Title: ETH_Sem3 / Physik_III / Woche_11

Rydberg formula:
The Rydberg formula is used to predict the spectral lines of Hydrogen
\(\nu=\frac{c}{\lambda} \)\(=cR_\infty\left(\frac{1}{n'^2}-\frac{1}{n^2}\right)\)
with \(R_\infty = 10973731.568527 \pm 73 m^{-1}\) the Rydberg constant.
More useful is the energy unit "Rydberg" \(R_e = hcR_\infty \approx 13.6eV\)
Back Extra: Funfact: The Rydberg constant is one of the most precicely known natural constants

Q: The [...] is [...]
[...]
A: visible part of the Rydberg series; the Balmer series:; \( \nu \propto \left( \frac{1}{4} - \frac{1}{m^2}\right), m = 3,4,5,\ldots\)

Solving the Hydrogen atom:
Part 1: Getting the energies
1. Write TISE with \(V(r) = -\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}\)
2. Only consider bound states with \(E<0\)
3. Define some constants
4. \(\dede{u}{\rho^2}=(1-\frac{\rho_0}{\rho} + \frac{l(l+1)}{\rho^2})u\)
5. Observe asymptotic behaviour as \(\rho \to \infty/0\)
6. Consider normalizability 
7. \(u(\rho) = \rho^{l+1}e^{-\rho}v(\rho)\)
8. Ansatz: \(v(\rho) = \sum_{j=0}^\infty c_j\rho^j\)
9. Find recursion for \(c_j\)
10. Because we already found the highest order term in step 7) we need the series to terminate
11. Define an upper bound \(N\) s.t. \(c_N =0\)
12. Solve for energy.
13. We now know the energy eigensvalues!

Q: [...] of [...]
[...]
A: Eigenenergies; the Hydrogen Atom:; \(E_n = -\left(\frac{m_e}{2\hbar^2}(\frac{e^2}{4\pi\varepsilon_0})^2\right)\frac{1}{n^2}=\frac{E_1}{n^2}, n=1,2,3,\ldots\)

Spatial wavefunction of Hydrogen components:
\(\Psi_{nlm} (r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)\)
with the quantum numbers:
\(n \in \mathbb{N}_{\geq0}\)
\(l = 0,1,2,\ldots, n-1\)
\(m=-l,\ldots,0,\ldots,l\)
where all eigenstates of \(m\) are degenerate (there are \(n^2\) of them)

Q: [...]
[...]
A: Ground state energy of H:; \(E_1 = -\left(\frac{m_e}{2\hbar^2}(\frac{e^2}{4\pi\varepsilon_0})^2\right)=-13.6eV=-R_e\)

Q: [...]: (Probably not relevant)
[...]
A: Normalized wavefunction of Hydrogen atom; \(\Psi_{nlm}(r,\theta,\phi) = \sqrt{(\frac{2}{na})^3\frac{(n-l-1)!}{2n(n+l)!} }e^{-r/na}(\frac{2r}{na})^l L^{2l+1}_{n-l-1}(\frac{2r}{na})Y_l^m(\theta,\phi)\)

Q: What is this?
A: Structure of energy level diagram of the hydrogen atom

Interpreting wavefunction plots:
Radial wavefunction has \(N-1\) zero-crossings
Angular wavefunction has \(l-m\)  zero-crossings in \([0,\pi]\)
(Angular (in xy plane) wavefunction has \(m\) zero-crossings in \([0,2\pi]\))
Example on back:

Commutator relations for angular momentum:
\([L_x,L_y] = i\hbar L_z\)
\([L_y,L_z] = i\hbar L_x\)
\([L_z,L_x] = i\hbar L_y\)
\([L^2,L_{xyz}]=0\)

Q: Ladder operator for angular momentum:
[...]
it's commtator relations are:
[...]
[...]
A: \(L_\pm \equiv L_x \pm iL_y\); \([L_z,L_\pm] = \pm\hbar L_\pm\); \([L^2,L_\pm]=0\)

Q: Why is it useful that the ladder operator doesn't commute with \(L_z\)?
A: From the commutation relation we can "gain" an additional \(L_\pm\) from switching with \(L_z\), which leads us to the next higher eigenstate.

Q: Angular momentum ladder operator on state \(\ket{l,m}\)?
A: \(L_\pm\ket{l,m} = A^m_{l,\pm}\ket{l,m\pm1}\)
with \(A^m_{l,\pm} = \hbar\sqrt{l(l+1)-m(m\pm1)}\)

Why and how can we calculate the maximum and minimum for the quantum number \(m\)?
- The component of the angular momentum in \(z\) direction cannot be larger than the total angular momentum.
- This means we know that at some point \(L_\pm\ket{l,\mu}=0\)
- We observe \(L_\pm L_\mp = L^2-L^2_z \pm \hbar L_z\)
- Rewriting \(L^2 \ket{l,\mu}\) with this relation we get \(L^2\ket{l,\mu} = \hbar^2\mu(\mu+1)\ket{l,\mu}\)
- but we also know \(L^2\ket{l,\mu} = \hbar^2l(l+1)\ket{l,\mu}\)
- so we find \(m_{max}=\mu=l\)
- equivalently \(m_{min}= \mu' = -l\)

Q: Why do we know the possible values for \(l\) by knowing the maximal and minimal \(m\)?
A: Because \(m_{max} = -m_{min} = l\) and because we know that they are separated by integer steps \(l\) can only take half-integer values.
\(l=0,1/2,1,\ldots\)

Q: Why can we never measure the maximal angular momentum if we measure along the \(z\) axis?
A: If we measured the maximal value \(x,y\) would be fully determined as \(0\) which violates the commutation relations (uncertainty).

Because the angular momentum problem is analogous to the central potential problem we find that:
Energy eigenstates for a central potential are also eigenstates of the Hamiltonian, \(L^2\) and \(L_z\) (for whole integer values)

Q: Expected magnetic moment of an atom in the bohr model:
A: \(\vec{\mu}=-\frac{e}{2m}\vec{L}\)

Q: [...] experiment:
A: Stern gerlach

Q: Why is the stern gerlach experiment important?
A: It both shows that the magnetic moment is quantized and that we have to allow half integer magnetic moments (So we can get two spots)

Properties for the spin:
(they are the same as for \(L\))
\([s_{xyz},s_{yzx}] = i\hbar s_{zxy}\)
\(s^2\ket{s,m_s}=\hbar^2 s(s+1)\ket{s,m_s}\)
\(s_z\ket{s,m_s} = \hbar m \ket{s,m_s}\)

Q: Vector representation of spin eigenstates:
[...]
[...]
[...]
A: \(\ket{l=\frac{1}{2},m_s=-\frac{1}{2} } \leftrightarrow = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\); \(\ket{l=\frac{1}{2},m_s=\frac{1}{2} } \leftrightarrow = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\); \(\ket{\psi} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}\)

Pauli matrices:
 \(\sigma_{x} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)
\(\sigma_{y} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\)
\(\sigma_{z} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)