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Basis Transformation for Wave Function

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data

    It's not a homework problem. I'm reading my textbook (Sakurai's Modern QM), and I'm not sure about a step (eq 3.6.6 through 3.6.8). Here it is:

    We start with a wave function that's been rotated:
    [tex]\langle x' + y' \delta \phi, y' - x' \delta \phi, z' | \alpha \rangle[/tex]

    Now, we change the coordinate basis from Cartesian to spherical. That is,
    [tex]\langle x', y', z' | \alpha \rangle \to \langle r, \theta, \phi | \alpha \rangle[/tex]

    I want to show that the previous wave function transforms to:
    [tex]\langle r, \theta, \phi - \delta \phi | \alpha \rangle[/tex]

    2. Relevant equations

    See above.

    3. The attempt at a solution

    Physically, I understand that we're rotating by angle [tex]\delta \phi[/tex] about the z-axis. So, it makes sense that the bra would acquire a [tex]-\delta \phi[/tex] (the ket would acquire a positive). But, I'm not sure how to mathematically show the steps (which is needed for more complicated transforms, such as Sakurai eq. 3.6.10 to 3.6.11).
     
  2. jcsd
  3. Dec 18, 2009 #2
    You are essentially acting the rotation operator, [itex]\mathcal{D}[/itex], on the bra in the spherical wavefunction:

    [tex]
    \langle r,\theta,\phi|\mathcal{D}|\alpha\rangle=\langle r,\theta,\phi\left|1-i\left(\frac{\delta\phi}{\hbar}\right)L_z\right|\alpha\rangle=\langle r,\theta,\phi-\delta\phi|\alpha\rangle
    [/tex]

    Mathematically, you are multiplying each term in [itex]\langle r,\theta,\phi|[/itex] by [itex]\mathcal{D}[/itex], but since [itex]L_zr=L_z\theta=0[/itex], you are left with

    [tex]
    \langle r,\theta,\phi|\mathcal{D}=\langle r\mathcal{D},\theta\mathcal{D},\phi\mathcal{D}|\rightarrow\left.\langle r,\theta,\phi\left(1-i\left(\frac{\delta\phi}{\hbar}\right)L_z\right)\right|=\langle r,\theta,\phi-\delta\phi|
    [/tex]

    Equations (3.6.10) and (3.6.11) use the "known result from wave mechanics" in Equation (3.6.9):

    [tex]
    \langle\mathbf{x}'|L_z|\alpha\rangle=-i\hbar\frac{\partial}{\partial\phi}\langle\mathbf{x}'|\alpha\rangle
    [/tex]

    and this is not the same as the second equation above because you do not have the [itex]\propto(1-\delta\phi L_z)[/itex] acting on the bra, you just have [itex]L_x[/itex], [itex]L_y[/itex], or [itex]L_z[/itex]. Converting [itex]L_x[/itex] and [itex]L_y[/itex] from Cartesian into spherical should give you Equations (3.6.10) and (3.6.11) when using the "known result" of Equation (3.6.9)
     
  4. Dec 18, 2009 #3
    Thanks for the help!

    How do you know that [itex]L_zr=L_z\theta=0[/itex]? The only thing I remember about [itex]L_z[/itex] is how it acts on an eigenstate: [itex]L_z|l,m\rangle = m \hbar | l,m \rangle[/itex]. How do you know how it acts on [itex]r[/itex], [itex]\theta[/itex], and [itex]\phi[/itex]?
     
  5. Dec 18, 2009 #4
    Well [itex]L_z[/itex] is an operator equivalent to

    [tex]
    L_z=-i\hbar\frac{\partial}{\partial\phi}
    [/tex]

    So since [itex]r[/itex] and [itex]\theta[/itex] have no [itex]\phi[/itex] component, [itex]L_z[/itex] acting on them would be zero.
     
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