Sunday, January 31, 2016

quantum mechanics - What is the usefulness of the Wigner-Eckart theorem?


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.



Answer




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:


dΩ(Y3m1)Y2m2Y1m3


Here Ylm -- are spherical harmonics and we integrate over the sphere dΩ=sinθdθdϕ.


This kind of integrals appear over and over in, say, spectroscopy problems. Let us calculate it for m1=m2=m3=0:


dΩ(Y30)Y20Y10=10532π3dΩcosθ(13cos2θ)(3cosθ5cos3θ)=


=10532π32πdθ(3cos2θsinθ14cos4θsinθ+15cos6θsinθ)=32335π


Hard work, huh? The problem is that we usually need to evaluate this for all values of mi. 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:


dΩ(Y3m1)Y2m2Y1m3=l=3,m1|Y2m2|l=1,m3=C321m1m2m3(3||Y2||1)


Cj1j2j3m1m2m3 -- are the Clebsch-Gordan coefficients



(3||Y2||1) -- is the reduced matrix element which we can derive from our expression for m1=m2=m3=0:


32335π=C321000(3||Y2||1)(3||Y2||1)=123π


So the final answer for our integral is:


dΩ(Y3m1)Y2m2Y1m3=34πC321m1m2m3


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.


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