# Spin 3/2 Rotation Matrices

## Main Question or Discussion Point

A long literature search has given me nothing, so I'm turning to this forum for help.

I have a spin-3/2 fermion field, and I want to find its wave functions corresponding to its 4 pure-spin states, +3/2, +1/2, -1/2, -3/2, which is normally done by finding the 4 eigenfunctions of its rotation operator (which has exactly these four eigenvalues). My problem is that I can't find this operator.

For a spin-1/2 particle, the operator is
$$R_{1/2} = \frac{1}{2} \left[ \begin{array}{ c c c c} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array} \right]$$
which has eigenvalues +1/2 and -1/2. For a spin-1 particle, the operator is
$$R_1 = \left[ \begin{array}{ c c c c} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array} \right]$$
having this time eigenvalues +1, 0 and -1.

I'm looking for the spin-3/2 version of these operators. The original paper by Rarita-Schwinger about the spin-3/2 field states that this operator should be the sum of the above two operators (one would rotate the spinor index, while the other would rotate the vector index), which gives eigenvalues +1/2, -1/2, +sqrt(5)/2 and -sqrt(5)/2; multiplying the second operator by a factor of sqrt(2) would give the correct eigenvalues, but I don't see how I could ever justify that.

The operator diag(3/2,1/2,-1/2,-3/2) would obviously give the right eigenvalues, but I strongly doubt it does what I want it to do, mostly because its equivalent representation for spin-1 is diag(1,0,-1) which does not even have the right dimensions.

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Well the Rarita-Schwinger field has a spinor index and a vector index, and they are not usually combined in a single matrix. However, (not having looked at the paper, my connection is quite bad at the moment), I think what R+S mean is that the operator is the $$\oplus$$ sum of the two operators, just like in quantum mechanics we write $$J_x = L_x + S_x$$, but the two operators on the RHS act on different spaces.

Thus, the Lorentz transformation acting on such a field would be of the form $$\Lambda = e^{i \theta^{ij} L_{ij} + i \omega^{ij} S_{ij}$$, where $$L_{12}$$ is a matrix that acts on Lorentz vectors, and $$S_{12}$$ is a matrix that acts on Lorentz spinors. The notation $$X_{12}$$ means a generator that rotates between the 1-2 planes, so in this case, the z axis. Likewise, $$X_{01}$$ would be a boost. Anyway, I hope the meaning is clear.

I've been adding them inside the exponential as you said, although I did have them all acting together, basically as matrices multiplying. I'll try to keep my various indices separate.

So basically, I'll be doing:
$$(\delta_\nu^\mu + \theta L_\nu^\mu)(\delta_b^a + (\omega \cdot \Sigma)^a_b)\Psi_{\mu a} = (1+\lambda) \Psi_{\nu b}$$.

Whenever I have some free time I'll plug that through and see what comes out

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After doing some quick calculations, this does not allow the +3/2 eigenvalue, at least not when using

$$L = \left[ \begin{array}{ c c c c} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array} \right]$$

$$\omega \cdot \Sigma = \frac{\theta}{2}\left[ \begin{array}{ c c c c} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array} \right]$$

Am I doing something wrong here, and if not, do you have any other ideas?

Great posts in that thread, but I still don't see how [6L] or [6R] transform under the Lorentz group. You go over in some detail on how [2L] transforms, but I don't see how I can use that or a similar approach for [6L].

Would you mind expanding on that, or giving me the name of a textbook or reference which would explain the transformations?

samalkhaiat
Great posts in that thread, but I still don't see how [6L] or [6R] transform under the Lorentz group. You go over in some detail on how [2L] transforms, but I don't see how I can use that or a similar approach for [6L].

Would you mind expanding on that, or giving me the name of a textbook or reference which would explain the transformations?
1)

On Minkowski space, we identify the irreducible representation $(j_{L},j_{R})$ by a spinorial tensor field

$$\left[(2j_{L}+1)(2j_{R}+1)\right] \equiv \Psi_{l_{1}...l_{2j_{L}}\dot{r}_{1}...\dot{r}_{2j_{R}}}$$

which is symmetric in the indices $(l_{1}l_{2}...l_{2j_{L}})$ and $(\dot{r}_{1}\dot{r}_{2}...\dot{r}_{2j_{R}})$; (otherwise it is reducible).

So, for (1,1/2) we have

$$[6_{R}] \equiv \Psi_{l_{1}l_{2}\dot{r}} = \Psi_{l_{2}l_{1}\dot{r}}$$

and for (1/2,1) we have

$$[6_{L}] \equiv \Psi_{l\dot{r}_{1}\dot{r}_{2}} = \Psi_{l\dot{r}_{2}\dot{r}_{1}}$$

Under Lorentz transformation [SL(2,C)], we have

$$\Psi^{'}_{st\dot{r}_{1}} = D^{l_{2}}{}_{s} D^{l_{1}}{}_{t} \bar{D}^{\dot{r}}{}_{\dot{r}_{1}} \Psi_{l_{2}l_{1}\dot{r}}$$

$$\Psi^{'}_{l_{1}\dot{s}\dot{t}} = \bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} D^{l}{}_{l_{1}} \Psi_{l\dot{r}_{1}\dot{r}_{2}}$$

The form of $D(\bar{D})$ is given in the other thread.
By contracting the above with $(\sigma^{\mu})^{st}$ and $(\bar{\sigma}^{\mu})^{\dot{s}\dot{t}}$ respectively, we find

$$(\sigma^{\mu})^{st}\Psi^{'}_{st\dot{r}_{1}} = \left( D^{l_{2}}{}_{s} D^{l_{1}}{}_{t}(\sigma^{\mu})^{st}\right) \bar{D}^{\dot{r}}{}_{\dot{r}_{1}}\Psi_{l_{2}l_{1}\dot{r}}$$

$$(\bar{\sigma}^{\mu})^{\dot{s}\dot{t}} \Psi^{'}_{l_{1}\dot{s}\dot{t}} = \left( \bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} (\bar{\sigma})^{\dot{s}\dot{t}}\right) D^{l}{}_{l_{1}} \Psi_{l\dot{r}_{1}\dot{r}_{2}}$$

One can show that $\sigma^{\mu}$ behaves on the one hand as a rank-2 spinor, on the other as a 4-dimensional Lorentz vector, i.e.,

$$D^{l_{1}}{}_{s} D^{l_{2}}{}_{t} (\sigma^{\mu})^{st} = \Lambda(D)^{\mu}{}_{\nu} (\sigma^{\nu})^{l_{1}l_{2}}$$

$$\bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} (\bar{\sigma}^{\mu})^{\dot{s}\dot{t}} = \Lambda(\bar{D})^{\mu}{}_{\nu} (\bar{\sigma}^{\nu})^{\dot{r}_{1}\dot{r}_{2}}$$

Thus

$$(\Psi^{\mu}_{\dot{s}})^{'} = \bar{D}^{\dot{r}}{}_{\dot{s}} \Lambda(D)^{\mu}{}_{\nu} \Psi^{\nu}_{\dot{r}}$$

$$(\Psi^{\mu}_{s})^{'} = D^{l}{}_{s} \Lambda(\bar{D})^{\mu}{}_{\nu} \Psi^{\nu}_{l}$$

where

$$\Psi^{(8)}_{\mu R} \equiv \Psi^{\mu}_{\dot{r}} = \frac{1}{2} (\sigma^{\mu})^{st} \Psi_{st\dot{r}}$$

$$\Psi^{(8)}_{\mu L} \equiv \Psi^{\mu}_{l} = \frac{1}{2} (\bar{\sigma}^{\mu})^{\dot{r}\dot{s}} \Psi_{l\dot{r}\dot{s}}$$

The S-R bispinor-vector is given by

$$\Psi^{(16)}_{\mu} = \Psi^{(8)}_{\mu L} + \Psi^{(8)}_{\mu R}$$

******
2)

Under Lorentz group, a multi-component field transforms according to

$$\Psi^{(16)} \rightarrow \exp ( \frac{i}{2} \omega_{\mu\nu}J^{\mu\nu}) \Psi^{(16)}$$

Now

$$(J^{\mu\nu})^{(16 \times 16)} \equiv (J^{\mu\nu})^{[4] \otimes ([2] \oplus [2])}$$

can be written as

$$J^{[4]}_{\mu\nu} \otimes 1^{[2] \oplus [2]} + 1^{[4]} \otimes J_{\mu\nu}^{[2] \oplus [2]}$$

$$(J_{\mu\nu})^{\rho a}_{\sigma b} = (J_{\mu\nu})^{\rho}_{\sigma}\delta^{a}_{b} + \delta^{\rho}_{\sigma}(J_{\mu\nu})^{a}_{b}$$

where

$$(J_{\mu\nu})^{\rho}_{\sigma} = \delta^{\rho}_{\nu} \eta_{\mu\sigma} - \delta^{\rho}_{\mu} \eta_{\nu\sigma}$$

and

$$(J_{\mu\nu})^{a}_{b} = - \frac{i}{4} [\gamma_{\mu}, \gamma_{\nu}]^{a}_{b}$$

***

3)

You can also use the nice but lengthy formulation of Hayward which leads to 6x6 spin matrix

$$\left( \begin{array}{cc} S^{3/2}_{(4 \times 4) & 0_{(4 \times 2)} \\ 0_{(2 \times 2)} & S^{1/2}_{(2 \times 2)} \end{array} \right)$$

See
Haward,R.W.,1976,"The dynamics of fields of higher spins",National Bureau of Standards, Monograph 154, Washington D.C.

On SL(2,C), see

Weinberg,S., in "Lectures on Particles and Field Theory", Brandies Summer Institute, 1964, Vol.2.

Moussa P. & Stora R., in "Methods in Subnuclear Physics", Vol.2, 1968, Gordon and Breach.

Lurie D.,1968, "Particles and Fields", Interscience, N.Y.

Nice and simple treatment can be found in