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Angular momentum operator acting on |j,m>

  1. Aug 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]e^{-i \pi J_x} \mid j,m \rangle =e^{-i \pi j} \mid j,-m \rangle[/itex]

    2. Relevant equations
    [itex]J_x \mid j,m \rangle =\frac{\hbar}{2} [\sqrt{(j-m)(j+m+1)} \mid j,m+1 \rangle +[/itex]
    [itex]\sqrt{(j+m)(j-m+1)} \mid j,m-1 \rangle][/itex]

    3. The attempt at a solution
    Expanding [itex]e^{-i \pi J_x}[/itex] to a power series and applying the equation in #2, I come up with an expression with coefficients for [itex]\mid j,m \rangle[/itex], [itex]\mid j,m+1 \rangle[/itex], [itex]\mid j,m-1 \rangle[/itex], [itex]\mid j,m+2 \rangle[/itex], [itex]\mid j,m-2 \rangle[/itex], and so on as far as the quantum number j allows. I will focus on the [itex]\mid j,m \rangle[/itex] term.

    [itex]\left\{ 1+ \sum _{n=1} ^\infty \frac{(i \pi \hbar)^{2n} \left[\sqrt{(j-m)^{2n} (j+m+1)^{2n}} +\sqrt{(j+m)^{2n} (j-m+1)^{2n}}\right]}{2^{n+1} (2n)!} +\sum _{n=1} ^\infty \frac{(i \pi \hbar)^{2n+2} \left[\sqrt{(j-m)^{2n} (j+m+1)^{2n} (j-m-1)^{2n} (j+m+2)^{2n}} +\sqrt{(j+m)^{2n} (j-m+1)^{2n} (j+m-1)^{2n} (j-m+2)^{2n}}\right]}{2^{n+3} (2n+2)!} +... \right\} \mid j,m \rangle[/itex]
    This coefficient goes on for as long as quantum number j allows.

    The right side of the equation I am trying to prove has no operators, so the only non-zero coefficient I should have is the one for [itex]\mid j,-m \rangle[/itex]. This means that the coefficient for [itex]\mid j,m \rangle[/itex] should be 0 for all cases except when [itex]m=0[/itex].

    The above series does converge, but to what it converges seems to depend on j and m. I have manipulated the series a couple different ways, but I have been unable to show that it equals 0 for all values of m except 0. I wonder if I am making this more difficult that it has to be. Any ideas?
     
  2. jcsd
  3. Aug 9, 2012 #2
    Well you could always cheat and notice that all you're doing is turning a spherical harmonic upside-down :smile:
     
  4. Aug 10, 2012 #3
    Your tip led me to realize that it would be simpler to think of it as a rotation applied to [itex]\mid j,m \rangle[/itex]. Then, I can use the Wigner D-matrix to simplify the expression:
    [itex]R(\alpha , \beta , \gamma) \mid j,m \rangle =\sum _{m'=-j} ^j D _{m'm} ^{(j)} (\alpha , \beta , \gamma) \mid j,m' \rangle[/itex]

    [itex]e^{-i \pi J_x}[/itex] can be written as [itex]e^{-i (-\frac{\pi}{2}) J_z} e^{-i \pi J_y} e^{-i \frac{\pi}{2} J_z}=R(-\frac{\pi}{2}, \pi , \frac{\pi}{2})[/itex].

    Now I can use the Wigner D-matrix equation:
    [itex]R(-\frac{\pi}{2}, \pi , \frac{\pi}{2}) \mid j,m \rangle =\sum _{m'=-j} ^j D _{m'm} ^{(j)} (-\frac{\pi}{2} , \pi , \frac{\pi}{2}) \mid j,m' \rangle[/itex]
    [itex]=\sum _{m'=-j} ^j e^{-i(-m' \pi /2 +m \pi /2)} d _{m'm} ^{(j)} \mid j,m' \rangle[/itex]
    [itex]=(-1)^{j-m} \sum _{m'=-j} ^j e^{-i(-m' \pi /2 +m \pi /2)} \delta _{m',-m} \mid j,m' \rangle[/itex]
    [itex]=(-1)^{j-m} e^{-i(m \pi /2 + m \pi /2)} \mid j,-m \rangle[/itex]

    So I'm close now, but it seems like they set [itex]m=j[/itex] to get to [itex]e^{-i \pi j} \mid j,-m \rangle[/itex], and I don't know why that step would be made.
     
  5. Aug 11, 2012 #4

    dextercioby

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    [tex] (-1)^{j-m} e^{-i(m\pi/2+m\pi/2)} = (-1)^{j-m} e^{-im\pi} = (-1)^{j-m} (e^{-i\pi})^m = (-1)^{j-m} (-1)^m = (-1)^{j-m+m} = (-1)^j = (e^{-i\pi})^j = e^{-i\pi j} [/tex]
     
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