Spin-orbital combination and the exclusion principle

[hokhani]

TL;DR

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.

 

[DrClaude]

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.

 

[hokhani]

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.

 

[DrClaude]

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 $M_J = -2, -1, 0, 1, 2$.

 

[hokhani]

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.

 

[vanhees71]

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 $m_{J}$.

 

[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.

 

[vanhees71]

Of course. You can combine $L=1$ and $S=1$ to $J=0,1,2$. Check this explanation of the helium-atom:

https://en.wikipedia.org/wiki/Helium_atom

 

[PeroK]

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.

 

[hokhani]

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 $J=3,m_j=0$?

 

[PeroK]

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.

 

[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?

 

[hokhani]

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.

 

[PeroK]

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.

 

[PeroK]

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.

 

[DrDu]

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.

 

[hokhani]

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.

 

[PeroK]

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.