# Is there an easy way to compute $\exp(i\pi J_2) |jm\rangle = (-1)^{j-m} |j,-m\rangle$?

Is there an algebraic way to compute $$\exp(i\pi J_2) |jm\rangle = (-1)^{j-m} |j,-m\rangle$$. I know this is basically the Wigner $$d$$-matrix (which I can just look up), but how is it derived in this special case where the rotation angle is $$\pi$$?

$$J_1,J_2,J_3$$ are just $$J_x,J_y,J_z$$

EDIT. Notice that I changed the question to $$\exp(i\pi J_2)$$ instead of its inverse $$\exp(-i\pi J_2)$$. Indeed, it should be noted that $$\exp(i\pi J_2)^2=(-1)^{2j}$$ and thus $$\exp(-i\pi J_2)|j,m\rangle =(-1)^{3j-m} |j,-m\rangle$$

• What is $J_2$ here? The total angular momentum associated with the quantum number $j$? Or is it the $J_y$ or $J_z$ or something? Oct 22, 2019 at 21:35
• As opposed to geometric, where a π rotation around the y axis reverses the z one? Oct 22, 2019 at 23:33

I think I figured out a simple (algebraic) way of proving the statement. For simplicity, let $$S = \exp{(i\pi J_2)}$$. Using the Adjoint/adjoint representation, it's clear that $$SJ_3 S^{-1} = -J_3$$ and that $$S J_+ S^{-1}=-J_-$$ Notice that \begin{align} SJ_3 S^{-1} S|jm\rangle &= SJ_3 |jm\rangle \\ -J_3 S|jm\rangle &= mS |jm\rangle \\ J_3 S|jm\rangle &= -mS |jm\rangle \end{align} Therefore, $$S|jm\rangle$$ is an eigenvector of $$J_3$$ with eigenvalue $$-m$$ and thus $$S|jm\rangle = c_m |j,-m\rangle$$ where $$|c_m| = 1$$ since $$S$$ is unitary. Apply a similar argument with $$J_+,J_-$$, i.e., \begin{align} SJ_+ S^{-1} S|j,m\rangle &= SJ_+ |j,m\rangle \\ -J_- S|j,m\rangle &= S \sqrt{j(j+1)-m(m+1)}|j,m+1\rangle \\ -c_m \sqrt{j(j+1)-m(m+1)} |j,-m-1\rangle &= c_{m+1} \sqrt{j(j+1)-m(m+1)}|j,-m-1\rangle \\ \end{align} Hence, if $$m, then $$c_{m}=-c_{m+1}$$. In particular, $$c_{j-m}=(-1)^m c$$ where I set $$c=c_j$$. Therefore, the key now is to show what $$c$$ is.

Notice that $$S^4=1$$. Hence, $$c^4=1$$ and thus $$c=\pm 1, \pm i$$. Notice that $$S$$ is real with respect to the orthonormal basis $$|j,m\rangle$$. Hence, $$c=\pm 1$$. However, I cannot tell whether $$c=1$$ or $$=-1$$.

EDIT: I realized that my original final step wasn't enough to prove the statement, so I changed it a little, but I'm still stuck between $$\pm 1$$.

EDIT2: Notice that if $$j=l$$ is an integer, then the character of $$SO(3)$$ is given by

$$\chi_l (\alpha) = \frac{\sin{((l+1/2)\alpha)}}{\sin{(\alpha/2)}}$$ where $$\alpha$$ denotes rotation about any axis of angle $$\alpha$$. In particular, if $$\alpha =\pi$$, then the right-hand-side $$=(-1)^l$$. Notice that the left-hand-side can be given by $$\sum_{m=-l}^l \langle lm| e^{i\pi J_2} |lm\rangle = c(-1)^l$$ since the only nonzero term in the summation is when $$m=0$$. Hence, in this case, we have $$c=+1$$.

EDIT 3: (Finally!) I think I have found a proof for the case where $$j=s$$ is a half-integer. Indeed, if $$j=1/2$$, it's clear that we should take $$c=1$$, since $$\exp (i\pi J_2) = \exp \left(\frac{\pi}{2} i\sigma_2 \right) = i\sigma_2$$ where $$\sigma_2$$ is the Pauli matrix. Now let $$j=s$$ be a half-integer and $$V_j$$ denote the corresponding irrep. Then we have $$V_j \otimes V_{1/2} = V_{j+1/2} \oplus V_{j-1/2}$$ More specifically, we should notice that \begin{align} \left|j+\frac12,j+\frac12\right\rangle &= |j,j\rangle \left|\frac12,\frac12\right\rangle \\ \left|j+\frac12,-j-\frac12\right\rangle &= |j,-j\rangle \left|\frac12,-\frac12\right\rangle \end{align} Indeed, since the basis is unique only up to a global phase, we can always choose $$|j,j\rangle \left|1/2,1/2\right\rangle$$ to be $$|j+1/2,j+1/2\rangle$$. By successively applying $$J_-$$, we see that each time we obtain a nonnegative coefficient in front of the basis and thus we arrive at the second equation, e.g., $$J_- |j,j\rangle \left|\frac12,\frac12\right\rangle = (\cdots) |j,j-1\rangle \left|\frac12,\frac12\right\rangle +(\cdots) |j,j\rangle \left|\frac12,-\frac12\right\rangle$$ where $$(\cdots)$$ represents some nonnegative coefficient. Notice that $$S |j,j\rangle \left|\frac12,\frac12\right\rangle = c_j |j,-j\rangle \left|\frac12,-\frac12\right\rangle$$ where $$c_j$$ is that corresponding to the action of $$S$$ on $$|j,j\rangle$$. Notice that $$j+1/2$$ is an integer and thus $$S$$ should act on $$|j+1/2, j+1/2\rangle$$ by $$+1$$, and thus $$c_j=+1$$.