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

[Fluffy86]

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

Homework Statement

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

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. Let's 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

 

[vela]

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. Let's 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?

Yup.

When yes then I get 7 additional cases
S=1
L=0: $$^3S_1$$
L=1: $$^3P_0, ^03P_1, ^3P_2$$
L=2: $$^3D_1, ^3D_2$$
L=3: $$^3F_2$$

I have really no idea if this is correct or not.

Looks good.

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?

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.

 

[Fluffy86]

Thx a lot!
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

 

[vela]

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?

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

 

[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?