Graduate Calculation a reduced matrix element using E-Wigner Th.

Click For Summary
The discussion focuses on calculating the reduced matrix element ⟨la∥Y(L)∥lb⟩ using the Eckart-Wigner theorem. The user encounters difficulty in understanding how the integral solution affects the Wigner 3j symbols in their calculations. They recognize that the Kronecker deltas from the integral play a crucial role in simplifying the Wigner symbols, leading to the survival of only specific terms. The Dirac delta function's properties clarify why only the term with ##l=l_a## and ##m=-m_a## remains after substitution. The conversation concludes with the user expressing understanding of the concept.
squareroot
Messages
76
Reaction score
0
Hello.

I fail to follow one step in the process of calculating ⟨la∥Y(L)∥lb⟩ .

The spherical harmonics Yma(L)(r) represent the 2L+1 components of the spherical tensor of rank L. Writing the Eckart-Wigner th. for M = 0 yields:
Screen Shot 2017-08-05 at 2.12.25 PM.png
(1)

Also one can write
Screen Shot 2017-08-05 at 2.13.17 PM.png
(2)

Coupling L and lb to l:
Screen Shot 2017-08-05 at 2.14.00 PM.png
(3)

Thus having
Screen Shot 2017-08-05 at 2.14.30 PM.png
(4)

Now solving the integral:
Screen Shot 2017-08-05 at 2.15.05 PM.png
(5)

So:

Screen Shot 2017-08-05 at 2.15.31 PM.png
(6)

Here is my problem! After solving the integral (5) and replacing it into (4) I don't understand how that changes the Wigner 3j symbols from
Screen Shot 2017-08-05 at 2.17.26 PM.png
(3) into
Screen Shot 2017-08-05 at 2.17.52 PM.png
(6)

Could anyone please help me with this step? I m guessing it has something to do with does kronecker deltas from solving the integral and they act on the wigner symbols after substitution... but i have no idea how!
 
Physics news on Phys.org
In equation (4), the right hand side is summed over ##l## and ##m##. However upon inserting the result from equation (5), only the term with ##l=l_a## and ##m=-m_a## survives. Yes it has to do with the property of Dirac delta symbol which is for any pair of integer ##i## and ##j##, the Dirac delta ##\delta_{ij}## will give zero if ##i\neq j##. If ##i=j##, ##\delta_{ij} = 1##.
 
Oww i get it! Thank you so much!
 

Similar threads

Replies
0
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K