb on trying to argue the same for how high-energy physicists treat majorana fermions, I would start from the Wikipedia definition of Majorana equation:

[tex]i\partial \psi-m\psi_c=0[/tex]
(note: [itex]\partial[/itex] should be replaced with \partialslash which does not seem to work)
where [itex]\psi_c=\gamma^2\psi^*[/itex] i.e. it is related to [itex]\psi[/itex] through operation which includes complex conjugation. This equation does not seem to necessarily imply that its solutions must be real? Rather the special case where the field is (pseudo-)real [itex]\psi=\psi_c[/itex] is the case when its solutions are majorana fermions. (please correct me if this interpretation is wrong)

One could consider the above majorana equation together with its complex conjugate and form an equation for a spinor

[tex]\begin{pmatrix}\psi\\\psi_c\end{pmatrix}[/tex]

this spinor is also (pseudo-) real meaning that we can transform it into a basis where it is real. This whole procedure is analogous to producing from one complex spinor two real ones, or equivalently from one complex eqm to two real ones.

So it seems to me that either you postulate that [itex]\psi=\psi_c[/itex] or you use the procedure above to combine [itex]\psi[/itex] and [itex]\psi_c[/itex] into two real spinors, i.e. replacing one complex equation for [itex]\psi[/itex] into two real equations.

Am I making sense to you? My point is simply that I don't think majorana equation gives you any new insight into the problem than what has been posted in terms of replacing complex schrödinger equation with two real ones.
 
  • #42
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0
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."
While I have not read this article and unfortunately don't have time (sorry for being lazy) I would still like to comment on this. This procedure seems very much like what is used in Higgs mechanism [note that I am in no way an expert on this]. An important point is that when you choose a gauge in this way it by no means removes the degree of freedom associated with the phase. Rather it is absorbed into the the gauge field. If you originally have a complex scalar field with two degrees of freedom (magnitude and phase) and a gauge field [itex]A^\mu[/itex] which, due to gauge invariance, has (only) three degrees of freedom, you will, after fixing the gauge in a way such that the scalar field becomes real, have one degree of freedom associated with the scalar field and 4! degrees of freedom of the gauge field.

So do I understand your point was that we could have taken the real field + non-gauge invariant EM field as a starting point? thus eliminating the need for complex fields at the "cost" of losing gauge invariance? This does seem quite appealing since gauge invariance is another one of those things which one might ask oneself "why do we need it?"
 
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  • #43
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1) The Dirac equation can be made to be real as the gamma matrices can be made to be pure imaginary: http://en.wikipedia.org/wiki/Gamma_matrices#Majorana_basis

So if the differential equation has real coefficients, does that mean the solution does too?

2) The solution of the Majorana equation does not necessarily describe a Majorona particle. Only if solution is its own conjugate is the solution a Majorana particle. The Majorana equation is a perfectly good Lorentz-invariant equation: however, the Dirac equation is the one that matches experiment.

3) Imposing the condition that the field is self-conjugate is different than adding a field and its conjugate or subtracting a field and its conjugate (and multiplying by i). In the former case, you're solving for 1 Majorana particle, and in the latter case you're creating 2 Majorana particles from 1 Dirac particle.
 
  • #44
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Consider the path integral approach:

[tex] \int e^{iS}=<x',t'|x,t>[/tex]

Suppose at time t, the wavefunction [tex]\psi(x,t)[/tex] is real. Then at a later time t':

[tex] \psi(x',t')=\int e^{iS} \psi(x,t) [/tex]

So if the Lagrangian is real, then the action ought to be real, and the imaginary part comes form the i in the exponential. The i in the exponential is needed to keep the path integral from being infinity?
 
  • #45
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So do I understand your point was that we could have taken the real field + non-gauge invariant EM field as a starting point? thus eliminating the need for complex fields at the "cost" of losing gauge invariance?
Yes.
 
  • #46
turin
Homework Helper
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My point is simply that I don't think majorana equation gives you any new insight into the problem than what has been posted in terms of replacing complex schrödinger equation with two real ones.
Yes, I'm sorry. At the time I posted, I hadn't read all of the posts, so I was merely providing a first order equation and not worrying about how many components were needed. I don't have that book that AEM mentioned, but, in defense of the need for multiple components, I would say that, if you require the solution to the equation to be a scalar, and you require space to have more than 1 dimension, then you must introduce also some vector into the equation. So, while this can happen when the medium in which the wave propagates has a flow, I don't consider this a wave equation in the same sense of a matter-wave equation, in which the matter is supposed to be described as a wave that propagates through free space. So, basically, I must be misunderstanding the point.
 
  • #47
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Can someone present a first order (in time) wave-equation which is not of the schrödinger form and does not have complex solutions?
What's wrong with just the Schroedinger equation, replacing i with an operator that (rather than performing complex multiplication, and also different from the differential operators) factors any function into a Fourier basis then advances each component by a quarter wave? Doesn't this have solutions [itex]\phi(x,t)[/itex] that are real and single scalar valued?
 
  • #48
strangerep
Science Advisor
3,172
1,016
I just wanted to remind people that the power of complex numbers
(vs 2 real numbers) lies in the theory of complex-analytic (holomorphic)
functions, I.e., Cauchy-Riemann equations, and all that.

Holomorphic functions of a complex variable are not equivalent
(in general) to functions of 2 real variables.

This leads to immensely powerful theorems, used in advanced
physics quite a lot. (E.g., contour integration, Wick rotation,
path integrals, etc, etc.) The success of these tools suggests that
complex-analyticity does indeed have something important to do
with real world physics.
 
  • #49
970
3
Holomorphic functions are indeed powerful, but have to do with certain complex functions of complex variables, while the wavefunction is a complex function of a real variable.

Wick's rotation seems to be just a mathematical trick to turn Minkowski space into Euclidean space so that you can integrate over hyperspheres, but in principle the rotation is not needed if you know how to do the integral. The path integral for scalar or vector fields gets the [tex] i\epsilon[/tex] from extracting the vacuum state, so I think it's just another trick (you could in principle integrate over the vacuum wavefunction to get your path integral instead of the epsilon prescription).
 
  • #50
samalkhaiat
Science Advisor
Insights Author
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992
Does anyone know a deeper reason why the quantum mechanical wavefunction has to be complex?

In the non-relativistic QM, the complex nature of the wave function is as fundamental as the Plank constant itself.

1) If the dynamical equation EXPLICITLY contains the imaginary number i, as does Schroginger's equation, then a PAIR of REAL functions ( or their equivalent in the form of a SINGLE COMPLEX function) are needed for complete physical description.
If you insert [itex]\psi = f + ig[/itex] into Schrodinger's equation ([itex]2m=\hbar =1[/itex]):

[tex]i \partial_{t}\psi = - \nabla^{2}\psi[/tex]

you end up with 2 COUPLED equations

[tex]
\partial_{t}f + \nabla^{2}g = 0, \ \ \ \partial_{t}g - \nabla^{2}f =0
[/tex]

Thus we see that f and g are not INDEPENDENT of each other,i.e., neither of them alone is a solution of Schrodinger's equation.
Therefore, a complete solution requires the REAL pair (f,g) or their equivalent COMPLEX [itex]\psi[/itex].
Notice that the probability current

[tex]j_{i} = 2(f \partial_{i}g - g\partial_{i}f)[/tex]

is a mutual property of f and g, which vanishes when either is identically zero.

2) If the imaginary (i) does not show up explicitly in your equation, then the USE of COMPLEX function is just an auxiliary device.
Take the wave equation

[tex]\frac{\partial^{2}\phi}{\partial t^{2}} = \nabla^{2}\phi[/tex]

Now put [itex]\phi = f +ig[/itex]. This leads to the following two independent equations

[tex](\frac{\partial^{2}}{\partial t^{2}} - \nabla^{2})f = 0[/tex]

[tex](\frac{\partial^{2}}{\partial t^{2}} - \nabla^{2})g = 0[/tex]

Thus, f and g remain independent of each other,i.e., a complete solution can be given in terms of f alone or g alone. Therefore, in this case, the wave function [itex]\phi[/itex] need not necessarily be complex function.

3)You might now say this: But the first-order Schrodinger's equation can be replaced by the "equivalent" second-order equation

[tex]\frac{\partial^{2}\psi}{\partial t^{2}} + \nabla^{4}\psi = 0 \ \ \ (X)[/tex]

Therefore, according to (2), the wave function [itex]\psi[/itex] does not need to be a complex function! Does this mean that the complex nature of Schrodinger's wave function is not FUNDAMENTAL after all?
This quetions is, of course, meaningless, because it is based on a wrong conclusion. Let me explain this.
It is true (as explained in (2)) that [itex]\psi[/itex] in the second-order equation (X) does not need to be complex function. It is also true that the equation (X) is "equivalent" to the first-order Schrodinger's equation. IT IS, however, a WRONG equation to use in QM! One way to see this is to prove that NO CONSERVED PROBABILITY can be set up which depends on [itex]\psi[/itex] only and not on [itex]\partial_{t}\psi[/itex].
To do this, write

[tex]P = P(\psi)[/tex]

Then

[tex]\partial_{t} \int dx \ P(\psi) = \int dx \frac{\partial P}{\partial \psi} \frac{\partial \psi}{\partial t}[/tex]

If this expression is to vanish for arbitrary [itex]\psi[/itex], it is necessary that [itex]\partial_{t}\psi[/itex] be given in terms of [itex]\psi[/itex]. But this implies a FIRST-ORDER wave equation. In a SECOND-ORDER differential (like eq(X)), [itex]\partial_{t}\psi[/itex] can be given an arbitrary initial value. Therefore, the above expression cannot vanish for all [itex]\psi[/itex]. For example, if we put

[tex]\frac{\partial \psi}{\partial t} = \frac{\partial P}{\partial \psi}[/tex]

then

[tex]\frac{\partial}{\partial t} \int dx \ P(\psi) = \int dx \ (\partial P / \partial \psi )^{2} > 0[/tex]

So, we see that EQ(X) can not be PHYSICALLY equivalent to Schrodinger's equation.


regards

sam
 

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