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Adding Three Spin 1/2s

  1. Oct 11, 2007 #1

    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:
    \left([1/2] \otimes [1/2]) \otimes [1/2]
    = ([1] \oplus [0]) \otimes [1/2]
    =[3/2] \oplus [1/2] \oplus [1/2]
    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?
    Last edited: Oct 11, 2007
  2. jcsd
  3. Oct 11, 2007 #2


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    There are two distinct spin 1/2 states, with different spin functions.
  4. Oct 11, 2007 #3
    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.
  5. Oct 12, 2007 #4
    Well lets work it out. Adding the first two angular momentums you get four states namely:

    [tex]\left| {1,1} \right\rangle = \left| {{1 \mathord{\left/ {\vphantom {1 {2,1/2}}} \right.
    \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 [/tex]

    [tex]\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.
    \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)[/tex]
    [tex]\left| {1, - 1} \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
    [tex]\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)[/tex]

    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 [tex]\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.
    \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 \uparrow - \downarrow \uparrow \uparrow } \right)[/tex] where I am using up and down arrows to represent spin up and spin down and simplify my notation. You also have [tex]\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.\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)[/tex]. These are the two 1/2 spins I think you missed in your counting
    Last edited: Oct 12, 2007
  6. Oct 12, 2007 #5

    Meir Achuz

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    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.
  7. Dec 9, 2007 #6
    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?


    Attached Files:

  8. Dec 10, 2007 #7

    Dr Transport

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    Using Clebsch-Gordan coefficients is the standard method to add three angular momenta.
  9. Dec 11, 2007 #8


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    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.
  10. Dec 11, 2007 #9
    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?
  11. Dec 11, 2007 #10


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    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 =)
  12. Dec 11, 2007 #11
    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.
  13. Dec 11, 2007 #12


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    "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.
    Last edited: Dec 11, 2007
  14. Dec 11, 2007 #13
    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.
  15. Dec 11, 2007 #14


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

    It is writing in TeX etc that takes time.
  16. Dec 11, 2007 #15
    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.
  17. Dec 11, 2007 #16


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    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 dont 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 :)
  18. Oct 26, 2008 #17


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    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 [tex]J^2[/tex] 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?
    Last edited: Oct 26, 2008
  19. Oct 27, 2008 #18


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    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
  20. Oct 27, 2008 #19

    Vanadium 50

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    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.
  21. Oct 27, 2008 #20
    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. Infact 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.
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