<aside> <img src="attachment:2bdd1692-42e8-4727-a5f4-5cfd37082b06:noethers_theorem.png" alt="attachment:2bdd1692-42e8-4727-a5f4-5cfd37082b06:noethers_theorem.png" width="40px" />
Noether’s theorem: If the Lagrangian or the equation of motion is symmetric under a given transformation then there will be a conserved quantity associated with this symmetry
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💃 Example: translational symmetry
Here out equation of motion is the wave equation as follows
$$ \psi'(\vec x)=\hat H \psi(\vec x) $$
We define a translation operator which turns $\vec x \to \vec x'$ where $\vec x'=\vec x + \delta \vec x$ where $\delta \vec x$ is the
$$ \hat D \psi(\vec x)=\psi(\vec x + \delta \vec x) $$
Using both our equations we can write
$$ \hat D \hat H\psi(\vec x)=\hat D \psi'(x)=\psi'(\vec x + \delta \vec x)=\hat H \psi(\vec x+ \delta \vec x)=\hat H \hat D \psi(x) $$
💎 Conclusion: we have $[\hat D , \hat H]=0$ commute
Now that we showed the commutation lets try and find $\hat D$ we expand $\psi(\vec x + \delta x)$ and compare
$$ \begin{aligned} \psi(\vec x+ \delta \vec x)&=\psi(x)+\delta \vec x \frac{\partial \psi}{\partial \vec x} \\ &= \psi(x) + i (\delta \vec x \cdot \vec p) \,\psi(x) \\ \Rightarrow \qquad \hat D&=1+i(\delta \vec x \cdot \vec p)
\end{aligned} $$
From this we see
$$ \boxed{[\hat p , \hat H]=0} $$
🧠 Remember: if an operator commutes with the Hamiltonian in quantum mechanics the quantity is conserved
💎 Conclusion: linear momentum is conserved!
🗒️ Note: the actual expression is $\hat D=\exp (i \vec a\cdot\vec p)$ where $\vec a$ is the displacement
| Symmetry | Conservation Law |
|---|---|
| Translation in space | Momentum |
| Translation in time | Energy |
| Rotation in space | Angular momentum |
<aside> <img src="attachment:d87119bd-6def-48c0-9b1e-d768e0d4f932:orbital_angular_momentum.png" alt="attachment:d87119bd-6def-48c0-9b1e-d768e0d4f932:orbital_angular_momentum.png" width="40px" />
Orbital angular momentum:
$$ \begin{aligned} \text{Operator:}& &\quad \hat{\bold L} &=(\hat L_x,\hat L_y,\hat L_z) \qquad \hat{\bold L}=\hat{\bold r} \times \hat{\bold p} \\ \text{Eigenvalues:}& & \hat{\bold L}^2 \psi(\vec r)&=l(l+1) \psi(\vec x) \qquad\, l\in \N \\ &&\hat L_z \psi(\vec x)&=m_l \psi(\vec x) \qquad \quad \, \, m_l\in[-l,l]\cap \N \\
\end{aligned} $$
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<aside> <img src="attachment:a70c4abb-24b9-4f11-8763-7f2855f2d845:spin_angular_momentum.png" alt="attachment:a70c4abb-24b9-4f11-8763-7f2855f2d845:spin_angular_momentum.png" width="40px" />
Spin angular momentum:
$$ \begin{aligned} \text{Operator:}&& \quad \hat{\bold S}&=(\hat S_x,\hat S_y , \hat S_z) \\ \text{Eigenvalues:}&& \quad \hat{\bold S}^2 \psi(\vec x)&=s(s+1)\psi(\vec x) \quad s\in \left \{ \tfrac{n}{2}\; | \; n\in \N \right \} \\ &&\hat S_z \psi(\vec x)&=m_s \psi(\vec x) \qquad \; \, m_s\in \{ s,s-1,\ldots ,-s\}
\end{aligned} $$
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💃 Example: electron is a fermion so $s=\frac 12$ and $m_s\in\{-\frac 12, \frac 12 \}$
<aside> <img src="attachment:532ef457-d2a0-4624-82b4-95e618434661:total_angular_momentum.gif" alt="attachment:532ef457-d2a0-4624-82b4-95e618434661:total_angular_momentum.gif" width="40px" />
Total angular moment:
$$ \begin{aligned} \text{Operator:}&& \quad \hat{\bold J}&=\hat{\bold L} + \hat {\bold S} \\ \text{Eigenvalues:}&& \quad \hat{\bold J}^2 \psi(\vec x)&=j(j+1)\psi(\vec x) \quad j\in \N \\ &&\hat J_z \psi(\vec x)&=m_j \psi(\vec x) \qquad \; \, m_j\in [-j,j]\cap \N
\end{aligned} $$
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Conservation of angular momentum
$$ \boxed{[\hat {\bold J}, \hat H]=0 \quad [\hat J_z, \hat H]=0 } $$
🗒️ Note: while $\hat {\bold J}$ is conserved $\hat {\bold L}$ and $\hat {\bold S}$ don’t have to be
For a composite system (ie a system in a bound state like a hadron)
$$ {\text{Spin of bound state } \hat {\bold S}B = \hat {\bold J}\text{constituents} \text{ total angular momentum of the constituents}} $$