Solving 3j Symbols Problem to Show Total Angular Momentum is 0

[dingo_d]

Homework Statement

Show that the state:

[itex]|\Psi\rangle=\sum\limits_{m_1\ m_2\ m_3}\begin{pmatrix}
j_1 & j_2 & j_3 \\
m_1 & m_2 & m_3 \\
\end{pmatrix}|j_1\ m_1\rangle |j_2\ m_2\rangle |j_3\ m_3\rangle[/itex]

has the total angular momentum equal to zero.

Homework Equations

There are bunch of formulas for Clebsch - Gordan coefficients and Wigner 3j symbols, but that can be found everywhere.

The Attempt at a Solution

I have absolutely no idea how to start this one :\

I could write explicitly the 3j symbols, but then what?

EDIT: I found in Landau & Lifgarbagez, that this is what you should get for system of 3 particles with total angular momentum 0, but how do I prove that? They left that part out...

 

[mark726]

Hello,

Thank you for your post. To show that the state |\Psi\rangle has total angular momentum equal to zero, we can use the fact that the total angular momentum operator, denoted by \hat{J}, is given by the sum of the individual angular momentum operators for each particle, denoted by \hat{J_1}, \hat{J_2}, and \hat{J_3}:

\hat{J} = \hat{J_1} + \hat{J_2} + \hat{J_3}

We can then use the commutation relations for angular momentum operators to show that the total angular momentum operator commutes with the state |\Psi\rangle:

[\hat{J}, |\Psi\rangle] = [\hat{J_1} + \hat{J_2} + \hat{J_3}, |\Psi\rangle] = [\hat{J_1}, |\Psi\rangle] + [\hat{J_2}, |\Psi\rangle] + [\hat{J_3}, |\Psi\rangle] = 0

This means that the state |\Psi\rangle is an eigenstate of the total angular momentum operator with eigenvalue 0. Therefore, the state has total angular momentum equal to zero.

To further understand why this is the case, we can look at the explicit form of the state |\Psi\rangle and see that it is a superposition of states with different angular momentum projections for each particle. Since the total angular momentum operator is a vector operator, it acts on each particle independently and does not change the relative magnitudes of the individual angular momenta. This means that the state |\Psi\rangle has the same angular momentum projection for each particle, resulting in a total angular momentum of zero.

I hope this helps to clarify the solution. Please let me know if you have any further questions.