Why is this C-G coefficient always zero?

[euphoricrhino]

From calculating a few CG-coefficient tables, it occurred to me that when we add two angular mometa $j_1=l$ and $j_2=1$ (with $l$ whole integer), the resulting $|j=l,m=0\rangle$ state always has zero C-G coefficient with $|j_1=l,j_2=1;m_1=0,m_2=0\rangle$ component, i.e., $\langle j_1=l,j_2=1;m_1=0,m_2=0,|j=l,m=0\rangle=0$.
But the usual C-G coefficient selection rule only dictates $|j=l,m=0\rangle$ may have components in 3 states $|j_1=l,j_2=1;m_1=-1,m_2=1\rangle, |j_1=l,j_2=1;m_1=0,m_2=0\rangle, |j_1=l,j_2=1;m_1=1,m_2=-1\rangle$. Why was the middle state $|j_1=l,j_2=1;m_1=0,m_2=0\rangle$ forbidden? Is there deeper insights?
Thanks

 

[DrClaude]

Think about classical vectors. You want to add $\ket{1,0}$ to $\ket{l,0}$ and still get $\ket{l,0}$, i.e., you want to add a vector of length $\sqrt{2}$ to a vector of length $\sqrt{l(l+1)}$ and recover a vector of length $\sqrt{l(l+1)}$, while all 3 vectors lie in the xy plane. Can't work.

 

[euphoricrhino]

Thanks for the reply, but I'm sorry I don't really understand this analogy.

Where I got lost is the part when you say "all 3 vectors lie in the xy plane", which 3 vectors are you referring to (it seems you are referring to the $|m_1=1,m_2=-1\rangle,|m_1=0,m_2=0\rangle,|m_1=-1,m_2=1\rangle$ as the 3 vectors in xy plane?)

I'm also not understanding what "add a vector of length $\sqrt{2}$" mean, since we are talking about projection of vector of length $\sqrt{l(l+1)}$ onto 3 vectors of length $\sqrt{2}$ while one of the projection has zero projected components.

 

[DrClaude]

I'm also not understanding what "add a vector of length $\sqrt{2}$" mean, since we are talking about projection of vector of length $\sqrt{l(l+1)}$ onto 3 vectors of length $\sqrt{2}$ while one of the projection has zero projected components.

Clebsch-Gordan coefficients are used in the addition of angular momenta, so one can use a the analogy of adding classical vectors. In this case, you have a state $\ket{j,m}$ that is the result of the addition of states $\ket{j_1,m_1}$ and $\ket{j_2,m_2}$, in particular
$$
\ket{j=l,m=0} = \sum_{m_1,m_2} \ket{l,m_1,1,m_2} \braket{l,m_1;1,m_2 | l, 0}
$$
and ask why $\braket{l,0;1,0 | l, 0} = 0$. The reason is that since $m=m_1=m_2 = 0$, the two vectors being added, namely $\ket{l,0}$ and $\ket{1,0}$, are in the xy-plane, but so is the resulting vector you are considering, and therefore the CG coefficient can only be $\neq 0$ is making a triangle out of the three vectors is possible, which is not the case if $j=l$ while $j_1=l$ and $j_2=1$.

This is similar to the fact that $\braket{j_1,m_1;j_2,m_2 | j, m} = 0$ if $j < |j_1-j_2|$ or $j > j_1+j_2$ (triangle rule) or if $m \neq m_1 + m_2$, as these conditions also lead to an arrangement of three vectors that is not a triangle.

 

[euphoricrhino]

Thanks a lot for the elaboration. I think I'm forming the mental picture now. You are suggesting to visualize any state $|j,m\rangle$ as a 3-D vector of length $\sqrt{j(j+1)}$ whose $z$-component is $m$, and claim CG coefficient to be non-zero only if the 3 vectors involved can form a triangle.
Am I understanding your analogy correctly? I can see how this can help prove my original observation.
Thanks a lot again!

 

[euphoricrhino]

Thinking about it a little more, I think I'm still not understanding the analogy (despite my earlier claim that I did). Take the example where we add $|l=1,0\rangle$ with $|l=1,0\rangle$, and we are asking why the resulting angular momentum cannot be $|l=1,0\rangle$. By the analogy, these are three vectors on xy plane, each of which has length $\sqrt{2}$, and the claim was they cannot form a triangle. But they can, as equilateral triangle, right?

 

[DrClaude]

Thinking about it a little more, I think I'm still not understanding the analogy (despite my earlier claim that I did). Take the example where we add $|l=1,0\rangle$ with $|l=1,0\rangle$, and we are asking why the resulting angular momentum cannot be $|l=1,0\rangle$. By the analogy, these are three vectors on xy plane, each of which has length $\sqrt{2}$, and the claim was they cannot form a triangle. But they can, as equilateral triangle, right?

You're making me doubt my answer! I'll have to think a bit more about it.

 

[euphoricrhino]

Now I think I finally come up with a proof for this claim using Wigner-Wickart theorem. The proof is simple but "mathematical", so some physical insights are still appreciated.

Here's the proof -
Consider the vector operator $\mathbf{J}$, and transform it into "spherical tensor of rank-1" form
$$
T^{(1)}_{\pm1} = \mp(J_x\pm iJ_y), \qquad T^{(1)}_0=J_z
$$
By Wigner-Eckart theorem,
$$
\left\langle j=l,m=0|T^{(1)}_q|j_1=l,m_1=0\right\rangle = \langle j_1=l,j_2=1;m_1=0,m_2=q|j=l,m=0\rangle\frac{\langle j=l||T^{(1)}||j_1=l\rangle}{\sqrt{2l+1}}
$$
But for $q=0$, the LHS vanishes because $T^{(1)}_0=J_z$, which means the CG coefficient on the RHS must vanish because the second factor on RHS, being the same for all $m$ and $q$ values, must be non-zero.