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:

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

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)

 

[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

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

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