⚛️ Atomic Physics
- Size: $\sim 10^{-10} \to 10^{-9} \,\rm m$
- Energies: $\sim 1\to 100 \, \rm eV$
☢️ Nuclear Physics
- Size: $\sim 10^{-15}\, \rm m$ or $1 \,\rm fm$ (Fermi or Femtometer)
- Energy: $\sim 10^6 \, \rm eV$ or $1 \rm \, MeV$
- mass: $\sim \rm \, GeV/c^2$
🗒️ Note: we will work in natural units $c=\hbar = 1$
$$ \begin{aligned} \hbar c &\approx 200 \rm\, MeV \, fm \\ \text{fine structure constant: } \quad\alpha &=\frac{1}{4\pi \epsilon_0}\frac{e^2}{\hbar c} = \frac{1}{137} \; \text{ (dimensionless) }\;\;
\end{aligned} $$
⚠️ Warning: don’t use SI units
Nuclei exhibit wave-like Phenomena, thus they follow the following relations:
De Broglie relation: $\lambda = \frac{h}{p}$
Heisenberg uncertainty principle: $\Delta x \Delta p_x \ge \frac{\hbar}{2}$
🗒️ Note: equality only holds for ideal situations
For a non-relativistic particle:
Kinetic energy: $E_\text{kin}=\frac{p^2}{2M}$
Schrodinger equation:
$$ -\frac{\hbar ^2}{2M} \frac{ \text d^2 \psi(x)}{\text dx^2} + V(x) \psi (x)=E \psi(x) $$
which can be rewritten using the Hamiltonian as
$$ \hat H \psi(x)=E\psi(x) $$
Wave function with a time dependence: $\Psi(x,t)= \psi(x) e^{iEt/\hbar}$
Must respect unitary:
$$ \int^\infin_{-\infin} P(x) \, \text dx = \int ^\infin _{-\infin} \Psi^* (x,t) \Psi(x,t) \, \text dx =1 $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/52196f78-16dc-4db1-9269-6ab934d44ae5/Antiparticles.gif" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/52196f78-16dc-4db1-9269-6ab934d44ae5/Antiparticles.gif" width="40px" />
Antiparticles: have the same properties but reversed charges. Postulated by Dirac they come from the fact that the total energy of a particle $E^2 =p^2 c^2+ m_0^2 c^4$ has a negative energy solution: $E =\pm \sqrt{p^2 c^2+ m_0^2 c^4}$
</aside>
Properties:
$$ \vec L = \vec r \times \vec p \Rightarrow \left \{ \begin{aligned} \hat L_x&=-i \hbar \left ( y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y} \right ) \\ \hat L_y&=i \hbar \left ( x\frac{\partial }{\partial z}-z\frac{\partial}{\partial x} \right ) \\ \hat L_z&=-i \hbar \left ( x\frac{\partial }{\partial y}-y\frac{\partial}{\partial x} \right ) \end{aligned} \right . $$