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

  • Thread starter Thread starter Nikratio
  • Start date Start date
  • Tags Tags
    Spin
Click For Summary
When adding three spin-1/2 particles, the total spin quantum number can yield values of s=3/2 or s=1/2, leading to a total of eight possible states. However, there appears to be a discrepancy with the eigenvector count, as only six eigenvectors are accounted for, raising questions about the degeneracy of the states. The discussion highlights the importance of distinguishing between the two distinct spin-1/2 states, which can be achieved by constructing an observable that commutes with S^2 and S_z. The standard method for adding angular momenta involves using Clebsch-Gordan coefficients, and while this approach is common, some participants express uncertainty about its effectiveness in this context. Ultimately, the conversation revolves around the complexities of identifying and verifying the states resulting from the combination of three spin-1/2 particles.
  • #31
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.
 
Physics news on Phys.org
  • #32
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).
 
  • #33
Gerenuk said:
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.
 
  • #34
malawi_glenn said:
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)
 
  • #35
malawi_glenn said:
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?
 
  • #36
Gerenuk said:
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)|-+++>


KFC said:
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>

KFC said:
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.
 
  • #37
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 :(
 
Last edited:
  • #38
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 :-P Not antisymmetric.
 
Last edited:
  • #39
malawi_glenn said:
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:

malawi_glenn said:
|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.
 
  • #40
Gerenuk said:
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...
 
  • #41
malawi_glenn said:
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

malawi_glenn said:
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).
 
  • #42
Gerenuk said:
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
 
Last edited by a moderator:
  • #43
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.
 

Similar threads

I Weakly interacting electrons in an atom
  • · Replies 2 ·
Replies
2
Views
2K
I Addition of angular momenta
  • · Replies 1 ·
Replies
1
Views
667
I How Does Spin-Orbit Coupling Affect Quantum Number Validity?
  • · Replies 14 ·
Replies
14
Views
2K
I Two identical spin 1/2 particles
  • · Replies 1 ·
Replies
1
Views
2K
I Fermi energy for a Fermion gas with a multiplicity function ##g_n##
  • · Replies 1 ·
Replies
1
Views
1K
I How are good quantum numbers related to perturbation theory?
  • · Replies 18 ·
Replies
18
Views
3K
A Why are there only two possible spin-eigenstates for deuterium's nucleus?
  • · Replies 20 ·
Replies
20
Views
4K
I Spin states for two identical 1/2 particles - Confused :s
  • · Replies 5 ·
Replies
5
Views
3K
I More doubts in perturbation theory
  • · Replies 8 ·
Replies
8
Views
1K
I Determining the Number of Points in the n-Space
  • · Replies 16 ·
Replies
16
Views
2K