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I am attempting to understand Weinberg's formulation of the spin-statistics theorem as presented in his book "The quantum theory of fields: foundations" pages 233-238. I have at my disposal all three of his Phys rev papers on "Feynman rules for any spin I-III," as well as Novoshilov's book on particle physics (1975, relevant pages 60-77 chapter 4), Streater and Wightman's "PCT, Spin and Statistics, and All That" (1989), Duck and Sudarshan's "Pauli and the spin-statistics theorem" (1998), and Pauli's 1940 paper "The connection between spin and statistics" (Phys rev 58, 716 1940).

Suffice to say either my interpretation of these references, or my understanding is getting stuck. My main problem is with the introduction of the $(-1)^{2j}$ term in the expression for the (anti)commutator relationship between fields: $$\left[ \psi_{ab}(x),\tilde{\psi}^\dagger_{\tilde{a}\tilde{b}}(y) \right]_{\mp}=\left[\kappa\tilde{\kappa}^*\mp (-)^{2A+2\tilde{B}}\lambda\tilde{\lambda}^*\right]P_{ab,\tilde{a}\tilde{b}}(-i\nabla)\Delta_+(\textbf{x}-\textbf{y},0) +\left[\kappa\tilde{\kappa}^*\pm (-)^{2A+2\tilde{B}}\lambda\tilde{\lambda}^*\right]Q_{ab,\tilde{a}\tilde{b}}(-i\nabla)\delta^3(\textbf{x}-\textbf{y})\tag{5.7.19* in Weinberg QtOF:I} $$

Or from Novoshilov page 77:

$$ \left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}= \frac{1}{(2\pi)^3}\int{\frac{d^3p}{2p_0}D^J_{\sigma\sigma'}}\left(\frac{p}{m}\right)\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y}\} $$

$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{-i\partial}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm(-1)^{2j}\eta\eta^*e^{-ip(x-y}\} $$

In this latter case, the explanation for the appearance of $(-1)^{2j}$ is given as "where we have used $m\alpha\alpha^+=p$ and $D^J\left(-1\right) = (-1)^{2j}$."

In Weinberg's case, the form of the fields $\psi_{\sigma}(x)$ requires that $\left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}$ include terms where the coefficient functions are multiplied by their complex conjugates (as $u_{ab}(\textbf{p},\sigma)\tilde{u}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma)$ below):

Ie: if $$\psi_l(x)^+ = \sum_{\sigma, n}(2\pi)^{-3/2}\int{d^3p*u_l(\textbf{p},\sigma,n)*e^{ip*x}*a(\textbf{p},\sigma,n)}$$ $$ \psi_l(x)^- = \sum_{\sigma, n}(2\pi)^{-3/2}\int{d^3p*v_l(\textbf{p},\sigma,n)*e^{-ip*x}*a^\dagger(\textbf{p},\sigma,n)} $$ $$ u_{ab}(\textbf{p},\sigma)=\frac{1}{\sqrt{2p^0}}\sum_{a',b'}\left(e^{-\hat{\textbf{p}}*\textbf{J}^{(A)}\theta}\right)_{aa'}\left(e^{\hat{\textbf{p}}*\textbf{J}^{(B)}\theta}\right)_{bb'}\times C_{AB}(j\sigma;a'b')\tag{5.7.14} $$ and $$ v_{ab}(\textbf{p},\sigma)=(-1)^{j+\sigma} u_{ab}(\textbf{p},-\sigma)\tag{5.7.15} $$

Then, we can write

$$ (2p^0)^{-1}\pi_{ab,\tilde{a}\tilde{b}}(\textbf{p}) \equiv \sum_{\sigma}u_{ab}(\textbf{p},\sigma)\tilde{u}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma) = \sum_{\sigma}v_{ab}(\textbf{p},\sigma)\tilde{v}^*_{\tilde{a}\tilde{b}}(\textbf{p},\sigma)\tag{5.7.20} $$ as $$ \pi_{ab,\tilde{a}\tilde{b}}(\textbf{p})=P_{ab,\tilde{a}\tilde{b}}(\textbf{p},\sqrt{\textbf{p}^2+m^2})\tag{5.7.22} $$

and regroup terms to turn this into a function of $\textbf{p}$ only:

$$ \pi_{ab,\tilde{a}\tilde{b}}(\textbf{p})=P_{ab,\tilde{a}\tilde{b}}(\textbf{p})+2\sqrt{\textbf{p}^2+m^2}Q_{ab,\tilde{a}\tilde{b}}(\textbf{p}) $$

Where

$$ P(-\textbf{p})=(-)^{2A+2\tilde{B}}P(\textbf{p}) $$ $$ Q(-\textbf{p})=-(-)^{2A+2\tilde{B}}Q(\textbf{p})\tag{5.7.26} $$

But in all these cases I do not see how we can preferentially multiply $(-1)^{2j}$ to the $e^{-ip(x-y)}$ term alone. In the case of Novoshilov, because

$$ \hat{p}\equiv-i\partial $$

His "page 77" simply reads to me as:

$$ \left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}= \frac{1}{(2\pi)^3}\int{\frac{d^3p}{2p_0}D^J_{\sigma\sigma'}}\left(\frac{p}{m}\right)\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y)}\} $$

$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{p}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y)}\} $$

$$ =\frac{1}{(2\pi)^3}D^J_{\sigma\sigma'}\left(\frac{-i\partial}{m}\right)\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm(-1)^{2j}\eta\eta^*e^{-ip(x-y)}\} $$

Wherein this $(-1)^{2j}$ term simply appears on the inverse exponential as if by magic. So too, does Weinberg's proof run into difficulty. The statement $(5.7.19*)$ only makes sense if the form of the integral in the (anti)commutator returns $Q(-\textbf{p})$ and $P(-\textbf{p})$ for the $Q(p)*e^{-ip(x-y)}$ and $P(p)*e^{-ip(x-y)}$ terms only. But I do not see how this occurs. Why would not both $e^{\pm ip(x-y)}$ terms simply act as $F(p)e^{\pm ip(x-y)}$ and not one preferentially as $F(-p)$?

In other words, why does the $(-1)^{2j}$ term only survive on one component of the commutator or anticommutator?

In Streater and Wightman's treatment, where as best I can tell the issue comes down to the number of dotted and undotted indices in the irreducible-Lorentz-representation spinors, this same sort of "preferential" action is expressed in $(4-51)$, where the authors write that "[...this result] is a consequence of the transformation law of [the holomorphic function] $\hat{W}$ under the group $SL(2,C)\otimes SL(2,C)$..." And this is borderline unintelligible to me.

Does anyone know why what seems to be a violation of the associative property is allowed here? I am likely missing something specific, and I would be greatful for any and all help towards the right direction.

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2 Answers 2

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I do not see how we can preferentially multiply $(-1)^{2j}$ to the $e^{-ip(x-y)}$ term alone

Because that's the origin of the spin-statistics theorem.
It comes from the requirement of the theory to be causal.
And the term that would cause a problem in this case is the space-like term $x-y$.

For a theory to be causal, the time ordering of physical events affecting the system's evolution cannot be reversed. This is especially problematic for space-like separations where a Lorentz boost may reverse the chronological order $t_{\mathrm{final}} - t_{\mathrm{initial}} < 0$. For causality to hold, any two space-like separated operators are required to commute: \begin{equation} [\mathcal{O}_1(x), \mathcal{O}_2(y)] = 0 \quad \text{if} \quad (x-y)^2 < 0, \quad g_{\mu \nu} = (+,-,-,-), \end{equation} to ensure their time ordering is irrelevant and not resulting in any physical consequence.

Because operators $\mathcal{O}(x)$ are usually just a product of $\prod_i \psi_i(x)$, requiring $[\mathcal{O}_1(x), \mathcal{O}_2(y)]=0$ is the same as requiring $\left [ \psi_A(x), \psi_B(y)\right ] = 0$.

The specific case for the space-like configuration is discussed on pag. 237 of Weinberg:

For $x-y$ space-like, we can adopt a Lorentz frame in which $x^0=y^0$, and write Eq. (5.7.19) as [...]. In order that this should vanish when $\mathbf{x}\neq\mathbf{y}$ we must have ...

and then Weinberg gets to the point of $2j \in \mathbb{N}$.

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  • $\begingroup$ Thanks for your answer SuperCiocia! It was my understanding from these sources that utilizing a Lorentz invariant representation of the field and proceeding through the analysis of $\left[ \psi_{\sigma}(x),\psi^+_{\sigma'}(y) \right]_{\mp}$ is all that is required to prove the theorem. In your answer, I am still unsure of why we should have, for instance: $\int\frac{d^3p}{2p_0}\{e^{ip(x-y)}\pm(-1)^{2j}\eta\eta^*e^{-ip(x-y)}\}$ instead of $\int\frac{d^3p}{2p_0}(-1)^{2j}\{e^{ip(x-y)}\pm\eta\eta^*e^{-ip(x-y)}\}$. In other words, where does this term $(-1)^{2j}$ arise from, explicitly. $\endgroup$ Jan 6, 2020 at 1:06
  • $\begingroup$ If the origin of the term $(-1)^{2j}$ is from the spin sum $\pi_{ab,\tilde{a}\tilde{b}}(\textbf{p})$ as I believe Weinberg to be implying, why does it only act on one exponential term? That it does act on only one brings us to the spin statistics theorem, but if it is a coefficient to both components of the field, I don't see how we can reach the same conclusion. $\endgroup$ Jan 6, 2020 at 1:12
  • $\begingroup$ Because the $(-1)^j$ term comes from the transformation of $\eta$. $\endgroup$
    – SuperCiocia
    Jan 6, 2020 at 6:56
  • $\begingroup$ It was my understanding via Weinberg and "Weinberg's proof of the spin-statistics theorem (Massimi, Redhead, 2003)" that $\eta$ was simply a constant from the linear combination of the particle annihilation and antiparticle creation fields. If this is so, it simply transforms as a scalar. In this case, I don't understand what you mean? Could you please elaborate? $\endgroup$ Jan 6, 2020 at 13:26
  • $\begingroup$ I am still very interested. If I am incorrect in any of my assumptions above that is perfectly ok, I just need to know where so I can learn. Could you elaborate at all on you answer? $\endgroup$ Jan 21, 2020 at 18:31
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So there hasn't been any development in four months or so, and I believe that I have the answer I was looking for. Just in case anyone else runs into the issue I did, I'm going to risk posting my own answer.

The main issue is that of the $(-1)^{2j}$ term coming from the Lorentz covariance algebra. For this I am relying on Weinberg's "Feynman Rules for Any Spin", Phys Rev 1964 B1318 1964, Streater and Wightman's "PCT, Spin, Statistics, and All That" Princeton University Press, 1980 pages 14-16, and Novozhilov's "Introduction to Elementary Particle Field Theory" Pergamon Press, 1975 pages 75-77.

Starting with Weinberg, we want to construct our creation and annihilation operators, and thus our fields, in a Lorentz covariant way. He does this by having the operators obey transformations in the proper homogeneous orthochronous Lorentz group.

$$ x^\mu\rightarrow \Lambda^\mu_{\space\space\nu}x^\nu = g_{\lambda\rho} $$ $$ g_{\mu\nu}\Lambda^\mu_{\space\space\nu}\Lambda^\nu_{\space\space\rho} \tag{Weinberg (W) 2.1} $$ $$ det\Lambda=1; \Lambda^0_{\space\space 0}>0 $$

In Einstein summation notation.

For each transformation $\Lambda$ there corresponds a unitary operator acting on Hilbert space with the group property $$ U[\Lambda_2]U[\Lambda_1]=U[\Lambda_2\Lambda_1] \tag{W 2.3} $$

We next describe the action of these $U[\Lambda]$ on one particle $|\textbf{p},\sigma\big>$ states. We first define these states as the result of a boost ($\Lambda = L(\textbf{p})$ which takes a particle of mass m at rest to momentum $\textbf{p}$) on a rest state $|\sigma\big>$ $$ |\textbf{p},\sigma\big> = [m/\omega(\textbf{p})]^{1/2}U[L(\textbf{p})]|\sigma\big> \tag{W 2.6} $$

This allows us to see how these states should transform under an arbitrary $\Lambda$

$$ \begin{align*} U[\Lambda]|\textbf{p},\sigma\big> &= [m/\omega(\textbf{p})]^{1/2}U[\Lambda]U[L(\textbf{p})]|\sigma\big> \\ &=[m/\omega(\textbf{p})]^{1/2}U[L(\Lambda\textbf{p})]U[L^{-1}(\Lambda\textbf{p})\Lambda L(\textbf{p})]|\sigma\big> \\ &= [m/\omega(\textbf{p})]^{1/2}\sum_{\sigma'}U[L(\Lambda\textbf{p})]|\sigma'\big>\times \big<\sigma'|U[L^{-1}(\Lambda\textbf{p})\Lambda L(\textbf{p})]|\sigma\big> \\ &=[\omega(\Lambda\textbf{p})/\omega(\textbf{p})]^{1/2}\sum_{\sigma'}|\Lambda\textbf{p},\sigma'\big>\times D_{\sigma',\sigma}^{(j)}[L^{-1}(\Lambda\textbf{p})\Lambda L(\textbf{p})] \end{align*} \tag{W 2.8} $$

$L^{-1}(\Lambda\textbf{p})\Lambda L(\textbf{p})$ is actually the pure rotation $R$, otherwise known as a "Wigner rotation," and so $D_{\sigma',\sigma}^{(j)}[R]$ are here the $2j+1$ dimensional unitary matrix representations of the rotation group.

To now assert the Lorentz covariance of the fields, we say their creation and annihilation operators transform as above:

$$ U[\Lambda]a^*(\textbf{p},\sigma)U^{-1}[\Lambda]=[\omega(\Lambda\textbf{p})/\omega(\textbf{p})]^{1/2}\sum_{\sigma'}D_{\sigma',\sigma}^{(j)}[L^{-1}(\Lambda\textbf{p})\Lambda L(\textbf{p})]a^*(\Lambda\textbf{p},\sigma')\tag{W 2.11} $$

And with the adjoint: $$ U[\Lambda]a(\textbf{p},\sigma)U^{-1}[\Lambda]=[\omega(\Lambda\textbf{p})/\omega(\textbf{p})]^{1/2}\sum_{\sigma'}D_{\sigma',\sigma}^{(j)}[L^{-1}(\textbf{p})\Lambda^{-1} L(\Lambda\textbf{p})]a(\Lambda\textbf{p},\sigma')\tag{W 2.12} $$

It is imperative that we get the forms of these to match, since the solution to my problem lies in the manipulation of these matrix coeffients $D_{\sigma',\sigma}^{(j)}[L(\textbf{p})]$ and their cross products in the eventual (anti)commutator. As such, we need make the following changes, using:

$$ \begin{align*} &D^{(j)}[R]^*=CD^{(j)}[R]C^{-1} \\ &D_{\sigma',\sigma}^{(j)}[R]=\{CD^{(j)}[R^{-1}]C^{-1}\}_{\sigma,\sigma'} \end{align*} \tag{W 2.13,2.15} $$

We transform $W 2.11$ into

$$ \begin{align*} &U[\Lambda]a^*(\textbf{p},\sigma)U^{-1}[\Lambda] \\ &=[\omega(\Lambda\textbf{p})/\omega(\textbf{p})]^{1/2}\sum_{\sigma'}\{CD^{(j)}[L^{-1}(\textbf{p})\Lambda^{-1} L(\Lambda\textbf{p})]C^{-1}\}_{\sigma,\sigma'}a^*(\Lambda\textbf{p},\sigma') \end{align*}\tag{W 2.16} $$

Now $b^*(\textbf{p},\sigma) $ transforms as $a^*(\textbf{p},\sigma)$, so we can use $W 2.16$ for the antiparticle creation operator, and $W 2.12$ for $a(\textbf{p},\sigma)$,the particle annihilation operator.

Weinberg next forms our $(j,0)$ representation from the standard sums of Lorentz $K$ and $J$ operators, leading us to the following useful identity:

$$ D^{(j)}[\Lambda]=\bar{D}^{(j)}[\Lambda^{-1}]^\dagger \tag{W 2.38} $$

Next we make use of the group property $\{W 2.3\}$ of the Lorentz group to split the Wigner rotations appearing in our formulae into three parts

$$ D^{(j)}[L^{-1}(\textbf{p})\Lambda^{-1} L(\Lambda\textbf{p})] = D^{(j)-1}[L(\textbf{p})]D^{(j)}[\Lambda^{-1}]D^{(j)}[L(\Lambda\textbf{p})]$$

Allowing us to write our previous transformation laws $\{W 2.12\}$ and $\{W 2.16\}$ as:

$$ \begin{align*} &U[\Lambda]\alpha(\textbf{p},\sigma)U^{-1}[\Lambda]=\sum_{\sigma'}D_{\sigma,\sigma'}^{(j)}[L(\Lambda^{-1})]\alpha(\Lambda\textbf{p},\sigma') \\ &U[\Lambda]\beta(\textbf{p},\sigma)U^{-1}[\Lambda]=\sum_{\sigma'}D_{\sigma,\sigma'}^{(j)}[L(\Lambda^{-1})]\beta(\Lambda\textbf{p},\sigma') \\ &\alpha(\textbf{p},\sigma)\equiv[2\omega(\textbf{p})]^{1/2}\sum_{\sigma'}D_{\sigma,\sigma'}^{(j)}[L(\textbf{p})]a(\textbf{p},\sigma') \\ &\beta(\textbf{p},\sigma)\equiv[2\omega(\textbf{p})]^{1/2}\sum_{\sigma'}\{D_{\sigma,\sigma'}^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}b^*(\textbf{p},\sigma') \end{align*} $$

Only one more Weinberg step left. We express our field as a fourier transform on the sum of lorentz invariant creation and annihilation operators $\alpha$ and $\beta$, and then substitute back in for $a$ and $b^*$:

$$ \psi_{\sigma}(x)=\frac{1}{(2\pi)^{3/2}}\int\frac{d^3\textbf{p}}{[2\omega(\textbf{p})]^{1/2}}\sum_{\sigma'}\left[\xi D_{\sigma,\sigma'}^{(j)}[L(\textbf{p})]a(\textbf{p},\sigma')e^{ip\cdot x}+\eta\{D^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}b^*(\textbf{p},\sigma')e^{-ip\cdot x}\right] $$

The (anti)commutator which we want: $[\psi_{\sigma}(x),\psi^\dagger_{\sigma'}(y)]_\pm$, will now return only those terms like: $D_{\sigma,\sigma'}^{(j)}[L(\textbf{p})]D_{\sigma,\sigma'}^{(j)}[L(\textbf{p})]^\dagger$ for the particle case "a," and terms like: $\{D^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}\{D^{(j)}[L(\textbf{p})]C^{-1}\}^\dagger_{\sigma,\sigma'}$ for the antiparticle case "b."

Returning to $\{W 2.15\}$, we have:

$$ \begin{align*} &D_{\sigma',\sigma}^{(j)}[R]=\{CD^{(j)}[R^{-1}]C^{-1}\}_{\sigma,\sigma'} \\ &\{C^{-1}D^{(j)}[R]C\}_{\sigma,\sigma'}=\{C^{-1}CD^{(j)}[R^{-1}]C^{-1}C\}_{\sigma,\sigma'} \\ &\{C^{-1}D^{(j)}[R]C\}_{\sigma,\sigma'}=D^{(j)}[R^{-1}]_{\sigma,\sigma'} \end{align*} $$

We can now group terms from the antiparticle case as so:

$$ \begin{align*} &\{D^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}\{D^{(j)}[L(\textbf{p})]C^{-1}\}^\dagger_{\sigma,\sigma'} \\ &=\{D^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}\{D^{(j)}[L(\textbf{p})]^\dagger C^{-1\dagger}\}_{\sigma,\sigma'} \\ &=\{D^{(j)}[L(\textbf{p})]C^{-1}\}_{\sigma,\sigma'}\{D^{(j)}[L(\textbf{p})]^\dagger C\}_{\sigma,\sigma'} \\ &=D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}\{C^{-1}D^{(j)}[L(\textbf{p})]^\dagger C\}_{\sigma,\sigma'} \\ &=D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}D^{(j)}[L(\textbf{p})^{-1}]^\dagger_{\sigma,\sigma'} \\ &=D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}\bar{D}^{(j)}[L(\textbf{p})]_{\sigma,\sigma'} \end{align*} $$ Where the last step was by applying $\{W 2.38\}$. This gives us now for the (anti)commutator a form like:

$$ [\psi_{\sigma}(x),\psi^\dagger_{\sigma'}(y)]_\pm=\frac{1}{(2\pi)^3}\int\frac{d^3\textbf{p}}{2\omega(\textbf{p})}\left[|\xi|^2 D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}e^{ip\cdot (x-y)}+|\eta|^2D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}\bar{D}^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}e^{-ip\cdot (x-y)}\right] $$

We now turn to Novozhilov, who indicates that in his nomenclature:

$$ D^J(\frac{p}{m})=e^{\frac{\beta(\textbf{J}\cdot\textbf{p})}{|\textbf{p}|}}, \theta^i=\beta\frac{p^i}{|\textbf{p}|} \tag{Novozhilov 4.80} $$ Which is the same form as in $\{W 2.39, 2.40\}$, where

$$ \begin{align*} &D^{(j)}[L(\textbf{p})] = e^{-\hat{p}\cdot\textbf{J}^{(i)}\theta} \tag{W 2.39} \\ &\bar{D}^{(j)}[L(\textbf{p})] = e^{+\hat{p}\cdot\textbf{J}^{(i)}\theta} \tag{W 2.40} \end{align*} $$

This implies, that by group property we can perform the following:

$$ D^{(j)}[L(\textbf{p})]\bar{D}^{(j)}[L(\textbf{p})]\equiv D^{J}(\frac{p}{m})D^{J}(\frac{-p}{m})=D^{J}(\frac{p}{m})D^{J}(\frac{p}{m})D^{J}(-1) $$

Leaving us with

$$ \begin{align*} &[\psi_{\sigma}(x),\psi^\dagger_{\sigma'}(y)]_\pm=\frac{1}{(2\pi)^3}\int\frac{d^3\textbf{p}}{2\omega(\textbf{p})}\Pi(\textbf{p})\left[|\xi|^2 e^{ip\cdot (x-y)}\pm|\eta|^2{D}^{(j)}[-1]_{\sigma,\sigma'}e^{-ip\cdot (x-y)}\right] \\ &\Pi(\textbf{p}) \propto D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'} \end{align*} $$

Novozhilov directly states that $D[-1]=(-1)^{2j}$ $\{page 77, in text\}$, but stops short of why. Which is where I turn to Streater and Wightman. In their book PCT, Spin and Statistics, and All that $(2000)$, page 15, they posit a form for these matrices $D^{(j)}$:

"Consider a set of quantities $\xi_{\alpha_{1}...,\alpha_{j},\dot{\beta}_1...\dot{\beta}_j}$, where the $\alpha$'s and $\dot{\beta}$'s take the values 1 and 2, and $\xi$ is symmetric under permutations of the $\alpha$'s and also under permutations of the $\dot{\beta}$'s. For each $A\in SL(2,C)$ we define a linear transformation of the $\xi$'s according to $$ \xi_{\alpha_{1}...,\alpha_{j},\dot{\beta}_1...\dot{\beta}_k} \longrightarrow \sum\limits_{(\rho)(\dot{\sigma})}A_{\alpha_1\rho_1}...A_{\alpha_j\rho_j}\bar{A}_{\dot{\beta}_1\dot{\sigma}_1}...\bar{A}_{\dot{\beta}_k\dot{\sigma}_k}\xi_{\rho_{1}...,\rho_{j},\dot{\sigma}_1...\dot{\sigma}_k} $$ [The dot over the index simply means that this index transforms according to $\bar{A}$ instead of $A$; the symbol ($\rho$) stands for $\rho_1...\rho_j$; the symbol ($\dot{\sigma}$) for $\dot{\sigma}_1...\dot{\sigma}_k$] This representation of SL(2,C) is usually denoted $\mathfrak{D}^{(\frac{j}{2},\frac{k}{2})}$. Every irreducible representation is equivalent to one of these."

From here, if we consider the case with $A\longrightarrow(-1)$ and $\mathfrak{D}^{(\frac{j}{2},0)}(A)\equiv D^{\frac{j}{2}}(A)$, then we can see that this transformation reduces to a multiplication by the inverse unit matrix $\textbf{-1}$ j times.

We're getting $\xi_{\alpha_{1}...,\alpha_{j},\dot{\beta}_1...\dot{\beta}_k} \longrightarrow \sum\limits_{(\rho)(\dot{\sigma})}-1_1\times...-1_j \xi_{\rho_{1}...,\rho_{j},\dot{\sigma}_1...\dot{\sigma}_k}$ or

$$\xi_{\alpha_{1}...,\alpha_{j},\dot{\beta}_1...\dot{\beta}_k} \longrightarrow \sum\limits_{(\rho)(\dot{\sigma})}(-1)^j \xi_{\rho_{1}...,\rho_{j},\dot{\sigma}_1...\dot{\sigma}_k} $$

At this point there is a difference in notation, with Streater and Wightman using $\frac{j_{integer}}{2}$ to label their representations, and Weinberg and Novozhilov using $j$ either whole or half integer. Since these are functionally equivalent, then $\mathfrak{D}^{(\frac{j}{2},0)}(-1)_{Streater}\equiv D^{j}(-1)_{Weinberg}\equiv (-1)^{2j}$.

And finally, this leads us to the result:

$$ \begin{align*} &[\psi_{\sigma}(x),\psi^\dagger_{\sigma'}(y)]_\pm=\frac{1}{(2\pi)^3}\int\frac{d^3\textbf{p}}{2\omega(\textbf{p})}\Pi(\textbf{p})\left[|\xi|^2 e^{ip\cdot (x-y)}\pm|\eta|^2(-1)^{2j}_{\sigma,\sigma'}e^{-ip\cdot (x-y)}\right] \\ &\Pi(\textbf{p}) \propto D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'}D^{(j)}[L(\textbf{p})]_{\sigma,\sigma'} \end{align*} $$

Leading us directly to the conclusion it always does: $|\xi|^2=(-1)\pm|\eta|^2(-1)^{2j}=\mp|\eta|^2(-1)^{2j}$, the spin statistic theorem.

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