Editing Matrix Elements and the Wigner Eckart Theorem Example

Let ${\vec {J}}_{1}$ and ${\vec {J}}_{2}$ be two angular momentum operators, ${\vec {J}}={\vec {J}}_{1}+{\vec {J}}_{2}$ is the sum of these two vectors, and $|JM\rangle$ denotes the eigen states of ${\vec {J}}^{2}$ and ${\mathbf {J}}_{z}$.

${\mathbf {(}}a)$ Show that the matrix elements of $J_{1}^{-},\langle JM|J_{1}^{-}|J'(M+1)\rangle$, vanish, unless ${\mathbf {J}}'={\mathbf {J}}$ or ${\mathbf {J}}'={\mathbf {J}}\pm 1$.

${\mathbf {(}}b)$ Show also that the following expressions are independent of ${\mathbf {M}}$:

$i){\frac {\langle JM|J_{1}^{-}|J(M+1)\rangle }{\sqrt {(J+M+1)(J-M)}}}$

$ii){\frac {\langle JM|J_{1}^{-}|(J-1)(M+1)\rangle }{\sqrt {(J-M+1)(J-M)}}}$

$iii){\frac {\langle JM|J_{1}^{-}|(J+1)(M+1)\rangle }{\sqrt {(J+M+2)(J+M+1)}}}$

${\mathbf {(}}a)$ According to the Wigner-Eckart Theorem we have: $\langle JM|J_{1}^{-}|J'(M+1)\rangle =\langle J',1,M+1,-1|J',1,J,M\rangle \langle J||{\vec {J}}_{1}||J'\rangle$ The Clebsch-Gordan coefficient unless the triangular relation among the vectors is satisfied, i.e. $|J'-1|\leq J\leq J'+1$

This implies that either ${\mathbf {J}}'={\mathbf {J}}$ or $J'=J\pm 1$

${\mathbf {(}}b)$ We use again the Wigner-Eckart Theorem

$i)\langle JM|J_{1}^{-}|J(M+1)\rangle =\langle J,1,M+1,-1|J,1,J,M\rangle \langle J||{\vec {J}}_{1}||J\rangle$

but we know that $\langle J,1,M+1,-1|J,1,J,M\rangle ={\sqrt {(J+M+1)(J-M)}}$

therefore ${\frac {\langle JM|J_{1}^{-}|J(M+1)\rangle }{\sqrt {(J+M+1)(J-M)}}}=\langle J||{\vec {J}}_{1}||J\rangle$ which does not depend on M.

$ii)\langle JM|J_{1}^{-}|(J-1)(M+1)\rangle =\langle J-1,1,M+1,-1|J-1,1,J,M\rangle \langle J||{\vec {J}}_{1}||J-1\rangle$

but we know that $\langle J-1,1,M+1,-1|J-1,1,J,M\rangle ={\sqrt {(J-M-1)(J-M)}}$

therefore ${\frac {\langle JM|J_{1}^{-}|(J-1)(M+1)\rangle }{\sqrt {(J-M-1)(J-M)}}}=\langle J||{\vec {J}}_{1}||J-1\rangle$ which does not depend on M.

$iii)\langle JM|J_{1}^{-}|(J+1)(M+1)\rangle =\langle J+1,1,M+1,-1|J+1,1,J,M\rangle \langle J||{\vec {J}}_{1}||J+1\rangle$

but we know that $\langle J+1,1,M+1,-1|J+1,1,J,M\rangle ={\sqrt {(J+M+2)(J+M)}}$

therefore ${\frac {\langle JM|J_{1}^{-}|(J+1)(M+1)\rangle }{\sqrt {(J+M+2)(J+M)}}}=\langle J||{\vec {J}}_{1}||J+1\rangle$ which does not depend on M.