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.

 

[malawi_glenn]

Iam not sure what you mean by "How would you add |1/2,+1/2> and another |1/2,+1/2> if both states are not actually fully specified."

Getting the internal structure of all states coupling three spin 1/2 using C.G would take me approx 5 minutes with pen and paper (and a CG table of course). It's not difficult.

 

[Gerenuk]

If you want to add 4 spin-1/2s. At one point you add 3 of them and get the two different |1/2,+1/2> states from above. Now you want to add the 4th spin.

What do you look up in the CG table respectively? You do not want to lose the identity of the different |1/2,+1/2> states (which are in fact a 3 spin-1/2 state).

 

[malawi_glenn]

If you want to add 4 spin-1/2s. At one point you add 3 of them and get the two different |1/2,+1/2> states from above. Now you want to add the 4th spin.

What do you look up in the CG table respectively? You do not want to lose the identity of the different |1/2,+1/2> states (which are in fact a 3 spin-1/2 state).

What do you want? Adding 4 spin 1/2?

Do you know how to construct all 8 states obtained from adding 3 spin 1/2 using C.G tables, yes or no? If no, tell me/us where the problem is, which states are you unable to construct?

The "trick" is to label them, you have one |1/2,+1/2>_symmetric and one |1/2,+1/2>_asymmetric, or call them whatever you like. Using C.G you don't care about the "iternal structure" of the states which you are adding to. e.g let's say j_1 = L_1 + S_1, a combination of angular and spin momentum. Then you want to add another j_2, which is j_2 = L_2 + S_1. Same thing, just use C.G for j_1 + j_2.

 

[Gerenuk]

What do you want? Adding 4 spin 1/2?
...
The "trick" is to label them, you have one |1/2,+1/2>_symmetric and one |1/2,+1/2>_asymmetric, or call them whatever you like. Using C.G you don't care about the "iternal structure" of the states which you are adding to.
...

But C.G. tables don't have entries for "_symmetric" and "_antisymmetric"? And if you just use |1/2,+1/2> you don't get enough states? And the parts of the full representation (i.e. "|+-+>") are not in a C.G. table either? Whereas simply saying one addition arises from a "_symmetric" state and the other from an "_antisymmetric" doesn't specify what the state really are (should 16 eigenstates for 4 spins)

 

[KFC]

Iam not sure what you mean by "How would you add |1/2,+1/2> and another |1/2,+1/2> if both states are not actually fully specified."

Getting the internal structure of all states coupling three spin 1/2 using C.G would take me approx 5 minutes with pen and paper (and a CG table of course). It's not difficult.

malawi_glenn, I read through this thread and I am totally lost now. Well, starting from the triplet states and consider coupling with another 1/2 spin. It is easily to write down all 6 states and I already done that. But, as you told, CG doesn't give combination coef. for the coupling of singlet state and spin 1/2.

1) You said it is trival to obtain those state, I don't see why and how. Would you please tell me how can you write down those symmetric and antisymmetric forms?

2) Obviously, from triplet states and singlet state, we get the same coupling state
|1/2, 1/2> and |1/2, -1/2>. What's the physical meaning for having two states of different form but have same quantum number?

 

[malawi_glenn]

But C.G. tables don't have entries for "_symmetric" and "_antisymmetric"? And if you just use |1/2,+1/2> you don't get enough states? And the parts of the full representation (i.e. "|+-+>") are not in a C.G. table either? Whereas simply saying one addition arises from a "_symmetric" state and the other from an "_antisymmetric" doesn't specify what the state really are (should 16 eigenstates for 4 spins)

you have two different |J = 1/2,M = +1/2>, where J is addition of 3 spin 1/2. For your information using C.G tables you DONT NEED TO KNOW ITS INTERNAL STRUCTURE, see post #33. Do you read what I post, and do you try for yourself BEFORE you posting here? Or do you just want me to do the work for you?

Just do this:
|J = 1/2,M = +1/2>*|+> = look in table for j_1 = 1/2, m_1 = 1/2, j_2 = 1/2, m_2 = 1/2 = |J_4 = 1, M_4 = +1> = now use what you know of state |J = 1/2,M = +1/2> , thera are two of them, see my post #25.

|J_4 = 1, M_4 = +1>_malawi = {sqrt(2/3)|++-> - sqrt(1/6)|+-+> - sqrt(1/6)|-++>}*|+> = sqrt(2/3)|++-+> - sqrt(1/6)|+-++> - sqrt(1/6)|-+++>

|J_4 = 1, M_4 = +1>_glenn = sqrt(1/2)|+-++> - sqrt(1/2)|-+++>

You said it is trival to obtain those state, I don't see why and how. Would you please tell me how can you write down those symmetric and antisymmetric forms?

Piece of cake:

|J = 1/2, M = 1/2> = |0,0> * |+> = ... (use what |0,0> is according to c.g table) ...= sqrt(2){|+-> - |-+>}*|+> = sqrt(2){|+-+> - |-+-> }

Now you try:
|0,0> * |->

Where |+> of course is the state |j=1/2, m= +1/2> and |-> is |j=1/2, m= -1/2>

2) Obviously, from triplet states and singlet state, we get the same coupling state
|1/2, 1/2> and |1/2, -1/2>. What's the physical meaning for having two states of different form but have same quantum number?

The most obviuos physical meaning is that the probability to get J = 1/2 by a measurment on a random coupled 3 spin 1/2 system is given as the sum of the probability to get the asymmetric state and the symmetric.

It matters in nuclear structure physics, when you also have other quantum numbers, such as the isospin. You demand the total wavefunction to be antisymmetric (you treat the nucleons as identical particles with spin-up and spin-down in Isospin space). So then it matters what angular-momentum part of the wavefunction a state of a collection of nucleons have.

 

[KFC]

malawi_glenn, please forgive me, but I don't really familiar with the symbol you post here. I guess what you mean "|0,0> * |-> " is the DIRECT PRODUCT of state |0,0> and |->, right? But I didn't learn direct product before, I am just reading a book and google about it. According to the definition, direct product seems only to change two column vectors into a bigger column vector by "concate them tail-to-head", is that right? If so, now, according to your example,

|0,0> * |+> = sqrt(1/2) (|+-> - |-+>) * |+> = sqrt(1/2)(|+-+> - |-++>)

It seems just insert the third spin into the two kets, isn't it? So why |0,0>*|-> didn't get this?

|0,0> * |-> = sqrt(1/2) (|+-> - |-+>) * |-> = sqrt(1/2)(|+--> - |-+->)

By the way, why you call |0,0> * |+> antisymmertic? If you exchange the first two spin, it appears to be antisymmetic; but if you exchange the last two or the first and the last one, it is not.

I knew it is quite annoying to keep asking the question. I feel very sorry about this, but I do want to learn something. If you feel to annoyed to this, just ignore my question :(

 

[malawi_glenn]

The |0,0> * |-> you did is correct. Now one can label kets as onw wishes, |-+-> Just means that first particle is spin down, second is spin up, third is spin sown, and that is just a short notation for the direct product:
|->*|+>*|-> = |-+->

You are right, one should call "|0,0> * |+> " Asymmetric, which I have done in all posts exept #25 Not antisymmetric.

 

[Gerenuk]

For your information using C.G tables you DONT NEED TO KNOW ITS INTERNAL STRUCTURE, see post #33. Do you read what I post, and do you try for yourself BEFORE you posting here? Or do you just want me to do the work for you?

Yes, I do read it. But then you do not refer to my posts fully. I did mention that it is possible to add with C.G. disregarding internal structure.
I do not try the calculation myself, because I see that it wouldn't work. You don't do it either but something different instead:

|J_4 = 1, M_4 = +1>_malawi = {sqrt(2/3)|++-> - sqrt(1/6)|+-+> - sqrt(1/6)|-++>}*|+> = sqrt(2/3)|++-+> - sqrt(1/6)|+-++> - sqrt(1/6)|-+++>

|J_4 = 1, M_4 = +1>_glenn = sqrt(1/2)|+-++> - sqrt(1/2)|-+++>

It's hard for me to check now, but why do you assume that simply adding the last spin |+> to the representation makes it a J^2 eigenstate? That might have worked for the trivial case |0,0>+|1/2,+1/2> but not in general (and you didn't use C.G. or do the contain "_malawi" and "_glenn" entries?). How would you add |J_3=1/2,M_3=-1/2>*|+>? For example |->*|+>=|-+> does not give a J^2 eigenstate.

It would be good to have a third persons opinion here.

 

[malawi_glenn]

Yes, I do read it. But then you do not refer to my posts fully. I did mention that it is possible to add with C.G. disregarding internal structure.
I do not try the calculation myself, because I see that it wouldn't work. You don't do it either but something different instead:It's hard for me to check now, but why do you assume that simply adding the last spin |+> to the representation makes it a J^2 eigenstate? That might have worked for the trivial case |0,0>+|1/2,+1/2> but not in general (and you didn't use C.G. or do the contain "_malawi" and "_glenn" entries?). How would you add |J_3=1/2,M_3=-1/2>*|+>? For example |->*|+>=|-+> does not give a J^2 eigenstate.

It would be good to have a third persons opinion here.

Of course I used C.G,

|J_4 = 1, M_4 = +1> = (when adding a j_1=1/2 with a j_2=1/2 state) is simply |m_1 = 1/2, m_2 = 1/2> .. there is only one term.

Maybe you don't know how to use C.G tables, but you look at what final state you want to construct FIRST, for example J = 1, M = 0 (when adding a j_1=1/2 with a j_2=1/2 state) then you look in the table what states it consists of and which c.g coefficient each state in the linear combination should have.

Now you try this, then rewrite is a linear combinations of |++--> 'like states.

I will give you "my" answer in approx 6h when I get back home from my office

can also, once and for all, write down all states you get when adding 3 spin 1/2's, and maybe all states you get if you add 4spin 1/2...

 

[Gerenuk]

Maybe you don't know how to use C.G tables, but you look at what final state you want to construct FIRST, for example J = 1, M = 0 (when adding a j_1=1/2 with a j_2=1/2 state) then you look in the table what states it consists of and which c.g coefficient each state in the linear combination should have.

Now you try this, then rewrite is a linear combinations of |++--> 'like states.

I need to think about that. Any particular reference for C.G. you could recommend? I only had a simplified table so far.
OK, the example from above had only only state and maybe that's how it was "trivial". Unfortunately I can't see the general principle if it's a trivial case

I will give you "my" answer in approx 6h when I get back home from my office

can also, once and for all, write down all states you get when adding 3 spin 1/2's, and maybe all states you get if you add 4spin 1/2...

No, don't worry about all the solution. I can find them by myself by other means. I'm more interested how C.G. does handle the degenerate states.

Maybe the single example how to add |J_3=1/2,M_3=+1/2>*|-> would help me understand (which I suppose is non-trivial).

 

[malawi_glenn]

I need to think about that. Any particular reference for C.G. you could recommend? I only had a simplified table so far.
OK, the example from above had only only state and maybe that's how it was "trivial". Unfortunately I can't see the general principle if it's a trivial case


No, don't worry about all the solution. I can find them by myself by other means. I'm more interested how C.G. does handle the degenerate states.

Maybe the single example how to add |J_3=1/2,M_3=+1/2>*|-> would help me understand (which I suppose is non-trivial).

I think this is where you'll encounter problems. Don't think as you should add a state j_1, m_1 with j_2 m_2. Instead, think of this:

You can add j_2 and j_1 to total J =>

|j_2-j_1| <= J <= j_2 + j_1

So beginning with maximum J, look at the CG table to find all |J,M> states in terms of |j_1,m_1> and |j_2,m_2> states.

For example, adding two spin 1/2 particles, the total J = 1, M = 0 state is given according to C.G table as:

|1,0> = 1/sqrt(2) { |+-> + |-+> }

And the J = 0, M = 0 state is:

|0,0> = 1/sqrt(2) { |+-> - |-+> }

This is how it works.

You CAN do it the other way around, using the table "inversely".

The state |+-> is then a linear combination of|1,0> and |0,0> states.

The "degenerate" states, are handled just as i showed you before:

(J here is addition of 3 spin 1/2 states, just as a reminder)

|J = 1/2,M = +1/2>*|+> = look in table for j_1 = 1/2, m_1 = 1/2, j_2 = 1/2, m_2 = 1/2 = |J_4 = 1, M_4 = +1> = now use what you know of state |J = 1/2,M = +1/2> , there are two of them, see my post #25.

|J_4 = 1, M_4 = +1>_malawi = {sqrt(2/3)|++-> - sqrt(1/6)|+-+> - sqrt(1/6)|-++>}*|+> = sqrt(2/3)|++-+> - sqrt(1/6)|+-++> - sqrt(1/6)|-+++>

|J_4 = 1, M_4 = +1>_glenn = sqrt(1/2)|+-++> - sqrt(1/2)|-+++>

Or another notation:

|J_4 = 1, M_4 = +1>_glenn = sqrt(1/2)|+>*|->*|+>*|+> - sqrt(1/2)|->*|+>*|+>*|+>

The notation |++> is just a short way of writing |j_a = 1/2, m_a = +1/2 > * |j_b = 1/2, m_b = +1/2 >

I use this as reference, it is attached to an old exam in QM:
http://www3.tsl.uu.se/thep/courses/QM/081021-exam.pdf

page 4

 

[Simplicio2]

I would like to answer Nikratio's question: "And with which operator is this degeneracy usually resolved? ". The answer is that the operator S_12**2 (being (S_1 + S_2)**2) is also needed as a member of the CSCO of three operators, alongside with S**2 (total spin squared) and S_z (z-component of total spin). As a consequence, not only the degeneracy is lifted, but also one sees clearly that the two spin-1/2 subspaces have different quantum numbers with respect to S_12, namely s_12=1 and s_12=0, respectively. An alternative choice of CSCO is of course the set of S_1z, S_2z and S_3z.