What is Clebsch-gordan: Definition and 25 Discussions
In physics, the Clebsch–Gordan (CG) coefficients are numbers that arise in angular momentum coupling in quantum mechanics. They appear as the expansion coefficients of total angular momentum eigenstates in an uncoupled tensor product basis. In more mathematical terms, the CG coefficients are used in representation theory, particularly of compact Lie groups, to perform the explicit direct sum decomposition of the tensor product of two irreducible representations (i.e., a reducible representation into irreducible representations, in cases where the numbers and types of irreducible components are already known abstractly). The name derives from the German mathematicians Alfred Clebsch and Paul Gordan, who encountered an equivalent problem in invariant theory.
From a vector calculus perspective, the CG coefficients associated with the SO(3) group can be defined simply in terms of integrals of products of spherical harmonics and their complex conjugates. The addition of spins in quantum-mechanical terms can be read directly from this approach as spherical harmonics are eigenfunctions of total angular momentum and projection thereof onto an axis, and the integrals correspond to the Hilbert space inner product. From the formal definition of angular momentum, recursion relations for the Clebsch–Gordan coefficients can be found. There also exist complicated explicit formulas for their direct calculation.The formulas below use Dirac's bra–ket notation and the Condon–Shortley phase convention is adopted.
I am looking at the point group C<sub>3v</sub> described shown here. I am trying to understand the block diagonalization process. The note says that changing the basis in the following way will result in the block diagonal form.
What is the rationale for choosing the new basis. Is it...
Hi,
I would like to know why a particle with spin=0 can't posses a magnetic dipole moment?
Using Wigner-Eckart theorem for ##\langle j,1,m,0|j,m \rangle## I get ##\langle j'|| \vec{J}|| j \rangle = \hbar \sqrt{j(j+1)} \delta_{jj'}##
It seems like the right hand side is the magnetic dipole...
Summary: Different sign in the combination of two ##\textbf{1/2}## isospins with opposite third component
Hello everybody!
I was doing an exercise regarding isospin and I noticed something from the Clebsch-Gordan coefficients that made me think.
For example, if I consider the combination...
Homework Statement
[/B]
I am trying to get the C-G Decomposition for 6 ⊗ 3.
2. Homework Equations
Neglecting coefficients a tensor can be decomposed into a symmetric part and an antisymmetric part. For the 6 ⊗ 3 = (2,0) ⊗ (1,0) this is:
Tij ⊗ Tk = Qijk = (Q{ij}k + Q{ji}k) + (Q[ij]k +...
Homework Statement
I am trying to improve my understanding of the Clebsch-Gordan coefficients. I am looking at page 5 of the following document https://courses.physics.illinois.edu/phys570/fa2013/chapter3.pdf
Homework Equations
I have derived the result for the I = 3/2 quadruplet but am...
Homework Statement
I was going through Sakurai's textbook and I tried to work out by myself the following expression for the Clebsch-Gordan coefficients (Addition of orbital angular momentum and spin 1/2)
\begin{equation}
\langle m-1/2,1/2|l+1/2,m \rangle = \sqrt{\frac{l+m+1/2}{2l+1}}...
Homework Statement
The question asks me to calculate the non-zero Clebsch Gordan coefficients
##\langle j_1+j_2, j_s-2|j_1,m_1;j_2,m_2\rangle##
Where ##j_s=j_1+j_2##.
Homework EquationsThe Attempt at a Solution
The ##j_s-2## part is the ##m## of a ##|j,m\rangle## and I know that m has to equal...
Homework Statement
So i have to calculate the Clebsch-gordan coefficients for the state j1=3/2 and j2=1/2[/B]Homework Equations
Recursiom formula, lowering and uppering operator[/B]The Attempt at a Solution
I have tried to calculate the first set of the coefficients stating that:
L- l2,2> =...
Why are ##<j_1j_2m_1m_2|jm>## and ##<j_2j_1m_2m_1|jm>## negative of each other when ##j_1+j_2-j## is odd as given below?
I would expect ##<j_1j_2m_1m_2|jm>## and ##<j_2j_1m_2m_1|jm>## to always have the same sign since nature doesn't care which particle we label as particle 1 and which as...
On one hand, in reading Georgi's book in group theory, I comprehend the invariant tensor as a special "tensor", which is unchanged under the action of any generators. On the other hand, CG decomposition is to decompose the product of two irreps into different irreps.
Now it is claimed that...
Homework Statement
The state of an electron is,
|Psi> =a|l =2, m=0> ⊗ |up> + Psi =a|l =2, m=1> ⊗ |down>,
a and b are constants with |a|2 + |b|2 = 1
choose a and b such that |Psi> is an eigenstate of the following operators: L2, S2, J2 and Jz.
The attempt at a solution
I am really not sure...
I'm doing a problem where I need to know the coefficients to change from the
\vec{J} = \vec{J}_1 + \vec{J}_2 + \vec{J}_3 to the {\vec{J}_1, \vec{J}_2, \vec{J}_3} for three spin-1 particles, but I'm having trouble finding a table or reference for this... surely every time someone needs to...
Hello everyone,
I'm reading a bit about Clebsch-Gordan coefficients and I found two things in their general description I didn't quite understand. Can anyone help me with these questions?
First, I read that the Clebsch-Gordan coefficients are zero unless the total angular momentum...
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)
The 1/sqrt term is clear but the (-1)^.. term not so much, is there a way to find this...
Hello,
I'm not sure whether I should have posted this in main Quantum mechanics thread because it's not really regarding homework, but I posted it in here just to be safe.
I was wondering, is there a way to derive the expression for these coefficients without the use of the general CGC...
Hi, for the deutron triplet state, I am trying to find the coefficient using the lowering operator but I kept getting 0 hope someone can help me.
Using
J-|j,m>=[(j+m)(j-m+1)]1/2|j,m-1>
I can get from |++> to 1/\sqrt{2}(|-+> + |+->) but from here I am stuck since j+m=0 for |+->.
What is...
Homework Statement
I have two spin-1/2 particles and I need to calculate their Clebsch-Gordan coefficients.Homework Equations
The Attempt at a Solution
I followed the procedure of applying J_- to |{j,m}\rangle
and J_{1-} and J_{2-} to |{m_1,m_2}\rangle and comparing them. I got correctly...
When adding the angular momenta of two particles, you use Clebsch-Gordan coefficients, which allow you, in fancy language, to decompose the tensor product of two irreducible representations of the rotation group into a direct sum of irreducible representations. (I'm not exactly clear on what...
Hello all,
I'm stuck on understanding part of a discussion of representations and Clebsch-Gordan series in the book 'Groups, representations and Physics' by H F Jones. I'd be grateful to anyone who can help me out.
For starters, this discussion is in the SU(2) case. I don't know how to...
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...
Trying to self-teach myself more quantum mechanics. Reading Zare's "Angular Momentum- Understanding Spatial Aspects in Chemistry and Physics". I don't really understand how to calculate eigenvalues of J using CG coefficients (I mean, I know the eigenvalues of J, just trying to calculate them a...
Clebsch-Gordan Theorem??
symmetric spinor tensors are IRR of SU(2), i.e., T_{\undergroup{\alpha_1\cdots\alpha_r}}
The Clebsch-Gordan theorem says,
{\{j_1\}}\otimes{\{j_2\}}={\{j_1+j_2\}}\oplus{\{j_1+j_2-1\}}\oplus\cdots\oplus{\{|j_1-j_2|\}}.
Can I prove this theorem by symmetrizing the...
Homework Statement
My textbook takes a look at the \Delta(1232) particle
It says that
\left|\pi p;\frac{3}{2},\frac{3}{2}\right>=\left|\pi;1,1\right>|N;\frac{1}{2},\frac{1}{2}\right>
where N stands for a nucleon and pi could be any of the three flavours of pion.
They then go on by applying...
Hey there, I'm new to this forum, but I just wanted to let you know that I have put out a clebsch-gordan calculator that is a little bit better than the ones I've found out there so far, check it out. http://phys.csuchico.edu/C-G/
-Fish