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Chapter 2 of the paper Symmetry of massive Rarita-Schwinger fields by T. Pilling mentions "the usual" spin projection operators. However, to me, they are not usual and I struggle with intuition and notation.

I understand that we find the correct Lorentz representation of the RS vector-spinor by taking the tensor product of a bispinor and vector representation (eq 1 in the paper):

$$\left[(1/2,0)\oplus(0,1/2)\right] \otimes (1/2, 1/2) \quad\quad\quad$$ $$\quad\quad\quad = (1, 1/2) \oplus (1/2, 1) \oplus (1/2, 0) \oplus (0, 1/2) \tag{1}$$

My main question is about the later mentioned projection operators. Clearly a RS-spinor $\psi_\mu$ has a mixture of 3/2 and 1/2 degrees of freedom. We are not interested in the 1/2 background, so we need a projection operator $P^{3/2}$ to get rid of them. Fine. Likewise, I can define an operator $P^{1/2}$ to get the spin-1/2 background. That's just a mathematical excercise. Now, what are the extra indices at $P^{1/2}_{11}$, $P^{1/2}_{12}$, $P^{1/2}_{21}$ and $P^{1/2}_{22}$? In the paper they are described as

the individual projection operators for the two different spin-1/2 components of the Rarita-Schwinger field

This is where I'm lost. What am I projecting? For what reason? Can someone explain this to me and give me some intution?

edit:

To clarify, consider the Projector

$$P_x = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$

The intution here is, that is projects the x-component of a three vector. Using this analogy, what is $P^{1/2}_{11}$ projecting?

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  • $\begingroup$ Your author defines very precisely these symbols in (4, unnumbered interstitial, 5) and you may check they have just the right projection properties and dimensionalities of subspace. You may afford to ignore the subscripts, related to polarizations and helicities, and further defined in gory detail in van Nieuwenhuizen's review cited. The author is condensing things to spare you further complications and asides, so take it as it is! $\endgroup$ Commented Aug 5, 2021 at 16:40
  • $\begingroup$ You are projecting out the last two of the r.h.side of (1), 4 out of the original 16 components, to be left with the first two terms, 12 components. $\endgroup$ Commented Aug 5, 2021 at 16:49
  • $\begingroup$ I don't see how "take it as it is" is helping me in understanding something I don't. $\endgroup$ Commented Aug 5, 2021 at 17:24
  • $\begingroup$ It's not clear what you don't understand. The exposition and logical flow are sensible, given the self-evident definitions. Give those indices other names, if they distracted you from the key points. $\endgroup$ Commented Aug 5, 2021 at 17:29
  • $\begingroup$ @Cosmas Zachos What is the difference between $P^{1/2}$ and $P^{1/2}_{ij}$? What is the difference between $P^{1/2}_{ij}$ and $P^{1/2}_{ji}$? $\endgroup$ Commented Aug 5, 2021 at 18:12

2 Answers 2

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It is apparent that the clear and explicit formal paper you are reading is not helping you systematize what you may have learned in your QFT theoretical course (this is distinctly not experimental HEP!), namely, the basic facts summed up in the paragraph following equation (12): at the end of the day, the independent conditions $\gamma \cdot \psi$ and $p\cdot \psi$ have no Lorentz indices anymore, and so they Lorentz- transform like plain spinors, which you know represent spin 1/2 s:

...as can be seen by multiplying the first equation in (10) on the left by $γ_μ$ and using the second equation. The condition $γ_\mu^{AB}ψ^μ_B = 0$ (where we now explicitly write the spinor indices A,B) represents a constraint equation for each value of the spin index A, whereas the condition $∂_μ ψ_B^μ = 0$ is an equation of motion for the spinor components $ψ_B^0 $. However, the Dirac equation (10) also gives an equation of motion for the same spinor components and when taken together, these result in another set of constraints. In four space–time dimensions, these two sets of equations each constitute four constraints; and serve to remove eight components of the 16 spinor components of the vector-spinor $ψ_A^μ$, leaving 2(2s + 1) = 8 physical degrees of freedom as required for a massive spin s = 3/2 particle.

That is, of the original 16 d.o.f. of the reducible field $\psi^\mu$, you prune out 4 d.o.f. by the first condition, and another 4 by the second, being left with 8 for the spin-3/2 block: a parity doublet of the 4 spin states of the spin quartet. (Remember: we have not gauged out the intermediate helicity $\pm 1/2$ states, since susy-gauge-invariance has not been assumed, this not being supergravity; it might as well be a Δ baryon.)

Now the author spends quite some time giving you a formal (Ogievetskian) implementation of these maneuvers, through the projector $(P^{3/2})^2=P^{3/2} $, the only operator you really need to appreciate, $$ P^{3/2}_{\mu\nu}= \frac{1}{6p^2} \bigl ( 4(p^2 g_{\mu\nu} - p^\mu p^\nu) \\ -p^2[\gamma_\mu,\gamma_\nu ]+p^\mu p^\kappa [\gamma_\kappa,\gamma_\nu] - p^\nu p^\kappa [\gamma_\kappa,\gamma_\mu] \bigr ) \tag{7.1} ~~~~\leadsto \\ \gamma^\mu P^{3/2}_{\mu\nu}=0=P^{3/2}_{\mu\nu} \gamma^\nu = 0=p^\mu P^{3/2}_{\mu\nu}=0=P^{3/2}_{\mu\nu} p^\nu. $$

As a result, after the two piecemeal projections of the two spin 1/2 s, the pure spin 3/2 field you only need consider is $\tilde \psi^\mu = P^{3/2}_{\mu\nu} \psi^\nu$, which, now, automatically satisfies the conditions, by above!

This is where I'm lost. What am I projecting? For what reason? Can someone explain this to me and give me some intuition?

You are projecting the spin 3/2 piece $P^{3/2}_{\mu\nu} \psi^\nu$ of the reducible field $\psi^\mu$ so you don't get distracted by the two irrelevant spin 1/2 spinors unfortunately also packaged in the vector-spinor; a demonstrably real risk, ipso facto. This is the intuition.

If, despite this, you really wished to look in the dross pile for the spin 1/2 pieces you discarded in two stages, 1.13 of van Nieuwenhuizen's review will furnish an overcomplete analysis of the polarization components of the two spin 1/2 s. In (1), $P^{1/2}_{22}$ of course projects onto the spin 1/2 of $(0,1/2)\oplus(1/2,0)$, while $P^{1/2}_{11}$ onto the remnant spin 1/2 spinor in the incompletely pruned $(1,1/2)\oplus (1/2,1)$, which still has 12, not 8, d.o.f.

For the simpler, massless, case, look at section II of van Nieuwenhuizen, P., Sterman, G., & Townsend, P. K. (1978): Unitary Ward identities and BRS invariance Phys Rev, 17 1501.

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  • $\begingroup$ Great answer +1. Can you please help with this one too? :) Thank you! physics.stackexchange.com/questions/652544 $\endgroup$
    – safesphere
    Commented Aug 8, 2021 at 2:19
  • $\begingroup$ Thanks; Anna's answer looks spot on: the calculated cross section for γγ->e+ e- is distinctly non-zero. The matrix element is identical to the time-reversed one in your picture, but, naturally, the phase-spec and cross section differ. $\endgroup$ Commented Aug 8, 2021 at 15:49
  • $\begingroup$ Thanks for your insight, but the essence of my question is different. In reality, the annihilation is self evidently non-reversible. Collide an election and positron and they practically always annihilate. Collide two photons in a vacuum and they won’t produce a pair - ever. So the non-reversibility of this reaction in reality is a given and undisputed fact. And yet the formulas are reversible, so everyone just blindly ignores the real facts and says, “it is reversible”. Thus my question is, what exactly is the theoretical justification or technicality that makes this reaction non-reversible? $\endgroup$
    – safesphere
    Commented Aug 8, 2021 at 17:44
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    $\begingroup$ I disagree with your experimental premise. e+ e- collisions have a non-vanishing cross-section, and so do γγ ones. I have no clue where your faulty "Collide two photons in a vacuum and they won’t produce a pair - ever" comes from. $\endgroup$ Commented Aug 8, 2021 at 17:47
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    $\begingroup$ I don’t believe the factor of a billion nonfact. If the cross sections bore out what you say, it should be phase space differences, not matrix elements (amps). With the advent of photon colliders, the process will be observed at the predicted level, unless something nonstandard happened… $\endgroup$ Commented Aug 8, 2021 at 19:27
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After hours of more research, I came to the following conclusion:

A representation of the inhomogeneous Lorentz group (ILG) has a spin space $S=\{s_\mathrm{max}, s_\mathrm{max}-1, ... , s_\mathrm{min}\}$, where each spin has a multiplicity $n(s)$, where $n(s_\mathrm{max})=1$.
A Lorentz vector has a spin decomposition of $1\oplus0$ and hence for a vector-spinor we find $$\mathrm{spin}\;\left( A_\mu \otimes \psi\right) = (1\oplus 0) \;\otimes\; 1/2 = 3/2 \oplus 1/2 \oplus 1/2\;.$$ This means $n(3/2)=1$ and $n(1/2)=2$. As mentioned in the question, for a spin-3/2 particle, we do not care about the lower spin background. This demands projection operators, such as $P(3/2)$ and $P(1/2)$. Now the question was, what are the indices at $P_{ij}(1/2)$?

To answer this, we will use a secondary notation. Let $P(s)$ be a projection operator, that selects the spin-$s$ representation of a wave function. Since $n(s_\mathrm{max})=1$, $P(s_\mathrm{max})$ always selects the irreducible representation, but if $n(s)>1$ we only find a reducible representation. Therefore, we introduce an additional projection operator $D_i(s)$ with $i = 1, ... , n(s)$ which precisely finds the irreducible representations. This means $P(s) = \bigoplus \limits_{i=1}^{n(s)} D_i(s)$.

Combining this with the above notation, $P_{ii}(s)=D_i(s)$. So the projection operators with two equal indices, find the irreducible representations of spin-$s$.

Finally, to switch between irreducible representations, we can introduce the transition (not projection!) operators $P_{ij}$ with $i\neq j$. They have the following properties:

$$P_{21}(s) P_{11}(s)P_{12}(s) = P_{22}(s)$$ $$P_{12}(s) P_{22}(s) P_{21}(s) = P_{11}(s)$$

For a more general description, see the above paper.

Further reading:

A. Aurilia and H. Umezawa, Theory of High Spin Fields, Phys. Rev. 182

Frits A. Berends et al., On Field Theory for Massive and Massless Spin 5/2 Particles, Nucl.Phys.B

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