Can a Rotation of 180º Leave a Spin-2 System Hamiltonian Unchanged?

[shetland]

For S = 2, I have the following hamiltonian,
$$H=a \left( 2\,{S}^{2}-4\,{S_{{z}}}^{2} \right) +4\,b \left( (\left { S_+}^{3}) +({S_-}^{3})$$I'm to show that a rotation of 180º,

$$e^{-i \pi S_x/ \hbar}$$

leaves the hamiltonian unchanged.

I started thinking I could use the baker-hausdorff formula, but now not so sure - you would get something like, H + commutation terms. The first would be obvious zero, [Sx, S^2], but the other ones get messy.
Is the way to do this to find the matrix rep for Sx, S+, S-, then you would have the hamiltonian in matrix rep, along with the operator, and then do the series expansion for the exp?
I guess this would satisfy another part to this quesiton, which is to get the rep for Dm'm matrix. But it seems to me that one should be able to answer the first part without directly calculating the matrix?

 

[Physics Monkey]

This is an easy task when you remember that the $$S_i$$ transform as vector under rotations. How do the components of a vector change under rotation by $$180^\circ$$ about the x-axis?

Edit: Ok, great. Now everything is fine.

 

[shetland]

Monkey,

I guess I'm tired. I looked a bit closer - the problem in the book is quite small. I have a misplaced parenthesis now - so you can do this by inspection, right?

We know the value of $$S_i$$ is 2$$\hbar$$, so that the rotation is then $$2\pi$$, which brings the system back to original.

 

[Physics Monkey]

You don't actually know what the "value" of the spin components are. In fact, they can't all be diagonalized so they can't all have definite values at the same time. However, you can definitely do it by inspection. Like I said, if I rotate by $$180^\circ$$ degress about the x-axis, how do vector components changes. Cleary the x component is unchanged so you have $$U S_x U^\dag = S_x$$. What happens to $$S_y$$ and $$S_z$$?

 

[shetland]

right. I'm not thinking. Of course they can't all be diagonalized. Thanks.

I thought about it as you suggested, clearly y and z are rotated by $$\cos{\pi}$$, z-> -z, y-> -y. For the first part of the hamiltonian, obviously S_z^2is unchanged, as is S^2, so its just a matter of the S± terms. I guess I can see since S± = S_x ± iS_y that the net result is to leave H unchanged (since the Sy terms cancel one another due to the ±?