Question 7.P.8: (a) Find the reduced matrix elements associated with the sph...

(a) Find the reduced matrix elements associated with the spherical harmonic Y_{kq} (\theta ,\varphi ).

(b) Calculate the dipole transitions 〈n^{\prime } l^{\prime } m^{\prime }|\vec{r} |nlm 〉.

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On the one hand, an application of the Wigner–Eckart theorem to Y_{kq} yields

〈l^{\prime },m^{\prime }|Y_{kq} |l,m 〉=〈l,k;m,q|l^{\prime }, m^{\prime } 〉〈l^{\prime }\parallel Y^{(k)} \parallel l 〉                (7.401)

and, on the other hand, a straightforward evaluation of

〈l^{\prime },m^{\prime }|Y_{kq}|l,m 〉=\int_{0}^{2\pi } d\varphi \int_{0}^{\pi } \sin \theta d\theta 〈l^{\prime },m|\theta \varphi 〉Y_{kq} (\theta ,\varphi )〈\theta \varphi |l,m 〉

 

=\int_{0}^{2\pi } d\varphi \int_{0}^{\pi } \sin \theta d\theta Y^{*}_{l^{\prime }m^{\prime }} (\theta ,\varphi )Y_{kq}(\theta , \varphi )Y_{lm} (\theta ,\varphi )                    (7.402)

can be inferred from the triple integral relation (7.244):

\int_{0}^{2\pi }d\alpha \int_{0}^{\pi }Y^{*}_{lm}(\beta ,\alpha )Y_{l_{1}m_{1} } (\beta ,\alpha )Y_{l_{2}m_{2} } (\beta ,\alpha ) \sin \beta d\beta =\sqrt{\frac{(2l_{1}+1)(2l_{2}+1)}{4\pi (2l+1)} } 〈l_{1},l_{2};0,0|l,0 〉\times 〈l_{1},l_{2};m_{1},m_{2}|l,m 〉.   (7.244)

 

〈l^{\prime },m^{\prime }|Y_{kq}|l,m 〉=\sqrt{\frac{(2l+1)(2k+1)}{4\pi (2l^{\prime }+1)} } 〈l,k;0,0|l^{\prime },0 〉〈l,k;m,q |l^{\prime },m^{\prime } 〉.      (7.403)

We can then combine (7.401) and (7.403) to obtain the reduced matrix element

〈l^{\prime }\parallel Y^{(k)}\parallel l 〉=\sqrt{\frac{(2l+1)(2k+1)}{4\pi (2l^{\prime }+1)} } 〈l,k;0,0|l^{\prime },0 〉.                 (7.404)

(b) To calculate 〈n^{\prime } l^{\prime } m^{\prime }|\vec{r} |nlm 〉 it is more convenient to express the vector \vec{r} in terms of the spherical components \vec{r}=(r_{-1}, r_{0},r_{1} ), which are given in terms of the Cartesian coordinates x, y, z as follows:

r_{1}=-\frac{x+iy}{\sqrt{2} } =\frac{r}{\sqrt{2} } e^{i\varphi } \sin \theta ,      r_{0}=z=r\cos \theta ,       r_{-1}=\frac{x-iy}{\sqrt{2} }=\frac{r}{\sqrt{2} } e^{-i\varphi }\sin \theta ,         (7.405)

which in turn may be condensed into a single relation

r_{q}=\sqrt{\frac{4\pi }{3} } rY_{1q} (\theta ,\varphi ),                           q=1, 0, -1.            (7.406)

Next we may write 〈n^{\prime } l^{\prime } m^{\prime }|r_{q} |nlm 〉 in terms of a radial part and an angular part:

〈n^{\prime } l^{\prime } m^{\prime }|r_{q} |nlm 〉=\sqrt{\frac {4\pi }{3} }〈n^{\prime } l^{\prime }|r_{q}|nl 〉〈l^{\prime }, m^{\prime }|Y_{1q}(\theta ,\varphi )|l,m 〉.               (7.407)

The calculation of the radial part, 〈n^{\prime } l^{\prime }|r_{q}|nl 〉=\int_{0}^{\infty } r^{3} R^{*}_{n^{\prime } l^{\prime }} (r)R^{*}_{nl} (r)dr,is straightforward and is of no concern to us here; see Chapter 6 for its calculation. As for the angular part 〈l^{\prime },m^{\prime }|Y_{1q} (\theta ,\varphi )|l,m 〉, we can infer its expression from (7.403)

〈l^{\prime },m^{\prime }|Y_{1q}|l,m 〉=\sqrt{\frac{3(2l+1)}{4\pi (2l^{\prime }+1)} } 〈l,1;0,0|l^{\prime },0 〉〈l,1;m,q|l^{\prime },m^{\prime } 〉.                 (7.408)

The Clebsch–Gordan coefficients 〈l,1;m,q|l^{\prime }, m^{\prime } 〉 vanish unless m^{\prime }=m+q and l-1\leq l^{\prime }\leq l+1 or \Delta m= m^{\prime }-m=q=1,0,-1 and \Delta l=l^{\prime }-l= 1,0,-1. Notice that the case \Delta l=0 is ruled out from the parity selection rule; so, the only permissible values of l^{\prime } and l are those for which \Delta l=l^{\prime }-l=\pm 1. Obtaining the various relevant Clebsch–Gordan coefficients from standard tables, we can ascertain that the only terms of (7.408) that survive are

〈l+1,m+1|Y_{11}|l,m 〉=\sqrt{\frac{3(l+m+1)(l+m+2)}{8\pi (2l+1)(2l+3)} } ,                      (7.409)

 

〈l-1,m+1|Y_{11}|l,m 〉=\sqrt{\frac{3(l-m-1)(l-m)}{8\pi (2l+1)(2l+3)} } ,                          (7.410)

 

〈l+1,m|Y_{10}|l,m 〉=\sqrt{\frac{3[(l+1)^{2}-m^{2} ]}{4\pi (2l+1)(2l+3)} } ,                  (7.411)

 

〈l-1,m|Y_{10}|l,m 〉=\sqrt{\frac{3(l^{2}-m^{2} )}{4\pi (2l+1)(2l-1)} } ,                     (7.412)

 

〈l+1,m-1|Y_{1-1}|l,m 〉=\sqrt{\frac{3(l-m+1)(l-m+2)}{8\pi (2l+1)(2l+3)} } ,                (7.413)

 

〈l-1,m-1|Y_{1-1}|l,m 〉=\sqrt{\frac{3(l+m)(l+m-1)}{8\pi (2l+1)(2l-1)} } .                (7.414)

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