Clebsch gordon coefficients

1. Aug 14, 2015

Wminus

1. The problem statement, all variables and given/known data

2. Relevant equations
https://en.wikipedia.org/wiki/Clebsch–Gordan_coefficients

I don't know how to calculate tensors though..

3. The attempt at a solution

OK so how do I even proceed? All I know about Clebsch-Gordon coefficients is that you can use them to calculate total spin of two spinors, and allegedly the total angular momentum too. My knowledge is basically limited to level of Griffith's intro to QM.

Help?

2. Aug 15, 2015

cpsinkule

Do you know about raising and lowering operators? It's mostly an algorithm you follow to calculate them. You start with either the maximum or minimum total z component ket. From the fact that the total m=m1+m2 and m1=j1 and m2=j2 (remember the maximum m value is equal to the j value) is the only possible way to build such a state of maximum total m, you automatically get a coefficient of 1 for the topmost and bottom-most states of total m which are both equal to the product kets of maximal\minimal z components. From there you use the total J- operator on one side of the equation and its definition (J1-+J2-) on the other side to generate the next set of coefficients. There are some rules that you also must follow. Every state must be normalized, every state must be orthogonal (you will have different states built of the same product kets, these must be orthogonal). is a decent video on the matter and should give you the general idea.

3. Aug 15, 2015

Wminus

But usually one writes these coefficients as $<j_1 j_2 m_1 m_2|j_1 j_2 m_1 m_2>$, yet my professor has invented his own notation (yet again). In the video (I watched 25 mins in), she deals with total angular momentums, but my prof. is asking for spins as well..

I'm quite confused to be honest. Would you mind giving me an example on how to calculate these things?

4. Aug 15, 2015

blue_leaf77

Clebsch-Gordan coefficients are coefficients used to represent the eigenstate of any total angular momentum constructed from two angular momentum operators in terms of the eigenstates of the separate angular momentum operators from which it's constructed. Both spin and orbital angular momentum operators are angular momentum operator, the way you work on this problem is exactly the same as you do in spinors case.

5. Aug 15, 2015

Wminus

What is my professor asking for when he says "Determine the clebsch gordan coefficient <......> from the l=1/2, s=1/2,j=1,m=1 to the m_l=1/2,m_s=1/2 state"?

Which angular momentums am I supposed to add?

6. Aug 16, 2015

blue_leaf77

The eigenstate of a total angular momentum $|l,s,j,m\rangle$ can be represented as a linear combination of the product between eigenstates of the respective individual angular momentum $|l,s,m_l,m_s\rangle$
$$|l,s,j,m\rangle = \mathop{\sum\sum}_{m_l+m_s=m} C_{lsjm}(m_l,m_s) |l,s,m_l,m_s\rangle$$
The coefficient $C_{lsjm}(m_l,m_s)$ is called Clebsch-Gordan coefficient. You are asked to compute this one for $l=1/2, s=1/2,j=1,m=1, m_l=1/2,m_s=1/2$.
$l$ and $s$.
If you do know this, then you should be able to work out your problem.

Last edited: Aug 16, 2015
7. Aug 16, 2015

Wminus

allright thanks ! i solved it now

8. Aug 16, 2015

Orodruin

Staff Emeritus
If he constructed the problem you are solving, what he has invented is his own historical physicists. In the pair of Clebsch and Gordan, the former generally seems to be the one people have less problems spelling although you might expect it to be the other way around ...

9. Aug 16, 2015

Wminus

Yea, his assignments really aren't all that good to be honest...

But anyways, insolence aside: I have no idea what a mixed state is to be honest, but I wrote the following paragraph down as my answer. Did I say something wrong?

"A mixed state is very similar to a superposition of states. In both cases, upon measurement you will observe some basis state with a predefined probability. The difference is that a superposition is a pure state that exists in the real world, while a mixed state in the real world is one of its basis states, but we don't know which.

So if a photon is in a superposition of right and left polarization, it can be linearly polarized. If it's in a mixed state of left and right, it has a 50% classical probability of being in either state, so it cannot be linearly polarized regardless of how its basis states are combined."

And what is a "perturbation diagonal"??? Does it have something to do with the density matrix? Please help, this is the last quantum mechanics problem in the assignment...

Last edited: Aug 16, 2015
10. Aug 16, 2015

Wminus

Can anybody help me with the last part of problem 4? Particularly what a "perturbation diagonal in the basis representation of the wave functions" is?

11. Aug 17, 2015

blue_leaf77

That means the new basis is the set of eigenstates of the perturbation matrix.
Diagonalizing the perturbation matrix becomes necessary when the original basis in which the perturbation matrix is not diagonal is degenerate. This is because, when doing the derivation for the perturbation formula you will encounter expression like
$$\sum_{n\neq m} \frac{V_{nm}}{E_n - E_m}$$
where $V_{nm}$'s are perturbation matrix elements and $E_n$ electronic energy level.

12. Aug 17, 2015

Wminus

ahh is that what he is talking about? I think I remember this from Griffiths book. thanks!