# Spin-orbital combination and the exclusion principle

• A
• hokhani

#### hokhani

TL;DR Summary
Disregarding some of spin-orbital states by exclusion principle
If we combine the two spin-orbital states with ##l=1, s=1/2##, we obtain the combined states with ##J=3/2## and ## J=1/2##. Also, we know that the exclusion principle forbids the two electrons with identical spin-orbital states ##|l,s,m_l,m_s>##. If we combine the two combined states with ##J=3/2##, we would have the resultant states with ##J=3,2,1,0##. Among these resultant states all the states with ##J=3## and ##J=1## are forbidden according to the exclusion principle. I am justified how the states with ##J=3## and ##m_J=3,2,1,-1,-2,-3## are constructed from the two identical spin-orbitals with ##|l=1,s=1/2,m_l,m_s>## and why they are forbidden. However, I can not understand why the state with ##J=3, m_J=0## is forbidden by exclusion principle. I appreciate any help.

However, I can not understand why the state with ##J=3, m_J=0## is forbidden by exclusion principle. I appreciate any help.
I don't understand. If you have a state with ##J=3##, then all possible values of ##M_J = -3, \ldots 3## are present.

I don't understand. If you have a state with ##J=3##, then all possible values of ##M_J = -3, \ldots 3## are present.
We note that the state with ##J=3## is constructed from the spin-orbitals of the two electrons, i.e., ##|l_1 =1,s_1=1/2, m_{l1}, m_{s1}>## and ##|l_2 =1,s_2=1/2, m_{l2}, m_{s2}>## with various combinations of ##m_l,m_s## for the two electrons, say, the combination of ##|{m_l}_1=1,{m_s}_1=1/2>## and ##|{m_l}_2=1,{m_s}_2=1/2>## (which are identical states) gives the state with ##J=3, m_j=3## which is forbidden by exclusion principle when for the two electrons we have ##n_1=n_2##. However, I don't know why the state with ##J=3, m_j=0## is forbidden.

You example doesn't work, because ##l_1=l_2=1## can give a maximum of ##J=2##. But using ##J=2## instead, you indeed have that the combination of ##|{m_l}_1=1,{m_s}_1=1/2 \rangle## and ##|{m_l}_2=1,{m_s}_2=1/2\rangle## is not possible, but that simply means that you cannot have ##J=2## and ##S=1## at the same time. You still have ##J=2## with ##S=0##, and thus get all the possible values of ##M_J = -2, -1, 0, 1, 2##.

• vanhees71
You example doesn't work, because ##l_1=l_2=1## can give a maximum of ##J=2##. But using ##J=2## instead, you indeed have that the combination of ##|{m_l}_1=1,{m_s}_1=1/2 \rangle## and ##|{m_l}_2=1,{m_s}_2=1/2\rangle## is not possible, but that simply means that you cannot have ##J=2## and ##S=1## at the same time. You still have ##J=2## with ##S=0##, and thus get all the possible values of ##M_J = -2, -1, 0, 1, 2##.
Thank for your answer. You are disregarding the spin states. For example, the state ##|l_1=1,s_1=1/2,{m_l}_1=1, {m_s}_1=1/2## is equivalent to ##|l_1=1,s_1=1/2,j_1=3/2, {m_j}_1=3/2>## and the combination of this state with the same state ##|l_2=1,s_2=1/2,j_2=3/2, {m_j}_2=3/2>## gives ##|j_1=3/2, j_2=3/2, J=3, M_j=3>##. I would like to know why the state with ##|j_1=3/2, j_2=3/2, J=3, M_j=0>## is forbidden by exclusion principle.

Isn't it easier to get by first coupling the spins? You have two spins 1/2, which you can couple either to total spin ##S=0## and ##S=1##. The former (singlet) is the antisymmetric combination and the latter (triplet) is the symmetric combination.

Further you have to orbital angular momenta ##l_1=l_2=1##. These you can couple to ##L \in \{ 0,1,2 \}##. Then to get a total angular momentum ##J=0## you must couple ##S=1## (symmetric) and ##L=2## (also symmetric, because the exchange of particles means a parity transformation for the states describing the relative motion, and the parity of orbital-angular momentum eigenstates is ##(-1)^{\ell}##). Now we discuss the relative motion of the two particles and if these two particles are two fermions the exchange of the two particles must be antisymmetric. Since for ##J=3## both the spin and the orbital part are symmetric, these states are excluded by the Pauli principle. It doesn't depend on ##m_{J}##.

• hokhani and PeroK
Since for ##J=3## both the spin and the orbital part are symmetric, these states are excluded by the Pauli principle. It doesn't depend on ##m_{J}##.
What about ##J = 1##? Can you get that from a combination of anti-symmetric ##L = 1## and symmetric ##S = 1## states?

The OP suggests otherwise:

...all the states with ##J=3## and ##J=1## are forbidden according to the exclusion principle.

Last edited:
Thank for your answer. You are disregarding the spin states. For example, the state ##|l_1=1,s_1=1/2,{m_l}_1=1, {m_s}_1=1/2## is equivalent to ##|l_1=1,s_1=1/2,j_1=3/2, {m_j}_1=3/2>## and the combination of this state with the same state ##|l_2=1,s_2=1/2,j_2=3/2, {m_j}_2=3/2>## gives ##|j_1=3/2, j_2=3/2, J=3, M_j=3>##. I would like to know why the state with ##|j_1=3/2, j_2=3/2, J=3, M_j=0>## is forbidden by exclusion principle.
If you consult the table of Clebsch-Gordan coefficients, you will see that:

1) The composition of two L-1 systems yields:

L-2 states, which are symmetric, regardless of the##L_z## component.

L-1 states, which are anti-symmetric, again regardless of the ##L_z## component.

L-0 state, which is symmetric

Likewise:

2) The composition of two s-1/2 systems yields:

S-1 states (triplet), which are symmetric, regardless of the ##S_z## component.

S-0 state, which is anti-symmetric

In your case, you are combining four angular momenta to give:

J-0 state, which must be anti-symmetric (as it is a combination of the symmetric L-0 state and the anti-symmetric S-0 state).

J-1 states, which may be either symmetric (e.g L-1 and S-0 or L-0 and S-1) or anti-symmetric (L-1 and S-1)

J-2 states, which may be anti-symmetric (e.g. L-2 and S-0 or L-1 and S-1) or symmetric (L-2 and S-1)

J-3 states, which must be symmetric (L-2 and S-1).

By this reasoning, only ##J=3## is not possible, as it must be a symmetric state. As confirmed by @vanhees71 above, it is possible to have an anti-symmetric ##J = 1## state.

• PeterDonis, hokhani and vanhees71
Many thanks for all the answers. I would like to find the root of my misunderstanding, from my main view to the problem (namely using the combinations ##J,m_j## instead of ##L,S##).

We note that the combination of spin and orbital for each electron is the following 6 states:

Four states with ##|l=1,s=1/2,j=3/2,m_j>## and two states with ##|l=1,s=1/2,j=1/2,m_j>## .

All the sates with ##|l=1,s=1,j=3/2,m_j>## are constructed from the states ##|l=1,s=1/2, m_l,m_s>## as follows:

##|l=1,s=1,j=3/2,3/2>=|l=1,s=1/2, m_l=1,m_s=1/2>##

##|l=1,s=1,j=3/2,1/2>=##
##\frac{1}{\sqrt(3)} [\sqrt(2)|l=1,s=1/s,m_l=0,m_s=1/2>+|l=1,s=1/2,m_l=1,m_s=-1/2>]##

##|l=1,s=1,j=3/2,-1/2>=##
##\frac{1}{\sqrt(3)} [|l=1,s=1/2,m_l=-1,m_s=1/2>+\sqrt(2)|l=1,s=1/2,m_l=0,m_s=-1/2>]##

##|l=1,s=1,j=3/2,-3/2>=|l=1,s=1/2,m_l=-1,m_s=-1/2>##

The states with ##J=3## are constructed from the two states of the form ##|l=1,s=1,j=3/2,m_j>## (with various combinations of the values of ##m_j##) . Now, the state with ##J=3,m_j=0## can only be constructed either from the combination of ##|l=1,s=1,j=3/2,3/2>## and ##|l=1,s=1,j=3/2,-3/2>## or from the combination of ##|l=1,s=1,j=3/2,1/2>## and ##|l=1,s=1,j=3/2,-1/2>##. None of these combinations includes the identical states. How does the exclusion principle forbid ##J=3,m_j=0##?

Remember that the states must be anti-symmetric. Let's take two spin 1/2 systems. We have a state:
$$|\psi \rangle = |\frac 1 2 \frac 1 2 \rangle \otimes |\frac 1 2 -\frac 1 2 \rangle$$
I think what you are saying is that could be a state with ##S = 1## and ##S_z = 0##. But, that state is not allowed for identical particles with a symmetric spatial wavefunction, as it is not anti-symmetric.

The only anti-symmetric spin state for two identical particles is the singlet:
$$|\psi \rangle = \frac 1 {\sqrt 2} (|\frac 1 2 \frac 1 2 \rangle \otimes |\frac 1 2 -\frac 1 2 \rangle - |\frac 1 2 -\frac 1 2 \rangle \otimes |\frac 1 2 \frac 1 2 \rangle)$$
And that has ##S = 0##. Two electrons in the ground state, therefore, must be in the singlet spin state. They cannot be in the triplet state and they cannot have ##S = 1##.

The Pauli exclusion principle is really about anti-symmetry. The heuristic "no two electrons can be in the same state" may be confusing you.

• hokhani and vanhees71
The reason, why it is more simple to first couple the orbital angular momentum to ##L## and then the spins to ##S## is that then you can read off the signs under exchange of the two particles easier.

• hokhani
According to your statements I guess we can not find directly that the state with ##J=3,m_j=0## is forbidden and we are forced to take help of the ##L, S## form to confirm it. Am I right?

The Pauli exclusion principle is really about anti-symmetry. The heuristic "no two electrons can be in the same state" may be confusing you.
You last statement was very helpful for me.

According to your statements I guess we can not find directly that the state with ##J=3,m_j=0## is forbidden and we are forced to take help of the ##L, S## form to confirm it.
If you have the patience you can write out the full form of the ##J = 3, m =0## state and you will see that it is symmetric in the two particles.

According to your statements I guess we can not find directly that the state with ##J=3,m_j=0## is forbidden and we are forced to take help of the ##L, S## form to confirm it. Am I right?
Here's how you would construct the ##J = 3, m = 0## state.

First, we must have ##L = 2## and ##S = 1##. Looking at the Clebsch-Gordan table we see that the ##3, 0## state as a composition of ##L = 2## and ##S = 1## is:
$$|3, 0 \rangle = \sqrt{\frac 1 5}|2, 1 \rangle \otimes |1, -1 \rangle + \sqrt{\frac 3 5}|2, 0 \rangle \otimes |1, 0 \rangle + \sqrt{\frac 1 5}|2, -1 \rangle \otimes |1, 1 \rangle$$
That gives us three combinations of ##L + S## states to look at. We need to confirm that each of these is symmetric in the two particles. Again, from the table we have:
$$|2, 1 \rangle = \sqrt{\frac 1 2}|1, 1 \rangle \otimes |1, 0 \rangle + \sqrt{\frac 1 2}|1, 0 \rangle \otimes |1, 1 \rangle$$ $$|2, 0 \rangle = \sqrt{\frac 1 6}|1, 1 \rangle \otimes |1, -1 \rangle + \sqrt{\frac 2 3}|1, 0 \rangle \otimes |1, 0 \rangle + \sqrt{\frac 1 6}|1, -1 \rangle \otimes |1, 1 \rangle$$ $$|2, -1 \rangle = \sqrt{\frac 1 2}|1, 0 \rangle \otimes |1, -1 \rangle + \sqrt{\frac 1 2}|1, -1 \rangle \otimes |1, 0 \rangle$$
Note that these are all symmetric states.

Finally, we look at the three ##S = 1## states, composed of two ##S = 1/2## particles. You should already know that these are symmetric, but in any case they are:
$$|1, -1 \rangle = |1/2, -1/2 \rangle \otimes |1/2, -1/2 \rangle$$ $$|1, 0 \rangle = \sqrt{\frac 1 2}|1/2, 1/2 \rangle \otimes |1/2, -1/2 \rangle + \sqrt{\frac 1 2}|1/2, -1/2 \rangle \otimes |1/2, 1/2 \rangle$$ $$|1, 1 \rangle = |1/2, 1/2 \rangle \otimes |1/2, 1/2 \rangle$$
The ##3, 0## state is, therefore, symmetric in the two particles, as all possible states are symmetric. For ##J = 3## to be allowed, at least one of the possible states be anti-symmetric.

You have a similar situation for the other ##J = 3, m = 1, 2, 3## states. These are all symmetric as well. Hence no ##J = 3## states are allowed.

• hokhani
Just a side note: Just because two states have the same value of l, doesn't necessary mean that they are identical. E.g. a helium atom in a 1p^12p^1 state.

• DrClaude
Here's how you would construct the J=3,m=0 state.

First, we must have L=2 and S=1. Looking at the Clebsch-Gordan table we see that the 3,0 state as a composition of L=2 and S=1 is:
|3,0⟩=15|2,1⟩⊗|1,−1⟩+35|2,0⟩⊗|1,0⟩+15|2,−1⟩⊗|1,1⟩
Many thanks again. Of course, without extra calculations, we could realize at this stage that the state ##J=3,m=0## is symmetric because both space function ##L=2## and spin function ##S=1## are symmetric as stated by vanhees in the post 6. Anyway, these calculations again utilise the ##L,S## form and not directly realize the symmetry without decomposing to ##L,S## states.

Many thanks again. Of course, without extra calculations, we could realize at this stage that the state ##J=3,m=0## is symmetric because both space function ##L=2## and spin function ##S=1## are symmetric as stated by vanhees in the post 6.
Post #16 was intended as an antidote to your post #10. • hokhani