Angular Momentum and Rotations

• What comes to mind when thinking about angular momentum?
  • Rotations
  • \(\vec{L}=\vec{r}\times \vec{p}\)
  • A quantity that is conserved in systems (with no external torque) with invariance under rotations

1. Geometrical Rotations
2. Rotations of state vectors \(|\Psi\rangle\Rightarrow|\Psi'\rangle=\mathcal{D}(R)|\Psi\rangle\)
3. \(\mathcal{D}(R)\Leftrightarrow J\), the generalized angular momentum

Rotate about \(z\) axis by \(\varphi\), we can use an active rotation (xyz coordinate system remain the same), \(\vec{r}\to\vec{r}'\) with an angle in the xy-plane of \(\varphi\). passive rotation (xyz coordinate system changes), \(\vec{r}\to\vec{r}'\) with the coordinate \(x',y'\) have an angle \(\varphi\) between them (\(-\varphi\) between \(\vec{r}\) and \(\vec{r}'\)).

We will use active rotations.

We accomplish this with \(\vec{r}' = R\vec{r}\) with \(R\) being a 3x3 orthogonal matrix. Orthogonal \((R^TR=RR^T=\mathbb{I})\) is due to \(|\vec{r}'|=|\vec{r}|\).

Then, \(R(\varphi)=\begin{pmatrix} \cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{pmatrix}\).

\(R_x(\varphi)=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\varphi & -\sin\varphi \\ 0 & \sin\varphi & \cos\varphi \end{pmatrix}\).

\(R_y(\varphi)=\begin{pmatrix} \cos\varphi & 0 & \sin\varphi \\ 0 & 1 & 0 \\ -\sin\varphi & 0 & \cos\varphi \end{pmatrix}\).

Recall that rotations are non-commutative \(R_i(\varphi)R_j(\varphi)\neq R_j(\varphi')R_i(\varphi)\).

For a small angle \([R_x(\varepsilon),R_y(\varepsilon)]=\begin{pmatrix} 0 & -\varepsilon^2 & 0 \\ \varepsilon^2 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}=R_z(\varepsilon^2)-\mathbb{I}\to0\).

\(R_i(\varphi+\varepsilon) = R_i(\varphi)R_i(\varepsilon) = R_i(\varepsilon)R_i(\varphi)\).

\(O(3)\) is the group of all 3x3 orthogonal matrices.

\(\det R=+1\) implies rotations. \(\det R=-1\) implies parity.

\(SO(3)\) gives the rotations in 3 dimensions.

Abelian groups are commutative. Non-abelian groups are non-commutative (SO(3)).

\(|\Psi\rangle \Rightarrow |\Psi'\rangle=\mathcal{D}(R)|\Psi\rangle\). Where \(\mathcal{D}\) stands for Drehung rotation.

\(\mathcal{D}\mathcal{D}^\dagger\). Remember generators for time, \(U_\varepsilon = 1-iG\varepsilon\) with \(G=\frac{H}{\hbar}\) \(\varepsilon=dt\). For space, \(G=\frac{p_x}{\hbar}\), \(\varepsilon=dx\). For rotations, \(G=\frac{J_z}{\hbar}=\frac{\vec{J}\cdot\vec{n}}{\hbar}\) \(\varepsilon=d\varphi\). For finite rotations, \(\mathcal{D}(\vec{n},\varphi)=\exp\left(\frac{-i}{\hbar}\vec{J}\cdot\vec{n}d\varphi\right)\)

Remember, \(A' = UAU^\dagger\). So, \(A' = \mathcal{D}(R)A\mathcal{D}^\dagger(R)\).

Remember, if a generator commutes with the observable, then \(A\) is conserved under the transformation. I.e. \(A'=\mathcal{D}(R)A\mathcal{D}^\dagger(R) \approx A-\frac{i}{\hbar}d\varphi[\vec{J}\cdot\vec{n},A]\). If this is \(A\) then \(A\) is conserved under the transformation and it is also called a scalar.

\(A'= A + \left[-\frac{i}{\hbar}(\vec{J}\cdot\vec{n})d\varphi,A\right] + \frac{1}{2!} \left[-\frac{i}{\hbar}(\vec{J}\cdot\vec{n})d\varphi,\left[-\frac{i}{\hbar}(\vec{J}\cdot\vec{n})d\varphi,A\right]\right] + \cdots\).

Let us consider the Hamiltonian. Let \(|\Psi(t_0)\rangle\) be transformed by \(\mathcal{D}(R)\) to \(|\Psi'(t_0)\rangle=\mathcal{D}(R)|\Psi(t_0)\rangle\) then be transformed by \(|\Psi'(t_0+dt)\rangle\).

Consider the opposite order, \(|\Psi(t_0)\rangle\) be time transformed \(|\Psi(t_0+dt)\rangle\) then rotated \(|\Psi''(t_0+dt)\rangle\).

\(\Psi'(t_0+dt)\rangle = |\Psi'(t_0)\rangle + dt\left.\frac{d|\Psi'\rangle}{dt}\right|_{t_0} = |\Psi'(t_0)\rangle + \left.\frac{dt}{i\hbar}\hat{H}|\Psi'\rangle\right|_{t_0} = \mathcal{D}(R)|\Psi(t_0)\rangle + \frac{dt}{i\hbar}\hat{H}\mathcal{D}(R)|\Psi(t_0)\rangle\).

\(|\Psi(t+dt)\rangle = |\Psi(t_0)\rangle + \frac{dt}{i\hbar}\hat{H}|\Psi(t_0)\rangle\). So, \(\mathcal{D}(R)|\Psi(t+dt)\rangle = \mathcal{D}(R)|\Psi(t_0)\rangle + \frac{dt}{i\hbar}\mathcal{D}(R)\hat{H}|\Psi(t_0)\rangle\). If we expect the states to be the same, then the Hamiltonian commutes with the rotation operator. If the states are different, then the Hamiltonian does not commute.

So, \([\hat{H},\vec{J}\cdot\vec{n}]=0\Rightarrow \hat{H}\) is a scalar.

Recall: \(\frac{d\langle A\rangle}{dt} = \langle\frac{\partial A}{\partial t}\rangle + \frac{1}{i\hbar}\langle[A,H]\rangle\).

If \(A=\vec{J}\) then \(\frac{d\langle J\rangle}{dt}=\frac{1}{i\hbar}\langle[\vec{J},H]\rangle=0\) implies that \(\vec{J}\) is a constant of the motion.

\(\mathcal{D}^{(s)}(\vec{n},\varphi) = \exp\left(-\frac{i}{\hbar}\varphi\vec{S}\cdot\vec{n}\right)\)

For \(s=\frac{1}{2}\), \(\vec{S}=\frac{\hbar}{2}\vec{\sigma}\), \(\mathcal{D}^{(s)}(\vec{n},\varphi) = \exp\left(-i\frac{\varphi}{2}\hat{\vec{\sigma}}\cdot\vec{n}\right) \approx \mathbb{I}\cos\frac{\varphi}{2}-i(\vec{\sigma}\cdot\vec{n})\sin\frac{\varphi}{2}\). (Not sure if actually approximate or exact). \(\hat{\vec{\sigma}}\cdot\vec{n}=\sigma_xn_x+\sigma_yn_y+\sigma_zn_z=\begin{pmatrix} n_z & n_x-in_y \\ n_x+in_y & -n_z \end{pmatrix}\). \(\exp(\cdots)=\begin{pmatrix} \cos\frac{\varphi}{2}-in_z\sin\frac{\varphi}{2} & -i\sin\frac{\varphi}{2}(n_x-in_y) \\ -i\sin\frac{\varphi}{2}(n_x+in_y) & \cos\frac{\varphi}{2}+in_z\sin\frac{\varphi}{2} \end{pmatrix}\)

For \(\varphi=2\pi\), \(\mathcal{D}^{(s)} = -\mathbb{I}\). So \(\mathcal{D}^{(s)}|\alpha\rangle=-|\alpha\rangle\). For \(\varphi=4\pi\), \(\mathcal{D}^{(s)} = \mathbb{I}\).

Thus, we get SU(2).

An arbitrary rotation of a rigid body can be accomplished in 3 rotations, Euler angles \(\alpha,\beta,\gamma\). \(R(\alpha,\beta,\gamma)=R_{z'}(\gamma)R_{y'}(\beta)R_z(\alpha)\) From Goldstein, \(R(\alpha,\beta,\gamma) = R_z(\alpha)R_y(\beta)R_z(\gamma)\)

For spin space, where \(s=\frac{1}{2}\), \(\mathcal{D}(\alpha,\beta\gamma) = \mathcal{D}_z(\alpha) \mathcal{D}_y(\beta) \mathcal{D}_y(\gamma)=\exp\frac{-i\sigma_z\alpha}{2}\exp\frac{-i\sigma_y\beta}{2}\exp\frac{-i\sigma_z\gamma}{2} = \begin{pmatrix} \exp\frac{-i\alpha}{2} & 0 \\ 0 & \exp\frac{i\alpha}{2} \end{pmatrix}\cdots = \begin{pmatrix} \exp\left(\frac{-i(\alpha+\gamma)}{2}\right)\cos\frac{\beta}{2} & \exp\left(\frac{-i(\alpha-\gamma)}{2}\right)\sin\frac{\beta}{2} \\ \exp\left(\frac{i(\alpha-\gamma)}{2}\right)\sin\frac{\beta}{2} & \exp\left(\frac{i(\alpha+\gamma)}{2}\right)\cos\frac{\beta}{2} \\ \end{pmatrix}\) called the \(j=\frac{1}{2}\) irreducible representation of the rotation operator.

\(\mathcal{D}_{m_s'm_s}^{(1/2)}(\alpha,\beta,\gamma)=\langle \frac{1}{2}m_s'|\exp(-iS_z\alpha/\hbar)\exp(-iS_y\beta/\hbar)\exp(-iS_z\gamma/\hbar)|\frac{1}{2}m_s\rangle\)

By replacing \(S_i\) with \(J_i\) we get a more general expression, \(\mathcal{D}_{m_j'm_j}^{(j)}=\langle jm_j'|\exp\left(\frac{-i}{\hbar}J_z\alpha\right)\exp\left(\frac{-i}{\hbar}J_y\beta\right)\exp\left(\frac{-i}{\hbar}J_z\gamma\right)|jm_j\rangle\).

Question:

• \(\mathcal{D}|j,m\rangle \Leftrightarrow c|j,m'\rangle\)
• Why do we not go to some \(j'\)?

Generalized rotation operator matrix element, \(\mathcal{D}_{m'm}^{(j)}(R) = \langle jm'|\exp\left(-\frac{i}{\hbar}\vec{J}\cdot\vec{n}\varphi\right)|jm\rangle\)

Remember that \([\vec{J}^2,\vec{J}] = 0\) so \([\vec{J}^2,\mathcal{D}(R)] = 0\). Thus, the \(\vec{J}^2\) eigenvalue is unchanged under \(\mathcal{D}(R)\).

I.e. \(\vec{J}^2\mathcal{D}(R)|jm\rangle = \vec{J}^2|\Psi\rangle\), \(\mathcal{D}(R)\vec{J}^2|jm\rangle = \mathcal{D}(R)\hbar^2j(j+1)|jm\rangle\) A zero commutator implies that \(|\Psi\rangle = \mathcal{D}(R)|jm\rangle\). Thus, \(m\) may change but not \(j\).

Inserting Identity, \(\mathcal{D}(R)|jm\rangle = \sum_{m'}|jm'\rangle\langle jm'|\mathcal{D}(R)|jm\rangle = \sum_{m'}\mathcal{D}_{m'm}^{(j)}(R)|jm'\rangle\). This gives the weights for each of the eigenstates. I.e. the probabilities to end up in the particular state.

Irreducible: The general rotation matrix has the sub rotation matrices on the diagonals, but due to the zeroes on the diagonals separating them, they cannot go between the \(j\) bases. I.e. there is no crossover.

\(\mathcal{D}_{m'm}^{(j)}(R)\) is a \((2j+1)\times(2j+1)\) matrix, \((2j+1)-\) dim irreducible representation of \(\mathcal{D}(R)\) implies that any \(|jm\rangle\) can be rotated by \(\mathcal{D}(R)|jm\rangle\) to yield \(|jm'\rangle\) which then span the entire \(\mathcal{E}_j\).

Returning to the matrix elements:

\begin{align*} \mathcal{D}_{m_j'm_j}^{(j)}(\alpha,\beta,\gamma) & =\langle jm_j'|\exp\left(-\frac{i}{\hbar}J_z\alpha\right)\exp\left(-\frac{i}{\hbar}J_y\beta\right)\exp\left(-\frac{i}{\hbar}J_z\gamma\right)|jm_j\rangle \\ & =\exp(-im'\alpha)\langle jm_j'|\exp\left(-\frac{i}{\hbar}J_y\beta\right)\exp(-im\gamma)|jm_j\rangle \\ &= \exp(-im'\alpha+m\gamma)d_{m'm}^{(j)} \end{align*}

Using the ladder operators, \(J_y = \frac{J_+-J_-}{2i}\), \(J_{\pm}=\hbar\sqrt{j(j+1)-m(m\pm 1)}|jm\pm1\rangle\), \(\langle 1m'|J_y|1m\rangle\Rightarrow \begin{pmatrix} 0 & -i\sqrt{2} & 0 \\ i\sqrt{2} & 0 & -i\sqrt{2} \\ 0 & i\sqrt{2} & 0 \end{pmatrix}\). Taylor expanding the exponential,

\begin{align*} \exp\left(-\frac{i}{\hbar}J_y\beta\right)&=\mathbb{I}-\frac{J_y^2}{\hbar^2}(1-\cos\beta)-i\frac{J_y}{\hbar}\sin\beta. \end{align*}

So,

\begin{align*} j = 1\Rightarrow d^{(1)}(\beta) &= \begin{pmatrix} \frac{1+\cos\beta}{2} & -\frac{1}{\sqrt{2}}\sin\beta & \frac{1}{2}(1-\cos\beta) \\ \frac{\sqrt{2}}{2}\sin\beta & \cos\beta & \frac{-\sqrt{2}}{2}\sin\beta \\ \frac{1}{2}(1-\cos\beta) & \frac{\sqrt{2}}{2}\sin\beta & \frac{1}{2}(1+\cos\beta) \end{pmatrix}. \end{align*}

Since the overall rotation matrix is unitary, we expect the sub matrix d to be orthogonal due to the fact that it is real. \(d^{(j)}(0) = \mathbb{I}\). Sum of the squares of the rows or columns is 1.

Recall: \(D(R) = D_z(\alpha)D_y(\beta)D_z(\gamma)\), \(d_{m'm}^{(j)}(\beta) = \langle jm'|\exp(-i\beta J_y/\hbar)|jm\rangle\).

How does \(d_{m'm}^{(j)}\Leftrightarrow d_{m'm}^{(j_1)}(\beta),d_{m'm}^{(j_2)}\).

\(d_{m'm}^{(j)} = \langle jm'|\exp(-i\beta J_y/\hbar)|jm\rangle = \sum_{m_1m_2}\sum_{m_1'm_2'}\langle jm'|j_1j_2m_1m_2\rangle\langle j_1j_2m_1m_2|\exp(-i\beta(J_{y_1}+J_{y_2})/\hbar)|j_1j_2m_1'm_2'\rangle\langle j_1j_2m_1'm_2'|jm\rangle = \sum_{m_1m_2}\sum_{m_1'm_2'}\langle jm'|j_1j_2m_1m_2\rangle\langle j_1j_2m_1'm_2'|jm\rangle\langle j_1j_2m_1m_2|\exp(-i\beta J_{y_1}/\hbar)|j_1j_2m_1'm_2'\rangle\langle j_1j_2m_1m_2|\exp(-i\beta J_{y_2}/\hbar)|j_1j_2m_1'm_2'\rangle = \sum_{m_1m_2}\sum_{m_1'm_2'}\langle jm'|j_1j_2m_1m_2\rangle\langle j_1j_2m_1'm_2'|jm\rangle d_{m_1'm_1}^{(j_1)}(\beta)d_{m_2'm_2}^{(j_2)}(\beta)\).

Inverting the expression, \(d_{m_1'm_1}^{(j_1)}(\beta)d_{m_2'm_2}^{(j_2)}(\beta) = \sum_{j=|j_1-j_2|}^{j_1+j_2}\sum_{mm'}\langle j_1j_2m_1m_2|jm\rangle\langle j_1j_2m_1'm_2'|jm'\rangle d_{m'm}^{(j)}(\beta)\).

We can replace the small \(d(\beta)\) with \(\mathcal{D}(\alpha,\beta,\gamma)\) for the same relationship.

Note that this is a reducible representation for \(\mathcal{D}^{(j_1)}\otimes\mathcal{D}^{(j_2)}\) since there are many different \(j\) s. We get a direct sum of irreducible: \(\mathcal{D}^{(j_1+j_2)}\oplus\cdots\oplus\mathcal{D}^{|j_1-j_2|}\)

Recall: \(\exp(A+B) = \exp(A)\exp(B)\) iff they commute.