Addition of Angular Momenta - Degeneracy

[unscientific]

Homework Statement

Show that when angular momenta ##j_1## and ##j_2## are combined, the total number of states is ##(2j_1 +1)(2j_1+2)##

Homework Equations

The Attempt at a Solution

For the two gyros in the box, there are ##2j_1 +1## possible orientations of the first gyro, and for each of these orientations, the second gyro can be oriented in ##2j_2+1## ways.

So shouldn't the total number of ways be ##(2j_1+1)(2j_2+1)##?

 

[CAF123]

Yes, there is an error in the question. While your reasoning is valid, I think the setter wants you to actually compute the sum on the LHS explicitly.

 

[unscientific]

Yes, there is an error in the question. While your reasoning is valid, I think the setter wants you to actually compute the sum on the LHS explicitly.

How would I compute it explicitly?

 

[CAF123]

Rewrite the sum like $$\sum_{J=j_1 - j_2}^{j_1 + j_2} 2J+1 = 2\sum_{J=j_1 - j_2}^{j_1 + j_2}J + \sum_{J=j_1 - j_2}^{j_1 + j_2} 1$$

Try to relabel the indices on each of these sums on the RHS so that you can use the well-known sum formula for the sum of the first ##n## positive integers. To use that, relabel the indices such that the sum starts at J=0.

The supposition is that ##j_1 \geq j_2## so ##j_1 - j_2## is non-negative which means rewriting the sums is a bit easier.

 

[unscientific]

Rewrite the sum like $$\sum_{J=j_1 - j_2}^{j_1 + j_2} 2J+1 = 2\sum_{J=j_1 - j_2}^{j_1 + j_2}J + \sum_{J=j_1 - j_2}^{j_1 + j_2} 1$$

Try to relabel the indices on each of these sums on the RHS so that you can use the well-known sum formula for the sum of the first ##n## positive integers. To use that, relabel the indices such that the sum starts at J=0.

The supposition is that ##j_1 \geq j_2## so ##j_1 - j_2## is non-negative which means rewriting the sums is a bit easier.

[tex]\sum_{J=j_1 - j_2}^{j_1 + j_2} 2J+1[/tex]
[tex] = 2\sum_{J=j_1 - j_2}^{j_1 + j_2}J + \sum_{J=j_1 - j_2}^{j_1 + j_2} 1[/tex]
[tex]= 2(\frac{j_2+1}{2})(2j_1) + 2j_2+1[/tex]
[tex] = (2j_1+1)(2j_2+1)[/tex]

I suppose the sum of states is a constant? Because degeneracy for each particle is constant, and degeneracy of a combined state depends on degeneracy of each particle. Hence sum of states before mixing = sum of states after mixing.

 

[CAF123]

Did you not expect ##(2j_1 + 1)(2j_2+1)##? Rewrite, for example, $$\sum_{J=j_1-j_2}^{j_1+j_2}\,J = \sum_{J=0}^{j_1+j_2}\,J - \sum_{J=0}^{(j_1-j_2)-1}\,J $$ and apply the sum of n integers formula.

 

[unscientific]

Did you not expect ##(2j_1 + 1)(2j_2+1)##? Rewrite, for example, $$\sum_{J=j_1-j_2}^{j_1+j_2}\,J = \sum_{J=0}^{j_1+j_2}\,J - \sum_{J=0}^{(j_1-j_2)-1}\,J $$ and apply the sum of n integers formula.

I did, and I got ##(2j_1+1)(2j_2+1)##.

Why would the sum of states before and after mixing be equal? Is my explanation correct?

 

[CAF123]

Why would the sum of states before and after mixing be equal? Is my explanation correct?

Yes. Effectively you can express any of your quantum states as a linear combination of states in the coupled (or in your language, mixed) basis or those in the uncoupled basis. Since both of these bases describe the same physical space of states, then the number of basis elements must be the same.