How Do You Find Clebsch-Gordan Coefficients for j_1 = 1 and j_2 = 1/2?

davesface
Messages
97
Reaction score
2

Homework Statement


Find the Clebsch-Gordan coefficients associated with the addition of two angular momenta j_1 = 1 and j_2 = \frac{1}{2}


Homework Equations


The table of coefficients.


The Attempt at a Solution



I think I am misunderstanding something important here. I can't see how it is possible to find the proper coefficients without more information since these two particles could be in a number of different states that satisfy J=\frac{3}{2}, \frac{1}{2}: \frac{3}{2}\frac{3}{2}, \frac{3}{2}\frac{1}{2}, \frac{1}{2}\frac{1}{2}, and \frac{1}{2}\frac{-1}{2}
 
Physics news on Phys.org
davesface said:

Homework Statement


Find the Clebsch-Gordan coefficients associated with the addition of two angular momenta j_1 = 1 and j_2 = \frac{1}{2}


Homework Equations


The table of coefficients.


The Attempt at a Solution



I think I am misunderstanding something important here. I can't see how it is possible to find the proper coefficients without more information since these two particles could be in a number of different states that satisfy J=\frac{3}{2}, \frac{1}{2}: \frac{3}{2}\frac{3}{2}, \frac{3}{2}\frac{1}{2}, \frac{1}{2}\frac{1}{2}, and \frac{1}{2}\frac{-1}{2}

Yes there are a number of states, but there is a Clebsch-Gordan coefficient for each of them. The CG coefficients are labeled as \langle j_1m_1j_2m_2| JM\rangle.
 
Right, but I am having trouble seeing how it's possible to move forward without knowing M, m_1 or m_2.

From what I can tell, the furthest I can go is to say
\sum_{m_1+m_2=M}C^{1,\frac{1}{2},J}_{m_1,m_2,M}\mid 1 m_1\rangle \mid \frac{1}{2} m_2\rangle
 
The way you generally compute CG coefficients is to identify the highest and lowest weight states and then act with the ladder operators :

J_\pm = J_\pm^{(1)} + J_\pm^{(2)}.

The highest weight state satisfies J_+ |hw\rangle =0, while the lowest weight state has J_- |lw\rangle =0. It's easy to work out CG for those. Then you use the ladder operators to work out the other coefficients.
 
While that does seem like an easier way to do it, the point of the exercise is to learn how to read off the values from a table (see attached picture). Clearly I'm looking for something in the 1x1/2 table (second picture), but beyond that I'm lost.
 

Attachments

  • CG table.jpg
    CG table.jpg
    67.5 KB · Views: 32,035
  • Subtable.jpg
    Subtable.jpg
    15.2 KB · Views: 3,970
First of all, write down the tensor product of spaces/representations. Then the tensor product of basis vectors.
 
I have no idea how to do what you just said.
 
The angular momentum 1 representation is 3-dim. The 1/2 representation is 2-dim. Their tensor product is 3x2=6 dim. and decomposes in a direct sum of 2 spaces, one of total ang. mom. 3/2 which is 4 dim. and the other of total ang. mom. 1/2 which is 2-dim.

This is stuff that you should know about. Now use the basis vectors and write down the tensor product of them using C-G coefficients.
 
Each state is labeled by m_J. For each of these states, you can determine what allowed values of m_{1,2} can appear in the linear combination you wrote in post #3. The coefficients in the linear combination are the CG coefficients you can read off the table. Remember that you usually need to take a square root of the number in the table to get the CG coefficient.
 
  • #10
Oh, so do I just find the coefficients corresponding to the 3/2 case and do a linear combination of them (using the appropriate coefficients from the table) and then do the same thing for the 1/2 case?
dextercioby said:
The angular momentum 1 representation is 3-dim. The 1/2 representation is 2-dim. Their tensor product is 3x2=6 dim. and decomposes in a direct sum of 2 spaces, one of total ang. mom. 3/2 which is 4 dim. and the other of total ang. mom. 1/2 which is 2-dim.

This is stuff that you should know about. Now use the basis vectors and write down the tensor product of them using C-G coefficients.

Yeah, all of that means nothing to me. I double checked in the index and neither "tensor product" nor "outer product" appears anywhere in the book (Griffiths), so I'm fairly certain that we will not ever cover that stuff in this class. I do appreciate the gesture, though.
 
  • #11
What you mean when you write \mid 1 m_1\rangle \mid \frac{1}{2} m_2\rangle is the tensor product of the j=1 and j=1/2 representations. Finding the linear combinations of those products that equal the |3/2 m_3/2\rangle and |1/2 m_{1/2}\rangle states amounts to the decomposition of the tensor product into irreducible representations.
 

Similar threads

Back
Top