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I am currently working on Angular Momentum part of Gasiorowicz's QM. The writer says that

\begin{align} L_z |l, m\rangle &= \hbar m|l, m\rangle \\ \Rightarrow \langle\theta,\phi|L_z|l, m\rangle&=\hbar m\langle\theta,\phi|l, m\rangle \\ L_z &=\frac{\hbar}{i}\frac{\partial}{\partial \phi} \\ \Rightarrow \langle\theta,\phi|L_z|l, m\rangle&= \frac{\hbar}{i}\frac{\partial}{\partial \phi}\langle\theta,\phi|l, m\rangle \end{align} The first two equality is easily understood but I don't understand why $\frac{\partial}{\partial \phi}$ can just move to the left of $ \langle\theta,\phi|$ though they were originally in the center.

Qmechanic
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Lee Seungwoo
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1 Answers1

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We know that $L_z$ is the generator of infinitesimal rotation about $z$-axis, Thus $$\langle r,\theta,\phi|\left[1-i\left(\frac{\delta \phi}{\hbar}\right)L_z\right]|lm\rangle =\langle r,\theta,\phi-\delta \phi|lm\rangle $$ $$=\langle r,\theta,\phi|lm\rangle -\delta \phi\frac{\partial}{\partial \phi}\langle r,\theta,\phi|lm\rangle $$ $$\langle r,\theta ,\phi|L_z|lm\rangle =-i\hbar \frac{\partial}{\partial\phi}\langle r,\theta,\phi|lm\rangle $$

Note that: $$L_z=-i\hbar \frac{\partial }{\partial\phi}\ \ \ \text{Abuse of notation}$$ $$L_z\rightarrow -i\hbar \frac{\partial }{\partial\phi}\ \ \ \ \ \ \ \text{In Position basis}$$

Young Kindaichi
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  • Ahh okay. So, the state where $|lm\rangle$ is rotated by $\delta \phi$ is $\left[1-i\left(\frac{\delta \phi}{\hbar}\right)L_z\right]|lm\rangle$. Hence in coordinate space, that state is the same as $\langle r,\theta,\phi-\delta \phi|lm\rangle$, if viewed as something like 'projection to the coordinate space'. Thank you. – Lee Seungwoo May 11 '21 at 05:51