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It is often claimed that spin-2 particles return to their previous state after $\pi$ rotation, just like spin-1/2 particles return after $4\pi$ rotation. But my calculation suggests otherwise.

Let z axis be the axis of rotation. The matrix form of $J_z$ under the basis $\{|2m\rangle\}(m=0,\pm 1,\pm 2)$ is $$J_z= \begin{pmatrix} 2& & & &\\ &1& & &\\ & &0& &\\ & & &-1&\\ & & & &-2 \end{pmatrix}.$$

The rotation operator of 180° rotation around z axis is $$e^{\mathrm{i} \pi J_z}= \begin{pmatrix} 1& & & &\\ &-1& & &\\ & &1& &\\ & & &-1&\\ & & & &1 \end{pmatrix},$$ which is not identity, nor scalar matrix.

So is the claim wrong, or do I make mistakes in my calculation?

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up vote 1 down vote accepted

That's only true when all the spin is aligned with the axis of rotation. When you have a state of a massive spin-2 particle at rest with 1 unit of spin in the z direction, a pi rotation around the z axis gives a -1, as you calculated. Your matrix is right--- only the action on the subspace of z-spin 2,0,-2 gives the identity. When there are 2 units or 0 units in the z-direction, it comes back to itself after $\pi$ rotation around the z axis.

For gravitons, the pi rotation around the direction of motion returns to the same state, because the helicity is always $\pm 2$, but a pi rotation around another axis doesn't produce the same state, simply because the direction of motion is rotated, the graviton is going in a different direction.

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