Clebsch-Gordan coefficients

This is a standard topic in any book on group theory.In summary, the conversation was about finding the Clebsch-Gordan coefficients associated with the addition of two angular momenta, j_1 = 1 and j_2 = \frac{1}{2}. The participants discussed the possibility of finding the proper coefficients without more information and the use of ladder operators to compute the coefficients. The conversation also touched on the idea of decomposing the tensor product of the two representations to find the coefficients.
  • #1
davesface
97
2

Homework Statement


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


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 [tex]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}[/tex]
 
Physics news on Phys.org
  • #2
davesface said:

Homework Statement


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


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 [tex]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}[/tex]

Yes there are a number of states, but there is a Clebsch-Gordan coefficient for each of them. The CG coefficients are labeled as [tex]\langle j_1m_1j_2m_2| JM\rangle[/tex].
 
  • #3
Right, but I am having trouble seeing how it's possible to move forward without knowing M, [tex]m_1[/tex] or [tex]m_2[/tex].

From what I can tell, the furthest I can go is to say
[tex]\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[/tex]
 
  • #4
The way you generally compute CG coefficients is to identify the highest and lowest weight states and then act with the ladder operators :

[tex]J_\pm = J_\pm^{(1)} + J_\pm^{(2)}.[/tex]

The highest weight state satisfies [tex]J_+ |hw\rangle =0[/tex], while the lowest weight state has [tex]J_- |lw\rangle =0[/tex]. It's easy to work out CG for those. Then you use the ladder operators to work out the other coefficients.
 
  • #5
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: 31,987
  • Subtable.jpg
    Subtable.jpg
    15.2 KB · Views: 3,866
  • #6
First of all, write down the tensor product of spaces/representations. Then the tensor product of basis vectors.
 
  • #7
I have no idea how to do what you just said.
 
  • #8
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.
 
  • #9
Each state is labeled by [tex]m_J[/tex]. For each of these states, you can determine what allowed values of [tex]m_{1,2}[/tex] 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 [tex]\mid 1 m_1\rangle \mid \frac{1}{2} m_2\rangle[/tex] is the tensor product of the [tex]j=1[/tex] and [tex]j=1/2[/tex] representations. Finding the linear combinations of those products that equal the [tex]|3/2 m_3/2\rangle[/tex] and [tex]|1/2 m_{1/2}\rangle[/tex] states amounts to the decomposition of the tensor product into irreducible representations.
 

What are Clebsch-Gordan coefficients?

Clebsch-Gordan coefficients are mathematical quantities that describe the relationship between two sets of quantum numbers in a system of two coupled angular momenta. They are used to determine the possible values of the total angular momentum for a given system.

How are Clebsch-Gordan coefficients calculated?

Clebsch-Gordan coefficients are calculated using the Clebsch-Gordan formula, which involves the use of the Wigner 3-j symbols and the Wigner 6-j symbols. These symbols are used to represent the coupling between angular momenta and are calculated using the quantum numbers of the two angular momenta involved.

What is the significance of Clebsch-Gordan coefficients in quantum mechanics?

Clebsch-Gordan coefficients are essential in understanding the behavior of systems with multiple angular momenta, such as atoms and particles. They help determine the allowed energy levels and transitions between them, which are crucial in explaining various physical phenomena in quantum mechanics.

Can Clebsch-Gordan coefficients be negative?

Yes, Clebsch-Gordan coefficients can be negative. This indicates an anti-symmetric coupling between the two angular momenta involved in a system. However, the overall value of the coefficient is always positive as it represents the probability amplitude of a particular coupling.

How are Clebsch-Gordan coefficients used in spectroscopy?

Clebsch-Gordan coefficients are used in spectroscopy to analyze the emission and absorption spectra of atoms and molecules. They help determine the allowed transitions between energy levels and the relative intensities of spectral lines, providing valuable information about the atomic and molecular structure.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
20
Views
4K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
2K
Replies
16
Views
524
  • Advanced Physics Homework Help
Replies
1
Views
778
  • Advanced Physics Homework Help
Replies
0
Views
269
  • Advanced Physics Homework Help
Replies
17
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Back
Top