View source for Editing Matrix Elements and the Wigner Eckart Theorem Example

Let <math> \vec J_1 </math> and <math> \vec J_2 </math> be two angular momentum operators, <math> \vec J = \vec J_1 + \vec J_2 </math> is the sum of these two vectors, and <math> |J M \rangle </math> denotes the eigen states of <math> \vec J^2 </math> and <math> \bold J _z </math>.

<math> \bold (a) </math> Show that the matrix elements of <math> J_1^- , \langle J M | J_1^- | J' (M + 1)\rangle </math>, vanish, unless <math> \bold J' = \bold J </math> or <math> \bold J' = \bold J \pm 1 </math>.


<math> \bold (b) </math>  Show also that the following expressions are independent of <math> \bold M </math>:

<math> i) \frac{\langle J M | J_1^- | J (M + 1)\rangle}{\sqrt{(J+M+1)(J-M)}} </math>

<math> ii) \frac{\langle J M | J_1^- | (J-1)(M + 1)\rangle}{\sqrt{(J-M+1)(J-M)}} </math>

<math> iii) \frac{\langle J M | J_1^- | (J+1)(M + 1)\rangle}{\sqrt{(J+M+2)(J+M+1)}} </math>

<math> \bold (a) </math> According to the Wigner-Eckart Theorem we have:
<math> \langle J M | J_1^- | J' (M + 1)\rangle = \langle J', 1,  M+1, -1 | J', 1, J, M\rangle \langle J || \vec J_1||J'\rangle</math>  The Clebsch-Gordan coefficient unless the triangular relation among the vectors is satisfied, i.e. <math> |J'-1| \leq J\leq J'+1 </math>

This implies that either <math> \bold J'= \bold J </math> or <math> J'=J \pm 1 </math>

<math> \bold (b) </math> We use again the Wigner-Eckart Theorem

<math>i) \langle J M | J_1^- | J (M + 1)\rangle = \langle J, 1,  M+1, -1 | J, 1, J, M\rangle \langle J || \vec J_1||J\rangle</math>

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

therefore <math> \frac{\langle J M | J_1^- | J (M + 1)\rangle}{\sqrt{(J+M+1)(J-M)}} = \langle J || \vec J_1||J\rangle</math> which does not depend on M.

<math> ii) \langle J M | 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</math>

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

therefore <math> \frac{\langle J M | J_1^- | (J-1) (M + 1)\rangle}{\sqrt{(J-M-1)(J-M)}} = \langle J || \vec J_1||J-1\rangle</math> which does not depend on M.

<math> iii) \langle J M | 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</math>

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

therefore <math> \frac{\langle J M | J_1^- | (J+1) (M + 1)\rangle}{\sqrt{(J+M+2)(J+M)}} = \langle J || \vec J_1||J+1\rangle</math> which does not depend on M.