Finding the Total Angular Momentum

[devd]

Say, I have two spin-1/2 particles in the states characterized by ##(n=2, l=1, m_l=1, m_s=1/2)##and##(n=2, l=1, m_l=1, m_s=-1/2)##. Now, using something like the jj coupling scheme, I first find out the (orbital+spin)angular momentum for the individual particles:(i) ##| 11\rangle |\frac{1}{2}-\frac{1}{2}\rangle =\sqrt{1/3}| \frac{3}{2}\frac{1}{2}\rangle+\sqrt{2/3}|\frac{1}{2}\frac{1}{2}\rangle##

(ii)##|11\rangle|\frac{1}{2}\frac{1}{2}\rangle=|\frac{3}{2}\frac{3}{2}\rangle##

How do i proceed to find the total angular momentum of the system?

I've tried to add like this:
##\Big(| 11\rangle |\frac{1}{2}-\frac{1}{2}\rangle\Big)\Big(|11\rangle|\frac{1}{2}\frac{1}{2}\rangle\Big)= \sqrt{1/6}|3,2\rangle+\Big(\sqrt{2/3}-\sqrt{1/6}\Big)|2,2\rangle##

But, the sum of the square of the coefficients don't add up to 1! So, where did i go wrong?

 

[blue_leaf77]

The reason it doesn't work is because the notation ##|\psi_1\rangle |\psi_2\rangle## does not really mean ##|\psi_1\rangle ## being algebraically multiplied with ##|\psi_2\rangle##. Notation like ##|\psi_1\rangle |\psi_2\rangle## merely means that, that state is a member of a composite vector space constituted from individual vector spaces in which ##|\psi_1\rangle## and ##|\psi_2\rangle## are respectively members.

 

[devd]

The reason it doesn't work is because the notation ##|\psi_1\rangle |\psi_2\rangle## does not really mean ##|\psi_1\rangle ## being algebraically multiplied with ##|\psi_2\rangle##. Notation like ##|\psi_1\rangle |\psi_2\rangle## merely means that, that state is a member of a composite vector space constituted from individual vector spaces in which ##|\psi_1\rangle## and ##|\psi_2\rangle## are respectively members.

We're basically taking the tensor product, right? The tensor product is associative. So, in the last step I've basically tried to take the tensor product of product spaces. How does one do that correctly?

 

[blue_leaf77]

You are basically summing 4 angular momenta ##\mathbf{L}_1##, ##\mathbf{S}_1##, ##\mathbf{L}_2##, and ##\mathbf{S}_2##. I can't think of any better way to do it other than using matrix operations. Since you want to express the state in the form ##|j,m \rangle##, you need to solve the eigenstates of both the total (squared) angular momentum and its z component:
$$
J^2 = ( \mathbf{L}_1\otimes\mathbf{1}_{S1}\otimes\mathbf{1}_{L2}\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes\mathbf{S}_1\otimes\mathbf{1}_{L2}\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes\mathbf{1}_{S1}\otimes\mathbf{L}_2\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes\mathbf{1}_{S1}\otimes\mathbf{1}_{L2}\otimes\mathbf{S}_2 )^2
$$
and
$$
J_z = L_{1z}\otimes\mathbf{1}_{S1}\otimes\mathbf{1}_{L2}\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes S_{1z}\otimes\mathbf{1}_{L2}\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes\mathbf{1}_{S1}\otimes L_{2z}\otimes\mathbf{1}_{S2} + \mathbf{1}_{L1}\otimes\mathbf{1}_{S1}\otimes\mathbf{1}_{L2}\otimes S_{2z}
$$
where the notation like ##\mathbf{1}_{V}## means an identity matrix whose dimension is the same as that of a space ##V##. Those expressions may look nasty, but if you have matrix based calculator, such as matlab, it can help your work considerably. Note that, both matrices ##J^2## and ##J_z## should be diagonalized by the same unitary matrix, indicating that they shared the same set of eigenstates. This means, you can, for example try to find the eigenstates (and hence, the eigenvalues) of ##J_z##, and then use the unitary matrix formed by these newly found eigenstates to diagonalize ##J^2##. Having done all these obviously tedious steps, you will have ##Dim[L_1]\times Dim[S_1]\times Dim[L_2]\times Dim[S_2]## equations, each representing an eigenstate ##|j,m\rangle## of ##J_z## (and ##J^2##) as a linear combination of the old eigenstates ##|l_1,m_{l1}\rangle |s_1,m_{s1}\rangle |l_2,m_{l2}\rangle |s_2,m_{s2}\rangle##. Then solve this system of linear equations for your original state in question. You should see that your original state is a linear combination of several ##|j,m\rangle##'s.