Here's another extra credit problem. If you do it, please turn it in to me separately.

The point of this problem is to convince you that those Clebsch-Gordon tables are not arcane magical black boxes - you're going to *derive* some nontrivial entries.

Suppose you have a spin 1/2 particle in an orbit with angular momentum L. Also suppose the system has a definite value of total , which we can call (in all generality!) m+1/2.

The general (angular) state of such a particle can always be written in the form , (where a and b are constants)

(Can you see why I can make that claim?)

i) Figure out what does to this state.

(Hint: The relation will help you here, as will Griffiths Eq. 4.136!)

In the end, your answer should be written in the form .

(Also, argue briefly why is not necessarily an eigenfn of )

ii) By construction, your had a fixed value of , namely (m+1/2). Suppose we want to also have a fixed value of . Let's pick J=L+1/2, for definiteness.

Use your results from part i) to find what a and b have to be in this case.

(You answer should involve the symbols m and L, only).

Note: you have just worked out some of the Clebsch Gordon coefficients for adding arbitrary L to spin 1/2! (There is some sign ambiguity in your answer, but Griffith's conventions are consistent with the naive positive square root in this case)

iii) Check your answer by picking some particular values. E.g, let L=2, m=1, (thus J = L+(1/2) = 5/2, and m_tot= m+ (1/2) = 3/2) and explicitly verify your result in the 2x(1/2) C-G table.

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Hint for part ii) You worked out what does to in part i, but if it's going to have a definite value of , namely J=L+1/2, that means we are insisting that

You can set this equation equal to your result from part i). Now use orthogonality: . Although it may look at this point like you've got 2 equations in the 2 unknowns a and b, the 2 equations are actually equivalent! So, you'll need to use the fact that is normalized, i.e. .