# Spin and angular momentum

1. Jan 25, 2010

### Brian-san

1. The problem statement, all variables and given/known data
1. Consider a deuterium atom (composed of a nucleus of spin I=1 and an electron. The total angular momentum of the atom is $\vec{F}=\vec{J}+\vec{I}$, the eigenvalues of $J^2$ and $F^2$ are $J(J+1)\hbar^2$ and $F(F+1)\hbar^2$ respectively.
a) What are the possible values of the quantum numbers J and F for a deuterium atom in the 1s ground state?
b) Same as a for deuterium in the 2p excited state.
c) Same as a for hydrogen in the 2p state.

2. Consider particle A of spin-3/2 which can disintegrate into two particles, B of spin-1/2 and C of spin-0. We place ourselves in the rest frame of A and total angular momentum is conserved during the disintegration.
a) What values can be taken on by the relative orbital angular momentum of the two final particles.? Show that there is only one possible value if the parity of the relative orbital state is fixed. Would this result remain valid if particle A had spin greater than 3/2?
b) Assume particle A is initially in the spin state characterized by the eigenvalue $m_A\hbar$ of it's spin component along the z axis. We know the final orbital state has definite parity. Is it possible to determine the parity by measuring the probabilities of finding particle B either in state $|+\rangle$, or $|-\rangle$?

2. Relevant equations
$\vec{J}=\vec{L}+\vec{S}$, for the electron.
The electron is a spin-1/2 particle.

3. The attempt at a solution
1.a) Since we are in the 1s state, the quantum number n=0, so l=0 and J=1/2 for the electron. This would lead to F=1/2, 3/2 for the atom.
b) In the 2p state n=2, so l=0, 1 and J=1/2, 3/2 for the electron. Then for the entire atom F=1/2, 3/2 (for J=1/2), and F=3/2, 5/2 (for J=3/2).
c) Again with quantum numbers, n=2 so l=0, 1. Thus J=1/2, 3/2. However this time since it is regular hydrogen, I=1/2, for the proton, so F=0, 1 (for J=1/2), and F=1, 2 (for J=3/2).

I'm pretty sure those 3 are correct, but I'm still confused with how I arrived at them. For example, in 1a, I used J=±1/2 (since the total angular momentum was equal to the spin and the electron could be spin up or down, it seemed logical at the time). A similar process was used for the rest. Is this the correct thinking, or was I simply lucky with my incorrect method?

2. a) In the rest frame of A the total angular momentum is J=3/2, so I have $\frac{3}{2}=L_B+\frac{1}{2}+L_C$, given the spin of the final particles, L being the orbital angular momentum of the respective particles. My first guess is the either $L_B=1, L_C=0$, or $L_B=0, L_C=1$. However, I don't see how parity would come into play with this answer, which leads me to believe it is either wrong or incomplete.

2. Jan 26, 2010

### vela

Staff Emeritus
Actually, your answer for 1b is slightly incorrect for the J=3/2 case. When you add angular momenta, say, J=L+S, the allowed values of quantum number J run from |L-S| to |L+S|. In 1b, you have F=J+I where J=3/2 and I=1, so the allowed values of F run from |3/2-1| to |3/2+1| in integer steps, i.e, F=1/2, 3/2, or 5/2.

Also, the p in 2p refers to the orbital angular momentum. It tells you L=1, so you don't have to bother with the L=0 cases.

3. Jan 26, 2010

### vela

Staff Emeritus
The particles don't each have their own orbital angular momentum; they have one relative orbital angular momentum L.

Orbital angular momentum states have definite parity, and it depends only on L.

4. Jan 26, 2010

### Brian-san

Thanks, everything makes more sense for part one now. It's always the sings on the numbers that would throw me off, when to use ±, when to only keep positive values, etc.

In 2, why would the resulting particles not have their own individual orbital angular momentum?
Assuming there is only one orbital angular momentum L, since particle A is spin 3/2, the value for J in it's rest frame would be 3/2. The spin 1/2 particle (B) can contribute a ±1/2. Based on this, L can take on values of L=1, 2. L can only have integer values, so the only way the two particles could have the same relative parity would be if L=2. Is this closer to being on the right track?

I don't think this would be the case for particles with spin greater than 3/2, simply because there are more possible combinations for the L value.

5. Jan 26, 2010

### vela

Staff Emeritus
I'm not sure how you concluded that L=2.

You also need to take into account the intrinsic parity of the particles. It's been too many years since I took particle physics, so I don't remember all the details. You probably have an expression which tells you how to calculate the parity of a particle or collection of particles based on the particles and their state. If the decay occurs via a parity-conserving process, the parity of the initial state and the final state should be the same. This will let you choose the value of L which results in the correct parity for the final state.

6. Jan 26, 2010

### Brian-san

I got L=2 from the fact that particle A is spin-3/2 and the total angular momentum for the resulting particles is some total orbital angular momentum L, plus the spin from the spin-1/2 particle. Since the total angular momentum is conserved, I went with 3/2=L±1/2, which gave L=1, 2. The I figured all L is split between particle B and C, since each particle could get some individual amount of angular momentum from the process.

If L=2, either LB=0 and LC=2, LB=2 and LC=0, or LB=1 and LC=1. If L=1, then either LB=0 and LC=1, or LB=1 and LC=0. I figured parity would be dependent on if the L values are odd/even, and the only time LB, LC, match in all cases was L=2. This is probably the wrong idea.