# Explicit calculation of $J_2$ induced rotation by $-\pi$ reversing the sign of $J_3$ operator

Let $\textbf{J}=(J_1,J_2,J_3)$ be the angular momentum QM operators in the x, y, z spacial directions respectively. These are also the generators of rotations around the x, y, z axes respectively. Intuitively, and according to literature I'm reading (Weinberg's QFT, bottom of page 78), a rotation by $\pi$ given by $e^{i\pi J_2}$ should reverse the sign of $J_3$. Mathematically I believe this means $$J_3 e^{i\pi J_2} = -e^{i\pi J_2} J_3 ,$$ where these operators are assumed to be acting on some state. I'm finding it hard to actually prove this. We have the commutation relations $[J_i,J_j]=i \epsilon_{ijk} J_k$. Writing

$$J_3 e^{i\pi J_2} =J_3 \: (1+i\pi J_2-\pi^2J_2^2 -i\pi^3J_2^3+...),$$ before even working at commuting $J_3$ through the terms it seems like there will be a problem since the first order term in the series will remain $J_3$ with a positive sign, so I don't see how I'd ever get an overall minus sign... I could be wrong anywhere here, any help would be appreciated.

It is hard to guess the answer for a series before actually computing it. You were on the right track, just needed to keep going.

Begin with

$$J_{3} e^{i \pi J_{2}} = e^{i \pi J_{2} } e^{-i \pi J_{2}} J_{3} e^{i \pi J_{2}}$$

and use what people call Baker-Hausdorrf lemma:

$$e^{i \lambda A} B e^{-i \lambda A} = B + i \lambda \left[ A, B \right] \, + \cdots \, + \frac{i^{n} \lambda^{n}}{n!} \left[ A, [A, [A, \,... [A, B] \, ... ]\right] + \cdots$$

This will give you

$$e^{i \pi J_{2}} \left( J_{3} - i \pi \left[ J_{2}, J_{3} \right] - \frac{\pi^{2}}{2!} [J_{2}, [ J_{2}, J_{3}]] + \cdots \right)$$

which by the commutation relations you provide,

$$\left[ J_{1}, J_{2} \right] = i J_{3}$$

and

$$\left[ J_{2} , J_{3} \right] = i J_{1},$$

leaves

$$e^{i \pi J_{2}} \left( J_{3} + \pi J_{1} - \frac{\pi^{2}}{2!} J_{3} + \cdots \right) = e^{i \pi J_{2}} J_{3} \left( 1 - \frac{\pi^{2}}{2!} + \frac{\pi^{4}}{4!} - \frac{\pi^{6}}{6!} + \cdots \right) = e^{i \pi J_{2}} J_{3} \cos \pi = e^{i \pi J_{2} }(- J_{3}) .$$

Two comments. I clevered ignored $J_{1}$'s because their coefficients sum to $\sin \pi$ (the previous thing that I wrote was pointed out to be incorrect).

The other thing is that you can work out the coefficients yourself, and thus verify that everytime you will get the correct factors of $\pm \pi^{n}$ in the cosine series.