Basis Transformation for Wave Function

[Proofrific]

Homework Statement

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:
\langle x' + y' \delta \phi, y' - x' \delta \phi, z' | \alpha \rangle

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

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

Homework Equations

See above.

The Attempt at a Solution

Physically, I understand that we're rotating by angle \delta \phi about the z-axis. So, it makes sense that the bra would acquire a -\delta \phi (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).

 

[jdwood983]

You are essentially acting the rotation operator, \mathcal{D}, on the bra in the spherical wavefunction:

<br /> \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<br />

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

<br /> \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|<br />

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

<br /> \langle\mathbf{x}&#039;|L_z|\alpha\rangle=-i\hbar\frac{\partial}{\partial\phi}\langle\mathbf{x}&#039;|\alpha\rangle<br />

and this is not the same as the second equation above because you do not have the \propto(1-\delta\phi L_z) acting on the bra, you just have L_x, L_y, or L_z. Converting L_x and L_y 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)

 

[Proofrific]

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

Thanks for the help!

How do you know that L_zr=L_z\theta=0? The only thing I remember about L_z is how it acts on an eigenstate: L_z|l,m\rangle = m \hbar | l,m \rangle. How do you know how it acts on r, \theta, and \phi?

 

[jdwood983]

Thanks for the help!

How do you know that L_zr=L_z\theta=0? The only thing I remember about L_z is how it acts on an eigenstate: L_z|l,m\rangle = m \hbar | l,m \rangle. How do you know how it acts on r, \theta, and \phi?

Well L_z is an operator equivalent to

<br /> L_z=-i\hbar\frac{\partial}{\partial\phi}<br />

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