Question about angular momentum

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Homework Help Overview

The discussion revolves around the angular momentum states of a deuteron, which consists of a proton and a neutron with a total angular momentum of J = 1. Participants also explore the implications of a wave function with azimuthal dependence, specifically regarding the measurement of the z component of orbital angular momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the possible angular momentum states for the deuteron, questioning the values of l based on the relationship between total angular momentum J and spin S. There is also exploration of the azimuthal dependence of the wave function and the corresponding outcomes for measurements of angular momentum.

Discussion Status

The discussion is active, with participants offering different interpretations of the angular momentum values and the implications of parity in the context of nuclear binding. Some participants express uncertainty about the mixing of even and odd angular momentum states, while others provide insights into the mathematical relationships involved.

Contextual Notes

There are indications of confusion regarding the application of quantum numbers and the mixing of states, as well as the specific conditions under which certain angular momentum states can coexist. The original poster and others are navigating these complexities without reaching a definitive conclusion.

stunner5000pt
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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
[tex]\psi(r,\theta,\phi) \propto \cos^2 \phi[/tex]
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 because suppose a particle had spin 3/2 then [itex]j=l\pm \frac{1}{2}[/itex] and [itex]j = l \pm \frac{3}{2}[/itex]

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 [itex]m_{l}=1[/itex]

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

Is that right??

Thanks for the help! it is greatly appreciated!
 
Last edited:
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"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.
 
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.
 
Meir Achuz said:
"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
 
stunner5000pt said:


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

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?
 
stunner5000pt said:
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.
 

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