Why are QM wave functions complex?

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735
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I don't think that prescription preserves expectation values, in general. Consider:
[tex]
\langle \psi | \hat{p} |\psi \rangle =\int dx \; \langle \psi | x \rangle \langle x | \hat{p} |\psi \rangle = \int dx \; \psi^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi(x)
[/tex]
then under the transformation
[tex]
\rightarrow \int dx \; \psi'^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi'(x) = \int dx \; \psi^{*}(x)e^{-i\theta(x)} \left( -i \frac{\partial}{\partial x} \right) e^{i \theta(x)}\psi(x) \neq \int dx \; \psi^{*}(x) \left( -i \frac{\partial}{\partial x} \right) \psi(x)
[/tex]
How is that accounted for?
Mind you, the wave function transform is just one part of the prescription. The other part is the transform of the electromagnetic four-potential. The latter enters into the generalized momentum. Remember, a gauge transform produces a physically equivalent solution. Do you really dispute this statement?
 
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What about the spin wavefunction for a spin-1/2 particle?
I thought I answered your question, but I cannot find my answer, so please forgive me if the following is a duplicate.

Quite surprisingly, the same procedure is valid for a Dirac spinor function. While you cannot make a Dirac spinor real by a gauge transform, you can algebraically eliminate three out of four components of the Dirac spinor function from the Dirac equation. The remaining component can be made real by a gauge transform. Please see the reference in my post 4 in this thread.
 
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Mind you, the wave function transform is just one part of the prescription. The other part is the transform of the electromagnetic four-potential. The latter enters into the generalized momentum. Remember, a gauge transform produces a physically equivalent solution. Do you really dispute this statement?
I certainly don't dispute that. What I dispute is that you have just referenced electrodynamics, and I am talking about normal non-relativistic QM, (there need not be any fields!). Under your prescription for transformation of the wave-function, expectation values are not preserved. Again, how do you account for this?
 
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I certainly don't dispute that. What I dispute is that you have just referenced electrodynamics, and I am talking about normal non-relativistic QM, (there need not be any fields!). Under your prescription for transformation of the wave-function, expectation values are not preserved. Again, how do you account for this?
You are certainly free to talk about anything you want, and I am certainly free to talk about anything I want (within the forum rules). I carefully defined what I was going to talk about in my post 4 in this thread. It is my understanding that relativistic QM fully accounts for non-relativistic QM. What can be described without any fields, can also be described using fields, moreover, the latter approach is more fundamental. If you're asking about non-electromagnetic forces, I have little to say about them at the moment. Let me just note that the non-electromagnetic forces (beyond gravity, and I have nothing to say about gravity) can be described by the Standard Model, which also has a gauge-invariant Lagrangian. So maybe what I am saying is valid for the Standard Model as well, maybe not, I just don't know (let me just note that it took 60 years after Schroedinger's 1952 article to find out that his approach is also valid for the Dirac equation. I don't know how much time it'll take to prove or disprove the same for the Standard Model). I am just saying (following Schroedinger) that charged particles do not necessarily require complex representation.
 

vanhees71

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This is definitely not true. Despite the fact that in relativistic quantum theory there is no known consistent physically interpretable scheme for that theory using wave functions in the same sense as the non-relativistic Schrödinger-wave function. The only (very!) successful formulation of relativistic quantum theory is local quantum field theory, because relativistic QT necessarily is a many-body theory with a non-fixed number of particles.

Further, within relativistic QT, particles carrying charge are described by non-hermitean operators in the most convenient way. Within canonical quantization this means to start with complex rather than real fields. The most simple example are spin-0 (scalar) fields, obeying the Klein-Gordon equation. For the complex free field there is one symmetry, namely the multiplication of a phase factor, leading to a conserved charge. For a real scalar field, there is no such charge. The real scalar field thus describes strictly neutral particles, i.e., such particles that are their own antiparticles.
 
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AA(bar) = 1 = 1 + eU + eU(bar) which implies U(bar) = -U so that U is pure imaginary which of course means the underlying vector space must be complex.
In order to make U hermitian a factor of i is inserted i.e. A=1+ieU,so it implies
AA+=1+ieU-ieU+=1,which implies U+=U
 
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In order to make U hermitian a factor of i is inserted i.e. A=1+ieU,so it implies
AA+=1+ieU-ieU+=1,which implies U+=U
U is a unitary operator so its not in general Hermitian. I think its Stones Theorem or something like that that says if U is unitary then A = e^iU is Hemitian so defines an observable - its used a lot in defining energy, momentum and other operators. And you are correct - your form is indeed Hermitian being the infinitesimal form of e^iU.

On second thought I think that part of my post is not that convincing - DrDu was correct it doesn't really imply a complex vector space. I like the argument I gave later based on Wigners Theorem better.

Thanks
Bill
 
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I thought I answered your question, but I cannot find my answer, so please forgive me if the following is a duplicate.

Quite surprisingly, the same procedure is valid for a Dirac spinor function. While you cannot make a Dirac spinor real by a gauge transform, you can algebraically eliminate three out of four components of the Dirac spinor function from the Dirac equation. The remaining component can be made real by a gauge transform. Please see the reference in my post 4 in this thread.
That's an interesting observation.

However, in the spirit of the original post, i was thinking of the non-relativistic "bog standard QM" formalism in which a single-particle wavefunction just consists of the product of a spatial part and a spin part (or a sum of such products). All i was pointing out in that case was that for a spin-1/2 particle the spin part has to be allowed to be complex because once the basis is fixed (Sz basis, say), there are spin states that require complex coefficients (eigenstates of Sy, say). I think this is a good enough motivation for the introduction of complex numbers to single-particle QM, regardless of what QFT has to say on the matter.

Also, even if the Dirac equation can be rewritten as a higher order equation in a single real component, if you then solve this equation and want to write down the field itself, you're going to presumably need complex numbers for the other components, so i'm not sure how much bearing this algebraic elimination has on the question at hand, but i might have misunderstood your point there.
 
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That's an interesting observation.

However, in the spirit of the original post, i was thinking of the non-relativistic "bog standard QM" formalism in which a single-particle wavefunction just consists of the product of a spatial part and a spin part (or a sum of such products). All i was pointing out in that case was that for a spin-1/2 particle the spin part has to be allowed to be complex because once the basis is fixed (Sz basis, say), there are spin states that require complex coefficients (eigenstates of Sy, say). I think this is a good enough motivation for the introduction of complex numbers to single-particle QM, regardless of what QFT has to say on the matter.
Let me note that all phenomena described by single-particle nonrelativistic QM with spin 1/2 can be correctly described (sometimes much better) by the Dirac equation.

Also, even if the Dirac equation can be rewritten as a higher order equation in a single real component, if you then solve this equation and want to write down the field itself, you're going to presumably need complex numbers for the other components, so i'm not sure how much bearing this algebraic elimination has on the question at hand, but i might have misunderstood your point there.
It is not quite clear why I would need "the field itself": I will know the real component and will be able to calculate the current or some average values of observables based just on this component. The component, the current, and the averages of observables are real. What else do I need to describe all the experiments that can be described by the Dirac equation? I do appreciate that complex numbers provide very important simplifications, but the OP asked: why wave functions must be complex? Mind you, "must", not "should".
 
735
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This is definitely not true.
Let me start with the following trivial remark. When you say that something "is definitely not true", it may be advisable to quote that "something", so that readers of your post do not have to guess what you had in mind. I guess that you aimed at "my" phrase "charged particles do not necessarily require complex representation." If my guess is wrong, please let me know.

I would also like to remark that this is not my phrase, but Schroedinger's (the precise quote is as follows: "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.") Of course, that does not necessarily mean that this phrase is correct.

vanhees71 said:
Despite the fact that in relativistic quantum theory there is no known consistent physically interpretable scheme for that theory using wave functions in the same sense as the non-relativistic Schrödinger-wave function. The only (very!) successful formulation of relativistic quantum theory is local quantum field theory, because relativistic QT necessarily is a many-body theory with a non-fixed number of particles.
First, I think the Dirac equation is also a very successful formulation of relativistic quantum theory, although it is approximate. Second, so far I don't see any proof that charged particles necessarily require complex representation.

vanhees71 said:
Further, within relativistic QT, particles carrying charge are described by non-hermitean operators in the most convenient way.
"Convenient" does not mean "necessary".

vanhees71 said:
Within canonical quantization this means to start with complex rather than real fields. The most simple example are spin-0 (scalar) fields, obeying the Klein-Gordon equation. For the complex free field there is one symmetry, namely the multiplication of a phase factor, leading to a conserved charge.
In Nature, there is no such thing as charged free field: as soon as you have charge, you have electromagnetic field.

vanhees71 said:
For a real scalar field, there is no such charge. The real scalar field thus describes strictly neutral particles, i.e., such particles that are their own antiparticles.
I respectfully disagree. If you think so, why don't you show me where exactly Schroedinger screwed up in his 1952 article? He showed that the Klein-Gordon-Maxwell equations describing a scalar charged field interacting with electromagnetic field can be re-written in terms of a real field interacting with electromagnetic field, and this real field definitely can have nonzero charge.
 
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It is not quite clear why I would need "the field itself"
As i understand it, you haven't shown that the other field components are zero, you've just eliminated them from the Dirac equation - they will still contribute to the current etc (i'd have thought... can't see why not).

A couple of other things: firstly, your paper appears to be on classical field theory... can any conclusions drawn from that paper necessarily be carried over to the quantum theory? Secondly, when vanhees71 says you need a complex field to have charged quanta, he means the field must have the the freedom to be complex, even if it happens to have been made real by a gauge transformation. Although it doesn't appear to realised in Nature, there's nothing inconsistent about a free field theory with charged particles, and the definition of charge as the Noether charge of a global phase symmetry is still good - no gauge fields needed.
 

vanhees71

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In answer to #35 (so that's clear what I'm referring too). Of course, you can translate everything formulated in terms of complex fields to coupled equations of its real and imaginary parts, giving real equations. That's not very convenient, but if you insist, you can do it.

Further, it's not clear to me in which context Schrödinger made this remark. I'm sure that he is right for his example, but it would be good to have a reference to check, what he was referring to, before discussing this issue.
 
735
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As i understand it, you haven't shown that the other field components are zero, you've just eliminated them from the Dirac equation - they will still contribute to the current etc (i'd have thought... can't see why not).
Of course, the other field components are not zero, but you can just forget about them: the theory can be completely rewritten in terms of just one real function (the former component of the Dirac spinor). We may declare that this "lonely" real function is in fact our wave function, and this new "wave function" is everything we need to describe all phenomena described by the Dirac equation.

psmt said:
A couple of other things: firstly, your paper appears to be on classical field theory... can any conclusions drawn from that paper necessarily be carried over to the quantum theory?
Yes. Please see the following articles: http://akhmeteli.org/akh-prepr-ws-ijqi2.pdf (published in the International Journal of Quantum Information; this paper is discussed in the following thread: https://www.physicsforums.com/showpost.php?p=3413745&postcount=741) and http://arxiv.org/pdf/1108.1588.pdf (accepted for publication in the European Physical Journal C, except for the last two paragraphs in Conclusion, which were written later). If you are interested, it's better to start with the second article, as it largely supersedes and improves the former. However, the part of the second article related to spinor electrodynamics does use complex numbers, but I am not sure it is strictly necessary.

psmt said:
Secondly, when vanhees71 says you need a complex field to have charged quanta, he means the field must have the the freedom to be complex, even if it happens to have been made real by a gauge transformation. Although it doesn't appear to realised in Nature, there's nothing inconsistent about a free field theory with charged particles, and the definition of charge as the Noether charge of a global phase symmetry is still good - no gauge fields needed.
Again, this does not mean that wave function must be complex.
 
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In answer to #35 (so that's clear what I'm referring too).
Thank you :-)

vanhees71 said:
Of course, you can translate everything formulated in terms of complex fields to coupled equations of its real and imaginary parts, giving real equations.
This is quite obvious, but this is not what I had in mind. There is just one real wave function after the gauge transform.

Further, it's not clear to me in which context Schrödinger made this remark. I'm sure that he is right for his example, but it would be good to have a reference to check, what he was referring to, before discussing this issue.
I gave the reference in post 4 in this thread.
 

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