# Talk:Spin–statistics theorem

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## Loophole

What is the proof of the theory?

"See Klein transformation on how to patch up a loophole in the theorem." Tell us more of this loophole of which you speak.GangofOne 05:02, 7 August 2005 (UTC)

the theorem states that the wavefunctions of a system of identical integer spins particals are symmetric under interchange of a pair of particle lables and anti-symetric under interchange of a pair of particle lables for a system of identical half-integer spin particals.

what is a "partical" and a "lable"? just misspellings i assume? each copy-pasted twice? but then again could be some spooky terms i haven't heared, so i wont be correcting this post. anti-symetric also is missing an "m"...

my head hurts from reading this kind of things even without weird words in the text.

## Proof needed

This article would need the proof for spin-statistics theorem. Proof would make clear to readers WHY cant fermions be in the same quantum state while bosons can. This seems to bother many laymans interested in this fields of physics. --83.131.77.136 10:16, 14 July 2007 (UTC)

I think I got the proof, but there it's written about "x" and "-x", while the theorem is described with "x" and "y" - so it just seems to be a proof of a special case. If I'm wrong, please make it more clear.--91.3.219.146 (talk) 12:40, 5 November 2008 (UTC)
x and -x are equivalent to x and y by translation invariance--- you just change coordinates so that the origin is halfway inbetween at (x+y)/2.Likebox (talk) 17:33, 12 March 2009 (UTC)

## Reference for the proof

Could anyone give a reference for the proof outlined in this article (and maybe a link to a full explanation)? —Preceding unsigned comment added by 91.21.204.16 (talk) 12:36, 5 April 2009 (UTC)

A very clear discussion is in the recent textbook of Banks. The original reference is the paper Quantum Theory of Fields I or II by Julian Schwinger. There is a slight difference--- he uses time reversal there, as opposed to CPT, but it is clear he is doing it from Euclidean considerations.Likebox (talk) 17:59, 14 April 2009 (UTC)

## Time Rotation

I didn't understand the statement in the proof: "The rotation plane includes time". Which rotation is reffered to? the inner (polarization rotation)? the outer x=>-x, or maybe the whole thing, :${\displaystyle R(\pi )\phi (Rx)=R\phi (-x)\ }$?
Both of them are just pi spatial rotation, so why any of them includes time? How can one connect any of them to CPT? Please make it more clear.

Yes--- this is the central point. It might not be clear enough in the article, please make it clearer if you can. The best way to explain is with examples.

Consider first a nonrelativistic scalar Schrodinger field. This is a pair of scalar fields ${\displaystyle \psi (x)}$, which annihilates a particle at x, and ${\displaystyle \psi ^{\dagger }(x)}$, which creates one. Let's try to see what the proof technique says for this case. We are supposed to consider the two point function

${\displaystyle \langle R\psi ^{\dagger }(x)\psi ^{\dagger }(-x)\rangle =G(2x)}$

Rotating this, we find that G(2x) = G(-2x), in other words that this part of the correlation function is odd even under rotations that interchange the two insertions. But this is absolutely no information at all, because this correlation function is actually identically zero.

${\displaystyle \langle \psi ^{\dagger }(x)\psi ^{\dagger }(y)\rangle =\langle \psi (x)\psi (y)\rangle =0}$

So it's symmetric, and also antisymmetric, and also whatever you want it to be. The two psi-daggers create a two particle state, which has zero overlap with the vacuum. Likewise, the two psi insertions annihilate two particles, which again gives zero in the vacuum. So the argument gives you absolutely nothing.

The field with a nonzero vacuum value:

${\displaystyle \langle R\psi ^{\dagger }(x)\psi (-x)\rangle }$

Doesn't give you any information about interchanging either, because the rotation of this gives

${\displaystyle \langle R\psi (x)\psi ^{\dagger }(-x)\rangle }$ (nope, it's ${\displaystyle \langle \psi ^{\dagger }(-x)R\psi (x)\rangle }$ but same difference)

Which is a different correlation function, rotation doesn't take it to itself. So this argument doesn't give a spin/statistics relation.

If you allow arbitrary states, rather than the vacuum, all that the rotation stuff says is that ${\displaystyle G_{\eta }(x)}$ has a certain symmetry under pi rotations, where ${\displaystyle G_{\eta }}$ is defined below:

${\displaystyle \langle \eta |R\psi ^{\dagger }(x)\psi ^{\dagger }(-x)|\eta \rangle =G_{\eta }(2x)}$

That if there is a component of eta which is linked by rotating ${\displaystyle G_{\eta }}$ by ${\displaystyle \pi }$, that the symmetry of this part of the wavefunction is determined by the spin of the field. That's some information, it's a set of selection rules for which spatial two-particle wavefunctions are allowed for the case where the spins are rotated by ${\displaystyle \pi }$, but it's the usual selection rules for angular momentum spatial symmetry. That's interesting, but it's not spin-statistics (unless you add extra assumptions about "naturalness of the Berry phase", or something else, but that's extra assumptions).

So far so good. Because there's no general spin-statistics theorem without relativity. Nonrelativistic fields can be consistently defined with any spin and any statistics independently.

What's the new thing in relativity? That's the fact that time mixes in with space, so that the rotations are extended to Lorentz transformations. Then every field acts not only to create a particle, but also to annihilate the corresponding antiparticle. The reason is that the field is local, so it creates a particle localized in a region smaller than the compton wavelength, so that the particle can fluctuate into an antiparticle. Another way of saying this is that the particle's propagator includes a component that propagates backwards in time. Another way of saying this is that the positive frequency in time Green's function is nonzero outside the lightcone. All these things are saying the same thing.

The most elegant way to describe this property is to extend the theory into imaginary time, so that the relativistic boosts become rotations that mix up space and time into each other. Then a rotation in time gives you new information about correlation functions that you don't have without relativity. You can extend to imaginary time in principle whenever the energies are positive definite, but the details of how to do this explicitly are a little tricky for chiral fermions. But it's not hard for tensor fields. The important thing to understand is that the imaginary time business is the analytic continuation of the Feynman propagator, the one which has only positive frequencies.

So next example: examine the correlation function for a real scalar field (the time-ordered one, the Feynman propagator, the one which extends to imaginary space-time):

${\displaystyle \langle R\phi (-x)\phi (x)\rangle }$

I shouldn't really be including the "R" factor, because for a scalar there's no polarization, so it doesn't do anything. But the difference in relativity, this correlation function is nonzero. The field creates a particle, and for a real scalar, it annihilates the same particle. When this correlation function is not identically zero, the argument tells you that it is symmetric under pi-rotations, which tells you that the scalars should be bosons, by doing the rotation.

Why does this work in relativity? It works because the same correlation function does three things simultaneously:

1. It creates two particles
2. It creates a particle then annihilates it
3. It annihilates two particles

So the statistics of creating two new particles (or annihilating two existing particles) is determined by the symmetry of creating a particle and then annihilating it, by the symmetry of the correlation function.

But this argument only works when the field R\phi annihilates the same particle that the field \phi creates. When does this happen? CPT suggests it should happen always, but I haven't explicitly checked this for spin 1/2 fields in all dimensions, because rotating them is a little harder to visualize (all the components mix up together no matter what you do, not like for tensor fields). I am only bringing this up because Sudarshan claimed that for fermions in 8 dimensions there is a counterexample, you can make bosonic fermions, and this argument doesn't immediately disprove this, because you have to check that the polarization of the field can be chosen so that the correlation function is not identically zero.

The important thing for the argument is to make sure that the correlation function R\phi \phi is nonzero. The next example will clarify this further. That's a scalar ghost, like in Yang-Mills theory. This is a field that acts very much like the non-relativistic scalar.

The scalar ghost ${\displaystyle \eta }$ is a scalar under rotations, but it's a Fermionic field. It doesn't have polarization. The propagator for this field is

${\displaystyle \langle {\bar {\eta }}(y)\eta (x)\rangle =G(x-y)}$

which is the usual Feynman propagator. So this field is just like the nonrelativistic field, the correlation functions which you want to use for the proof are identically zero:

${\displaystyle \langle {\bar {\eta }}{\bar {\eta }}\rangle =\langle \eta \eta \rangle =0}$

just like the nonrelativistic example, the correlation functions which are nonzero don't give you a spin-statistics relation.

But this is where CPT comes in to tell you that the ${\displaystyle \eta }$ is creating states of negative norm. If you start with the vacuum and act with ${\displaystyle \eta }$, you create scalar fermions going forward/backwards in time. If you rotate this in the time direction by 180 degrees, you get the same field, because it's a scalar, except now the scalar fermions are going backward/forwards in time. So the correlation function

${\displaystyle \langle \eta (x)\eta (-x)\rangle =0}$

is not zero for the normal reason, that the two ${\displaystyle \eta }$'s are not bringing you back to the vacuum. That's not what is happening. What's happening is that ${\displaystyle \eta }$ is making a positive norm state and also a negative norm state, and then instead of adding up like they normally would, they cancel each other!

To see this in detail for the scalar fermion ghost, remember that the fields ${\displaystyle \eta }$ and ${\displaystyle {\bar {\eta }}}$ are independent Grassman variables, and you can diagonalize the quadratic part by defining new fields:

${\displaystyle \alpha =\eta +{\bar {\eta }}}$
${\displaystyle \beta =\eta -{\bar {\eta }}}$

up to square roots of two. The kinetic part of the Lagrangian is now:

${\displaystyle \alpha \partial ^{2}\alpha -\beta \partial ^{2}\beta }$

So that the cancellations between the alpha and beta states is obvious now.

CPT suggests that this is always true, because when you can create a particle, the same field should annihilate it, by rotating by pi in imaginary time. The only quibble is if the polarization business works out. The argument goes through whenever the correlation function is not identically zero. The next example is 2d Fermions.

Look at a free chiral majorana Weyl Fermion in 1+1d with coordinates (t,x). This is a single real grassman field \psi with equation of motion

${\displaystyle (\partial _{t}-\partial _{x})\psi =0}$

This particle only propagates to the "left", towards negative x. The Feynman propagator satisfies the equation of motion and when it has positive frequencies only (positive energy states), it continues to imaginary time and satisfies the equation

${\displaystyle {\partial _{z}}G=0}$

meaning that the propagator is holomorphic. You can check this case works out too. I'll finish this later.Likebox (talk) 16:46, 6 June 2009 (UTC)

Thanks alot for the blessed effort. I read everything, but I afraid that I lost you very early: you said "Rotating this, we find that G(2x) = G(-2x), in other words that this part of the correlation function is odd under rotations that interchange the two insertions." Didn't you mean "even"? And also, "rotating this" means what? Do you mean that in a thoery that is invariant under spatial rotation, G(2x) = G(-2x)? Also, later you said " The field with a nonzero vacuum value:

${\displaystyle \langle R\psi ^{\dagger }(x)\psi (-x)\rangle }$

Doesn't give you any information about interchanging either, because the rotation of this gives

${\displaystyle \langle R\psi (x)\psi ^{\dagger }(-x)\rangle }$"

How come spatial rotation gives you this? —Preceding unsigned comment added by 96.237.14.196 (talk) 03:32, 7 June 2009 (UTC)

Ooops! Yes, clearly, it's even. Sorry.
In a theory that is rotationally invariant, G(2x)=G(-2x). But that's not news, because in the particular theory G(2x)=0! So it vanishes, period. It's odd and even, and everything inbetween, because it's identically zero. That's why you don't have a spin-statistics theorem without relativity. I was just using that case as a warmup, to show you what a bogus argument looks like.
The argument you are missing is the main argument in the article. If you rotate the correlation function
${\displaystyle R\phi (-x)\phi (x)}$
by 180 degrees, you have to rotate each factor separately. You get
${\displaystyle RR\phi (x)R\phi (-x)}$
where R is the matrix that rotates the polarizations by 180. RR rotates by 360, so it's -1 for spin 1/2 3/2 5/2, and 1 for spin 0,1,2,3. So this is equal to
${\displaystyle \pm \phi (x)R\phi (-x)}$
which is the same thing, with the order of the operators interchanged. A good way to visualize it is this: (in ascii art, sorry )

<- x ->
x is the origin, and the arrow is supposed to represent polarization of the field. When you rotate this by 180 around the middle x, you get the same state back, except the order is reversed. The trick is to turn this argument into a real proof. This requires rotating in time.
Then you learn from relativity (time rotation) that certain correlation functions are even for integer spin and odd for spin half-integer spin. That's it. So far, I have to check each case with the same argument each time, but that's the argument, but you need to take the time to cross the t's and dot the i's.Likebox (talk) 04:15, 7 June 2009 (UTC)

Again me. Maybe you missed my point. What bothered me in your example above is that when you pi rotated ${\displaystyle \langle R\psi ^{\dagger }(x)\psi (-x)\rangle }$ you switched also the order of the operators from ${\displaystyle \psi ^{\dagger }\psi }$ to ${\displaystyle \psi \psi ^{\dagger }}$, for no particular reason. Is it a mistake? Is it importent?

Also, That makes me be confused about another statement: (I read further) it is not true that "This argument only works when the field R\phi annihilates the same particle that the field \phi creates. When does this happen? CPT suggests it should happen always". For any charged particle it is not the case, no? It is still not clear.

Yes, sorry about the order screwup in rotating the nonrelativistic field, but it doesn't change the argument that follows or the conclusion: if you rotate the correlation function of Rpsi-dagger(x) psi(-x) you get psi-dagger(-x) Rpsi(x), not the same correlation function, so the argument doesn't work. It's obvious why: in the nonrelativistic case, creation and annihilation are done by separate fields. In the relativistic case, the same field both creates and annihilates.
For the charged particle case, you are introducing two separate notions:
1. Multiple component fields with a symmetry.
2. Coupling to electromagnetic fields, gauging a part of this symmetry.
The second is a red-herring. It just makes it the complications introduced by the first case more difficult to understand, because superpositions of different charges start to look weird. Spin statistics is a property of free field theories, depending on the positive-definite metric of the hilbert space of free particles, it doesn't have anything to do with the interactions. So turn the electromagnetism off.
For a "charged scalar field", just a complex scalar， consider the correlation function
<R\phi phi>
This correlation function is nonzero for the real and imaginary parts separately. When you combine them for formal reasons into a complex scalar, the nonzero function is phi-conjugate phi, but that's because you put in an "i" by hand in defining the complex field, so that the phi-phi correlator is the real-real correlator minus the imaginary-imaginary correlator, and they cancel out (but both have the same sign). Rotating this, you see that the real and imaginary parts have to be bosonic.
Since CPT rotates each representation of the Lorentz group separately, you should ignore any internal symmetries when doing the rotation in this argument, gauged or not.Likebox (talk) 23:22, 9 June 2009 (UTC)
BTW, I hope it's clear that the "R" in the scalar case is just nothing. It's just for consistencyLikebox (talk) 23:31, 9 June 2009 (UTC)

What you did is to take a complex field, and extract its real operator by ${\displaystyle \xi =\phi +\phi ^{\dagger }}$. then the correlator ${\displaystyle \langle R\xi \xi \rangle }$ is not zero. But you can do that also in the nonrelativistic case, by defining ${\displaystyle \xi =\psi +\psi ^{\dagger }}$ and again the correlation function ${\displaystyle \langle R\xi \xi \rangle }$ would not be zro. I fail to see the difference. How anti-particles/time rotation enters here at all? please give the simplest example. I really want to understand. —Preceding unsigned comment added by 140.247.241.200 (talk) 19:37, 11 June 2009 (UTC)

These really are the simplest examples. You can't do what you want in the nonrelativistic case--- the field psi is just an annihilation operator, and it's conjugate psi-dagger is just a creation operator. If you take real part, you get the correlation function:
${\displaystyle \langle (\psi (x)+\psi ^{\dagger }(x))(\psi (y)+\psi ^{\dagger }(y)\rangle =\langle \psi (x)\psi ^{\dagger }(y)\rangle =\delta (x-y)}$
it's only nonzero when x and y are right on top of each other. That's like the ultralocal field in relativity, which can have any spin and any statistics independently, because it has infinite mass. Since there is no dynamics, you can check that the Lagrangian
${\displaystyle S=\int m^{2}\phi ^{2}}$
makes sense for any spin and any statistics (phi squared for any spin means the appropriate Lorentz invariant quadratic function of the field variables, but be careful writing it down for integer spins, for spin 1 it should be three ultralocal fields, not four, for spin 2, five ultralocal fields etc). For example, if phi is a scalar, it could be a scalar Fermion with no problem. But for ultralocal fields, the propagator is stupid--- there's no propagation. The correlation function for phi with itself is only nonzero when they're at the same point, just like the nonrelativistic case. That's the reason for the "finite, not infinite, mass" condition in the article.
If you look at the time dependent free nonrelativistic fields, you can examine the same correlation function at different times:
${\displaystyle \langle (\psi (x,t+a)+\psi ^{\dagger }(t+a))(\psi (y,t)+\psi ^{\dagger }(y,t))\rangle =G(x-y,a)=e^{i}{(x-y)^{2} \over 2a}}$
but, rotating this thing doesn't tell you squat about operator commutations, because the psi-dagger is always later than the other, so you never can rotate it to be the same thing. If the operators are at different times, in order for a rotation to bring them back to the same thing, the rotation plane must include time.
The path integral makes it clearer. In a path integral, the order of the operators is always the time order, so you should think of the operator to the right as a tiny bit later than the operator to the left. A nonrelativistic rotation will always keep this property, you can't flip the order of operators by rotation. In relativity, you can rotate in time by 180, and this will flips the time order of operators, which means it flips their order in the correlation function. This is why you get relations on orders of operators from relativity.
What happens when you take the nonrelativistic limit? I'll show you explicitly in a second, but intuitively, it's like this: a local field in the relativistic case makes particles and antiparticles which propagate forward and backward in time. You can make it real in relativity by taking real and imaginary parts and then you are identifying particle and antiparticle. But the process of taking the nonrelativistic limit separates a real field into a complex field which creates and a complex field which annihilates which are not related by rotations.
Mathematically, when you take the nonrelativistic limit, you are throwing away fourier modes of the field which are far above the frequency m, and shifting the frequency by m to subtract the mass-energy. In terms of fields Fourier transformed in space but not time,
${\displaystyle \psi (k,t)=e^{imt}\phi (k,t)\,}$
For (k<<m) and up to some stupid relativistic normalization factor or other. You can then see that the low energy dynamics is given by the Schrodinger field ${\displaystyle \psi }$ and it's conjugate, and that the scalar Lagrangian dynamics are given by the Schrodinger field lagrangian in this limit:
${\displaystyle S=\int \psi ^{\dagger }i{d \over dt}\psi -\psi ^{\dagger }\nabla ^{2}\psi }$
In this Lagrangian, psi and psi-dagger are conjugate operators, and they don't commute (their commutator is a delta function, it's just the canonical commutation relations for this Lagrangian). If you take real and imaginary parts, the imaginary part is conjugate to the real part, and they don't disentangle.
Why is it that the low energy limit of a real field becomes complex? It's because the annihilation and creation operators for a real relativistic field are both mixed up together inside the field at different frequencies. The creation operators are near frequency m, while the annihilation operators are near frequency -m. In the nonrelativistic limit, the two regions are infinitely far from each other in frequency, and they don't talk to each other, and you might as well call them different fields, psi and psi-dagger. Relativistically, you can see that they are the same because when the two points x and y are closer than the compton wavelength, the field phi has nonzero propagator outside the light cone. The operators that create are mixed up with those that annihilate.
I am sorry for being so verbose, but I am just saying the same thing again and again. I hope it is clear now. If not, please tell me what is confusing.Likebox (talk) 00:12, 12 June 2009 (UTC)

I thought about what you wrote above, and also reread to proof in the article. I feel that we all dispersed a bit, and want to be more focused, Let me state what I understood is the outline of the proof, then state what is understood and not, regarding this proof. So the proof goes as follows

1) The relativistic correlator G(x,t=0)=<Rphi(-x,0)phi(x,0)> is strictly invariant under pi spatial rotations (among others): G(x,t=0)=G(-x,0).

2) A relativistic correlator G(x,t=0) is non-zero.

3) As a local Field theory that must maintain causality, for any two local fields phi and xi at equal time xi(y)phi(x) = phi(x)xi(y)*z is true as an operator identity, where z is a phase factor.

4) Putting xi(y)=Rphi(-x) then points 1+2 restrict z to be 1 for integer spin field and -1 for half integer spin field.

5) Local transformation can't change R. So z is constant for all polarizations.

Now the time rotation, CPT arguments etc. are only for explaining point 2. This point 2 marks the only difference between relativistic and non- relativistic particles. What I don't understand is why we need all that argumetns. Following Feynman's simple argument, point 2 is true because a) for large t – G(x,t) is non-zero, somewhere inside the light cone (x<ct). b) Because G(x,t) is constructed only using positive energies, It has to be nonzero for any other t's, except for a zero-measure sections. This is a property of a Fourier transform: if f(w)=0 in a finite measure section (in our case, the section is w<0), f(t) can't be zero apart for zero-measure section. c) Then G(x,t) is non zero somewhere *outside* the light cone. d) Lorentz invariant of G(x,t) will get you the same non-zero value at t=0 G(x,t)=G(x',t=0), qed.

The last sentence "d" is the only thing which doesn't hold for non-relativistic particles, and this is what allowing G(x,t=0)=0 for them.

So why do we need CPT, and why you need to call the pi rotation "rotation in time"? Can't we write the article much simpler, as the above suggestion? —Preceding unsigned comment added by 96.237.14.196 (talk) 03:58, 15 June 2009 (UTC)

I see that you got the main point, because you are restating it in Feynman's language. But this argument is originally due to Schwinger, and I wanted to say it in Schwinger's language. That means euclidean fields.
But I will try to explain why I don't like the Feynman description. Feynman's argument is fine to show that the propagator is nonzero outside the light-cone, but to go from that to spin-statistics requires the connection between rotation and swapping. That requires the particular correlation function in the article. This correlation function is nonzero in the scalar case, but for example for the gauge ghost it is zero again. To see why it should be nonzero in general you should think about it Schwinger's way, with imaginary time.Likebox (talk) 04:38, 15 June 2009 (UTC)
I should add that Feynman has a perfectly valid spin-statistics demonstration using loop "-1"'s and perturbation by external fields. This might be a good addition. The source is an appendix to Feynman's 1950s lectures on Quantum Electrodynamics.Likebox (talk) 04:47, 15 June 2009 (UTC)

I understand you tendency to go fully with Schwinger - Because he was the first he connect rotation and swapping (which is that point 1 can fix z in point 3). But I still don't see why going to Euclidean 4d space makes your life simpler then Feynmann agrgument + your "heuristic argumet" (which is exectly the connection between spatial rotation ans swapping). Can you explain that? What 4d Euclidean view adds to the spin-statistics theorm?

About the Gauge Gohst - I think that the argument fails there because, as field with negative norm, G(x,t) can have negative frequencies, while real particle can't. So point b doesn't hold. Is that right? Then Feynmann argument is still much simpler... —Preceding unsigned comment added by 140.247.251.139 (talk) 14:42, 15 June 2009 (UTC)

The point of 4d is that it gives you a good mental picture of how to connect forward in time propagation to backward in time propagation: it's just rotation. It's mostly equivalent to what Feynman does, but it shows you that there is an extra symmetry which is not completely obvious in real space. In real space, you can do Lorentz transformations, but the boost parameter can never reverse the direction of time. The only new thing Euclidean space gives you really is a geometric interpretation which shows you why boosts actually tell you something about reversing the direction of time, because they analytically continue to Euclidean rotations.
What's clearer or less clear is different for different readers. I do tend to agree that Feynman's argument is somewhat more intuitive if you don't have a good feeling for Euclidean fields. But he gives a bunch of arguments, some not at all correct. For example, if you read the 1986 Dirac lecture, at the very end, he says the rotation-swapping business in an intuitive way that seems to rotate only in space, not time. This is incorrect, you need to do "time reversals" (or time rotations--- same thing) to get a good argument, otherwise the theorem would apply nonrelativistically too. But it's mostly correct, you just have to be comfortable with Euclidean space. But I think everyone should be made more comfortable with transforming to Euclidean space. It's not trivial for spinors. For example, in 1+1 dimensions, there is a 1-real-component spinor, but there is no 1-real-component spinor in 2 Euclidean dimensions. So how does the continuation work? In higher dimensions, this problem lead to a lot of hand wringing and gnashing of teeth. But it must be done. So maybe it would be good to add Feynman's 1950's argument (which he reproduces, mixing in some Schwinger, in the 1986 lecture).Likebox (talk) 17:42, 15 June 2009 (UTC)

## Removed this stuff

(I copedited)

The proof is based on the notion of causality, which means that two events in space-time cannot influence each other if light does not have time to travel from one event to the other. Two such events are Space-like separated.
The statement that such two event cannot influence each other is formulated by stating that for any two measurements, one taken at each event, the result of one the measurement does not depend on the preforming of the other measurement. In quantum mechanics we say that the operators related to these measurements commute (for bosonic operators) or anticommute (for fermionic ones).

The problem with this is that it is not true that the result of measuring two anticommuting fermionic operators is independent. Only commuting operators are independent. If it were possible to measure fermionic operators by a local measurement, then this would violate causality. But this is impossible. The only quantities which can be measured locally are those which are bosonic, and so the bosonic bilinears need to commute with each other. The bilinears commute with either statistics, so causality is not a clue about the connection between spin and statistics.Likebox (talk) 04:21, 8 January 2010 (UTC)

I guess that I was indeed misleading, because I meant quite another thing. You are of course right that one cannot make a local fermionic measurement, because a fermionic operator cannot be Hermitian. My mistake. I was trying to explain why two spacelike operators must either commute or anticommute. I was not trying to suggest (at this point) that it has anything to do with spin. Two spacelike bilinears must commute, and for this to happen would require two spacelike fermionic operators to anticommute. I will write this part again, with your correct remarks taken into consideration.

The proof requires the notion of causality, which means that two events in space-time cannot influence each other if light does not have time to travel from one event to the other. Two such events are Space-like separated.
The statement that such two event cannot influence each other is formulated by stating that for any two measurements, one taken at each event, the result of one the measurement does not depend on the preforming of the other measurement. In quantum mechanics we say that the measurement operators related to these measurements commute. Since such operators may be compound of either bosonic of fermionic operators, it follows that two spacelike separated bosonic operators must commute, and two spacelike separated fermionic operators must anticommute. From this argument, the two preceding equations follow.

(these are the quations in the preceding setions) Dan Gluck (talk) 10:33, 8 January 2010 (UTC)

This again is not precisely correct: the problem is that there is no general requirement for commutation/anticommutation in order for bilinears to commute. You can make bilinears commute with some other rules, but these are not realized in 4 dimensions in nature. It is not good to state that causality is the reason for spin/statisics, because causality without relativistic invariance is not enough, and even causality with relativistic invariance leaves loopholes for anyons and ghosts, so you need other assumptions. The assumptions listed in the "proof" section are sufficient.Likebox (talk) 18:59, 9 January 2010 (UTC)

## More on time rotation

What historically has been called time rotation by π, is actually a misleading term, and the more accurate "time reversal" should be used. This is because unlike the Euclidean 4-dimensional rotation group SO(4), which is connected, the Lorentz group SO(3,1) is not connected, but rather has 4 connected components. In other words, in a Euclidean space (whose rotational symmetry group is SO(4)) you can reach a rotation by π by successive small rotations, while in a Minkowsky space (whose rotational symmetry group is SO(3,1), the Lorentz group) you cannot reach time reversal by successive small steps.

From this it follows that R should actually be defined so that it includes both rotation of the internal state, plus time reversal.

Moreover, it is not true that a in quantum field theory (= relativistic quantum mechanics) every field includes both creation and annihilation operators of the same particles. In fact, the fermionic field ψ contains the creation operators of electrons and the annihilation operators of positrons (which is the same as the creation operators of negative-energy electrons), while its charge-conjugate ${\displaystyle {\bar {\psi }}}$ contains the annihilation operators of electrons and the creation operators of positrons. Therefore, for the proof to work, R must actually also transform ψ to ${\displaystyle {\bar {\psi }}}$ (amnd vice versa). This means that R also includes the charge conjugation operator, C. This point has been left unnoted in the above discussion because for real bosonic fields, the charge conjugation does nothing. However, for complex bosonic fields, it takes the field to its complex conjugate, and this solves the problem that has been mentioned above.

Thus actually R = CPT except for the spatial part of P, or in other words Rψ(-x) = CPT ψ(x). This also explains why the proof relies on "the CPT invariance of the correlation function" as written.

I am going to add explanations of this and of the previous discussion about the so-called "time rotation". Dan Gluck (talk) 19:02, 8 January 2010 (UTC)

The essential point is that a relativistic quantum field theory with a positive definite energy can be analytically continued to a Euclidean field theory with rotational invariance. In this theory, the analog of CPT is a rotation by 180 degrees around two axes one of which is time. So "time rotation" (meaning Euclidean time rotation) is CPT transformation in the Lorentzian theory.
This is the key to the proof of the CPT theorem, and is also important for the spin-statistics theorem. The operator "R" is the polarization transformation for a 180 degree rotation, and includes the particle/antiparticle business. I thought that the original version was pretty clear about this.
But the statements about causality, that fermionic fields anticommute because of causality, are incorrect and should be removed.Likebox (talk) 18:56, 9 January 2010 (UTC)
About the "heuristic argument". This argument doesn't work at all, because it does not involve a CPT transformation (time rotation). It is completely incorrect as stated, but it can be made into a proof when the rotation plane involves time in a theory with a Euclidean continuation. This heuristic argument sometimes is claimed in the literature to be a proof, and we must not endorse this.Likebox (talk) 19:01, 9 January 2010 (UTC)

## Remove Causality stuff

If you want to say something about causality, do not say that causality implies that spacelike separated operators commute or anticommute. This is false. You can say that causality implies that measurable spacelike operators commute, and that for reasonable theories, this means that certain bilinears commute. But this does not imply that the basic field operators commute or anticommute. They could have all sorts of different types of commutation relations.Likebox (talk) 19:38, 9 January 2010 (UTC)

I second Likebox. Moreover, the link "microcausality" goes to the page causality (physics), which mentions neither microcausality nor commutation. — Preceding unsigned comment added by 128.111.9.44 (talk) 21:47, 19 July 2011 (UTC)

## Streater/Whitman vs. Path integerals

The actual arguments in Streater Whitman are obvious results from the literature, entombed in ponderous difficult to decipher outdated mathematical language. Their book is where I first saw the spin/statistics proof, and it took forever to decipher what they are saying, because they say it like pompous fools.

They do emphasize the analytic continuation of Green's functions to Euclidean space, and then they point out that the rotation of vacuum expectation values of operators restrict the allowed statistics according to the spin/statistics rule. But they don't actually understand what they are doing very deeply.

The proof here, based on Schwinger's original arguments (incrementing on Pauli/Fierz) is superior in every respect. It does the exact same thing that Streater and Whitman does, except without the ridiculous pompous language.Likebox (talk) 17:50, 10 January 2010 (UTC)

I removed the following text:

The formal proof differs from the heuristic argument by several points:

1. The fields are defined by field operators acting on the vacuum state.
2. Rotation by 2π in the x-y or x-z plane is replaced by rotation by π in the y-z plane plus a "rotation" by π in the x-t plane. Thus all space-time directions are reversed. The "rotation" in the x-t plane cannot be done in real spacetime, and a mathematical concept known as analytic continuation is used.
3. Not only the field but also its hermitian conjugate is considered. Very loosely speaking, this is the field of antiparticles conrresponding to the field of particles at hand (though sometimes they are just the same field).
Why this is no good
1. The fields can act on any state in either argument. You can take the vacuum expectation value in the first argument, and its no different from the second. In fact, they are the same argument. The actual difference is physical. Without relativity, there is no vaccuum to vacuum two-point function, because the nonrelativistic two field product creates two particles. The relativistic two field product either creates two particles, annihilates two (anti)particles, or creates a particle and then annihilates it.
2. The rotation only needs to be in the x-t plane. There is no need for the y-z plane rotation at all. In a rotationally symmetric theory, either operation produces a CPT transformation of some kind.
3. The hermitian conjugate is NOT considered (at least not for a scalar field). This disease is caused by reading Streater and Whitman. The two field product is the same field at two points, with different polarization. The reason you get confused on this point is that the rotation in the time plane does (when you make it into a Minkowski operation) turn particles into antiparticles, but this is not exactly the same as hermitian conjugation.

Please read the exposition in a "Quantum Fields and Strings: A course for Mathematicians" from 1999 (although this is also entombed in overly mathematical language), or the exposition in Banks recent field theory textbook.

The statement that Schwinger provided the obviously wrong heuristic argument (which should not be called that at all--- it's actually a bogus argument) is a slur on Schwinger and must not be repeated. Schwinger gave a correct proof based on time reversals, which is essentially equivalent (for CP invariant theories, which were all that anyone considered back then) to the proof given here.Likebox (talk) 17:58, 10 January 2010 (UTC)

### Some remarks

After some thought, I've realized that this proof is indeed equivalent to Streater & Wightman's. I wouldn't call them fools, though. I think they were the first to prove this rigorously. Also, they prove it with less assumptions (an extremely minimal set of assumptions, actually), which makes them quite smart.

Regarding your 3 "no goods", with 1 I agree - I just meant that this is how we may define fields, nothing more.

With 2 I don't agree - I'll elaborate on it.

Regarding 3, the hermitian conjugation is just a consequence of the CPT transformation. So it's just a pedagogical matter whether to write about it or not. I'm fine with not writing about it here.

I'm fine with the current form of the proof as it appears in text, because it is probably easier to understand than the version which I wrote (I also had an error in my version - see in what follows). But, there are two corrections that I think should be made:

1. The CPT transformation must be reversal of all directions (x, y, z and t). However, I was wrong in saying that such a transformation is like a 2π rotation. I will explain all this.
2. The plus/minus sign in the proof you gave comes in the wrong place.

First - why the CPT transformation must reverse all directions:

Lets us try the two options: P1 reverse only x, and P2 reverses x, y, and z. Therefore, CP1T makes a π rotation in the x-t plane, while CP2T makes also a π rotation in the y-z plain.

We want to claim that :${\displaystyle G(x)=\langle 0|R\phi (-x)\phi (x)|0\rangle \,}$ is positive definite. For scalar fields, it doesn't matter which CPT transformation we take. But for this to be true for non-scalar fields as well, we must take CP2T. Indeed, this is the transformation taken by Streater & Wightman (excpet that they call it Λ = -1, which means that the Lorentz transformation is minus unity, i.e. takes every direction to minus itself). As a simple example, Let us consider a vector field Aμ.

Look at how the CPT transformations act on the ingredients of this vector field:

P1 reverse the x direction, but not the y and z. Therefore, it must act with an opposite sign on Ax relative to Ay and Az. So no matter how we define C and T, CP1T gives a different extra minus sign when acting on Ax relative to Ay and Az. Therefore ${\displaystyle G(x)=\langle 0|CP_{1}T\phi (-x)\phi (x)|0\rangle \,}$ cannot be positive definite for both φ = Ax and φ = Ay (or φ = Az).

On the other hand, P2 reverses both x, y and z, and so there is no such sign difference with ${\displaystyle G(x)=\langle 0|CP_{2}T\phi (-x)\phi (x)|0\rangle \,}$.

This is why the CPT transformation has to act so that it reverses all directions.

Now let's turn to where the plus/minus sign (for integer/half-integer spin) should appear.

First, note that if the CPT transformation acts on a state as a linear transformation S, then it acts on an operator O as: ${\displaystyle SOS^{-1}}$. So, instead of writing ${\displaystyle R\phi }$, we should write ${\displaystyle R[\phi (-x)]=S\phi (x)S^{-1}}$. CPT invariance holds if S in a unitary operator, but it can also hold if S is an antiunitary operator. In fact, the latter is correct.

Now, let's try to follow the proof as it's written now. If we assume S to be unitary (which is actually wrong), and use the invariance of the vacuum under S, we get (by replacing the vacuum with S times the vacuum):

${\displaystyle G(x)=\langle 0|R[\phi (-x)]\phi (x)|0\rangle \,=\langle 0|S\phi (x)S^{-1}\phi (x)|0\rangle \,=\langle 0|S^{-1}S\phi (x)S^{-1}\phi (x)S|0\rangle \,=\langle 0|\left(S^{-1}(S\phi (x)S^{-1})S\right)\left(S^{-1}\phi (x)S\right)|0\rangle \,=\langle 0|R^{-1}[R[\phi (x)]]R[\phi (-x)]|0\rangle \,}$

(the 1st, 2nd and 5th equalities are by definition, in the 3rd we multiply both vacua by S and in the 4th we multiply in the middle by SS-1 =1) So instead of getting RR we get R-1R which is identically one. Therefore the result always has the same sign, no matter what the spin is:

${\displaystyle ...=\langle 0|\phi (x)R[\phi (-x)]|0\rangle \,}$

This is much as if we'ld take two subsequent π rotations, with one clockwise and one anticlockwise, rather than two such rotations clockwise. The result will be as if we did nothing, no matter what the spin is.

However, this spin-statistics result (that everything is bosonic) is wrong because the CPT transformation needs not be unitary. In fact, it is antiunitary, so we actually get the two operators switching places. This just gives what we started with, and hence no information (in fact the following proves that S is antiunitary if and only if R = R-1 i.e. R2; we'll encounter this later again):

${\displaystyle G(x)=\langle 0|R[\phi (-x)]\phi (x)|0\rangle \,=\langle 0|S\phi (x)S^{-1}\phi (x)|0\rangle \,=\langle 0|S^{-1}\phi (x)S\phi (x)S^{-1}S|0\rangle \,=\langle 0|S^{-1}\phi (x)S\phi (x)|0\rangle \,=\langle 0|R^{-1}[\phi (-x)]\phi (x)|0\rangle \,}$

The correct way to progress is not to act with S on the vacuum, but directly on the operators themselves. Here we'll have to use their spin properties. First, remember that R[φ]φ allways have integer spin. Now we use the fact that the CPT transformation reverses all directions. Therefore when it acts on an operator with an integer spin, i.e. a tensor, it gives a factor (-1) for every space-time index it has. Thus, the overall factor is (-1)J for an (integer) spin J operator. Since the operator R[φ(-x)]φ(x) has twice the spin φ(x) has, we get an overall factor of 1 for a half-integer spin φ and an overall factor (-1) for an integer spin φ. Therefore we have:

${\displaystyle G(x)=\langle 0|R[\phi (-x)]\phi (x)|0\rangle \,=\pm \langle 0|R\left[R[\phi (x)]\phi (-x)\right]|0\rangle \,=\pm \langle 0|SS\phi (x)S^{-1}\phi (x)S^{-1}|0\rangle \,=\pm \langle 0|SS\phi (x)S^{-1}S^{-1}S\phi (x)S^{-1}|0\rangle \,=\pm \langle 0|R\left[R[\phi (x)]\right]R[\phi (-x)]|0\rangle \,}$

Now, we could say that a double CPT transformation is just the unity, so the result is:${\displaystyle \pm \langle 0|\phi (x)R[\phi (-x)]|0\rangle \,}$. Instead, we may also use the antiunitarity of S:

${\displaystyle G(x)=...=\pm \langle 0|R\left[R[\phi (x)]\right]R[\phi (-x)]|0\rangle \,=\pm \langle 0|S^{-1}R[\phi (-x)]R\left[R[\phi (x)]\right]S|0\rangle =\pm \langle 0|S^{-1}S\phi (x)S^{-1}SS\phi (x)S^{-1}S^{-1}S|0\rangle =\pm \langle 0|\phi (x)S\phi (x)S^{-1}|0\rangle =\pm \langle 0|\phi (x)R[\phi (-x)]|0\rangle }$.

Which completes the proof. Dan Gluck (talk) 15:11, 16 January 2010 (UTC)

Thank you for the thoughtful remarks--- I'll think about the positive definite business. You have discussed it for spin-1, but you can always choose the polarization of a spin-1 field (or any tensor field) so that it is unchanged by the x-t rotation, and then the positive definiteness of at least some components follows.
For spinor fields, it is not obvious to me that either CP1T or CP2T will produce positive definite Green's functions. In simple examples, you can see that it ends up being positive definite with some polarizations, but spinor components all mix up with one another under any rotation, and I have not been able to see why it must be positive definite in all circumstances, I always have to try it out case-by-case.
If you have an argument for why an extra y-z rotation proves that the relevant correlation function for spinor fields is positive definite, then I'll fix up the text. Right now, it's not rigorous, as you say, precisely because you have to worry about whether the polarizations match up between the annihilating operator and the creating one.Likebox (talk) 17:00, 16 January 2010 (UTC)
Also, I did not mean to imply that Streater and Whitman are fools--- they aren't. But they write in this foolish overly formal language which is obfuscatory, and this is something that was encouraged in the sixties. Maybe it's just been too long since I looked at their book.Likebox (talk) 17:20, 16 January 2010 (UTC)
OK, I will have to think about CP2T and the spinors, or see how Streater and Whightman prove this point (they do not do that directly, because they follow a somewhat different - though parallel - line of argument). PS I agree that their book is written with conventions which are hard to follow, but I guess it's just because these were acceptable conventions at the time. Maybe in 50 years people will say the same thing about today's books. Dan Gluck (talk) 18:12, 16 January 2010 (UTC)
I will also correct soon the plus/minus thing of which I wrote, unless you find some fault in what I've written.Dan Gluck (talk) 18:17, 16 January 2010 (UTC)

### Remarks on "Some remarks"

I read the technical stuff you wrote just now--- some of it is not right. You do NOT get that everything is bosonic by rotating the correlation function--- you get exactly spin/statistics independent of whether you use "P1" or "P2". The operator "R" is not what you think it is. It's just a polarization rotation matrix. It could be neither unitary nor antiunitary, because it acts on fields, not on states. It is a very simple polarization rotation matrix, not an operator acting on the Hilbert space of field states.

This is why I don't talk about the Minksowski stuff at all. I don't want to talk about anti-unitary or unitary business, which is completely avoidable if you stay in Euclidean space. It becomes all that in Minkowski space, but that is not necessary for understanding spin/statistics. The argument is simple: if you have the correlation function "Rphi phi", then just by rotation it equals -" phi R phi" for half integer spin.

I am well aware that Streater and Whitman use the operator "lambda" which does a full parity reversal on all spatial dimensions, but I don't like it. It doesn't generalize to even number of space dimensions, and secondly it does a second rotation in the y-z plane which is completely irrelevant for the argument. The only possible relevance is to ensure that the proper polarization operator is positive definite, but they don't discuss this point in there, as far as I remember.Likebox (talk) 22:47, 16 January 2010 (UTC)

If this is so, then the proof you wrote just isn't right, because for spinors you get that G(x) is zero. Indeed, you don't give a proof that G(x) is positive definite. You have to assume a CPT transformation because this is the proper way to rotate things through the x-t plane (AFAI understand, P and T make a "naive" rotation plus chaning the 3d spin representation, while C also changes representation under boosts). Also, if you reverse the time direction, you have to assume anti-unitarity (see Peskin and Schroeder, section 3.6, the pasrt on the T symmetry). You cannot ignore operators because even putting x and t as arguments in φ are formulated as translation operators acting on the operator φ. This is quantum mechanics.
Also, the y-z plane IS relevant because you may define two different R's, one which makes a rotation in that plane and the other that doesn't, and potentially there is a sign difference in the result for spinors if you use one and not the other. This shows that your proof as it stands now has some problem. By the way, I am no longer sure which type of CPT is correct (with or without y-z rotation), because it seems (I still have to check calculations) that G(x) the way I define it is non-zero for fermions only if we take CP1T (this also shows that the y-z rotation does matter, but I am not sure yet if it should be there or not).
Also, "positive definiteness" is not good, because a CPT transformation on fermions have an overall phase ambiguity. You should use "non-vanishing" instead. S&W do prove something equivalent to this non-vanishing, because they show that the anticommutator of some quantity is non-vanishing, from which the non-vanishing of what we want follows, but I am not sure about their proof anyway.
Dan Gluck (talk) 19:36, 19 January 2010 (UTC)
Actually I did have a mistake in my argument, because while using the antiunitarity of S, I didn't switch the places of the S-s and S-1-s as well. Therefore, for example, it is not necessarily true that RR = 1. I think I am beginning to suspect what's going on, but I'll write when I am more sure about it. Dan Gluck (talk) 20:56, 19 January 2010 (UTC)
Hello, thank you for checking everything over--- that's very valuable especially for the spinor case, which I didn't do very carefully. But I can clarify some points that might be confusing at the moment.
Hilbert Space Operators vs. Field Transformations
This is about whether the "R" matrix acting as an "operator". It doesn't, and this shouldn't be confusing, it's just fine. This is a point which I am not confused about and I am pretty sure you aren't either, but its good to get it down for the benefit of other editors which might come by in the future.
When you have a transformation like moving a field from point 0 to point x, you can implement it in the quantum theory by using a unitary transformation on the Hilbert space, but you do it this way:
${\displaystyle \phi (x)=T(x)\phi (0)T(x)^{-1}}$
Given the unitary hilbert space matrix T(x), you need to use T(x) inverse on one side, and T(x) on the other, and then it does the translation on the field operator. Similarly, to spatially rotate the field phi, you can use a unitary Hilbert space operator, lets call it U(R):
${\displaystyle R\phi =U(R)\phi U(R)^{-1}}$
The U(R) is the unitary operator, the R is the simple matrix that mixes up the components of phi. The unitary operator is completely different from the R-matrix, which is just a polarization matrix. When I say glibly "rotate the correlation function", one way of thinking of it is as rotating the state
${\displaystyle R\phi \phi |0\rangle }$
so that it gets multiplied by U(R)
${\displaystyle U(R)R\phi \phi |0\rangle =U(R)R\phi U(R)^{-1}U(R)\phi U(R)^{-1}U(R)|0\rangle }$
and this interspersings of U(R) and its inverse just say that rotating the state is the same as rotating the fields, and rotating the initial state. Rotating the fields can be done in the obvious way--- just by multiplying by the R matrix, and this is the reason that the proof works.
The action of the rotation hilbert space operator U(R) acting in this sandwiched way on the field is to mix up the components according to the rotation matrix R.
Why don't I say it this way? It's because in Euclidean space, the hilbert space of states is sort of remote. You should think of the Euclidean theory more like a monte-carlo simulation with a probability space, or as some sort of signed probability space for Fermions, and the Hilbert space can be reconstructed from the correlation functions after analytic continuation.
Each rotation becomes a unitary operator once you reconstruct the hilbert space, and if the rotation plane involves time, the corresponding operator is antiunitary in the analyic continuation back to Minkowski space. I didn't want to get involved with this silliness, because this is where you can get endlessly bogged down. You can sidestep any mention of Hilbert spaces and quantum mechanics by staying Euclidean, and knowing that the rotation properties of the field are given by the spin, while ignoring the precise implementation of these rotations/time-reversals as quantum mechanical operators. I know they have a unitary/antiunitary version, but who cares what it is exactly.
The precise implementation of the unitary operator corresponding to a rotation depends on whether the rotation is in a plane involving the dimension which you analytically continue to be time, or whether it involves only two space dimensions. These annoyances get S&W so bogged down in technicalities, and I don't want to have the same problem here.
If you stay in Euclidean space, you can just say "rotate this correlation function", and it means just that--- consider rotating the fields in a plane of rotation in the obvious way--- just rotate the field representation. This rotation will get implemented as some sort of unitary or antiunitary transformation once you construct the quantum theory from the Euclidean theory, but you can ignore all the details. The only thing that is essential is that you get a plus or minus sign from the rotation according to the spin representation of the field.
Many components
For anything except scalar fields, there are many components in the correlation function: consider \phi_a where a is the representation index (say a spinor index), and the R matrix acts on this index. So Rphi phi is actually a bunch of different correlation functions, which can be annotated:
(R\phi)_a \phi_a
If any of these diagonal correlation functions is not identically zero, then the proof goes through. As I said, you can check it case by case, and it is clearly true for bosons in more than two space-time dimensions whose polarizations can be perpendicular to the plane of the R rotation, and so unrotated, but I didn't check it in general. It shouldn't be too hard to do, I'll think about it actively (sorry for being lazy).Likebox (talk) 01:29, 20 January 2010 (UTC)

#### Missing part?

OK, I think I've finally understood what you mean. R, by the way, cannot be simply a rotation matrix in the spinoric case (I'll consider this later), but this is of no importance. Let me write it in a more rigorous way, and tell me if that's really what you meant:

Consider the correlation function:

${\displaystyle G(x)=\langle 0|R[\phi (-x)]\phi (x)|0\rangle }$

Where R is a CPT transformation, which is equivalent by analytic continuation to a rotation in the x-t plane for a Euclidean 4-dimensional space.

Because the theory and the vaccum state are Lorentz-invaiant, G(x) is invariant under any Lorentz transformation Λ, meaning that:

${\displaystyle G(x)=\langle 0|\Lambda [R[\phi (-\Lambda ^{-1}x)]]\Lambda [\phi (\Lambda ^{-1}x)]|0\rangle }$

We may now use analytic continuation to Euclidean space and holomorphicity in the parameters of Λ (I'm not sure that this is allowed, because correlation functions is holomorphic in x, but are they in any parameters as well? seems correct but I'm not sure). We then take Λ to be a π rotation in the x-t plane. This is just the analytic continuation of R, which we shall denote R'. so we have:

${\displaystyle =\langle 0|R'[R'[\phi (x)]]R'[\phi (-x)]0|\rangle }$

Since the first term is twice a rotatin by π in the x-t plane, it brings φ to plus or minus itself, depending on spin, Thus we have:

${\displaystyle =\pm \langle 0|\phi (x)R'[\phi (-x)]|0\rangle }$

We now make analytic continuation back to Minkowsky space and get:

${\displaystyle G(x)=\pm \langle 0|\phi (x)R[\phi (-x)]|0\rangle }$

This consitute a proof if, and only if, G(x) can be shown to be non-zero. This, however, is not trivial to show. This is a part missing in your proof. Dan Gluck (talk) 13:44, 22 January 2010 (UTC)

R is just a matrix which performs a rotation on the components of phi. It is not just a rotation matrix (except for the case of spin 1), it's a matrix which, given a rotation, shows you how to mix up the components of phi in order to implement the rotation. This is what is called a "group representation", and it is the definition of what spin is. In the spinor case, it is the matrix which mixes up the spinor components according to the rotation.
For a dirac spinor, the matrix R is the exponential of the product of two gamma matrices. For spin 3/2 in the Rarita-Schwinger form, considered as a vector of spinors, R will be do a reflection on the x-t components of the vector, while it will do the spinor thing on the spinor index. For each representation, you construct the matrix R.
R is not a CPT transoformation. It is a simple matrix. CPT transformations acts on the hilbert space of states in the Minkowski theory. R is a simple matrix that acts on the field components in the Euclidean theory. They are two different things. Once you know the Euclidean rotation properties, you can figure out the Minkowski-space CPT properties, but this requires some annoying work to reconstruct the states and to figure out what each Euclidean rotation analytically continues to.
While I agree that it is slightly non-trivial to show that G(x) is nonzero, you can explicitly check that it is nonzero for any of the propagators you find in a field theory textbook, and it is pretty obvious that this is because of positive definite metric.Likebox (talk) 16:58, 22 January 2010 (UTC)
To expand a little: you analytically continue to Euclidean space by considering all correlation functions as functions of "x,y,z,it". It is then true that any Lorentz invariant field theory with a positive definite energy (so that the analytic continuation works) becomes a rotationally invariant Euclidean theory. This is what Streater and Whitman spend a large part of their book doing. I just assumed this is already known.
The argument here uses only the Euclidean theory. I never said anything about the Lorentzian theory, because I didn't want to get involved in the reconstruction of the Hilbert space. Again, the only gap in the proof is the demonstration that G(x) is always nonzero, and you can explicitly check that this is true for any special case.Likebox (talk) 17:03, 22 January 2010 (UTC)

## Just in case I disappear

There are people pissed off at me over a long-standing dispute at Godel's incompleteness theorems. I gave a proof of the theorem which uses non-standard terminology, and there are a few editors that oppose it strongly. These editors are annoyed enough to ask that I be blocked. If I am blocked, this page should still get completed. I have not finished thinkng about positive definiteness yet, but I can tell you the following:

1. For tensors, the operation "R" just reverses the sign of two Euclidean directions, x and t. It is easy to see that the relevant correlation functions are nonzero, from the Lehman spectral representation. In the bosonic case, everything follows from positivity of the Hilbert space metric.
2. For spinors, the operation "R" acts like multiplication by "i", in other words, it mixes up real and imaginary components of spinors into each other. This is always true, but in some presentations of the gamma matrices it won't be multiplication by "i", but by some anti-symmetric matrix which acts like i. To see that the resulting correlation function should be nonzero, you need to consider the reflection positivity condition for spinors. This isn't done here (and it isn't fully discussed in S&W if I recall correctly).
3. By combining spinor and tensor indices, you should be able to do any representation of the Lorentz group, in any dimension.
4. The spinor case needs to be worked out very carefully, because the spinor representations in Euclidean space are different than the spinor representations in Minkowski space. That doesn't mean that Euclidean continuation doesn't work--- it does. It just requires careful thinking.
5. Sudarshan recently claimed that in eight dimensions, there is a counterexample to spin-statistics. He claims that there is a spinor representation which can be consistently bosonic in 8d.
6. Sudarshan's claim is logically possible, but I am not sure if it is correct. But it is important to realize that this means that for this 8d spinor, the relevant correlation function would have to be identically zero. If there is a spinor representation where all the correlation functions of the form Rpsi psi are zero, then you could give it bosonic statistics. I didn't read Sudarshan's paper in detail, since he does everything is sort of old fasioned ways. But is might be worthwhile to do so.Likebox (talk) 12:20, 23 January 2010 (UTC)

1. You cannot prove this for a Euclidean theory, because such a theory does not posess "microcausality", which is defined as the property that every two spacelike separated fields either commute or anticommute. Without such a property, the whole spin-statistics theorem is meaningless. Indeed, microcausality is one of the few assumptions which S&W take in their proof. By the way, as I get deeper into their book, I must admit that I agree with your comments about them :(
2. In a Minkowsky space, R must be a CPT transformation. Otherwise, for a spinor, G(x) is zero. To show this, let's assume it's z instead of x - the math is easier. Then, in a Euclidean space a π rotation in the z-t plane is:
${\displaystyle e^{(i\pi /2)i\gamma ^{0}\gamma ^{3}}=e^{(\pi /2)\gamma ^{0}\gamma ^{z}}=e^{(i\pi /2)\left({\begin{array}{cc}\sigma ^{3}&0\\0&-\sigma ^{3}\end{array}}\right)\,},}$

Since the argument of the exponent is the diagonal {1,-1,-1,1}, the result is simply the diagonal

${\displaystyle \{e^{i\pi /2},e^{-i\pi /2},e^{-i\pi /2},e^{i\pi /2}\}}$ = {i,-i,-i,i} = -iγ0γ3


So if R is just this, we have:

${\displaystyle G(z)=-i\langle 0|\gamma ^{0}\gamma ^{3}\psi (-z)\psi (z)|0\rangle =0}$

It is very easy to show that this is indeed zero, because the ψs include only annihilation operators of type "a" ("electrons") and only creation operators of type "b" ("positrons").

It is easy to see that the reason is that we had to take R to be a CPT transformation.

Let's see that:

ψ is a sum of annihilation/creation operators multiplying both the usual factor of exp(ipμxμ) and also a spin-half representation of the Lorentz group, which is of two types - either: ${\displaystyle u^{s}(p)=\left({\begin{array}{cc}{\sqrt {p\cdot \sigma }}\eta ^{s}\\{\sqrt {p\cdot {\bar {\sigma }}}}\eta ^{s}\end{array}}\right)\,,}$

or: ${\displaystyle v^{s}(p)=\left({\begin{array}{cc}{\sqrt {p\cdot \sigma }}\eta ^{s}\\-{\sqrt {p\cdot {\bar {\sigma }}}}\eta ^{s}\end{array}}\right)\,,}$

where ${\displaystyle p\cdot \sigma =p^{0}-p^{i}\sigma ^{i}}$ and ${\displaystyle p\cdot {\bar {\sigma }}=p^{0}+p^{i}\sigma ^{i}}$. Also, the η-s are 2-vectors representing spin, with η1,2 being two unit-vectors expanding the 2-dimensional space they live in.

It is u and v that are multiplied by the γ-s we had earlier.

Because the v-s actually have an opposite energy related to that of the u-s, we have: ${\displaystyle \psi (x)=\int {d^{3}p \over (2\pi )^{3}}{1 \over {\sqrt {2E_{p}}}}e^{ip\cdot x}\sum _{s,s^{\prime }}\left(a_{p}^{s}u^{s}(p)+b_{-p}^{s^{\prime }}v^{s^{\prime }}(-p)\right)}$ where a are annihilation operators and the b-s are creation operators, though before knowing the result of the spin-statistics theorem we may erronously think the b-s should be annihilation operators as well. Eventually, it is a custom to call "b" by "b-dagger" for convinience.

Let us call the states of the type ${\displaystyle a^{\dagger }|0\rangle }$ "electrons", and states of the type ${\displaystyle b^{\dagger }|0\rangle }$ "positrons". The former multiply u-s, and the latter multiply v-s. i.e., they belong to different representations.

The main point is, that if we take ψ and multiply it by γ0γ3, the u-s are turning into v-s and vice versa. For example, if we have a spin state in the up z-direction, then we have: ${\displaystyle u(p)=\left({\begin{array}{cc}{\sqrt {p^{0}-p^{z}}}\\0\\{\sqrt {p^{0}+p^{z}}}\\0\\\end{array}}\right)\,,}$

${\displaystyle \gamma ^{0}\gamma ^{3}u(p)=\left({\begin{array}{cc}-{\sqrt {p^{0}-p^{z}}}\\0\\{\sqrt {p^{0}+p^{z}}}\\0\\\end{array}}\right)\,=-v(p),}$

So we have an electorn state of the wrong representation: ${\displaystyle a_{p}^{\dagger }v(p)|0\rangle }$. There is no such electron state in our Hilbert space. In order for the z-t rotation to give us a state which belongs to our theory, we have to take "a" operators to "b" and vice versa. The meaning of this is simply to change the Lorentz representation we are working with, in accordance with the rotation we have made. This is precisely a C transformation, and as far as I understand it, it can be understood without the need to really explore all your Hilbert space - you just have to remember to change your Lorentz representation.

More generally, when you make a π z-t rotation in a Euclidean space and want to analytically continue to Lorentz space, you get from one copy of Lorentz group (the "orthochronous proper" one) to another copy (the "antichronous antiproper" one or something like that), and your representation may change - in particular, left-handed chiralities and right-handed chiralities change sign with respect to one another (in the Dirac fermion example, the first halves of u and v are the left-handed chiralities and the second halves are the right-handed). The C transformation is what changes the representation properly. Dan Gluck (talk) 13:03, 23 January 2010 (UTC)

## Yet again, why the missing part is non-trivial

Continuing the Dirac Fermion example, We have the freedom to take R to be a π rotationboth in the z-t plane and in the x-y plane. In this case, it can be shown (in a similar manner to what I did) that this rotation is made by γ0γ1γ2γ3 (up to a factor). But in this case, the double rotation in the proof brings you back to ψ with no minus sign! This means that we must have in this case G(x) = 0, or we get a wrong spin-statistics theorem, stating that spinors are fermions!

Indeed, G(x) = 0 in this case and it is non-zero in the other case (if you work properly, with CPT in Minkowsky space):

${\displaystyle \langle 0|CP2T[\psi (z)]\psi (z)|0\rangle =\langle 0|\psi ^{\dagger }(-z)\gamma ^{0}\gamma ^{1}\gamma ^{2}\gamma ^{3}\psi (z)|0\rangle =\langle 0|\psi ^{\dagger }(-z)\gamma ^{5}\psi (z)|0\rangle =\langle 0|{\bar {\psi }}(-z)\gamma ^{0}\gamma ^{5}\psi (z)|0\rangle =0}$
${\displaystyle \langle 0|CP1T[\psi (z)]\psi (z)|0\rangle =\langle 0|\psi ^{\dagger }(-z)\gamma ^{0}\gamma ^{3}\psi (z)|0\rangle =\langle 0|{\bar {\psi }}(-z)\gamma ^{3}\psi (z)|0\rangle \neq 0}$

The equalities are up to factors of i and so which I didn't want to follow.

The last equalities in both cases are not so trivial and follow (for a free fermion, which is what we're studying anyway) from Psekin&Schroeder (3.115). In the second case:

${\displaystyle \langle 0|{\bar {\psi }}(-z)\gamma ^{3}\psi (z)|0\rangle =-Trace\left(\gamma ^{3}(m+i\gamma ^{\mu }{d \over dx^{\mu }})\right)\int {d^{3}p \over (2\pi )^{3}}{1 \over 2E_{p}}e^{2ip_{z}z}=-i{d \over dz}\int {d^{3}p \over (2\pi )^{3}}{1 \over 2E_{p}}e^{2ip_{z}z}\neq 0}$

While in the first case the trace is over ${\displaystyle \gamma ^{0}\gamma ^{5}(m+i\gamma ^{\mu }{d \over dx^{\mu }})}$ which is always zero.

It is still not clear to me whether this non-trivial calculation can be avoided by arguments from symmetries. Even if this is so, you should now show this, case by case, for all types of fermions in all possbile representaion. Since there are infinite such representaions, I think you should find some general argument instead. Dan Gluck (talk) 13:23, 23 January 2010 (UTC)

PS, good luck with not disappearing, but I'm not sure I would like you to win the Godel's case, because I was the original writer of the article about the proof :) . Yet, at that time it was relatively simple and since then it has been expanded and made rigorous by mathematicians experting on the subject, and I got two children born, so I no longer follow it, and I don't know what they've made out of it. Anyway, hope to continue our correspondence. Dan Gluck (talk) 13:27, 23 January 2010 (UTC)

Heh--- the Godel article is not too bad. The issue is that there is a much simpler way to present the proof, if you are willing to argue about the properties of computer programs. The basic idea is that if you have an axiom system "S" which proves enough things about the integers, you write the program GODEL to do the following
1. print its own code into a variable R
2. deduce all theorems of S looking for "R does not halt"
3. if it finds this theorem, it halts
Then it is easy to see that Godel does not halt is the Godel sentence for the theory, but good luck convincing the people there that this is equivalent to Godel's argument. I gave proofs of the Godel-Rosser theorem, and of Lob's theorem, and the simplification is tremendous. All the nontrivial work is in the first step--- showing that a program can print its own code. I discussed it at length there. I probably won't get banned. I was just getting intimidated by the strong-arm tactics.
Anyway, about this page, I think that you studied Minkowski quantum field theory, but you are not completely comfortable with Euclidean theory. The Euclidean theory is not a quantum field theory in Euclidean space, it's a completely different thing. It's a mathematical construction which can be used to reconstruct the Lorentzian quantum theory. It lives in another world altogether.
You shouldn't talk about "microcausality" in the Euclidean theory--- these concepts don't make sense. This is a Euclidean theory, it doesn't have a light cone, and energy states don't get multiplied a phase when you move forward in time like in quantum mechanics, they shrink according to the energy. The Euclidean theory is a rotationally invariant statistical field theory, and the only thing that makes sense for this statistical field theory is the path integral (which tells you the probability weight of different states) and the rotation properties of the fields. It is not trivial to translate physical intuition from Minkowski to Lorentzian theories.
Because you can reconstruct the Lorentzian from the Euclidean theory, there are direct analogs for every Minkowski space concept: there is a dictionary.
Positive definite metric (Lorentzian) is Reflection positivity (Euclidean)
Micro-causality is (implied by) a local action function
Spin (Lorentzian) is Field representation (Euclidean)
Quantum State (Lorentzian) is a superposition of Euclidean configurations, but like probability
The statement that you make that "R" needs to be a CPT operation is not true exactly the way you say: there is a CPT operator, let's call it U, such that UphiU^{-1} is equal to Rphi, so U implements R in the Lorentzian theory, and indeed U is an antiunitary operator. But R itself is just a stupid matrix. U acts on the Hilbert space, while R acts on the fields. The property of spin-statistics is just a consequence of the symmetry properties of certain correlation functions, and it is easiest to demonstrate this symmetry in Euclidean space, then use this symmetry to establish spin-statistics after translating to Minkowksi space using the dictionary.
If you stay Euclidean, the symmetry property of the correlation function is the analog of spin-statistics. It translates to to spin-statistics by going to Lorentzian quantum field theory.
What you wrote about spinors is correct only if you do normal rotations. If you rotate the Euclidean space version of a spinor field by 180 degrees, you change the spinor field to its conjugate in the Lorentzian continuation. Your arguments are doing rotations in the Lorentzian theory, not rotations in the Euclidean theory, and they can be stated much more easily than you say: rotating a spinor in Lorentzian space never mixes up psi with its conjugate psi-bar, so that for a non-Majorana spinor the correlation function is always zero. That's true, but irrelevant.
When you rotate in the Euclidean theory, the rotation does mix up psi and psi-bar when you reverse the direction of time. It corresponds to an antiunitary operator in the Lorentzian theory. It's not zero in the case of an ordinary spinor in 4d.Likebox (talk) 13:40, 23 January 2010 (UTC)
Hi.
The Godel proof looks really cool if it's true, I have to think about it.
Also, I agree with everything you say, except that I think it's better to write things in the Lorentzian theory with which most people are more familiar. If you insist on the Euclidean one, you should write the dictionary which you mentioned here. Also, I think that the Lorentzian version should be mentioned anyway, i.e. either go through the Lorentzian option or go through both. Otherwise it is very confusing (it was for me, at least). I also agree that what I showed about R needing to include C in the Lorentz version is trivial, I just didn't know if you accept this.
Could you please explain "Positive definite metric (Lorentzian) is Reflection positivity (Euclidean)"?
Last but not least, you still have to prove this missing part, whichever signature you choose. What about it?
Dan Gluck (talk) 20:03, 23 January 2010 (UTC)
OK, I got the Godel idea. It is really non-trivial to show that step 1 is possible. I like the direct connection with the halting problem. Is this proof due to yourself? Dan Gluck (talk) 20:05, 23 January 2010 (UTC)
Thinking about it yet again, do you save any effort in this way of proving Godel's theorem? Well, I guess I'll have to read the discussion there some day.Dan Gluck (talk) 20:24, 23 January 2010 (UTC)
Unfortunately, the Godel proof is basically due to me. It is basically identical to Godel's original argument--- it proves no new theorems without further work--- so I think it is unpublishable as is. It is not accepted here because it talks about computer programs instead of recursive functions. That's Wikipedia for you.
Reflection positivity is discussed in Streater and Whitman. Their discussion of reconstruction of the Hilbert space is one of the best parts of the book. Unfortunately, they do not discuss reconstruction details for chiral spinors, which is where the headaches lie. I'll try to clear it up and write it out. I completely agree with you about the gap in the proof (I noticed it myself, and the same gap appears in all published proofs). I am still trying to fix it, but its annoying to do. There's also Sudarshan's example, which needs to be understood in detail (I don't).Likebox (talk) 21:12, 23 January 2010 (UTC)
Its pretty easy to show that a program can print its own code by construction. But the construction requires writing code! In order to avoid this, mathematicians use what they call "recursive function theory", and then the relevant lemma is the recursion theorem. This says that instead of feeding a program its own code as one of the inputs, you can make the program generate its own code internally. It's a corrolary of what they call the s-m-n theorem, which is a triviality in terms of real computer code. I discussed it ad nauseum at the relevant page. Hope to see you there.

I agree that S&W's proof also lacks a corresponding part (there it's in their assumption that the comination of two fermions give a vector, while in principle it can also give a scalar, depends on the combination), which is one reason why I don't insist on giving it here, thought technically the proof you gave here is original work. In fact so is your proof in the Godel article. I think these particular things are good to have in Wikipedia, but since the Wiki standards are good as general guidelines, you cannot resist someone who insist on giving Godel's original proof instead of yours. Without reading what's there right now, why don't you accept this as the main proof, and add yours as "relation of the proof to the halting problem" or so? I guess this will bring up much less resistance, and anyone who's interested will get to read your more intuitive proof as well. Dan Gluck (talk) 15:21, 24 January 2010 (UTC)

That's exactly what I did. The arguments came from people who specificially did not want to see a computer program in the proof, and these people colluded with people who did not understand the mathematics.Likebox (talk) 18:39, 24 January 2010 (UTC)

## Remove the "bogus proof"

In any mathematical or physical discussion, the inclusion of an incorrect line of argument is highly discursive. Although I am not remotely equipped to critique the content of any argument in this article, nevertheless, speaking as one attempting to learn something, I can say that the inclusion of the bogus proof adds significant confusion.

If it is somehow too much of a loss to remove the entire bogus argument, I strongly suggest condensing it down to, say, two or three sentences. This is already a long and complex article anyway!

## Typography in the title

Should it become the Spin–statistics theorem (with a true dash)? Incnis Mrsi (talk) 12:42, 22 August 2011 (UTC)

## Final section on Lorentz group representations

What exactly is meant by the last section on how this relates to representations of the Lorentz Group? How exactly does Fermi-Dirac statistics solve the 'naive' problem described? I have looked at the given reference but it is a large book and I don't know which part is meant.

Tobycrisford (talk) 20:25, 21 September 2013 (UTC)

## The spin can be uncertain, but bosons never mix with fermions

This is partially a continuation of talk:Identical particles #Composite particles, partially some thoughts parallel to WT:WikiProject Physics‎‎ #Quantum particle and subsequent threads, and partially a sequel to my [1] edit.

I am largely a self-educated physicist (my main education is mathematical). That’s why I am not so interested in the proof, but interested in physically reasonable generalizations. The statement that all particles have spin is hypocritical. Of course, for an elementary particle the spin is certain. In other words, for a particle at rest: J2 = const for all internal (a.k.a. spin) states; this is possibly related to the “total angular momentum” thread at WT:PHYS. For such composite particles as hadrons it was observed that their rest mass eigenstates also have certain spins. I mean:
π+
/
ρ+
,
π0
/
ρ0
,
π
/
ρ
are essentially different states of the same thing, but they have different masses and different spins. Note that all of these are bosons.

Is the situation with an 1H atom different, in principle, from π and ρ mesons? It isn’t: there is a spin-0 singlet and a spin-1 triplet. Obviously, the s–s is applicable to each of these separately, but the bosonic statistics also applies to the 1H1 as a whole, including mixtures between the singlet and the triplet. Similar (but with more intrinsic states) should be true for 4
He
. I am sure the practical Bose condensate state of its atoms does not have a certain spin; it is a superposition of several atomic spin states. Maybe I say trivial things for advanced physicists but, remember, I am self-educated and recovered this piece of theory from my own meditation.

We see: the s–s theorem explains why bosons and fermions never mix. So, the problem is whether a formulation exists that permits for an uncertain spin, for particles where S2 is not constant. Note there was some confusion about whether atoms can be bosons – now I understand why. Incnis Mrsi (talk) 17:36, 21 January 2014 (UTC)

## Importance

Why is this article "rated as Mid-importance on the project's importance scale"? The spin-statistics relation is a major aspect of quantum mechanics with implications ranging from the existence of atoms with a variety of chemical properties to the pressure countering collapse of neutron stars. Perhaps if the importance were acknowledged, an expert in this field would become inspired to re-work the article into a form with improved presentation more readily available to both physicist and layperson. David in Cincinnati (talk) 15:34, 17 February 2016 (UTC)

The operator

${\displaystyle \iint \psi (x,y)\phi (x)\phi (y)\,dx\,dy}$

(with ${\displaystyle \phi }$ an operator and ${\displaystyle \psi (x,y)}$ a numerical function) creates a two-particle state with wavefunction ${\displaystyle \psi (x,y)}$, and depending on the commutation properties of the fields, either only the antisymmetric parts or the symmetric parts matter.

This makes little sense.

How is ${\displaystyle \phi }$ an operator? It appears to be a function of one variable. How can the whole expression be an operator? It appears to be a number. At best, it is a functional that assigns a number to... something, but the author should then say what.

How can an expression "create" a state? Maybe it computes an amplitude. I think it computes the amplitude for the joint state ${\displaystyle \psi }$ to actually be the pure tensor product state ${\displaystyle \phi \otimes \phi }$, where the two particles are in the same state ${\displaystyle \phi }$.

That is the only sense I can make of this, but as stated, it is word salad. 198.129.64.60 (talk) 17:21, 3 May 2016 (UTC)