# Explanation of the sign of Clebsch-Gordan coefficients These are the Clebsch-Gordan coefficients when the orbital and spin-angular momenta of a single spin 1/2 particle are added.

I'm not able to understand the explanation. What I can understand is that:

Since the elements of $$J_-$$ are positive, for $$-l-\frac{1}{2}\leqslant m\leqslant l+\frac{1}{2}$$, the elements of all kets $$\mid {j=l+\frac{1}{2},m}\rangle$$ will have same sign as the ket $$\mid {j=l+\frac{1}{2},m=l+\frac{1}{2}\rangle}$$. Also, since $$\mid m_l=l,m_s=\frac{1}{2}\rangle = \mid {j=l+\frac{1}{2},m=l+\frac{1}{2}\rangle}$$, all the elements of the kets $$\mid m_l=m+\frac{1}{2},m_s=-\frac{1}{2}\rangle$$ and $$\mid m_l=m-\frac{1}{2},m_s=\frac{1}{2}\rangle$$ will also have same sign as $$\mid {j=l+\frac{1}{2},m=l+\frac{1}{2}\rangle}$$. So, $$\langle m_l=m+\frac{1}{2},m_s=-\frac{1}{2}\mid {j=l+\frac{1}{2},m}\rangle$$ will be positive.

But $$\langle m_l=m-\frac{1}{2},m_s=\frac{1}{2}\mid {j=l-\frac{1}{2},m}\rangle$$ is negative, that means $$\mid {j=l-\frac{1}{2},m=l-\frac{1}{2}}\rangle$$ and $$\mid {j=l+\frac{1}{2},m=l+\frac{1}{2}\rangle}$$ have elements of opposite sign.

How is it so? Is my explanation correct?

• Can you link to a source for, or provide a citation for, the screenshotted text? – rob Dec 28 '18 at 20:13
• fisica.net/quantica/… Pg-214 of Modern Quantum Mechanics by Sakurai – Asit Srivastava Dec 29 '18 at 5:38

I'm not completely sure where the problem is but the issue of the sign is best seen with an explicit example. I will pick $$\ell=4$$ for the purpose of the example and you can generalize to any $$\ell$$. We have, using the notation $$\vert JM_J\rangle$$ for the arguments of kets: \begin{align} \textstyle\vert\frac{9}{2},\frac{9}{2}\rangle = \vert 4,4\rangle\vert\frac{1}{2},\frac{1}{2}\rangle\, ,\tag{1} \end{align} and, as a result, all the CGs for the $$J=\frac{9}{2}$$ will be positive because the matrix element of $$L_-$$ is always positive. Thus \begin{align} \textstyle\vert\frac{9}{2},\frac{7}{2}\rangle = \frac{1}{3} \vert 4,4\rangle\vert\frac{1}{2},-\frac{1}{2}\rangle +\frac{2\sqrt{2}}{3}\vert 4,3\rangle\vert\frac{1}{2},\frac{1}{2}\rangle \, .\tag{2} \end{align} The state $$\vert\frac{7}{2},\frac{7}{2}\rangle$$ must be orthogonal to (2) so one of the coefficient must be negative, i.e. one must have by orthogonality \begin{align} \textstyle\vert\frac{7}{2},\frac{7}{2}\rangle= \pm\left(\frac{2\sqrt{2}}{3}\vert 4,4\rangle\vert\frac{1}{2},-\frac{1}{2}\rangle-\frac{1}{3} \vert 4,3\rangle\vert\frac{1}{2},\frac{1}{2}\rangle\right)\, . \tag{3} \end{align} The convention is the choose the coefficient of the form $$\vert \ell,\ell\rangle \vert s m_s\rangle$$ to be positive, so that in (3) we keep the $$+$$ sign in front of the whole state. Once you have this, the relative minus sign in front of the second term carries through so that \begin{align} \textstyle\vert\frac{7}{2},\frac{5}{2}\rangle=\frac{\sqrt{7}}{3} \vert 4,3\rangle\vert \frac{1}{2},-\frac{1}{2}\rangle -\frac{\sqrt{2}}{3}\vert 4,2\rangle\vert \frac{1}{2},\frac{1}{2}\rangle\, .\tag{4} \end{align} The sign of the $$m_s=1/2$$ component does not change because $$L_-$$ acting on the second term in (3) gives $$L_-\left(-\frac{1}{3}\right)\vert 4,3\rangle\vert \textstyle\frac{1}{2},\frac{1}{2}\rangle=-\frac{1}{3}\sqrt{14} \vert 4,2\rangle\vert \textstyle\frac{1}{2},\frac{1}{2}\rangle -\frac{1}{3} \vert 4,3\rangle\vert \frac{1}{2},-\frac{1}{2}\rangle \tag{5}$$ Now, $$L_-$$ acting on $$\vert 4,4\rangle\vert \frac{1}{2},-\frac{1}{2}\rangle=+2\sqrt{2}\vert 4,3\rangle\vert \frac{1}{2},-\frac{1}{2}\rangle$$ is positive, but the corresponding term in (5) has a negative sign so it could be, in some circumstance, the pieces of the $$\vert 4,2\rangle\vert \frac{1}{2},-\frac{1}{2}\rangle$$ come out with unknown sign. In fact, since we know that all the coefficients of $$\vert \frac{9}{2}, M_J\rangle$$ are positive; by orthogonality, the coefficients for $$\vert \frac{7}{2},M_J\rangle$$ must have opposite signs or else there is no chance they will be orthogonal to the $$J=\frac{9}{2}$$ states. Since we know from (4) that the coefficient for the $$m_s=1/2$$ state is negative, it follows that the one for the $$m_s=-1/2$$ state must be positive.