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Trivial spin problem

  1. Apr 19, 2005 #1
    A trivial problem, but I am stuck.

    Prove that
    [tex]e^{i\pi S_y}|S\ 0\rangle = (-1)^S |S\ 0\rangle[/tex]

    I proved the S = 1 case, by expanding [tex]|S\ 0\rangle[/tex] in the basis of [tex]S_y[/tex]'s eigenvectors. How to do for general case?
  2. jcsd
  3. Apr 19, 2005 #2
    i think its something like this
    you have Sy|S k> = -S+k for k=0,1,...,2S+1
    exp(i*pi*Sy)|S k> = exp(i*pi*(-S+k))|S k>


    exp(i*pi*(-S+k)) = exp(-i*pi*S)*exp(i*pi*k) if k = 0 then

    exp(i*pi*Sy)|S 0> = exp(i*pi*(-S))|S 0> = cos(Pi*S)|S 0> and cos(Pi*S) = (-1)^S
  4. Apr 19, 2005 #3
    What did you mean?

    [tex]|S\ k\rangle[/tex] is eigenstate of [tex]S_z[/tex] not [tex]S_y[/tex]
  5. Apr 19, 2005 #4


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    U can write

    [tex]\hat{S}_{y}=\frac{1}{2i}\left(\hat{S}_{+}-\hat{S}_{-}\right) [/tex]

    and u know the action of the lowering & rising ladder operators on the standard basis [itex] |s,m_{s}\rangle [/itex]

  6. Apr 20, 2005 #5
    your correct, didnt think about that :)
  7. Apr 22, 2005 #6
    Sure, I used that trick and proved the S=1 case (3 dimensional matrices)

    For general case. The problem is, we are dealing with the exponentiation of the operator, and the dimension of the matrix can be very large.
  8. Apr 22, 2005 #7


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    The dimension of the matrices will always be 2...

  9. Apr 23, 2005 #8
    :confused: :confused: :confused:
    d = 2 is spin half case.
  10. Apr 24, 2005 #9


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    I'm sorry.It can't be 2.The dimension of the matrix is [itex] \left(2s+1,2s+1\right) [/itex],where "s" is an integer (because of the "0" value of [itex] m_{s} [/itex]).

    I'm getting 0.I dunno why.Here's what i do.

    [tex] e^{i\pi\hat{S}_{y}}=\mathcal{D}(0,-\pi,0) [/tex]

    [tex] \mathcal{D}(0,-\pi,0)|s,0\rangle =d^{(j)}_{0,0}(-\pi) |s,0\rangle [/tex]

    Now,using Wigner's formula

    [tex] d^{(j)}_{m,m'}(\beta)=\left[\frac{(j+m)!(j-m)!}{(j+m')!(j-m')!}\right]^{1/2} \sum_{t}\left(\begin{array}{c} j+m'\\j-m-t\end{array}\right)\left(\begin{array}{c} j-m'\\t\end{array}\right) (-1)^{j-m'-t}\left(\cos\frac{\beta}{2}\right)^{2l+m+m'}\left(\sin\frac{\beta}{2}\right)^{2j-2t-m-m'} [/tex]

    for the case [itex] j=s,m=m'=0,\beta=-\pi [/itex]

    ,i get 0...:surprised:

    Tell me what u did...And how.

  11. Apr 24, 2005 #10


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    I found a solution and it looks as something rather fishy happens here...I thought i applied the Wigner formula correctly,but there are books which prove me wrong...Galindo & Pascual,first volume,page ~210 make a discussion on this issue...

    I can post it,if u don't have the book.

    Something is really weird.I saw that the Wigner functions in Sakurai are different than the ones taken by Constantinescu & Magyari,but they both would lead to 0 instead of [itex] (-1)^{j} [/itex]...

  12. Apr 25, 2005 #11
    OMG, it's an awesome formula, I had never seen it before

    I have exam tomorrow, will give you a reply after that.
  13. Apr 25, 2005 #12


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    It's the form i found in Constantinescu & Magyari book.They're both Romanian & i'm a patriot.:approve:

  14. May 3, 2005 #13
    :grumpy: I checked the formula in the book by Constantine you said
    okay, cosine 90 is zero, so the only surviving term in the binomial series is when the power of the cosine is zero.

    I have
    [tex]\langle s\ m | D(0, -\pi, 0) | s\ 0 \rangle = \frac{\sqrt{(s+m)!(s-m)!}}{s!}\ ^sC_{s-m/2}\ ^sC_{-m/2}(-1)^{s+m/2}[/tex]

    [tex]\ ^sC_{s-m/2}[/tex] makes sense only if [tex]0 \leq m \leq 2s[/tex]
    [tex]\ ^sC_{-m/2}[/tex] makes sense only if [tex]-2s \leq m \leq 0[/tex]

    so the only non-zero term is when [tex]m = 0[/tex].

    and the coefficient is [tex](-1)^s[/tex]

    Is that the correct way to use Wigner's formula?
  15. May 5, 2005 #14


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    How did u get that [itex] m/2 [/itex] in the exponent and the binomial coefficients...?

    Last edited: May 5, 2005
  16. May 7, 2005 #15
    I used Wigner's formula to compute
    [tex]\langle s\ m | D(0, -\pi, 0) | s\ 0 \rangle[/tex]

    there is a non-zero term in the series (the term without any cosine)
  17. May 7, 2005 #16


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    [tex] \langle s,m|\mathcal{D}\left(0,-\pi,0\right)|s,0\rangle=d^{(s)}_{m,0}(-\pi)\delta_{m,0} [/tex]

    [tex] d^{(s)}_{m,0}(-\pi)=\frac{\sqrt{(s+m)!(s-m)!}}{s!}\sum_{t}\left(\begin{array}{c}s\\s-m-t\end{array}\right)\left(\begin{array}{c}s\\t\end{array}\right)(-1)^{s-t}\left[\cos\left(-\frac{\pi}{2}\right)\right]^{2s+m}\left[\sin\left(-\frac{\pi}{2}\right)\right]^{2s-2t-m} [/tex]

    And you impose that the power of the 0 (of the cosine) be 0.

    [tex] 2s+m=0\Rightarrow m=-2s [/tex]

    But on the other hand,[itex]m=0 [/itex] from the states orthonormalization...

    Hmmm...I'm getting that [itex] (-1)^{s-t} [/itex] with a summation after "t" and an "s=0"...Which gives "+1".

    What do you think?

  18. May 8, 2005 #17
    something is wrong with your formula
    your power of the cosine should contain t, the dummy index
  19. May 8, 2005 #18


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    Yes,you're right,there was no typo in my book,just in my formulas...:redface:

    So here goes.To prevent a [itex] 0^{0}[/itex],i'll write it again,for an arbitrary [itex] \beta [/itex] and then take it [itex] -\pi [/itex].

    [tex] \langle s,0|\mathcal{D}\left(0,\beta,0\right)|s,0\rangle=d^{(s)}_{0,0}(\beta) [/tex] (1)


    [tex] d^{(s)}_{0,0}(\beta)=\sum_{t}\left(\begin{array}{c}s\\s-t\end{array}\right)\left(\begin{array}{c}s\\t\end{array}\right)\left(-1\right)^{s-t}\left(\cos\frac{\beta}{2}\right)^{2t}\left(\sin\frac{\beta}{2}\right)^{2s-2t} [/tex] (2)

    Okay.Now,i'm interested in canceling the exponent of cosine,since that "cosine",when evaluated on a rotation of [itex] -\pi [/itex],would yield 0,and so i need no cosine term in the sum after "t".


    [tex] 2t=0\Leftrightarrow t=0 [/tex] (3)

    and the sum is reduced to only one term,the one corresponding to (3).

    [tex] d^{(s)}_{0,0}(-\pi)=\left(\begin{array}{c}s\\s\end{array}\right)\left(\begin{array}{c}s\\0\end{array}\right)(-1)^{s}(-1)^{2s}=(-1)^{s} [/tex] (4)


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