Adding 3 Spin 1/2s: Total Spin Quantum Number

[Nikratio]

Hello,

I am considering the case of the total spin when adding three spin 1/2s. The combined system has dimension 2x2x2=8. The possible values for the total spin quantum number are:
<br /> \left([1/2] \otimes [1/2]) \otimes [1/2]<br /> = ([1] \oplus [0]) \otimes [1/2] <br /> =[3/2] \oplus [1/2] \oplus [1/2]<br />
so either s=3/2 or s=1/2.

s=3/2 is 4 fold degenerate and s=1/2 is 2 fold degenerate. But this sums up to only 6 eigenvectors.

Where are the missing two eigenvectors? It seems to me that S^2 and S_z no longer form a C.S.C.O in this case, since for s=1/2 both m=1/2 and m=-1/2 still have to be 2-fold degenerate. Is this correct? And with which operator is this degeneracy usually resolved?

 

[Meir Achuz]

There are two distinct spin 1/2 states, with different spin functions.

 

[Nikratio]

There are two distinct spin 1/2 states, with different spin functions.

Yes, that's what I expected. But is there a common way to distinguish between these two states? I could probably construct an observable that commutes with S^2 and S_z and distinguishes between these states, but I'd like to know whether there is a standard way of choosing this observable.

 

[Jim Kata]

Well let's work it out. Adding the first two angular momentums you get four states namely:

\left| {1,1} \right\rangle = \left| {{1 \mathord{\left/ {\vphantom {1 {2,1/2}}} \right.<br /> \kern-\nulldelimiterspace} {2,1/2}}} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right. \kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle

\left| {1,0} \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}}\right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2, -{1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.<br /> \kern-\nulldelimiterspace} {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.<br /> \kern-\nulldelimiterspace} 2}}}} \right\rangle + \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.<br /> \kern-\nulldelimiterspace} {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.<br /> \kern-\nulldelimiterspace} 2}}}} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle } \right)
\left| {1, - 1} \right\rangle = \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.<br /> \kern-\nulldelimiterspace} {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.<br /> \kern-\nulldelimiterspace} 2}}}} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2, - {1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle <br />
\left| {0,0} \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2, - {1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle - \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2, - {1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle }\right)

Now we add the last 1/2 angular momentum. I'm not going to write out all of the states because I'm getting bored of this, but I'll show you the two 1/2 ones I think your missing in your count. Consider \left| {{1 \mathord{\left/{\vphantom {1 {2,{1\mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle = \left| {0,0} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.<br /> \kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/<br /> {\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle = \frac{1}<br /> {{\sqrt 2 }}\left( { \uparrow \downarrow \uparrow - \downarrow \uparrow \uparrow } \right) where I am using up and down arrows to represent spin up and spin down and simplify my notation. You also have \left| {{1 \mathord{\left/{\vphantom {1 {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.<br /> \kern-\nulldelimiterspace} {2, - {1 \mathord{\left/{\vphantom {1 2}} \right.<br /> \kern-\nulldelimiterspace} 2}}}} \right\rangle = \left| {0,0} \right\rangle \otimes \left| {{1 \mathord{\left/{\vphantom {1 {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right.\kern-\nulldelimiterspace} {2,{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right\rangle = \frac{1}{{\sqrt 2 }}\left( { \uparrow \downarrow \downarrow - \downarrow \uparrow \downarrow } \right). These are the two 1/2 spins I think you missed in your counting

 

[Meir Achuz]

Yes, that's what I expected. But is there a common way to distinguish between these two states? I could probably construct an observable that commutes with S^2 and S_z and distinguishes between these states, but I'd like to know whether there is a standard way of choosing this observable.

One spin 1/2 state has the first two spins in a symmetric spin one state.
The other spin 1/2 state has the first two spins in an antisymmetric spin zero state.
In the quark model, a spin 1/2 baryon is composed of three spin 1/2 quarks in the symmetric spin one state. For instance |p>=uud, with the two u quarks in a spin one state.

 

[Gerenuk]

Hi!
I was asking myself exactly the same question. People said use Clebsch coefficients, but that seemed to me an inadequate approach. I instead simply diagonalized the corresponding matrix.

See attachment for my results.

I found some asymmetric states, which I symmetrized as to make them commute with the cyclic permutation operator 1->2->3->1

I just guessed. But is this operator always sufficient to distinguish between degenerate states?

Now there was something that puzzled a lot of students who never thought beyond the university examples:

Is it OK, that it is not possible to symmetrize or antisymmetrize the total spin 1/2 states? (you get zero... and P_cycl is not 1 or -1)

Or does it just follow that the total wavefunction won't separate into position part and spin part?

Anton

 

[Dr Transport]

Hi!
I was asking myself exactly the same question. People said use Clebsch coefficients, but that seemed to me an inadequate approach.
Anton

Using Clebsch-Gordan coefficients is the standard method to add three angular momenta.

 

[malawi_glenn]

First, construct all states you get by adding two s=½. Then each of these states you couple to s=½. So it is like adding an ensamble of s=1 states to s=½. And one uses CG for this, as Dr Transport mentioned. This should be the easiest way to work em out.

 

[Gerenuk]

I admit everyone else also suggested that to me first, but none of them actually did the calculation to know that it's easiest. I guess you have to put some thinking in in order to get the 8 states. You probably have to try adding different pairs initially and before combining them with the third spin.
Or do you see another way to get 8 states?

 

[malawi_glenn]

why think? Just first combine to get all states for two s=½, (you get 4 s=1 states), then coupling those with s=½ again, you'll get 8states. It is very straightforward in my opinion =)

 

[Gerenuk]

You may be right. Actually if you combine two spin you get 3 s=1 states and 1 s=0 state. The question is, if you combine the (s=1 m=0) or the (s=0 m=0) state with the third spin, do you get different state functions? Note that the solution has two different orthogonal state functions for the same momentum eigenvalues s=1/2 m=1/2.

 

[malawi_glenn]

"may be right" ?

I have done excerceise of this kind several times, and also our professor did it on the black board once. If I had more time, I would be delighted to demonstrate the whole procedure for you all.


"The question is, if you combine the (s=1 m=0) or the (s=0 m=0) state with the third spin, do you get different state functions? "

Yes, you do.

 

[Gerenuk]

I'm just not sure how to verify it myself.
Please post the full answer, if it's as easy as looking up Clebsch Gordon. I don't think it's as direct as the common examples. You need to get two different wavefunction for a state with equal eigenvalues in you representation.

 

[malawi_glenn]

After my final exams the 17th Dec i might have time.

It is writing in TeX etc that takes time.

 

[Gerenuk]

OK. I really would like to know.
Graduate theoreticians said it must be straightforward, but then actually they got stuck. Hope Clebsch Gordan ist really enough and one doesn't need any of the more complicating momentum formalism.

 

[malawi_glenn]

i would take my 10min to work em out on a paper, but how to show it on the computer takes more time;)

The thing is that adding s_3 = ½ to the (s=0, m=0) where s = s_1 + s_2, states you don't even need GC, adding a spin ½ to spin 0 can only give you two states.

(s=0, m=0) + (s_3=½, m_3 = +½) and (s=0, m=0) + (s_3=½, m_3 = -½)

Maybe this is the most difficult thing, it is so easy that you think its hard :)

 

[KFC]

Using Clebsch-Gordan coefficients is the standard method to add three angular momenta.

I just consider the combination of two spin 1/2 first, get four states and then add the third one in. Consider one singlet and three triplet states, each of them I take the third spin 1/2 up and down, this will give me 8 states, but is it that easy? I even don't have to do calculation! I don't know if this is wrong or not. By the way, how do I know if my result is correct or not? Should I have J^2 operate on each of them?

I do have interest to know how can we use Clebsch-Gordan coefficients to add the third momenta to two-spin1/2 coupled system, can you show us an example?

 

[malawi_glenn]

Just treat the constructed states from two spin 1/2 as "black boxes" i.e as spin 1 with M = 1. Then spin 1 with M = 0 etc. Just do the procedure you would have done if you where told to coulpe J = 1 with J = 1/2, and J = 0 with J = 1/2

 

[Vanadium 50]

There are something called 6-j symbols which let you add three angular momenta in one step. (Clebsch-Gordon coefficients are equivalent to 3-j symbols). It's worth learning how to use them (even if one never uses them again), as one sees how one has to label the individual states.

 

[Gerenuk]

Hmm, is the initial question answered? Maybe I'm missing something in the notation here, but it seems everyone is adding two spins and then claiming without prove that the rest is easy?
Do CG always give definite J^2 and J_z? Because then there are only 6 possibilities for adding 3 spins. And note that in the direct low-level answer there are 4 symmetric states and 4 states which are neither symmetric nor antisymmetric. In fact it is impossible to generate antisymmetric spin-states with 3 spin-1/2 particles.

Could someone please write down the final solution? My proposition is found in one of the earlier answers - however no CG used.

 

[malawi_glenn]

It is easy?

Start with adding a J_c = 1/2 particle to the J_1 = 1, M_1 = 1 state:

The J = 3/2 M = 3/2 state must be given as:

|J_1 = 1, M_1 = 1 > * |J_c= 1/2 , M_c = 1> = |J = 3/2, M = 3/2 >

but

|J_1 = 1, M_1 = 1 > = |J_a = 1/2, M_a = 1/2> * |J_b= 1/2, M_b= 1/2>

So

|J = 3/2, M = 3/2 > = (|J_a = 1/2, M_a = 1/2> + |J_b= 1/2, M_b= 1/2> )* |J_c= 1/2 , M_c = 1>

or in another notation:

|J = 3/2, M = 3/2 > = | + + + >

a,b,c denotes electron (spin 1/2) particles.
J_1 is momenta of a coupled to b,
J_2 is J_1 coupled to c.

Now use for example ladder operators to get the J=3/2 M = 1/2, M = -1/2 , M = -3/2 states.

Then when you construct the state J = 1/2, M = 1/2, use orthogonality of state, in perticular, use that:
< J = 1/2, M = 1/2 | * |J = 3/2, M = 1/2 > = 0.

Then repete.

Dr. Transport maybe shows his way do to this soon.

 

[Gerenuk]

...
or in another notation:

|J = 3/2, M = 3/2 > = | + + + >

a,b,c denotes electron (spin 1/2) particles.
J_1 is momenta of a coupled to b,
J_2 is J_1 coupled to c.

Now use for example ladder operators to get the J=3/2 M = 1/2, M = -1/2 , M = -3/2 states.

Then when you construct the state J = 1/2, M = 1/2, use orthogonality of state
...

Well, that's again only part of the answer and in fact the part that I already understand. I'm more curious about the J=1/2 M=+1/2 states(!) since there are two different ones!
As mentioned in the initial question J=3/2 gives 4 states and J=1/2 apparently only 2 which sums to 6 possible outcomes, whereas in fact there are 8 eigenvectors?! What's wrong here?

 

[malawi_glenn]

in CG table:

j_1 = 1, j_2 = 1/2, adding to for state J = 3/2, M =1/2 is:

|J = 3/2, M =1/2 > = sqrt(2/3)|m_1=1,m_2= -1/2> - sqrt(1/3)|m_1=0,m_2= 1/2>

Where:
| m_1=1,m_2= -1/2> is found by looking in CG table for coupling two spin 1/2.
you need to find out what two 1/2 spins j_a and j_b coupled to j_1 = 1 with m_1 = 1, which is trivial.

Rule: Full solutions is not given here
Another rule: In order to get help, attempt to solution must be shown.

Can you do your part of this "game" ?

Remember that you also must add a spin 1/2 to the J = 0 state, this is not mentioned in CG tables.

 

[Gerenuk]

Rule: Full solutions is not given here
Another rule: In order to get help, attempt to solution must be shown.

Can you do your part of this "game" ?

Please look at my first post where I give the full solution of all 8 eigenstates without using CG. From looking at the result it seems CG couldn't give the two different J=1/2 m=+1/2 states. So I actually would like this one part of the result only:

Two different states corresponding to J=1/2 M=+1/2 in the simplified notation (i.e. like "|+++>")

 

[malawi_glenn]

No you posted the answer in that post.

|J=1/2 M=+1/2 >_symmetric = sqrt(2/3)|++-> - sqrt(1/6)|+-+> - sqrt(1/6)|-++>


|J=1/2 M=+1/2 >_antisymmetric = sqrt(1/2)|+-+> - sqrt(1/2)|-++>

The antisymmetric state is done by coupling J = 0 with spin 1/2 and is thus not covered in C.G, since it is trivial.

 

[Gerenuk]

No you posted the answer in that post.

|J=1/2 M=+1/2 >_symmetric = sqrt(2/3)|++-> - sqrt(1/6)|+-+> - sqrt(1/6)|-++>
|J=1/2 M=+1/2 >_antisymmetric = sqrt(1/2)|+-+> - sqrt(1/2)|-++>

The antisymmetric state is done by coupling J = 0 with spin 1/2 and is thus not covered in C.G, since it is trivial.

Now that's is exactly the form of answer I was looking for. Both states are correct results if I compare with my calculations. However, neither of the states is fully symmetric or antisymmetric. I think it is not possible to find a fully antisymmetric 3 spin-1/2 state.

So is the answer to the initial question basically that
|1,0>+|1/2,+1/2> and |0,0>+|1/2,+1/2> both give a |1/2,+1/2> state however in each case a different one?

 

[malawi_glenn]

No it is not possible to find such.

Yes, it is two different states, with same observables.

 

[Gerenuk]

OK, but am I right to conclude that plain C.G. fails in adding more spin-1/2 since there are intrinsically different J=1/2 m=+1/2 states which you wouldn't find in (basic?) tables?

Anyone has a suggestion what other observable should distinguish the degenerate states?

 

[malawi_glenn]

OK, but am I right to conclude that plain C.G. fails in adding more spin-1/2 since there are intrinsically different J=1/2 m=+1/2 states which you wouldn't find in (basic?) tables?

Anyone has a suggestion what other observable should distinguish the degenerate states?

It depends on how you look at it, adding a spin j_1 = J to j_2 = 0 is so trivial that only a person with knowledge at all in quantum angular momentum will have troubles to find the correct results. It's like evaluating a hard integral, and someone says "hey its all in integral tables", but then one encounters one term: \int dx and that is not covered in the tables of integral. What is the conclusion?

So it depends on how you see it ;-)

 

[Gerenuk]

It depends on how you look at it, adding a spin j_1 = J to j_2 = 0 is so trivial that only a person with knowledge at all in quantum angular momentum will have troubles to find the correct results. It's like evaluating a hard integral, and someone says "hey its all in integral tables", but then one encounters one term: \int dx and that is not covered in the tables of integral. What is the conclusion?

So it depends on how you see it ;-)

Hmm? That's not quite what I meant. How would you add |1/2,+1/2> and another |1/2,+1/2> if both states are not actually fully specified.
From the example above one can see that the states are intrinsically different. OK, one would probably get the right quantum numbers with C.G., but this time the intrinsic structure to get all eigenstates is non-trivial.