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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?

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

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  • $\begingroup$ Thanks. I have understood. But for eq.3 if I choose minus sign in front of the whole state then what's the problem? Only the order of terms will be reversed to have positive in the front. $\endgroup$ – Asit Srivastava Dec 29 '18 at 5:51
  • $\begingroup$ @AsitSrivastava There is a conventional choice, and that’s just convention. What is known for sure is that one of the terms will be + and the other will be -. There is no problem in choosing the - sign in (3): that would simply be a different convention, and it would ripple through as any other convention. $\endgroup$ – ZeroTheHero Dec 29 '18 at 5:53

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