Chapter 6. Symmetries and Conservation Laws - (5)

6.5 Rotational Symmetry

6.5.1 Rotations about the z-axis.

A rotation operator, as we can see, it is an operator that rotates a function about the z-axis by an angle \[\varphi\] 

:\[\hat{R}\psi(r,\theta,\phi)=\phi'(r,\theta,\phi)=\phi'(r,\theta,\phi-\varphi)\]

 

And we have already defined z-component rotation operator 

Rotation operator \[\hat{R]_{z}\] is closely related to the z component of angular momentum 

\[L_{z}=-i\hbar \frac{\partial }{\partial \phi}\]

 

For the same reason that led to \[\hat{R}_{z}(\phi)=exp\left [ \frac{-i\varphi}{\hbar} \hat{L}_{z} \right ]\]

\[\hat{L}_{z}\]

So we say that \[L_{z}\] is the generator of rotations about the z-axis. (Let's compare this with \[\hat{T}(a)=exp[\frac-{ia}{\hbar}\hat(p}]-(6.3)\] (We called this "generator" of translations in 6.2)

 

How does the operator \[\hat{r}\] transform under rotations? 

Consider the infinitesimal form of the operator. \[\hat{T}(a)\simeq \frac{-ia}{\hbar}\hat{p}\]

How do the operators \[\hat{r}\] and \[\hat{p}\]

 

Then the operator \[\hat{x}\] transforms as \[\hat{R}^{\dagger}\hat{x}\hat{R}\simeq\left ( 1+\frac{ia}{\hbar}\hat{L}_{z} \right )\hat{x}\left ( 1-\frac{ia}{\hbar}\hat{L}_{z} \right )\]

\[\hat{R}^{\dagger}\hat{x}\hat{R}\simeq\left ( 1+\frac{ia}{\hbar}\hat{L}_{z} \right )\hat{x}\left ( 1-\frac{ia}{\hbar}\hat{L}_{z} \right )\simeq\hat{x}+\frac{i\delta}{\hbar}\left [ \hat{L}_{z},\hat{x} \right ]\simeq\hat{x}-\delta\hat{y}\]

\[\hat{z}'=\hat{z}\]

 

\[\begin{pmatrix} \hat{x}'\\ \hat{y}' \\ \hat{z}' \end{pmatrix}=\begin{pmatrix}1 & -\delta & 0\\ \delta & 1 & 0\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} \hat{x}\\ \hat{y} \\ \hat{z} \end{pmatrix}\]

 

And we fix this matrix for rotation

\[\begin{pmatrix} \hat{x}'\\ \hat{y}' \\ \hat{z}' \end{pmatrix}=\begin{pmatrix}cos\varphi & -sin\varphi & 0\\ sin\varphi & cos\varphi & 0\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} \hat{x}\\ \hat{y} \\ \hat{z} \end{pmatrix}\]

 

6.5.2 Rotations in 3D

In 3D, we generalize 

We can combine these results into a matrix equation \[\hat{R}_{z}(\phi)=exp\left [ \frac{-i\varphi}{\hbar} \hat{L}_{z} \right ]\]
into

\[\hat{\mathbf{R}}_{n}(\phi)=exp\left [ \frac{-i\varphi}{\hbar} \mathbf{n}\cdot \hat{\mathbf{L}}\ \right ]\]

 

As we mentioned, angular momentum is the generator of rotations.

Any operator that transforms the same way as the position operator under rotations is called a vector operator.

\[\mathbf{\hat{V}}'=\mathrm{D}\mathbf{\hat{V}}\]

 

We rewrite this in matrix form

\[\begin{pmatrix} \hat{V_{x}}'\\ \hat{V_{y}}' \\ \hat{V_{z}}' \end{pmatrix}=\begin{pmatrix}cos\varphi & -sin\varphi & 0\\ sin\varphi & cos\varphi & 0\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} \hat{V_{x}}\\ \hat{V_{y}} \\ \hat{V_{z}} \end{pmatrix}\]

 

The transformation rule shows this commutation relations between \[\hat{L}_{i}\] and \[\hat{V}_{j}\]

and we use Levi-civita notation.

\[\left [ \hat{L}_{i},\hat{V}_{j} \right ]=i\hbar\epsilon_{ijk}\hat{V}_{k}\]

 

Now we classify operators.

We can classify operators as either scalar or vectors. It is based on their commutation relations with \[L\] and as true or pseudo quantities, based on their commutators with \[\hat{\Pi}\]

The scalar operator is a single quantity that is unchanged by rotations (=commutes with angular momentum operator L)

 

\[\left [ \hat{L}_{i},\hat{f} \right ]=0 \]

Remind that infinitesimal rotation operator is

\[\hat{R}_{\mathbf{n}}(\delta)=1- \frac{-i\delta}{\hbar} \hat{\mathbf{n}} \hat{\mathbf{L}}\]

 

From the commutation relation between hamiltonian and Angular momentum, we use the Ehrenfest theorem again.

\[\frac{\mathrm{d} \left \langle \mathbf{L} \right \rangle}{\mathrm{d} x}=\frac{i}{\hbar}\left \langle \left [ \hat{H},\mathbf{\hat{L}} \right ] \right \rangle=\mathbf{0}\]

Used the Ehrenfest theorem for central potential

And we conclude "Angular momentum is conserved due to rotational invariance"

 

Since the Hamiltonian for a central potential commutes with all three components of angular momentum. it also commutes with \[\hat{L}\] too.

Operator \[\hat{H},\hat{L}^2,\hat{L}_z\] form a complete set of compatible observables for bound states of a central potential. We rewrite this with this.

 

\[\left [ \hat{H},\hat{L}^2 \right ]=0\] \[\left [ \hat{H},\hat{L}_z \right ]=0\] \[ \left [ \hat{L}_z,\hat{L}^2 \right ]=0\]

 

So that the eigenstates of \[\hat{H}\] can be chosen to be simultaneous eigenstates of \[\hat{L}^2,\hat{L}_z\]

 

\[\hat{H}\psi_{nlm}=E_{n}\psi_{nlm}\] \[\hat{L}_{z}\psi_{nlm}=m\psi_{nlm}\] \[\hat{L}^{2}\psi_{nlm}=l(l+1)\hbar^2\psi_{nlm}\]

 

'They are complete' means that the quantum numbers n,l, and m uniquely specify a bound state of the Hamiltonian with any central potential.

 

 

Reference: Introduction to quantum mechanics (David J. Griffith) 3rd