How Do Quantum States Combine in Spin and Orbital Angular Momentum?

[positron98]

The following linear combination of states is considered in almost all quantum mechanics textbooks when they try to explain the addition of spin 1/2 and orbital angular momentum. The thing I don't understand is how the left hand side is equal to the right. Please, if you can, explain how.

|j,m + 1/2> = a|lm,+> + b|lm,-> where |+> is spin up and |-> is spin down.


Thanks,

Sam

 

[quantumdude]

Write the spin and orbital states separately, and the total as a direct product of the two, as follows:

|J,M_J>=|L,M_L>*|1/2,1/2>

Then use the ladder operators to work your way down.

Example:

L=1, S=1/2

|3/2,3/2>=|1,1>*|1/2,1/2>

Apply the ladder operator J_-=L_-+S_-, noting that on the RHS the operator only acts on its respective ket.

J_-|3/2,3/2>=(L_-|1,1>)*|1/2,1/2>+|1,1>*(S_-|1/2,1/2>)

and keep working down.

Try to work out the combination for |J=3/2,M_J=1/2>. If you get stuck, post what you came up with and I'll help you through it.

 

[quantumdude]

What the hell, I'm feeling ambitious!

J_-|3/2,3/2>=(L_-|1,1>)*|1/2,1/2>+|1,1>*(S_-|1/2,1/2>)

RHS:
J_-|3/2,3/2>=3^1/2(hbar)|3/2,1,2>

LHS:
(L_-|1,1>)*|1/2,1/2>+|1,1>*(S_-|1/2,1/2>)=
2^1/2(hbar)|1,0>*|1/2,1/2>+(hbar)|1,1>*|1/2,-1/2>

Putting them together and solving for |3/2,1/2> yields:

|3/2,1/2>=(2/3)^1/2|1,0>*|1/2,1/2>+(1/3)^1/2|1,1>*|1/2,-1/2>

Try the rest, and let me know if you need help.

Remember: Once you get to J=1/2, you will have a parity change. See me if you need help on that, too.

 

[positron98]

Thanks Tom for your time and help.

Sam