Question about angular momentum

[stunner5000pt]

Homework Statement

The deuteron consists of a proton and a neutron in a state of spin 1 and total angular momentum J = 1. What are the possible angular momentum states for this system.

Another question: Consider a wave function with azimuthal dependence
\psi(r,\theta,\phi) \propto \cos^2 \phi
What are the possible outcomes of a measurement of the z component of the
orbital angular momentum and what are the probabilities of these outcomes?

Homework Equations

J = L + S

The Attempt at a Solution

since J =1 and S = 1, and thus -1 and 0. (this becuase suppose a particle had spin 3/2 then j=l\pm \frac{1}{2} and j = l \pm \frac{3}{2}

so then the possible values for l are 2,0,1??

seemingly trivial but i don't quite udnerstand this completely yet soo...

For the second question - it appears that m_{l}=1

and thus we can calcculate the probability of this measurement using
c_{n}=\frac{1}{\sqrt{2\pi}} \int_{0}^{2\pi} e^{-i\phi} \cos^2\phi d\phi
For ml=1, then
\left<L_{z}\right>=\hbar

Is that right??

Thanks for the help! it is greatly appreciated!

 

[Meir Achuz]

"so then the possible values for l are 2,0,1??
seemingly trivial but i don't quite udnerstand this completely yet soo.."

L could be either a mix of 0 and 2 (the actual case) or L=1.
Even L and odd L cannot mix in the same state because parity is a good quantum number for nuclear binding.
L=1 is ruled out if Isospin is introduced.

 

[Meir Achuz]

cos^2\phi=[1+cos(2\phi)]/2.
This mean m_L= a combiination of +2, -2, 0.
For the probabilties, do your integrals using the appropriate Y_LM.
The \Psi in your question should not depoend on phi.

 

[stunner5000pt]

"so then the possible values for l are 2,0,1??
seemingly trivial but i don't quite udnerstand this completely yet soo.."

L could be either a mix of 0 and 2 (the actual case) or L=1.
Even L and odd L cannot mix in the same state because parity is a good quantum number for nuclear binding.
L=1 is ruled out if Isospin is introduced.

how are we 'mixing' even and odd L if we took L = 0, and L =1 and L = 2 separately??

i don't quite see how parity is involved here
also i edited the second question

 

[cgw]

since J =1 and S = 1, and thus -1 and 0. (this becuase suppose a particle had spin 3/2 then j=l\pm \frac{1}{2} and j = l \pm \frac{3}{2}

so then the possible values for l are 2,0,1??

I think your reasoning is a little off.
j = l+s or l-s

If the particle had a spin of 3/2 then j= l +or- 3/2. There is no 1/2 spin.

Right?

 

[Meir Achuz]

how are we 'mixing' even and odd L if we took L = 0, and L =1 and L = 2 separately??

i don't quite see how parity is involved here

Parity implies that the ground state can be a miixture of L=0 and L=2
(the actual case), but if L=1, no even L could be mixed in.