Basis Transformation for Wave Function

1. Dec 18, 2009

Proofrific

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:

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

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:
$$\langle r, \theta, \phi - \delta \phi | \alpha \rangle$$

2. Relevant equations

See above.

3. 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).

2. Dec 18, 2009

jdwood983

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

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

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

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

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

$$\langle\mathbf{x}'|L_z|\alpha\rangle=-i\hbar\frac{\partial}{\partial\phi}\langle\mathbf{x}'|\alpha\rangle$$

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)

3. Dec 18, 2009

Proofrific

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 $\phi$?

4. Dec 18, 2009

jdwood983

Well $L_z$ is an operator equivalent to

$$L_z=-i\hbar\frac{\partial}{\partial\phi}$$

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