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Homework Help: Applying S_ operator on composite state |5/2 3/2>

  1. Feb 28, 2010 #1
    Original question:

    A system contains two particles: the spin of particle 1 is 3/2 and the spin of particle 2 is 1. Motion of particles can be ignored.

    Part A asked to find the total spin state |5/2 3/2> using Clebsch-Gordon coefficients.

    I did so, and came up with,

    [tex] \mid 5/2\ \ \ 3/2\ \rangle = \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle [/tex]

    Had no trouble with that.

    Part D asks to apply the operator:

    [tex]S_- = S_-^{(1)} + S_-^{(2)}[/tex],

    on the composite state constructed in part A.

    Now I have,

    [tex] S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle [/tex]

    and

    [tex]S_- = S_x - iS_y[/tex]

    Kinda don't know where to go from here. I understand that since we are using minus operator, spin will be decreasing. But how do values of s and m change accordingly?
     
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2

    vela

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    The individual S- operators act on the respective parts of the state. If |s m>=|s1 m1>|s2 m2>, you'd have

    [tex]S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)[/tex]
     
  4. Feb 28, 2010 #3
    So for,

    [tex]S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)[/tex]

    it would be,

    [tex]S_-|5/2\ \ 3/2\rangle = (S_-^{(1)}|3/2\ \ 3/2\rangle)|1\ \ 0\rangle+|3/2\ \ 1/2\rangle(S_-^{(2)}|1\ \ 1\rangle)[/tex]
     
  5. Feb 28, 2010 #4
    hi! i'm a new member. just got a problem of getting the integral of e to the negative x squared from a to b. not the typical error function from 0 to infinity. please help me out with this. thank you.
     
  6. Feb 28, 2010 #5
    Try the homework help forum, for calculus.
     
  7. Feb 28, 2010 #6
    And if my above attempt is true, do I keep the sqrt(2/5) and sqrt(3/5)?
     
  8. Feb 28, 2010 #7

    vela

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    Yes, you have to keep the coefficients. Sorry, I should have included them in the line I wrote above.
     
  9. Feb 28, 2010 #8
    Awesome! I will post my result in a few min, if you can just check it that'd be great!
     
  10. Feb 28, 2010 #9
    So,

    [tex]S_-|s\ m\rangle = (S_-^{(1)}|s_1\ m_1\rangle)|s_2\ m_2\rangle+|s_1\ m_1\rangle(S_-^{(2)}|s_2\ m_2\rangle)[/tex]

    so,

    [tex]S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ (S_-^{(1)}|3/2\ \ 3/2\rangle)|1\ \ 0\rangle+\sqrt{3/5}\ |3/2\ \ 1/2\rangle(S_-^{(2)}|1\ \ 1\rangle)[/tex]

    where,

    [tex](S_-^{(1)}|3/2\ \ 3/2\rangle) = \sqrt{3/2(3/2+1) - 3/2(3/2-1)}\ \hbar\mid 3/2\ \ 3/2-1 \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle[/tex]

    and,

    [tex](S_-^{(2)}|1\ \ 1\rangle) = \sqrt{1(1+1) - 1(1-1)}\ \hbar\mid 1\ \ 1-1 \rangle = \sqrt{2}\ \hbar\mid 1\ \ 0 \rangle[/tex]

    Substituting in the previous two, we get:

    [tex]S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle+\sqrt{3/5}\ |3/2\ \ 1/2\rangle\sqrt{2}\ \hbar\mid 1\ \ 0 \rangle[/tex]

    which equals:

    [tex]S_-|5/2\ \ 3/2\rangle = \sqrt{2/5}\ \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle+\sqrt{3/5}\sqrt{2}\ \hbar\ |3/2\ \ 1/2\rangle\mid 1\ \ 0 \rangle[/tex]

    To conclude:

    [tex]S_-|5/2\ \ 3/2\rangle = (\sqrt{2/5}\ \sqrt{3}+\sqrt{3/5}\sqrt{2}\ )\ \hbar\\ \mid 3/2\ \ 1/2 \rangle|1\ \ 0\rangle[/tex]

    Done correctly ? :)
     
  11. Feb 28, 2010 #10
    You are forgetting to expand the whole equation out. It is similar to:

    (s1+s2)*(v1+v2) = s1*v1+s1*v2+s2*v1+s2*v2

    So you should have 4 parts and not just 2.
     
  12. Feb 28, 2010 #11
    OK. I get what you're saying, but little los at the same time.

    In this case, what would my "v1" and "v2" be?
     
  13. Feb 28, 2010 #12

    vela

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    They'd be [itex]\sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle[/itex] and [itex]\sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle[/itex].
     
  14. Feb 28, 2010 #13
    So you have:

    [tex] S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \left( S_{-}^{1}+S_{-}^{2}\right) \cdot \left( \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle \right) [/tex]

    [tex]= \sqrt{2/5}(S_{-}^{1}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{1}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle+....[/tex]
     
  15. Feb 28, 2010 #14
    Oooo. For some reason when I saw (s1+s2)*(v1+v2) = s1*v1+s1*v2+s2*v1+s2*v2, I was thinking s1*(v1+s1)*(v2+s2)*(v1+s2)*v2 and it made me think in circles. Silly mistake on my part.

    Thanks to the both of you! I will be back in a couple to make sure I got it! :)
     
  16. Feb 28, 2010 #15
    Continuing with this,

    [tex] S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \left( S_{-}^{1}+S_{-}^{2}\right) \cdot \left( \sqrt{2/5} \mid 3/2\ \ \ 3/2\ \rangle\mid 1\ \ \ 0\ \rangle + \sqrt{3/5} \mid 3/2\ \ \ 1/2\ \rangle\mid 1\ \ \ 1\ \rangle \right) [/tex]

    [tex]= \sqrt{2/5}(S_{-}^{1}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{1}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}(S_{-}^{2}\mid 3/2\ \ \ 3/2\ \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(S_{-}^{2}\mid 3/2\ \ \ 1/2\ \rangle)\mid 1\ \ \ 1\ \rangle[/tex]

    Now applying:

    [tex]S_-\mid s\ \ m \rangle = \sqrt{s(s+1) - m(m-1)}\ \hbar\mid s\ \ m-1 \rangle[/tex]

    First,

    [tex]S_-\mid 3/2\ \ 3/2 \rangle = \sqrt{3/2(3/2+1) - 3/2(3/2-1)}\ \hbar\mid 3/2\ \ 3/2-1 \rangle = \sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle[/tex]

    second,

    [tex]S_-\mid 3/2\ \ 1/2 \rangle = \sqrt{3/2(3/2+1) - 1/2(1/2-1)}\ \hbar\mid 3/2\ \ 1/2-1 \rangle = \sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle[/tex]

    Substituiting:


    [tex] S_{-}\mid 5/2\ \ \ 3/2\ \rangle = \sqrt{2/5}(\sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(\sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle)\mid 1\ \ \ 1\ \rangle+ \sqrt{2/5}\sqrt{3}\ \hbar\mid 3/2\ \ 1/2 \rangle)\mid 1\ \ \ 0\ \rangle+\sqrt{3/5}(\sqrt{4}\ \hbar\mid 3/2\ \ -1/2 \rangle)\mid 1\ \ \ 1\ \rangle[/tex]

    Now, simplifying, I want to combine to two terms:

    [tex] S_{-}\mid 5/2\ \ \ 3/2\ \rangle = 2\sqrt{6/5}\ \hbar\mid 3/2\ \ 1/2 \rangle\mid 1\ \ \ 0\ \rangle+2\sqrt{3/5}\ \hbar\mid 3/2\ \ -1/2 \rangle\mid 1\ \ \ 1\ \rangle[/tex]

    Crossing my fingers. . .
     
  17. Feb 28, 2010 #16

    vela

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    You don't have [itex]S_-^{(2)}[/itex] acting on the right part of the states.
     
  18. Feb 28, 2010 #17
    Yes and no.

    Don't S_^1 |3/2 3/2> and S_^2 |3/2 3/2> essentially equal the same thing?
     
  19. Feb 28, 2010 #18
    Wait are you saying that im not acting on |1 0> for say in the first term?
     
  20. Feb 28, 2010 #19

    vela

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    Yeah. S1 only acts on the first ket in a product, while S2 only acts on the second ket.
     
  21. Feb 28, 2010 #20
    Ok. So let me get this. . .

    S1 acts on the |3/2 3/2> and |3/2 1/2> and the S2 acts on the |1 0> and |1 1>.
     
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