Phy5645/Angular Momentum Problem 1

(a)

${\hat {R}}_{\Delta \phi }f=\left[\exp \left(\Delta \phi {\frac {\partial }{\partial \phi }}\right)\right]f$

$=f(\phi )+\Delta \phi {\frac {\partial f}{\partial \phi }}+{\frac {\left(\Delta \phi \right)^{2}}{2}}{\frac {\partial ^{2}f}{\partial \phi ^{2}}}+\cdots =f\left(\phi +\Delta \phi \right).$

(b) Let $\delta \phi \!$ be an infinitesimal angle so that $\Delta \phi =n\delta \phi \!$ in the limit that $n>>1\!$. For the infinitesimal rotation

$\mathbf {r} '=\mathbf {r} +\delta \mathbf {r} =\mathbf {r} +\delta {\vec {\phi }}\times \mathbf {r}$

so that

$f\left(\mathbf {r} +\delta \mathbf {r} \right)=f(\mathbf {r} )+\delta {\vec {\phi }}\times \mathbf {r} \cdot \mathbf {\nabla } f(\mathbf {r} )$

$=f(\mathbf {r} )+\delta {\vec {\phi }}\cdot \mathbf {r} \times \mathbf {\nabla } f(\mathbf {r} )$

$=f(\mathbf {r} )+{\frac {i}{\hbar }}\delta {\vec {\phi }}\cdot \mathbf {r} \cdot \mathbf {\hat {p}} f(\mathbf {r} )$

$=f(\mathbf {r} )+{\frac {i}{\hbar }}\delta {\vec {\phi }}\cdot \mathbf {\hat {L}} f(\mathbf {r} )$.

In the Taylor series expansion of $f\left(\mathbf {r} +\delta \mathbf {r} \right)\!$ above we have only kept terms of $O\left(\delta \phi \right)\!$. [The expression $\delta \mathbf {r} =\delta {\vec {\phi }}\times \mathbf {r} \!$ is valid only to terms of $O\left(\delta \phi \right)\!$.] In this manner we obtain

$f\left(\mathbf {r} +\delta \mathbf {r} \right)=\left({\hat {I}}+{\frac {i}{\hbar }}\delta {\vec {\phi }}\cdot \mathbf {\hat {L}} \right)f(\mathbf {r} )={\hat {R}}_{\delta {\vec {\phi }}}f(\mathbf {r} )$

For a finite rotational displacement through the angle $\mathbf {\Delta } {\vec {\phi }}=n\delta {\vec {\phi }}\!$, we apply the operator ${\hat {R}}_{\delta {\vec {\phi }}}\!$, $n\!$ times:

${\hat {R}}_{n\delta {\vec {\phi }}}=\left({\hat {R}}_{\delta {\vec {\phi }}}\right)^{n}=\left({\hat {I}}+{\frac {i}{\hbar }}\delta {\vec {\phi }}\cdot \mathbf {\hat {L}} \right)^{n}$

and pss to the limit $n\rightarrow \infty \!$ or, equivalently, $\Delta \phi /\delta \phi \rightarrow \infty \!$.

${\hat {R}}_{\Delta \phi }=\lim _{\Delta \phi /\delta \phi \rightarrow \infty }\left({\hat {I}}+{\frac {i}{\hbar }}\delta {\vec {\phi }}\cdot \mathbf {\hat {\mathbf {L} }} \right)^{\Delta \phi /\delta \phi }=e^{i\Delta {\vec {\phi }}\cdot \mathbf {\hat {\mathbf {L} }} \hbar }$.

The operator ${\hat {R}}_{\delta {\vec {\phi }}}\!$ rotates $\mathbf {r} \!$ to $\mathbf {r} +\delta {\vec {\phi }}\times \mathbf {r} \!$ with respect to a fixed coordinate frame. If, on the other hand, the coordinate frame is rotated through $\delta {\vec {\phi }}\!$ with $\mathbf {r} \!$ fixed in space, then in the new coordinate frame this vector has the value $\mathbf {r} -\delta {\vec {\phi }}\times \mathbf {r} \!$. Thus, rotation of coordinates through $\delta {\vec {\phi }}\!$ is generated by the operator ${\hat {R}}_{-\delta {\vec {\phi }}}.$

Back to Angular Momentum as a Generator of Rotations in 3D