Homework Help: Quantum Numbers J,S and L in p+n and n+n system

1. Feb 24, 2010

Fluffy86

Hey,
I have some problems concerning one of the exercises I got last week

1. The problem statement, all variables and given/known data
List the quantum numbers J,S,L (in spectroscopic notation $$^{2S+1}L_J$$ for partial waves of the n + p systemup to total angular momentum J=2
The quantum numbers J, S and L represent the total angular momentum, the total spin, and the relativ orbital momentum, repectively.
b) Repeat for the n+n system
c) Use the above information, together with the known spectroscopy of the deuteron, to explain why there are no bound states of two neutrons

3. The attempt at a solution

Ok first of all we can have S=0 or S=1. Then we go through the different L states.
With L=0: S L=1 : P L=2: D
S=0:
L=0: $$^1S_0$$
L=1: $$^1P_1$$
L=2: $$^1D_2$$

Now I have some problems with the S=1 states. Lets assume we have L=1 and S=1. What values can J have. Is it the same as I learned for the electron orbits $$J=|L-S|..L+S$$? So can J be in the example be 0,1 and 2?
When yes then I get 7 additional cases
S=1
L=0: $$^3S_1$$
L=1: $$^3P_0 ^3P_1 ^3P_2$$
L=2: $$^3D_1 ^3D_2$$
L=3: $$^3F_2$$

I have really no idea if this is correct or not. But when this is true I am not sure about the second part. Now we have a n+n system, this means we have to consider the Pauli principle. Which of the combination is than not allowed any longer? Of course the $$^3S_1$$ one, because then both have same spin. But are all S=1 states forbidden, even when they have a relative angular momentum?

In the last part the question about a bound state of 2 neutrons. I know that the ground state in the deuteron is the $$^3S_1$$ state. This one is not allowed in the 2 neutron system, but why not the next higher state is the ground state for the 2 neutrons?(e.g. the first excited state of the deuteron)

Would be really nice if you could help me
Best regards
Fluffy

2. Feb 24, 2010

vela

Staff Emeritus
Yup.
Looks good.
The symmetry of the entire state needs to be antisymmetric for the two-neutron system. The S=1 spin state is symmetric, so the orbital part of the state needs to be antisymmetric so that the state as a whole is antisymmetric.

3. Feb 24, 2010

Fluffy86

Thx alot!
Do you mean with an antisymmetric orbital part that the L has to be 1 or 3 or how do I know what symmetry the orbital part has?

Fluffy

4. Feb 24, 2010

vela

Staff Emeritus
Yes, that's what I meant. I'm not sure how you'd subsequently rule those states out for the n-n system though.

5. Feb 25, 2010

Fluffy86

hmm maybe it is possible to argue with the parity, the relative orbital momentum gives $$(-1)^l$$ so in the n+n system and S=1 l has to be an odd number and in the S=0 and even number?
Or am I thinking totally wrong?