# Angular Momentum and Rotations

### Motivating Questions

• 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

### Geometrical Rotations

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

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

#### The set of rotations R constitutes a group

An abstract group is a set of elements $$G$$ and a binary operation $$\cdot$$ such that for $$a,b\in G$$ satisfies:

1. $$a\cdot b\in G$$
2. $$\exists e.a\cdot e = e\cdot a= a$$
3. $$\exists e.\forall a.\exists a^{-1}.a\cdot a^{-1} = e$$
4. The binary operation is associative: $$(a\cdot b)\cdot c=a\cdot(b\cdot c)$$

A group can be finite if the set is finite and infinite if there are infinitely many. The infinite case can be countably infinite or uncountably infinite.

$$\mathbb{Q}$$ is denumerable since it gives an injection to $$(\mathbb{N},\mathbb{Z})$$ which is equivalent to $$\mathbb{N}$$?

• Examples
• $$\mathbb{Z}$$ under multiplication? No, the inverse element does not exist!
• $$\mathbb{Z}$$ under addition? Yes
• Rotations Yes. $$R_n(\varphi)R_m(\varphi')\in$$ rotations. $$R^TR=RR^T=R_n(\varphi)R_m(\varphi')R_m(\varphi')R_n(\varphi)=\mathbb{I}=R_o(0)$$. $$R^T=R^{-1}$$. $$(R_n(\varphi)R_m(\varphi'))R_o(\varphi'')=R_n(\varphi)(R_m(\varphi')R_o(\varphi''))$$.

### Rotation Groups

$$O(3)$$ 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)).

### Rotations in State Space

$$|\Psi\rangle \Rightarrow |\Psi'\rangle=\mathcal{D}(R)|\Psi\rangle$$. 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.

## Rotations in Spin Space

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

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).

## Representations of the Rotation Operator

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.

### Spherical Harmonics

The connection between the Spherical Harmonics and the rotation matrices.

Recall: $$\mathcal{D}(\alpha,\beta,\gamma)|jm\rangle = \sum_{m'=-j}^j \mathcal{D}_{m'm}^{(j)}(\alpha,\beta,\gamma)|jm'\rangle = \sum_{m'=j}^j\exp\left(-i(m'\alpha+m\gamma)\right)d_{m'm}^{(j)}(\beta)|jm'\rangle$$

#### Wigner Formula

$$d_{m'm}^{(j)} = \sum_{k}(-1)^{k+m'-m}\frac{\sqrt{(j+m)!(j-m)!(j+m')!(j-m)!}} {(j-m'-k)!(j+m-k)!(k+m'-m)!k!} \left(\cos\frac{\beta}{2}\right)^{2j+m-m'-2k} \left(\sin\frac{\beta}{2}\right)^{m'-m+2k}$$ The $$k$$ s are such that the factorials in the denominator are positive.

$$|\ell m\rangle \doteq Y_\ell^m(\Omega)$$. $$\langle\vec{n}|\ell m\rangle = Y_\ell^m(\Omega)$$.

Start with some $$|\vec{z}\rangle$$ and rotate it on $$|\vec{n}\rangle$$. Hence, $$\mathcal{D}(R)|\vec{z}\rangle = |\vec{n}\rangle = \sum_m\mathcal{D}(R)|\ell m\rangle\langle \ell m|\vec{z}\rangle$$.

Multiply by $$\langle \ell m'|$$ so, $$\langle \ell m'|\vec{n}\rangle = \sum_m\langle \ell m'|\mathcal{D}(R)|\ell m\rangle\langle \ell m|\vec{z}\rangle$$. Thus, $$Y_\ell^{m*}(\Omega) = \sum_m \mathcal{D}_{m'm}^{(\ell)}(\alpha,\beta,\gamma)\langle \ell m|\vec{z}\rangle$$.

Ansatz: $$\varphi = \alpha+\gamma$$, $$\theta = \beta$$.

Since rotation about $$z$$ leaves $$\vec{z}$$ unchanged, we can set $$\gamma=0$$ without loss of generality. Then, $$\theta = \beta$$. So, $$\varphi=\alpha$$. Thus, $$Y_\ell^{m'* }(\Omega) = \sum_m \mathcal{D}_z(\theta)\mathcal{D}_y(\varphi)\langle \ell m|\vec{z}\rangle$$. $$\langle \ell m|\vec{z}\rangle = Y_{\ell}^{m*}(\theta=0,\varphi) = Y_\ell^{0*}(0)\delta_{m,0} = \sqrt{\frac{2\ell+1}{4\pi}}P_\ell(\cos 0)\delta_{m,0} = \sqrt{\frac{2\ell+1}{4\pi}}\delta_{m,0}$$

Inserting this back into the sum, $$Y_\ell^{m'*}(\theta,\varphi) = \mathcal{D}_{m'0}^{(\ell)}(\alpha=\varphi,\beta=\theta,\gamma=0)\sqrt{\frac{2\ell+1}{4\pi}}$$. So, $$\mathcal{D}_{m'0}^{(\ell)}(\alpha,\beta,\gamma=0)=\left.\sqrt{\frac{4\pi}{2\ell+1}}Y_{\ell}^{m'*}(\theta,\varphi)\right|_{\alpha=\varphi,\beta=\theta}$$.

$$\mathcal{D}_{m'm}^{(j)} = \exp(-im'\alpha+m\gamma)d_{m'm}^{(j)}(\beta)$$.

For $$j=1$$. Recall we set $$\gamma=0$$. Then, $$\mathcal{D}_{m'm}^{(1)}=\exp(-im'\alpha)\begin{pmatrix} \frac{1}{2}(1+\cos\beta) & \frac{-\sin\beta}{\sqrt{2}} & \frac{1}{2}(1-\cos\beta) \\ \cdots & \cos\beta & \cdots \\ \cdots & \frac{\sin\beta}{\sqrt{2}} & \cdots \end{pmatrix}$$

The rest of the matrix is on the worksheet.

The middle column is then our spherical harmonics.

So, $$\mathcal{D}_{m'0}^{(1)}=\exp(-im'\varphi)\begin{pmatrix} \frac{-\sin\theta}{\sqrt{2}} \\ \cos\theta \\ \frac{\sin\theta}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \exp(-i\varphi)\frac{-\sin\theta}{\sqrt{2}} \\ \cos\theta \\ \exp(i\varphi)\frac{\sin\theta}{\sqrt{2}} \end{pmatrix} = \sqrt{\frac{4\pi}{3}}\begin{pmatrix} Y_1^{1*} \\ Y_1^{0*} \\ Y_1^{-1*} \end{pmatrix}$$.

If $$m'=0$$, then, $$\mathcal{D}_{00}^{(\ell)}(\varphi,\theta,0) = d_{00}^{(\ell)}(\theta) = \sqrt{\frac{4\pi}{2\ell+1}}Y_\ell^{0*}(\theta) = P_\ell(\cos\theta)$$.

## Rotation Matrices for Coupling Two Angular Momentum

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.

### Spherical Harmonics

$$j_1=\ell_1,j_2=\ell_2,m=0$$. Recall: $$\mathcal{D}_{m'0}^{(\ell)}(\alpha,\beta,\gamma) = \sqrt{\frac{4\pi}{2\ell+1}}Y_{\ell}^{m' *}(\theta,\varphi)$$.

$$\mathcal{D}_{m_1'0}^{(\ell_1)}\mathcal{D}_{m_2'0}^{(\ell_2)} = \sum_{\ell,m'}\langle \ell_1\ell_200|\ell0\rangle\langle\ell_1\ell_2m_1'm_2'|\ell m'\rangle\mathcal{D}_{m'0}^{(\ell)}$$ $$Y_{\ell_1}^{m_1'}Y_{\ell_2}^{m_2} = \sum_{ell,m'}\sqrt{\frac{(2\ell_1+1)(2\ell_2+1)}{4\pi(2\ell+1)}}\langle\ell_1\ell_200|\ell0\rangle\langle \ell_1\ell_2m_1'm_2'|\ell m'\rangle Y_\ell^{m'}$$

Applying $$Y_\ell^{m*}$$ and integrating over a solid angle gives an orthogonality relationship, $$\int Y_\ell^{m*}Y_{\ell_1}^{m_1}Y_{\ell_2}^{m_2}\:d\Omega = \sqrt{\frac{(2\ell_1+1)(2\ell_2+1)}{4\pi(2\ell+1)}}\langle\ell_1\ell_200|\ell0\rangle\langle\ell_1\ell_2m_2m_2|\ell m\rangle$$.

#### Tangent

$$|\ell_2m_2\rangle \to |\ell m\rangle$$, $$\mathcal{P}\sim|\langle\ell m|H_{int}|\ell_2m_2\rangle|^2$$, $$H_{int} = -e\vec{r}\cdot\vec{E} = -eE(r\cos\theta) = -\alpha eE Y_{1}^0$$. So, $$\int Y_{\ell}^{m*}Y_{1}^{0}Y_{\ell_2}^{m_2}d\Omega$$

### Aside

#### Supplemental Paper

For two sources with different phases and energies, $$|E_1\exp(i\varphi_1) + E_2\exp(i\varphi_2)|^2$$

$$|\alpha,t=0\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle)$$ time translated gives $$|\alpha,t\rangle = \exp\left(\frac{-i}{\hbar}Ht\right)|\alpha,0\rangle$$

Two pathways possible.

Path 1: $$H=H_0=\frac{p^2}{2m}$$

Path 2: $$H=H_0-\vec{\mu}\cdot\vec{B} = H_0 -\frac{g_ne}{m_nc}S_zB$$

Let $$\omega = \frac{g_n e}{m_n c}B$$. So, $$|\alpha,t\rangle = \exp\left(\frac{i}{\hbar}S_z\omega t\right)|\alpha,0\rangle$$ On our initial state, $$|\alpha,t\rangle = \frac{1}{\sqrt{2}}\exp\frac{i\omega t}{2}|+\rangle + \frac{1}{\sqrt{2}}\exp\frac{-i\omega t}{2}|-\rangle$$

The detector detects $$\left||\alpha,0\rangle + |\alpha,t\rangle\right|^2$$. Note that the standard hamiltonian phases are equal and so they get washed out, thus we can express path 1 as only the time unevolved state. So, $$I\sim||\alpha,0\rangle + \cdots|^2 = 2+2\Re\langle\alpha,0|\alpha,t\rangle\sim\left(1+\cos\frac{\omega}{2}t\right)$$.

$$g_n = -1.91$$

### Noether’s Theorem

For every continuous symmetry there exists a corresponding conservation law. I.e. there exists a conserved variable.

Example, for a group, Wigner’s theorem states for a unitary transformation U there is a generator such that $$U=\exp(i\varepsilon G)$$ which implies that if $$[H,G]=0$$ then $$\frac{dG}{dt}=0$$. Further, $$[G_i=G_j]=c_{ijk}G_k$$.

Created: 2023-06-25 Sun 17:19

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