Clebsch-Gordan coefficients for spin-one particle decomposition

[Marty]

I started working on this topic and thought I would post it for other to contribute their ideas. The Clebsch-Gordan coefficients basically tell you how to move between two different representations of spin. I am trying to work them out logically. For starters I am doing the very simplest case: how to represent a spin-one particle as two spin one-half particles.

The +1 and -1 spin states of course can only be up-up and down-down. It's the other two states that are interesting. We have in the two-particle representation:

up-down (call it A>) and down-up (call it B>)

We have to combine these to get the states l=1,m=0 (call it 10>) and l=0,m=0 (call it 00>)

I think I know the answer: A> + B> gives 10>, and A> - B> gives 00>

I wonder how we would argue that it has to work this way? I would find it convincing if there was an argument using an external magnetic field, and showing how the respective states behaved properly under its influence. In particular, the state 00> should do absolutely nothing under an external field, so I wonder how we would show that the difference of the A and B states behaves this way?

 

[Marty]

I'm working on the helium atom with two electrons, and I'm trying to account for the electron spin states (ignoring the orbital spin). There should be four spin states with l and m restricted to the interval from -1 to +1:

1, 1>
1, 0> and 0,0>
1,-1>

We create these states by different combinations of up-up, up-down, down-up, and down-down. It seems clear to me that if the atom is in the ground state, we can't have either the up-up or the down-down cases, because the two electrons must have opposite spin to share the ground state orbital. So looking at the electron spin states, only the middle two which I've listed should be possible.

Now I have a paradox which I'd like some help to resolve. I believe I can create an electron spin state of l=1, m=0 using combinations of up-down and down-up; so the helium atom is in the ground state. Here is the paradox: since there is nothing special about the z-axis, I can create analagous states with respect to the x-axis and y axis. But the superposition of the x- and y- axis 10> states gives me the z-axis 11> state (you can see that this is true by looking at the superpositions of spherical harmonics corresponding to the p orbitals). But I already concluded that the 11> spin state could not exist for helium in the ground state (because up-up violates the Pauli exclusion principle).

Help?

 

[per.sundqvist]

I'm working on the helium atom with two electrons, and I'm trying to account for the electron spin states (ignoring the orbital spin). There should be four spin states with l and m restricted to the interval from -1 to +1:

1, 1>
1, 0> and 0,0>
1,-1>

We create these states by different combinations of up-up, up-down, down-up, and down-down. It seems clear to me that if the atom is in the ground state, we can't have either the up-up or the down-down cases, because the two electrons must have opposite spin to share the ground state orbital. So looking at the electron spin states, only the middle two which I've listed should be possible.

Now I have a paradox which I'd like some help to resolve. I believe I can create an electron spin state of l=1, m=0 using combinations of up-down and down-up; so the helium atom is in the ground state. Here is the paradox: since there is nothing special about the z-axis, I can create analagous states with respect to the x-axis and y axis. But the superposition of the x- and y- axis 10> states gives me the z-axis 11> state (you can see that this is true by looking at the superpositions of spherical harmonics corresponding to the p orbitals). But I already concluded that the 11> spin state could not exist for helium in the ground state (because up-up violates the Pauli exclusion principle).

Help?

Its a question about that you have the total spin constant, i.e., an eigen problem of S_tot^2. A general linear combination of Stot=0 (up-down, down-up) is

$$\vec{\Psi}(\vec{r}_1,\vec{r}_2)= \left( \begin{array}{c} \psi_{\uparrow,\uparrow} \\ \psi_{\uparrow,\downarrow} \\ \psi_{\downarrow,\uparrow} \\ \psi_{\downarrow,\downarrow} \\ \end{array} \right)= c_1\cdot\left( \begin{array}{c} 0 \\ \psi_A \\ -\psi_A \\ 0 \\ \end{array} \right)+ c_2\cdot\left( \begin{array}{c} 0 \\ \psi_B \\ -\psi_B \\ 0 \\ \end{array} \right)$$

You could clearly not flip into a up-up with this combination.

 

[per.sundqvist]

oh i see what your paradox is. The state with l=1, m=0 is valid both for both S=0 and S=1, and if S=1, you could have the z-projection Sz=-1,-1/2,1/2,+1/2, hence a term in up-down terms, but a Sz=+/-1/2 state for S=1 has to look like:
$$\left( \begin{array}{c} 0 \\ \psi \\ \psi \\ 0 \\ \end{array} \right)$$
note the difference in sign with S=0 case in message above.

 

[reilly]

Two things:

The ground state of helium is a 1 s^^2 state, the electrons are in a l=0 (S wave) and a Mz=0, S =1 state. See any book covering atomic physics, orbitals and all that stuff.

This is a sort of abstract answer to your paradox, which is not at all a paradox. The angular momentum operator for two electrons, J=J1 + J2, is invariant under interchange of electrons. Hence the rotation taking one from the z axis to the x axis, that is taking Mz to Mx, is invariant under exchange of two electrons. Thus the unitary transformation that changes the direction of the z axis cannot change the symmetry characteristics of the orbital or spin angular momentum.

Marty -- put in your magnetic interaction, and grind it out. This is a very standard, basic QM problem, which will yield by means of lots of algebra. This will also help you explain your paradox=error.
Regards,
Reilly Atkinson

 

[Marty]

So can I ask what exactly was wrong with my logic?

 

[reilly]

You might start be examining what you mean by superimposing the x and y axes -- what does this mean?
Regards,
ReillyAtkinson

 

[Marty]

I hope it was clear what I meant. I wrote that one could have a spin state which would be 10> with respect to the x axis; you could also have a state that was 10> with respect to the y axis. Certainly with the correct phasing these two states can be combined to form the 11> state with respect to the z axis.

 

[Marty]

Its a question about that you have the total spin constant, i.e., an eigen problem of S_tot^2. A general linear combination of Stot=0 (up-down, down-up) is

$$\vec{\Psi}(\vec{r}_1,\vec{r}_2)= \left( \begin{array}{c} \psi_{\uparrow,\uparrow} \\ \psi_{\uparrow,\downarrow} \\ \psi_{\downarrow,\uparrow} \\ \psi_{\downarrow,\downarrow} \\ \end{array} \right)= c_1\cdot\left( \begin{array}{c} 0 \\ \psi_A \\ -\psi_A \\ 0 \\ \end{array} \right)+ c_2\cdot\left( \begin{array}{c} 0 \\ \psi_B \\ -\psi_B \\ 0 \\ \end{array} \right)$$

You could clearly not flip into a up-up with this combination.

Your refer to the "general linear combination" but there doesn't appear to be a need
for the c2 term in your expression: can you not write out the same physics if you set c2=0?
Or do your A and B subscripts refer to something that I'm not aware of?

But this is probably not the main point. Clearly, in setting up the helium atom, I have allowed for the case of (in your representation)

$$\left( \begin{array}{c} 0 \\ \psi_B \\ \psi_B \\ 0 \\ \end{array} \right)$$

which you have excluded. I don't see the physical constraint which sets this term to zero.

 

[reilly]

Marty --
1. Think about the symmetry of the problem. And, just grind it out -- compute your CG coefficients.
2. Your problem is a good candidate for a first year QM exam. That is, it provides a simple way to look at and compute how two angular momenta combine. You'll find it discussed in many books on QM and on Angular Momentum.

Yours is a good question, but a very old one.
Regards,
Reilly Atkinson

 

[Marty]

Marty --
1. Think about the symmetry of the problem. And, just grind it out -- compute your CG coefficients.
2. Your problem is a good candidate for a first year QM exam. That is, it provides a simple way to look at and compute how two angular momenta combine. You'll find it discussed in many books on QM and on Angular Momentum.

Yours is a good question, but a very old one.
Regards,
Reilly Atkinson

I wonder what my question is exactly, in your terms? Maybe it would help if you would rewrite in just the way you might phrase it if you wanted to include it on one of your own first-year exams.

 

[reilly]

I wonder what my question is exactly, in your terms? Maybe it would help if you would rewrite in just the way you might phrase it if you wanted to include it on one of your own first-year exams.

OK. Assume that the electronic states of a Helium atom are appropriately antisymmetrized hydrogen orbitals, with no spin dependent terms involved. Look at the first two shells, with n=1 l=0, s=0 (l and s are total orbital and spin angular momenta), and n=2 with both S and P waves.

Write out the Hamiltonian for the Helium electrons in a static magnetic field, B. Evaluate the possible transitions generated by B among the n=1 and n=2 shells, Work in a basis that includes total angular and spin momenta.

Regards,
Reilly Atkinson

 

[Marty]

Reilly,

Are you using "l" (lower-case L) the same way I am? Maybe I am not following standard terminology. I see in Wikipedia that L is defined (sometimes?) as ORBITAL spin, which I'm trying to avoid getting into here...I'm trying to work out electron spin. I thought I was pretty clear in my very first post in which I talk about putting two electrons together to create a spin-one particle. I attempted to designate the four possible spin states as

l=1, m=1
l=1, m=0 and l=0, m=1
l=1, m=-1

and then tried to line these up with the up-up, up-down, etc. representation.

Am I incorrect in using "small-L" here? Because there is no orbital angular momentum. How else do I designate these states?

In your problem statement (previous post) you say "l and s are total orbital and spin momenta". Do you mean "l is total orbital and s is spin..." or do you mean to include spin in your L?

Marty

 

[reilly]

Sorry to be so late with my response. Bad notation on my part. I mean by l, l1+l2, the total orbital angular moment, and s1+s2=s is the total spin angular momentum. Whether you keep l or no, simply evaluate (s1+s2)*(s1+s2) on the various spin states of interest. (This problem is done in countless QM texts.)
Regards,
Reilly

 

[TriTertButoxy]

I thought that the state $$|\Psi\rangle=|1s,\uparrow\rangle\,|1s,\downarrow\rangle-|1s,\downarrow\rangle\,|1s,\uparrow\rangle$$ would be the correct ground state of the helium atom. It is properly antisymmetric with respect to the interchange of the electrons, and both electrons are sitting in the 1s orbital.

 

[per.sundqvist]

Your refer to the "general linear combination" but there doesn't appear to be a need
for the c2 term in your expression: can you not write out the same physics if you set c2=0?
Or do your A and B subscripts refer to something that I'm not aware of?

But this is probably not the main point. Clearly, in setting up the helium atom, I have allowed for the case of (in your representation)

$$\left( \begin{array}{c} 0 \\ \psi_B \\ \psi_B \\ 0 \\ \end{array} \right)$$

which you have excluded. I don't see the physical constraint which sets this term to zero.

The thing is that you are mixing states/super posing with different total spin, but where the wave function is the same:

$$\vec{\Psi}(\vec{r}_1,\vec{r}_2)= \left( \begin{array}{c} \psi_{\uparrow,\uparrow} \\ \psi_{\uparrow,\downarrow} \\ \psi_{\downarrow,\uparrow} \\ \psi_{\downarrow,\downarrow} \\ \end{array} \right)= c_1\cdot\left( \begin{array}{c} 0 \\ \psi_{l1,m0} \\ -\psi_{l1,m0} \\ 0 \\ \end{array} \right)_{S=0}+ c_2\cdot\left( \begin{array}{c} 0 \\ \psi_{l1,m0} \\ \psi_{l1,m0} \\ 0 \\ \end{array} \right)_{S=1}$$

so you get the mean value of total spin to be in between 0 and 1, like: 0 < <S> < 1. You mix singlet with triplet.

 

[Marty]

The thing is that you are mixing states/super posing with different total spin, but where the wave function is the same:

$$\vec{\Psi}(\vec{r}_1,\vec{r}_2)= \left( \begin{array}{c} \psi_{\uparrow,\uparrow} \\ \psi_{\uparrow,\downarrow} \\ \psi_{\downarrow,\uparrow} \\ \psi_{\downarrow,\downarrow} \\ \end{array} \right)= c_1\cdot\left( \begin{array}{c} 0 \\ \psi_{l1,m0} \\ -\psi_{l1,m0} \\ 0 \\ \end{array} \right)_{S=0}+ c_2\cdot\left( \begin{array}{c} 0 \\ \psi_{l1,m0} \\ \psi_{l1,m0} \\ 0 \\ \end{array} \right)_{S=1}$$

so you get the mean value of total spin to be in between 0 and 1, like: 0 < <S> < 1. You mix singlet with triplet.

Yes, I kind of think this is what I am doing but I am still confused by your notation (although it makes more sense now that you are showing a combination of symmetric and antisymmetric). I take it that you are using my convention where L and M refer to electron spin and not orbital (phi being symmetric anyhow)...then shouldn't the first column (antisymmetric) have the subscripts L0,M0?? (uppercase used for clarity here). I think I'm understanding that your subscripts here are intended simply to clarify, and that we can in any case "read" the L,M states off the column just by looking at the symmetry state.

I'm only asking this because I think you're showing me something that's helpful here.

Marty