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Finding the magnetic field based on vector potential

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    If you have |[itex]\psi[/itex]>=cos([itex]\theta[/itex]/2)ei*l*[itex]\varphi[/itex]/2+sin([itex]\theta[/itex]/2)e-i*l*[itex]\varphi[/itex]/2ei[itex]\varphi[/itex]
    and A=<[itex]\psi[/itex]|[itex]\partial[/itex][itex]\varphi[/itex]|[itex]\psi[/itex]>[itex]\hat{r}[/itex]
    find B in polar coordinates
    2. Relevant equations

    B=[itex]\nabla[/itex]xA


    3. The attempt at a solution

    So far I got B=[itex]\frac{1}{r}[/itex][-iei[itex]\varphi[/itex]([itex]\frac{l(l-1)}{4}[/itex]e(l-1)-[itex]\frac{(1-l)(1-\frac{l}{2})}{2}[/itex]e(1-l))[itex]\hat{\theta}[/itex]-([itex]\frac{l}{2}[/itex]cos([itex]\theta[/itex]/2)sin([itex]\theta[/itex]/2)+[itex]\frac{l}{4}[/itex]sin2([itex]\theta[/itex]/2)ei[itex]\varphi[/itex](l-1)-[itex]\frac{l}{4}[/itex]cos2([itex]\theta[/itex]/2)ei[itex]\varphi[/itex](l-1)+([itex]\frac{1}{2}[/itex]cos2([itex]\theta[/itex]/2)-[itex]\frac{1}{2}[/itex]sin2([itex]\theta[/itex]/2)-[itex]\frac{l}{4}[/itex]cos2([itex]\theta[/itex]/2)+[itex]\frac{l}{4}[/itex]sin2([itex]\theta[/itex]/2)ei[itex]\varphi[/itex](1-l)-cos([itex]\theta[/itex]/2)sin([itex]\theta[/itex]/2)+[itex]\frac{l}{2}[/itex]sin([itex]\theta[/itex]/2)cos([itex]\theta[/itex]/2))[itex]\hat{\varphi}[/itex]]

    but I was told that the answer is l+1. How do I simplify this equation?
     
  2. jcsd
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