I am doing some self-study in between undergrad and grad school and I came across the beastly Wigner-Eckart theorem in Sakurai's Modern Quantum Mechanics.  I was wondering if someone could tell me why it is useful and perhaps just help me understand a bit more about it.  I have had two years of undergrad mechanics and I think I have a reasonably firm grasp of the earlier material out of Sakurai, so don't be afraid to get a little technical.
I will not get into theoretical details -- Luboš ad Marek did that better than I'm able to.
Let me give an example instead: suppose that we need to calculate this integral:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3}$

Here $Y_{lm}$ -- are spherical harmonics and we integrate over the sphere $d\Omega=\sin\theta d\theta d\phi$.

This kind of integrals appear over and over in, say, spectroscopy problems. Let us calculate it for $m_1=m_2=m_3=0$:  

$\int d\Omega (Y_{30})^*Y_{20}Y_{10} = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\int d\Omega \cos\theta\,(1-3\cos^2\theta)(3\cos\theta-5\cos^3\theta)=$

$ = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\cdot 2\pi \int d\theta\,\left(3\cos^2\theta\sin\theta-14\cos^4\theta\sin\theta+15\cos^6\theta\sin\theta\right)=\frac{3}{2}\sqrt{\frac{3}{35\pi}}$

Hard work, huh? The problem is that we usually need to evaluate this for all values of $m_i$. That is 7*5*3 = 105 integrals. So instead of doing all of them we got to exploit their symmetry. And that's exactly where the Wigner-Eckart theorem is useful:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3} = \langle l=3,m_1| Y_{2m_2} | l=1,m_3\rangle = C_{m_1m_2m_3}^{3\,2\,1}(3||Y_2||1)$

$C_{m_1m_2m_3}^{j_1j_2j_3}$ -- are the Clebsch-Gordan coefficients  

$(3||Y_2||1)$ -- is the reduced matrix element which we can derive from our expression for $m_1=m_2=m_3=0$:  

$\frac{3}{2}\sqrt{\frac{3}{35\pi}} = C_{0\,0\,0}^{3\,2\,1}(3||Y_2||1)\quad \Rightarrow \quad (3||Y_2||1)=\frac{1}{2}\sqrt{\frac{3}{\pi}}$

So the final answer for our integral is:  

$\int d\Omega(Y_{3m_1})^*Y_{2m_2}Y_{1m_3}=\sqrt{\frac{3}{4\pi}}C_{m_1m_2m_3}^{3\,2\,1}$

It is reduced to calculation of the Clebsch-Gordan coefficient and there are a lot of, tables, programs, reduction and summation formulae to work with them. 
