Fining Clebsch-Gordan coeffs in special cases, alternate method

In summary, the Clebsch-Gordan coefficients can be derived using stepping operators and involve arbitrary phase factors that can depend on the angular momentum states. These conventions are known as the Condon and Shortley phase conventions.
  • #1
Jesssa
51
0
Hello,

I was wondering, is there a way to derive the expression for these coefficients without the use of the general CGC formula?

For example, the J=0 case (this is taken from wikipedia)

65a3e02b0a4b4e80f8d539efb5c908aa.png


The 1/sqrt term is clear but the (-1)^.. term not so much, is there a way to find this coefficient?

I have read it requires the use of spinors and so on but I was wondering if there was a more simple way?

I found a derivation of the CGC using only binomial expansions, but it was difficult to see exactly how they saw the starting point, it seemed like it must have been educated guesses until worked. Other derivations of the coefficients rely on the J=0 coefficients or more advanced mathematics.

Thanks,
 
Physics news on Phys.org
  • #2
The usual way to derive Clebsch-Gordan coefficients in special cases is to use stepping operators. For the example you give, if you let |0,0> = ∑ Cm|j,m>|j,-m>, you know that J+|0,0> = 0.

J+(∑ Cm|j,m>|j,-m>) = ∑ Cm√(j-m)(j+m+1) (|j,m+1>|j,-m> + |j,m>|j,-m+1>) = ∑√(j-m)(j+m+1)(Cm + Cm+1)|j,m+1>|j,-m>

In order for this sum to vanish, the Cm's must all have the same value and alternate in sign.
 
  • #3
Oh I see,

Thanks for your answer!

However I am still having trouble figuring out how the negative sign of the coefficients depends on j1 and m1 in this example.

(-1)j1 + m1

From application of either of the ladder operators you get

Cm = -Cm+1

But from this it's not clear which coefficient is actually negative, only that they have opposite signs.

Is it possible to simply find things like this? What I mean is, in the case of this example, the dependence of the negative sign on j1 and m1?
 
  • #4
Some of the phase factors involved in the definition of the angular momentum states and the Clebsch-Gordan coefficients are not uniquely determined, and must be arbitrarily assigned. For example the state with the highest m-value is conventionally chosen to have a plus sign. The standard choices are known as the Condon and Shortley phase conventions.
 

Related to Fining Clebsch-Gordan coeffs in special cases, alternate method

1. What is the purpose of finding Clebsch-Gordan coefficients in special cases?

The Clebsch-Gordan coefficients are used to calculate the coupling of angular momenta in quantum mechanics. In special cases, such as when dealing with specific symmetries or restrictions, alternate methods may be necessary to find these coefficients.

2. What are some common special cases in which alternate methods are needed for finding Clebsch-Gordan coefficients?

Some common special cases include dealing with particles with a spin of 1/2, particles with multiple symmetries, and particles with specific angular momentum values.

3. What are some alternate methods for finding Clebsch-Gordan coefficients in special cases?

Some alternate methods include using Wigner's 3-j symbols, Racah coefficients, and recoupling coefficients. These methods can be more efficient and effective in certain special cases.

4. How do these alternate methods differ from the traditional method of finding Clebsch-Gordan coefficients?

The traditional method of finding Clebsch-Gordan coefficients involves using the Clebsch-Gordan formula, which can become cumbersome and complex in special cases. Alternate methods offer simpler and more elegant solutions for these cases.

5. Are there any drawbacks to using alternate methods for finding Clebsch-Gordan coefficients?

While alternate methods may be more efficient in certain special cases, they may not always provide the most accurate results. It is important to carefully consider the specific case and choose the most appropriate method for finding Clebsch-Gordan coefficients.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
556
  • Quantum Physics
Replies
3
Views
4K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
2K
Replies
6
Views
3K
  • Advanced Physics Homework Help
Replies
6
Views
2K
Replies
0
Views
128
Replies
16
Views
6K
  • Calculus and Beyond Homework Help
Replies
4
Views
745
Replies
2
Views
1K
Back
Top