Clebsch-Gordan coefficients (l+1/2)

In summary, the conversation discusses the attempt to derive the Clebsch-Gordan coefficients for the addition of orbital angular momentum and spin 1/2. The individual steps and equations used in the attempt are outlined, and the mistake is identified to be in the use of the ladder operator, resulting in an answer upside down. The mistake is eventually realized and the conversation ends with a thank you.
  • #1
Daviddc

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}}
\end{equation}

Homework Equations



\begin{equation}
J_{\pm} |jm \rangle=\sqrt{(j \mp m)(j \pm m +1)} \\
J_{i,\pm} |m_1 m_2 \rangle=\sqrt{(j_i \mp m_i)(j_i \pm m_i +1)}
\end{equation}

The Attempt at a Solution



\begin{equation}
|l+1/2,m \rangle = \sum_{m'} \left[|m'-1/2,1/2\rangle \langle m'-1/2,1/2|l+1/2,m \rangle + |m'+1/2,-1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle \right]
\end{equation}

Aplying J_+

\begin{equation}
\sqrt{(l+1/2-m)(l+1/2+m+1)}|l+1/2,m+1 \rangle = \sum_{m'} \left[\sqrt{(l-m'+1/2)(l+m'-1/2+1)}|m'+1/2,1/2\rangle \langle m'-1/2,1/2|l+1/2,m \rangle + \sqrt{(1/2+1/2)(1/2-1/2+1)} |m'+1/2,+1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle +\sqrt{(l-m'-1/2)(l+m'+1/2+1)}(|m'+3/2,-1/2\rangle \langle m'+1/2,-1/2|l+1/2,m \rangle \right]
\end{equation}

Taking the inner product \begin{equation} \langle m+1/2, 1/2 | \end{equation}

\begin{equation}
\sqrt{(l+1/2-m)(l+1/2+m+1)} \langle m+1/2,1/2 | l+1/2, m+1 \rangle = \sqrt{ (l-m+1/2)(l+m+1/2)} \langle m-1/2,1/2 | l+1/2, m \rangle
\end{equation}

\begin{equation}
\langle m-1/2,1/2 | l+1/2, m \rangle=\sqrt{\frac{l+m+3/2}{l+m+1/2}} \langle m+1/2,1/2 | l+1/2, m+1 \rangle
\end{equation}

Aplying the ladder operator until m reaches its maximum value (i.e. l+1/2),

\begin{equation}
\langle m-1/2,1/2 | l+1/2, m \rangle=\sqrt{\frac{2l+1}{l+m+1/2}}
\end{equation}

As you can see, I obtain the answer upside down. It is obviously wrong since you get values larger than the unity. In Sakurai's textbook he starts with this:

\begin{equation}
\sqrt{(l+1/2-m)(l+1/2+m+1)} \langle m-1/2,1/2 | l+1/2, m \rangle = \sqrt{ (l-m+1/2)(l+m+1/2)} \langle m+1/2,1/2 | l+1/2, m+1 \rangle
\end{equation}

But I don't understand why it is like that instead of the one I derived. I'm sure that my mistake is something stupid, but I can't find it and it is pissing me off since it's 1 am and I didn't have dinner yet because of this. If any of you can point out where I am failing I will be thankful.

Thank you for your time.
 
Physics news on Phys.org
  • #2
Oh, I have just realized where my mistake is...

Thank you anyways.
 

1. What are Clebsch-Gordan coefficients (l+1/2)?

Clebsch-Gordan coefficients (l+1/2) are mathematical quantities used in quantum mechanics to describe the coupling of angular momenta in a system.

2. How are Clebsch-Gordan coefficients (l+1/2) calculated?

Clebsch-Gordan coefficients (l+1/2) are calculated using a mathematical formula that takes into account the individual angular momenta (l) and the total angular momentum (l+1/2) of the system.

3. What is the significance of Clebsch-Gordan coefficients (l+1/2) in quantum mechanics?

Clebsch-Gordan coefficients (l+1/2) play a crucial role in calculating transition probabilities and predicting the outcomes of quantum mechanical experiments.

4. How do Clebsch-Gordan coefficients (l+1/2) relate to the quantum mechanical concept of spin?

Clebsch-Gordan coefficients (l+1/2) are used to describe the coupling of spin states in a system. They allow us to understand how different particles with different spin values can interact with each other.

5. Can Clebsch-Gordan coefficients (l+1/2) be negative?

Yes, Clebsch-Gordan coefficients (l+1/2) can have both positive and negative values. The sign of the coefficient depends on the relative orientation of the angular momenta in the system.

Similar threads

  • Advanced Physics Homework Help
Replies
17
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
13
Views
1K
Replies
16
Views
545
  • Advanced Physics Homework Help
Replies
2
Views
758
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
14
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
1K
Back
Top