$\renewcommand\bm[1]{\mathbf{#1}}$ $\renewcommand\h{\hbar}$ $\renewcommand\ket[1]{|#1\rangle}$ $\renewcommand\mean[1]{\langle #1 \rangle}$ $\renewcommand\norm[1]{||#1||}$ 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 \begin{equation} J+ = J_1 + iJ_2 \quad \text{and } \quad J_- = J_1 - iJ_2 \end{equation} Using these, one writes \begin{equation} J_1 = \frac{1}{2}\left(J_-+J_+\right) \quad J_2 = \frac{i}{2}\left(J_--J_+\right)\tag{1}\label{1} \end{equation}

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.

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

1 Answer 1


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.

  • $\begingroup$ … in which case the uncertainty relation is saturated for the “rotated” observables $R L_kR^{-1}$. $\endgroup$ Jan 4, 2022 at 22:06

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