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## 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

[tex]

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]

[/tex]

which has eigenvalues +1/2 and -1/2. For a spin-1 particle, the operator is

[tex]

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]

[/tex]

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.

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

[tex]

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]

[/tex]

which has eigenvalues +1/2 and -1/2. For a spin-1 particle, the operator is

[tex]

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]

[/tex]

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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