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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?

infinitezero
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  • 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! – Cosmas Zachos Aug 05 '21 at 16:40
  • 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. – Cosmas Zachos Aug 05 '21 at 16:49
  • I don't see how "take it as it is" is helping me in understanding something I don't. – infinitezero Aug 05 '21 at 17:24
  • 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. – Cosmas Zachos Aug 05 '21 at 17:29
  • @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 $P^{1/2}_{ji}$? – infinitezero Aug 05 '21 at 18:12
  • You may subtract them from their explicit definitions (7) and (8) in the published version, here for d=4. So, for instance, you may see from elementary γ matrix algebra that $(P^{1/2}{ij}){\mu\nu}=(P^{1/2}{ji}){\nu\mu}$. – Cosmas Zachos Aug 05 '21 at 18:53
  • @Cosmas Zachos The definition does not help me. It's just an equation. I added an example to my post. – infinitezero Aug 05 '21 at 22:14

2 Answers2

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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.

Cosmas Zachos
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  • Great answer +1. Can you please help with this one too? :) Thank you! https://physics.stackexchange.com/questions/652544 – safesphere Aug 08 '21 at 02:19
  • 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. – Cosmas Zachos Aug 08 '21 at 15:49
  • 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? – safesphere Aug 08 '21 at 17:44
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    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. – Cosmas Zachos Aug 08 '21 at 17:47
  • It comes from the fact that photons colliding in a vacuum have never been observed to produce a pair despite massive efforts. (E.g. the Atlas experiment was not in a field-free vacuum.) However, this point is not important. This is not about the “non-vanishing cross section”. Let’s assume two photons do produce a pair once in a blue moon. So let me put it this way. Collide an electron and positron a billion times. They annihilate approximately a billion times. Collide two photons a billion times. They produce a pair may be once. So in reality the reaction is practically non-reversible. Why? – safesphere Aug 08 '21 at 18:56
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    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… – Cosmas Zachos Aug 08 '21 at 19:27
  • A billion or trillion or million or even a thousand - it doesn’t matter. There is no practical symmetry anyway. “Reversible” means, I reverse time and the reaction happens every time, not one in a million cases. My question is why. You say, the phase space. OK, it’s a good start. Perhaps even correct, but a layman has no idea what this means and especially why this “phase space” is “non-reversible”. Can you possibly post an answer to make this clear for non-specialists? Thank you! – safesphere Aug 08 '21 at 20:43
  • I'm sorry, but I don't appreciate the condescending tone. My question was about the indices at the projection operators, but you answered a question I didn't ask. Only in your latest edit, you mentioned something, but that was not really comprehensive. In my opinion, this attitude is the main reason, a lot of people are afraid to ask questions, because they get brushed with the impression, they're too stupid too understand. Thank you for the effort. – infinitezero Aug 09 '21 at 16:42
  • @infinitezero Well, your title suggests you wish to understand how the RS vector spinor is projected into spin 3/2, not how the dross is discarded/ignored: the two zeros in your edit matrix. It is assumed you have mastered the Dirac equation and how spin 1 and 0 projections work in standard bosonic gauge fields ; but, if not, you are reading a very unsuitable reference. Perorating on spin projections in all generality takes much longer than the paper you are reading; I gave you a trail-map on how to understand it. I am sorry you perceive condescension: you might focus on the trail map. – Cosmas Zachos Aug 09 '21 at 16:54
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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

infinitezero
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