# Spin-orbital combination and the exclusion principle

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• 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