Angular momentum coherent states

$$\renewcommand\bm[1]{\mathbf{#1}}$$ $$\renewcommand\h{\hbar}$$ $$\renewcommand\ket[1]{|#1\rangle}$$ $$\renewcommand\mean[1]{\langle #1 \rangle}$$ $$\renewcommand\norm[1]{||#1||}$$ Let $$\bm{J}$$ be an angular momentum operator, meaning that it follows the usual definition $$[J_i,J_j] = i\h\epsilon_{ijk}J_k$$. We define the Casimir elements $$J^2 = \sum_i J_i^2$$, such that $$[J^2,J_i] = 0$$ for any $$i=1,2,3$$. Furthermore, we have that $$\ket{l,m,n}$$ are the eigenvalues of the Casimir element and one of the $$J_i$$, take $$J_3$$. The associated eigenvalues are $$\h j\left(j+1\right)$$ and $$\h m$$ (respectively). Note that we have $$-j\leq m\leq j$$. The litterature usually now introduces the operators $$$$J+ = J_1 + iJ_2 \quad \text{and } \quad J_- = J_1 - iJ_2$$$$ Using these, one writes $$$$J_1 = \frac{1}{2}\left(J_-+J_+\right) \quad J_2 = \frac{i}{2}\left(J_--J_+\right)\tag{1}\label{1}$$$$

The uncertainty principle (UP) states that for any operators $$\bm{A}$$ and $$\bm{B}$$, $$\Delta \bm{A}\Delta\bm{B}\geq \frac{1}{2}\norm{i\mean{[\bm{A},\bm{B}]}}$$.

Let us assume that we find ourselves in one of the eigenstates of $$J_3$$. In that eigenspace, one can show the following facts:

• $$\mean{J_1} = 0 = \mean{J_2}$$.
• $$\mean{J_3} = \h m$$.
• $$\mean{J_1^2} = \frac{\h^2}{2}\left(j(j+1)-m^2\right)$$
• $$\mean{J_2^2} = \frac{\h^2}{2}\left(j(j+1)+m^2\right)$$

In particular, from there one has that $$\Delta J_1 = \sqrt{\frac{\h^2}{2}\left[j(j+1)-m^2\right]}$$, $$\Delta J_2 = \sqrt{\frac{\h^2}{2}\left[j(j+1)+m^2\right]}$$ and $$\Delta J_3 = \mean{J_3^2} - \mean{J_3}^2 = 0$$. It is here that my problems arise. We can explicitly check that the UP is followed:

• $$\Delta J_1\Delta J_2 = \frac{\h^2}{2}\left(j(j+1)-m\right) = \frac{\h^2}{2}\sqrt{j^2(j+1)^2-m^2} \leq \frac{\h^2}{2}\norm{m}$$ is obviously verified because of the values $$m$$ can take.
• $$\Delta J_2\Delta J_3 = 0 = \Delta J_3\Delta J_1$$.

Notice that UP is saturated (all three inequalities are saturated) iff $$\norm{m} = j$$, meaning iff $$m = \pm j$$.

Let us now now assume we are in the eigenspace $$\ket{j,\pm j}$$. For simplicity, I will write this $$\ket{j,\pm}$$. I am asked to compute $$\left(\Delta J^2\right)^2 = \mean{\vec{J}^2}-\mean{\vec{J}}^2$$, which I do. It is equal to $$\h^2j(j+1)-\h^2j^2 = \h^2j$$. I am now asked "how can we deduce all the other states following this property (coherent states of the angular momentum?". However, I do not truly understand the question, and am therefore unable to answer it.

• how is $j(j+1)-m=\sqrt{j^2(j+1)^2-m^2}$? moreover should it not be $j(j+1)-m^2$? Jan 4, 2022 at 22:04

I think that they mean you to apply a general rotation to $$|j,j\rangle$$. Factor $$R=e^{\zeta J_-}e^{ \xi J_3} e^{\eta J_+}$$ and as $$J_+ |j,j\rangle=0$$ and acting by $$J_3$$ just goves a number, you can write $$R |j,j\rangle =e^{\zeta J_-}e^{ \xi J_3} e^{\eta J_+} |j,j\rangle\propto e^{\zeta J_-}|j,j\rangle$$
to define an (unormalized) spin coherent state] $$|\zeta\rangle=e^{\zeta J_-}|j,j\rangle.$$ Here $$\zeta$$ is the sterographic coodinate on the unit sphere in which $$\zeta=0$$ is the north pole (spin up) See page 661 and following here.
• … in which case the uncertainty relation is saturated for the “rotated” observables $R L_kR^{-1}$. Jan 4, 2022 at 22:06