When we transpose particles in the non-relativistic quantum mechanics we write down T \psi(t, x_1, x_2) = a \psi(t, x_2, x_1).
Of course, we cannot write this down in another frame, since there is only one time here, but the time in different spatial points would be different in a new frame. Each of the particles would have it's own time.
Since Einsten is one of the authors of the Einstein-Podolski-Rosen article, he must have thought about this. Evidently, his thoughts remained unknown to the most like me.
I think this is best to do in a second quantized formalism, i.e. relativistic quantum field theory. Then your natural objects are functions of two spacetime variables (which contain information of equal-time "wave-functions" in any given frame). In this formalism you get Bose or Fermi statistics for free. I have no idea how Einstein-Podolski-Rosen thought about it.
In my view the greatest problem is that you must be able to do time evolution (exactly) in order to do Lorentz transformations (exactly). This is possible for noninteracting theories. There one can define a Lorentz invariant integration measure in momentum space. This corresponds to a nonlocal measure in position space, so the non-relativistic results only emerges when one constructs states out of momenta p where |p|
Luiz> The symmetrization or anti symmetrization must be put by hand in the final scattered wave functions
I would not say that the (anti-)symmetrization must be put in by hand --- if you do all the algebra of the S-matrix expansion from scratch. But nobody human would want to do that, so one jumps directly to the Feynman diagrams. And they must be added up with the correct relative signs, according to the statistics of the quantum fields.
And it is not enough to take into account the correct symmetrization of the initial and final state wave functions: Two diagrams which are related by the interchange of a fermion in the final state with an anti-fermion in the initial state must be added with a relative minus sign; just do, or look up, the calculation of e+ e- -> e+ e- for an example. It is really the statistics of the quantum fields which matter, not only the symmetrization of states.
This is discussed here: http://ejde.math.txstate.edu/conf-proc/04/w1/wightman.pdf
I'm afraid, a pointlike particle on the light cone makes no sense: eventually a delta function type of Weyl particle? This has now spin, but never statistics.
I see the way how the spin mechanism makes sense: the spin could be a global cyclic property of the vacuum actualising quanta locally.
Though spin do not require any mechanism as a quantity, interactions in vacuum do require a mechanism and spin could be due to that. My point is that vacuum could be a siqnificant factor for the spin, for example by cyclic coherences.
Charles, the spin-statistics theorem refers to observable properties (pairwise exclusion rules such as forbidding L+S odd in the CM frame) not to (ambiguously defined) operators of field theory. The use of the word "statistics" refers to the behavior of large ensembles of identical particles obeying those exclusion rules. It is a historical accident that people describe the theorem in terms of operator commutation relations. The theorem does not depend on field theory in any way.
The (anti)commutation relation you refer to, Charles, depends on treating a two-valued quantity as single-valued by using a notation that does not distinguish frames of reference related by a 2pi rotation.[ a+(p) ->- a+(p) under a 2pi rotation for fermions.] Eliminate the 2pi ambiguity and you get pure permutation symmetry for all particles (a requirement for uniquely-defined state vectors that account for all physically significant observables since nature does not care about particle ordering -- or "creation" ordering). This has nothing to do with field theory perturbations. The field theory commutation relations (or anti-symmetrizations of state vectors) depend on ignoring the two-valuedness in a way that accidentally has no observable consequences (at least not for pairwise states).
The simplest way to understand the theorem is permutation symmetry requires exclusion rules.
Charles, I am fully aware of the traditional errors that you repeat. I am not going to waste time arguing it with you any more unless you read my paper cited above where you will see that sign changes occur only because of a convention that contains an ambiguity. It is very easy to see that you can get rid of the sign change by resolving the ambiguity. Just do a 2pi rotation on the spin quantization frame of one particle and you change the sign for fermions.. Since the usual spatial (or momentum) vector and "third component of spin" notation does not distinguish the rotated state from the unrotated state the state vector is ambiguous. When you remove the ambiguity you can get either symmetry or antisymmetry according to your choices, but the physics is the same for particle pairs. For states of more than two particles it is more complicated, Many other physicists have pointed this out, but many more are, like yourself, unaware and keep on repeating the error. The fact remains that the content of the spin-statistics theorem is simply that the fact that nature does not care about the order in which we describe identical particle states results in easily determined exclusion rules. That's all there is to it. If you cannot understand this, then I can't help you any more.
Mike> That's all there is to it.
Please Mike, demonstrate in detail how you compute lowest order Bhaba scattering (e+ e- -> e+ e-) with your symmetrization prescription.
No. Why are you asking? The thread is about identical particle states. Everything in QED concerning identical pairs follows through in just the same way as the usual convention. It's just that the statement that creation operators MUST anticommute is wrong. You can define commuting operators and get the same result. You just have to handle the spin quantization frames correctly. Why is this so hard to understand?
Mike> The thread is about identical particle states.
in local relativistic theories.
Mike> Why are you asking?
I just wanted to check if you had the slightest understanding of what Charles is talking about; you have just clearly demonstrated that you don't. For Bhaba scattering you do not have identical particles in the initial or final state; i.e. there is nothing to symmetrize. Nevertheless, the anticommuting property of the Dirac field describing electrons and positrons dictates a relative minus-sign between two contributions to the scattering amplitude, with physical effects. This is but one example where your proposal will lead to wrong results.
In your paper you are only referring to wave functions, not considering dynamics, or the requirements behind the spin-statistics theorem at all.
There are, to some extent, ways to change the symmetry of wavefunctions describing systems with fixed number of particles. This is f.i. being used to describe anyonic excitations in systems exhibiting the Fractional Quantum Hall Effect (systems of effectively two spatial dimensions). However, it seems impossible to transfer this to a quantum field theoretic description without introducing non-local operators, which is a no-no for relativistic local (i.e. causal) theories.
Why is this so hard to understand?
Wrong statements are often hard to understand (unless one is doing the mistakes oneself). I am not saying that the math you have done is wrong (I didn't check carefully), it is the somewhat sweeping conclusions you are making from this math.
-- Exactly, which is why I asked why you were asking since it appeared to have nothing to do with the issue. However, some folks might bring in crossing symmetry to invoke identical particle states. Nevertheless, it is totally irrelevant as I pointed out.
-- Which clearly shows you have not understood anything from this discussion.
-- Nonsense. My link is to a paper which discusses relativistic state vectors and is the first complete proof of the theorem. Apparently you didn't read it at all.
-- As you have ably demonstrated.
-- I see no evidence that you made any attempt to understand it. Which is strange because it is all very standard stuff.
Apparently you do not understand the difference between a tensor product and a symmetrized or anti-symmetrized two-particle state vector. So I guess it is useless to expect you to understand any of my papers. I will not waste any time on the rest of your arrogant and comical posturing.
Your Fermion notation is only possible if |f> and |g> are two-valued due to each undergoing a possible sign change due to arbitrary 2pi rotations. When you explicitly remove those rotations you will get your Fermion expression only if you do so in an order-dependent way.
Clearly it is you who does not understand your own notations. But, of course, that is beyond your understanding. Please feel free to enjoy your ignorance, Charles.
Mike> Which clearly shows you have not understood anything from this discussion.
Mike> Which is strange because it is all very standard stuff.
Radio: To everyone driving southwards on E6, north of Gardermoen: Alert! Alert! There is a car driving in the wrong direction on the motorway!
Driver: What twaddle! There are dozens of cars here driving in the wrong direction!
I note that it is about 40 years since you worked in physics, based on your publication record. And about 18 years since you wrote the (unpublished) preprint with your proof. Don't you think there is a small probability that you have forgotten a little of what is very standard stuff? Or actually never learned it, since you worked with nuclear physics, not relativistic quantum field theory? You have probably encountered several referees and editors who also have not understood anything.
Apparently you, Charles, are unaware that rotating the reference frame of a half-integer spin particle about the quantization axis changes the phase of the state vector (without changing any of the physical properties of f) and a 2pi rotation changes the sign. because, according to your misunderstanding of the significance of your arbitrary factor "a' you can simply wipe away that phase factor by changing 'a'. But if that is the case, you can never normalize your Hilbert space or calculate any interference effects. No wonder you can't get your expression right. It's time you learned some basic techniques for handling spin.
FYI Kare, my PhD thesis was on the relativistic spin structure of hadron collisions. I think I know something about the subject. It's one thing to be driving a car and quite another to be riding a bicycle and thinking it is a car.
By your own admission you don't understand the math (dealing with spin, boosts and rotations) in my paper, so you tell me I don't understand! Such arrogant stupidity!
Oh, and BTW, my work was published in the proceedings of the conference at which I presented it and it has been cited by other authors working in the field. Or perhaps you would like to check with Berry, Sudarshan, Duck, Goldberg, Jabs, Balachandran, Broyles, all of whom have published similar work that recognizes the importance of a unique choice of spin quantization frame.
Hey Charles, what's up? Have you discovered Hilbert space normalization and unitary transformations yet? Or are you still insisting that you don't have to use unique state vectors for any fully described physical state? The fact that you can choose any vector on a ray is irrelevant -- you still have to choose a unique state vector if you want your math to have a physical meaning. This is particularly true with spin for the reasons I pointed out. The coventions are well established. Waiting with bated breath for your next red herring.....
-- Wrong again! There has to be a unique state vector only for states that are fully defined. A rotation of the spin quantization frame about the spin quantization axis changes the phase because it has physically significant consequences. This happens to be true for fermions even in the case of a 2pi rotation. The key to understanding the spin-statistics theorem lies with the relative orientation of one particle's frame to the other. It just happens that the anti-symmetry case occurs when this relative orientation is chosen in an order-dependent way. The reason this occurs (yet was not historically recognized because people just assume there is no such relative orientation in a common frame) is because of a subtle bit of geometry that shows that the relative orientation cannot be a null rotation without introducing a geometric asymmetry between the particles' momentum (or position) vectors somewhat akin to the distinction between a left-handed frame and a right-handed frame.
Edit: Corrected "order-independent" to "order-dependent".
Mike> abuse
I advise you to carefully read through your own posts, and edit your most aggressive and abusive statements. Or, you may suddenly find yourself to again be No Longer Affiliated. With ResearchGate this time.
@Charles Francis: "It is a fundamental result that the (anti) commutation relation linked to spin is required for consistency of the perturbation expansion. I would regard this as both much more fundamental, and easier to understand, than the behaviour of large ensembles".
1) this may be a matter of taste! why "large" ensembles? take |2) and |1,1). Plain statistics is for sure more "fundamental" than sophistcated commutation rules.
2) these spin statistic rules are valid only for (massive) Dirac- or Klein-Gordon states. Beyond this, anything goes.
dear Dr. Francis, are you serious? "Nothing can be more fundamental than a single elementary particle, ... , We know nothing more fundamental or more elementary than Dirac particles, and fundamental Bosons." . the "philosophy" of a plain free Dirac state is quite mysterious, like states of negative energy, "Dirac see", naked electron, ... and more. Classical mechanics is not worse than quantum mechanics and much more "fundamental"
"In normal usage, a statistic is a property of a large ensemble" ? not really: we have measure theory and probability functions which do not depend of the number of the possible events. f.i., the most basic answer is "tru or false", what is "more basic"? Next, by introducing a "RG probability distribution" I state, the probability that you agree to anything is zero, a so called "strong result"
Hi, Dr. York, "I am curious as to why you think spin-statistics applies only to massive states", is not my "thinking". I agree with Dr. Francis that (massive) Dirac states need "anticommutator" rules, which lead to higher states like |1, 0, ... 1,0, >, with n symbols "1" and similar bosons commutators leading to |n>. that's all! Further, I "believe" that on the light cone there is no frame at all. This light cone is a singularity in Minkowsky space separating the time like- from the space like domain and physics on the light cone is a kind of static.
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