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

Geometrical Rotations
Rotations of state vectors $| Ψ ⟩ \Rightarrow | Ψ^{'} ⟩ = D (R) | Ψ ⟩$
$D (R) \Leftrightarrow J$ , the generalized angular momentum

Geometrical Rotations

Rotate about $z$ axis by $φ$ , 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 $φ$ . passive rotation (xyz coordinate system changes), $\vec{r} \to {\vec{r}}^{'}$ with the coordinate $x^{'}, y^{'}$ have an angle $φ$ between them ( $- φ$ 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 $(R^{T} R = R R^{T} = I)$ is due to $| {\vec{r}}^{'} | = | \vec{r} |$ .

Then, $R (φ) = (\begin{matrix} \cos φ & - \sin φ & 0 \\ \sin φ & \cos φ & 0 \\ 0 & 0 & 1 \end{matrix})$ .

$R_{x} (φ) = (\begin{matrix} 1 & 0 & 0 \\ 0 & \cos φ & - \sin φ \\ 0 & \sin φ & \cos φ \end{matrix})$ .

$R_{y} (φ) = (\begin{matrix} \cos φ & 0 & \sin φ \\ 0 & 1 & 0 \\ - \sin φ & 0 & \cos φ \end{matrix})$ .

Recall that rotations are non-commutative $R_{i} (φ) R_{j} (φ) \neq R_{j} (φ^{'}) R_{i} (φ)$ .

For a small angle $[R_{x} (ε), R_{y} (ε)] = (\begin{matrix} 0 & - ε^{2} & 0 \\ ε^{2} & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = R_{z} (ε^{2}) - I \to 0$ .

$R_{i} (φ + ε) = R_{i} (φ) R_{i} (ε) = R_{i} (ε) R_{i} (φ)$ .

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:

$a \cdot b \in G$
$\exists e . a \cdot e = e \cdot a = a$
$\exists e . \forall a . \exists a^{- 1} . a \cdot a^{- 1} = e$
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.

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

Examples
- $Z$ under multiplication? No, the inverse element does not exist!
- $Z$ under addition? Yes
- Rotations Yes. $R_{n} (φ) R_{m} (φ^{'}) \in$ rotations. $R^{T} R = R R^{T} = R_{n} (φ) R_{m} (φ^{'}) R_{m} (φ^{'}) R_{n} (φ) = I = R_{o} (0)$ . $R^{T} = R^{- 1}$ . $(R_{n} (φ) R_{m} (φ^{'})) R_{o} (φ^{″}) = R_{n} (φ) (R_{m} (φ^{'}) R_{o} (φ^{″}))$ .

Rotation Groups

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

$det R = + 1$ implies rotations. $det R = - 1$ implies parity.

$S O (3)$ gives the rotations in 3 dimensions.

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

Rotations in State Space

$| Ψ ⟩ \Rightarrow | Ψ^{'} ⟩ = D (R) | Ψ ⟩$ . Where $D$ stands for Drehung rotation.

$D D^{†}$ . Remember generators for time, $U_{ε} = 1 - i G ε$ with $G = \frac{H}{ℏ}$ $ε = d t$ . For space, $G = \frac{p_{x}}{ℏ}$ , $ε = d x$ . For rotations, $G = \frac{J_{z}}{ℏ} = \frac{\vec{J} \cdot \vec{n}}{ℏ}$ $ε = d φ$ . For finite rotations, $D (\vec{n}, φ) = \exp (\frac{- i}{ℏ} \vec{J} \cdot \vec{n} d φ)$

Remember, $A^{'} = U A U^{†}$ . So, $A^{'} = D (R) A D^{†} (R)$ .

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

$A^{'} = A + [- \frac{i}{ℏ} (\vec{J} \cdot \vec{n}) d φ, A] + \frac{1}{2!} [- \frac{i}{ℏ} (\vec{J} \cdot \vec{n}) d φ, [- \frac{i}{ℏ} (\vec{J} \cdot \vec{n}) d φ, A]] + \dots$ .

Let us consider the Hamiltonian. Let $| Ψ (t_{0}) ⟩$ be transformed by $D (R)$ to $| Ψ^{'} (t_{0}) ⟩ = D (R) | Ψ (t_{0}) ⟩$ then be transformed by $| Ψ^{'} (t_{0} + d t) ⟩$ .

Consider the opposite order, $| Ψ (t_{0}) ⟩$ be time transformed $| Ψ (t_{0} + d t) ⟩$ then rotated $| Ψ^{″} (t_{0} + d t) ⟩$ .

$Ψ^{'} (t_{0} + d t) ⟩ = | Ψ^{'} (t_{0}) ⟩ + d t {\frac{d | Ψ^{'} ⟩}{d t} |}_{t_{0}} = | Ψ^{'} (t_{0}) ⟩ + {\frac{d t}{i ℏ} \hat{H} | Ψ^{'} ⟩ |}_{t_{0}} = D (R) | Ψ (t_{0}) ⟩ + \frac{d t}{i ℏ} \hat{H} D (R) | Ψ (t_{0}) ⟩$ .

$| Ψ (t + d t) ⟩ = | Ψ (t_{0}) ⟩ + \frac{d t}{i ℏ} \hat{H} | Ψ (t_{0}) ⟩$ . So, $D (R) | Ψ (t + d t) ⟩ = D (R) | Ψ (t_{0}) ⟩ + \frac{d t}{i ℏ} D (R) \hat{H} | Ψ (t_{0}) ⟩$ . 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: $\frac{d ⟨ A ⟩}{d t} = ⟨ \frac{\partial A}{\partial t} ⟩ + \frac{1}{i ℏ} ⟨ [A, H] ⟩$ .

If $A = \vec{J}$ then $\frac{d ⟨ J ⟩}{d t} = \frac{1}{i ℏ} ⟨ [\vec{J}, H] ⟩ = 0$ implies that $\vec{J}$ is a constant of the motion.

Spin

Rotations in Spin Space

$D^{(s)} (\vec{n}, φ) = \exp (- \frac{i}{ℏ} φ \vec{S} \cdot \vec{n})$

For $s = \frac{1}{2}$ , $\vec{S} = \frac{ℏ}{2} \vec{σ}$ , $D^{(s)} (\vec{n}, φ) = \exp (- i \frac{φ}{2} \hat{\vec{σ}} \cdot \vec{n}) \approx I \cos \frac{φ}{2} - i (\vec{σ} \cdot \vec{n}) \sin \frac{φ}{2}$ . (Not sure if actually approximate or exact). $\hat{\vec{σ}} \cdot \vec{n} = σ_{x} n_{x} + σ_{y} n_{y} + σ_{z} n_{z} = (\begin{matrix} n_{z} & n_{x} - i n_{y} \\ n_{x} + i n_{y} & - n_{z} \end{matrix})$ . $\exp (\dots) = (\begin{matrix} \cos \frac{φ}{2} - i n_{z} \sin \frac{φ}{2} & - i \sin \frac{φ}{2} (n_{x} - i n_{y}) \\ - i \sin \frac{φ}{2} (n_{x} + i n_{y}) & \cos \frac{φ}{2} + i n_{z} \sin \frac{φ}{2} \end{matrix})$

For $φ = 2 π$ , $D^{(s)} = - I$ . So $D^{(s)} | α ⟩ = - | α ⟩$ . For $φ = 4 π$ , $D^{(s)} = I$ .

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 $α, β, γ$ . $R (α, β, γ) = R_{z^{'}} (γ) R_{y^{'}} (β) R_{z} (α)$ From Goldstein, $R (α, β, γ) = R_{z} (α) R_{y} (β) R_{z} (γ)$

For spin space, where $s = \frac{1}{2}$ , $D (α, β γ) = D_{z} (α) D_{y} (β) D_{y} (γ) = \exp \frac{- i σ_{z} α}{2} \exp \frac{- i σ_{y} β}{2} \exp \frac{- i σ_{z} γ}{2} = (\begin{matrix} \exp \frac{- i α}{2} & 0 \\ 0 & \exp \frac{i α}{2} \end{matrix}) \dots = (\begin{matrix} \exp (\frac{- i (α + γ)}{2}) \cos \frac{β}{2} & \exp (\frac{- i (α - γ)}{2}) \sin \frac{β}{2} \\ \exp (\frac{i (α - γ)}{2}) \sin \frac{β}{2} & \exp (\frac{i (α + γ)}{2}) \cos \frac{β}{2} \end{matrix})$ called the $j = \frac{1}{2}$ irreducible representation of the rotation operator.

$D_{m_{s}^{'} m_{s}}^{(1 / 2)} (α, β, γ) = ⟨ \frac{1}{2} m_{s}^{'} | \exp (- i S_{z} α / ℏ) \exp (- i S_{y} β / ℏ) \exp (- i S_{z} γ / ℏ) | \frac{1}{2} m_{s} ⟩$

By replacing $S_{i}$ with $J_{i}$ we get a more general expression, $D_{m_{j}^{'} m_{j}}^{(j)} = ⟨ j m_{j}^{'} | \exp (\frac{- i}{ℏ} J_{z} α) \exp (\frac{- i}{ℏ} J_{y} β) \exp (\frac{- i}{ℏ} J_{z} γ) | j m_{j} ⟩$ .

Question:

$D | j, m ⟩ \Leftrightarrow c | j, m^{'} ⟩$
Why do we not go to some $j^{'}$ ?

Generalized rotation operator matrix element, $D_{m^{'} m}^{(j)} (R) = ⟨ j m^{'} | \exp (- \frac{i}{ℏ} \vec{J} \cdot \vec{n} φ) | j m ⟩$

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

I.e. ${\vec{J}}^{2} D (R) | j m ⟩ = {\vec{J}}^{2} | Ψ ⟩$ , $D (R) {\vec{J}}^{2} | j m ⟩ = D (R) ℏ^{2} j (j + 1) | j m ⟩$ A zero commutator implies that $| Ψ ⟩ = D (R) | j m ⟩$ . Thus, $m$ may change but not $j$ .

Inserting Identity, $D (R) | j m ⟩ = \sum_{m^{'}} | j m^{'} ⟩ ⟨ j m^{'} | D (R) | j m ⟩ = \sum_{m^{'}} D_{m^{'} m}^{(j)} (R) | j m^{'} ⟩$ . 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.

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

Returning to the matrix elements:

\begin{aligned} D_{m_{j}^{'} m_{j}}^{(j)} (α, β, γ) & = ⟨ j m_{j}^{'} | \exp (- \frac{i}{ℏ} J_{z} α) \exp (- \frac{i}{ℏ} J_{y} β) \exp (- \frac{i}{ℏ} J_{z} γ) | j m_{j} ⟩ \\ = \exp (- i m^{'} α) ⟨ j m_{j}^{'} | \exp (- \frac{i}{ℏ} J_{y} β) \exp (- i m γ) | j m_{j} ⟩ \\ = \exp (- i m^{'} α + m γ) d_{m^{'} m}^{(j)} \end{aligned}

Using the ladder operators, $J_{y} = \frac{J_{+} - J_{-}}{2 i}$ , $J_{\pm} = ℏ \sqrt{j (j + 1) - m (m \pm 1)} | j m \pm 1 ⟩$ , $⟨ 1 m^{'} | J_{y} | 1 m ⟩ \Rightarrow (\begin{matrix} 0 & - i \sqrt{2} & 0 \\ i \sqrt{2} & 0 & - i \sqrt{2} \\ 0 & i \sqrt{2} & 0 \end{matrix})$ . Taylor expanding the exponential,

\begin{aligned} \exp (- \frac{i}{ℏ} J_{y} β) & = I - \frac{J_{y}^{2}}{ℏ^{2}} (1 - \cos β) - i \frac{J_{y}}{ℏ} \sin β . \end{aligned}

So,

\begin{aligned} j = 1 \Rightarrow d^{(1)} (β) & = (\begin{array}{c} \frac{1 + \cos β}{2} & - \frac{1}{\sqrt{2}} \sin β & \frac{1}{2} (1 - \cos β) \\ \frac{\sqrt{2}}{2} \sin β & \cos β & \frac{- \sqrt{2}}{2} \sin β \\ \frac{1}{2} (1 - \cos β) & \frac{\sqrt{2}}{2} \sin β & \frac{1}{2} (1 + \cos β) \end{array}) . \end{aligned}

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

Spherical Harmonics

The connection between the Spherical Harmonics and the rotation matrices.

Recall: $D (α, β, γ) | j m ⟩ = \sum_{m^{'} = - j}^{j} D_{m^{'} m}^{(j)} (α, β, γ) | j m^{'} ⟩ = \sum_{m^{'} = j}^{j} \exp (- i (m^{'} α + m γ)) d_{m^{'} m}^{(j)} (β) | j m^{'} ⟩$

Wigner Formula

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

$| ℓ m ⟩ ≐ Y_{ℓ}^{m} (Ω)$ . $⟨ \vec{n} | ℓ m ⟩ = Y_{ℓ}^{m} (Ω)$ .

Start with some $| \vec{z} ⟩$ and rotate it on $| \vec{n} ⟩$ . Hence, $D (R) | \vec{z} ⟩ = | \vec{n} ⟩ = \sum_{m} D (R) | ℓ m ⟩ ⟨ ℓ m | \vec{z} ⟩$ .

Multiply by $⟨ ℓ m^{'} |$ so, $⟨ ℓ m^{'} | \vec{n} ⟩ = \sum_{m} ⟨ ℓ m^{'} | D (R) | ℓ m ⟩ ⟨ ℓ m | \vec{z} ⟩$ . Thus, $Y_{ℓ}^{m *} (Ω) = \sum_{m} D_{m^{'} m}^{(ℓ)} (α, β, γ) ⟨ ℓ m | \vec{z} ⟩$ .

Ansatz: $φ = α + γ$ , $θ = β$ .

Since rotation about $z$ leaves $\vec{z}$ unchanged, we can set $γ = 0$ without loss of generality. Then, $θ = β$ . So, $φ = α$ . Thus, $Y_{ℓ}^{m^{'} *} (Ω) = \sum_{m} D_{z} (θ) D_{y} (φ) ⟨ ℓ m | \vec{z} ⟩$ . $⟨ ℓ m | \vec{z} ⟩ = Y_{ℓ}^{m *} (θ = 0, φ) = Y_{ℓ}^{0 *} (0) δ_{m, 0} = \sqrt{\frac{2 ℓ + 1}{4 π}} P_{ℓ} (\cos 0) δ_{m, 0} = \sqrt{\frac{2 ℓ + 1}{4 π}} δ_{m, 0}$

Inserting this back into the sum, $Y_{ℓ}^{m^{'} *} (θ, φ) = D_{m^{'} 0}^{(ℓ)} (α = φ, β = θ, γ = 0) \sqrt{\frac{2 ℓ + 1}{4 π}}$ . So, $D_{m^{'} 0}^{(ℓ)} (α, β, γ = 0) = {\sqrt{\frac{4 π}{2 ℓ + 1}} Y_{ℓ}^{m^{'} *} (θ, φ) |}_{α = φ, β = θ}$ .

$D_{m^{'} m}^{(j)} = \exp (- i m^{'} α + m γ) d_{m^{'} m}^{(j)} (β)$ .

For $j = 1$ . Recall we set $γ = 0$ . Then, $D_{m^{'} m}^{(1)} = \exp (- i m^{'} α) (\begin{matrix} \frac{1}{2} (1 + \cos β) & \frac{- \sin β}{\sqrt{2}} & \frac{1}{2} (1 - \cos β) \\ \dots & \cos β & \dots \\ \dots & \frac{\sin β}{\sqrt{2}} & \dots \end{matrix})$

The rest of the matrix is on the worksheet.

The middle column is then our spherical harmonics.

So, $D_{m^{'} 0}^{(1)} = \exp (- i m^{'} φ) (\begin{matrix} \frac{- \sin θ}{\sqrt{2}} \\ \cos θ \\ \frac{\sin θ}{\sqrt{2}} \end{matrix}) = (\begin{matrix} \exp (- i φ) \frac{- \sin θ}{\sqrt{2}} \\ \cos θ \\ \exp (i φ) \frac{\sin θ}{\sqrt{2}} \end{matrix}) = \sqrt{\frac{4 π}{3}} (\begin{matrix} Y_{1}^{1 *} \\ Y_{1}^{0 *} \\ Y_{1}^{- 1 *} \end{matrix})$ .

If $m^{'} = 0$ , then, $D_{00}^{(ℓ)} (φ, θ, 0) = d_{00}^{(ℓ)} (θ) = \sqrt{\frac{4 π}{2 ℓ + 1}} Y_{ℓ}^{0 *} (θ) = P_{ℓ} (\cos θ)$ .

Addition of Angular Momenta

Spectroscopic Notation

Rotation Matrices for Coupling Two Angular Momentum

Recall: $D (R) = D_{z} (α) D_{y} (β) D_{z} (γ)$ , $d_{m^{'} m}^{(j)} (β) = ⟨ j m^{'} | \exp (- i β J_{y} / ℏ) | j m ⟩$ .

How does $d_{m^{'} m}^{(j)} \Leftrightarrow d_{m^{'} m}^{(j_{1})} (β), d_{m^{'} m}^{(j_{2})}$ .

$d_{m^{'} m}^{(j)} = ⟨ j m^{'} | \exp (- i β J_{y} / ℏ) | j m ⟩ = \sum_{m_{1} m_{2}} \sum_{m_{1}^{'} m_{2}^{'}} ⟨ j m^{'} | j_{1} j_{2} m_{1} m_{2} ⟩ ⟨ j_{1} j_{2} m_{1} m_{2} | \exp (- i β (J_{y_{1}} + J_{y_{2}}) / ℏ) | j_{1} j_{2} m_{1}^{'} m_{2}^{'} ⟩ ⟨ j_{1} j_{2} m_{1}^{'} m_{2}^{'} | j m ⟩ = \sum_{m_{1} m_{2}} \sum_{m_{1}^{'} m_{2}^{'}} ⟨ j m^{'} | j_{1} j_{2} m_{1} m_{2} ⟩ ⟨ j_{1} j_{2} m_{1}^{'} m_{2}^{'} | j m ⟩ ⟨ j_{1} j_{2} m_{1} m_{2} | \exp (- i β J_{y_{1}} / ℏ) | j_{1} j_{2} m_{1}^{'} m_{2}^{'} ⟩ ⟨ j_{1} j_{2} m_{1} m_{2} | \exp (- i β J_{y_{2}} / ℏ) | j_{1} j_{2} m_{1}^{'} m_{2}^{'} ⟩ = \sum_{m_{1} m_{2}} \sum_{m_{1}^{'} m_{2}^{'}} ⟨ j m^{'} | j_{1} j_{2} m_{1} m_{2} ⟩ ⟨ j_{1} j_{2} m_{1}^{'} m_{2}^{'} | j m ⟩ d_{m_{1}^{'} m_{1}}^{(j_{1})} (β) d_{m_{2}^{'} m_{2}}^{(j_{2})} (β)$ .

Inverting the expression, $d_{m_{1}^{'} m_{1}}^{(j_{1})} (β) d_{m_{2}^{'} m_{2}}^{(j_{2})} (β) = \sum_{j = | j_{1} - j_{2} |}^{j_{1} + j_{2}} \sum_{m m^{'}} ⟨ j_{1} j_{2} m_{1} m_{2} | j m ⟩ ⟨ j_{1} j_{2} m_{1}^{'} m_{2}^{'} | j m^{'} ⟩ d_{m^{'} m}^{(j)} (β)$ .

We can replace the small $d (β)$ with $D (α, β, γ)$ for the same relationship.

Note that this is a reducible representation for $D^{(j_{1})} \otimes D^{(j_{2})}$ since there are many different $j$ s. We get a direct sum of irreducible: $D^{(j_{1} + j_{2})} \oplus \dots \oplus D^{| j_{1} - j_{2} |}$

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

Spherical Harmonics

$j_{1} = ℓ_{1}, j_{2} = ℓ_{2}, m = 0$ . Recall: $D_{m^{'} 0}^{(ℓ)} (α, β, γ) = \sqrt{\frac{4 π}{2 ℓ + 1}} Y_{ℓ}^{m^{'} *} (θ, φ)$ .

$D_{m_{1}^{'} 0}^{(ℓ_{1})} D_{m_{2}^{'} 0}^{(ℓ_{2})} = \sum_{ℓ, m^{'}} ⟨ ℓ_{1} ℓ_{2} 00 | ℓ 0 ⟩ ⟨ ℓ_{1} ℓ_{2} m_{1}^{'} m_{2}^{'} | ℓ m^{'} ⟩ D_{m^{'} 0}^{(ℓ)}$ $Y_{ℓ_{1}}^{m_{1}^{'}} Y_{ℓ_{2}}^{m_{2}} = \sum_{e l l, m^{'}} \sqrt{\frac{(2 ℓ_{1} + 1) (2 ℓ_{2} + 1)}{4 π (2 ℓ + 1)}} ⟨ ℓ_{1} ℓ_{2} 00 | ℓ 0 ⟩ ⟨ ℓ_{1} ℓ_{2} m_{1}^{'} m_{2}^{'} | ℓ m^{'} ⟩ Y_{ℓ}^{m^{'}}$

Applying $Y_{ℓ}^{m *}$ and integrating over a solid angle gives an orthogonality relationship, $\int Y_{ℓ}^{m *} Y_{ℓ_{1}}^{m_{1}} Y_{ℓ_{2}}^{m_{2}} d Ω = \sqrt{\frac{(2 ℓ_{1} + 1) (2 ℓ_{2} + 1)}{4 π (2 ℓ + 1)}} ⟨ ℓ_{1} ℓ_{2} 00 | ℓ 0 ⟩ ⟨ ℓ_{1} ℓ_{2} m_{2} m_{2} | ℓ m ⟩$ .

Tangent

$| ℓ_{2} m_{2} ⟩ \to | ℓ m ⟩$ , $P \sim | ⟨ ℓ m | H_{i n t} | ℓ_{2} m_{2} ⟩ |^{2}$ , $H_{i n t} = - e \vec{r} \cdot \vec{E} = - e E (r \cos θ) = - α e E Y_{1}^{0}$ . So, $\int Y_{ℓ}^{m *} Y_{1}^{0} Y_{ℓ_{2}}^{m_{2}} d Ω$

Aside

Supplemental Paper

For two sources with different phases and energies, $| E_{1} \exp (i φ_{1}) + E_{2} \exp (i φ_{2}) |^{2}$

$| α, t = 0 ⟩ = \frac{1}{\sqrt{2}} (| + ⟩ + | - ⟩)$ time translated gives $| α, t ⟩ = \exp (\frac{- i}{ℏ} H t) | α, 0 ⟩$

Two pathways possible.

Path 1: $H = H_{0} = \frac{p^{2}}{2 m}$

Path 2: $H = H_{0} - \vec{μ} \cdot \vec{B} = H_{0} - \frac{g_{n} e}{m_{n} c} S_{z} B$

Let $ω = \frac{g_{n} e}{m_{n} c} B$ . So, $| α, t ⟩ = \exp (\frac{i}{ℏ} S_{z} ω t) | α, 0 ⟩$ On our initial state, $| α, t ⟩ = \frac{1}{\sqrt{2}} \exp \frac{i ω t}{2} | + ⟩ + \frac{1}{\sqrt{2}} \exp \frac{- i ω t}{2} | - ⟩$

The detector detects ${| | α, 0 ⟩ + | α, t ⟩ |}^{2}$ . 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, $I \sim | | α, 0 ⟩ + \dots |^{2} = 2 + 2 ℜ ⟨ α, 0 | α, t ⟩ \sim (1 + \cos \frac{ω}{2} t)$ .

$g_{n} = - 1.91$

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 $U = \exp (i ε G)$ which implies that if $[H, G] = 0$ then $\frac{d G}{d t} = 0$ . Further, $[G_{i} = G_{j}] = c_{i j k} G_{k}$ .