Title: ETH_Sem3 / Physik_III / Woche_12

Q: Assumptions for small pertubations with a Hamiltonian \(\hat{H} = \hat{H}_0 + \delta\hat{H}\) so that \(\hat{H}\ket{\psi_n} = E_n\ket{\psi_n}\) solves the TISE:
A: \(\hat{H_0}\ket{\psi^0_n} = E^0_n\ket{\psi^0_n}\)
\(E_n = E^0_n + \delta E_0, |\delta E_n|\ll E^0_n, \forall{n}\)
\(\ket{\psi_n} = \ket{\psi^0_n} + \ket{\delta\psi_n}, |\ket{\delta\psi_n}|\ll \ket{\psi^0_n}, \forall{n}\)
note: this needs to be fullfilled in any basis

Assuming a time independent small pertubation. One can write \(\hat{H} \ket{\psi_n} = E_n\ket{\psi_n}\)as \(\hat{H}_0\ket{\delta\psi_n} + \delta\hat{H}\ket{\psi^0_n} = E^0_n\ket{\delta\psi_n} + \delta E\ket{\psi_n}(*)\)
Since {\(\ket{\psi^0_n}\)} is a complete basis we can write \(\ket{\delta\psi_n} = \sum_l c_l\ket{\psi^0_l}\) Note that \(c_n = 0\) because we defined \(\ket{\psi_n} = \ket{\psi^0_n} + \ket{\delta\psi_n} \) which allows us to write \(\hat{H}_0\ket{\delta\psi_n} = \) \(\hat{H}_0\sum_l c_l\ket{\psi^0_l} = \sum_lE^0_lc_l\ket{\psi^0_l}\)
Taking the inner product of \((*)\) with \(\bra{\psi^0_m}\) now gives us \(E^0_mc_m + \bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = \)  \(E^0_nc_m + \delta E_n\braket{\psi^0_m|\psi^0_n}\)

Q: Using \(E^0_mc_m + \bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = E^0_nc_m + \delta E_n\braket{\psi^0_m|\psi^0_n}\) solve for \(n = m\) :
A: \(\bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = \delta E_n\)
So one simply gets the expectation value of \(\delta\hat{H}\) for the original Eigenstates

Using \(E^0_mc_m + \bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = E^0_nc_m + \delta E_n\braket{\psi^0_m|\psi^0_n}\) solve for \(n \neq m\) :
\(\bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = c_m(E^0_n - E^0_m)\)
This can be used to write \(\ket{\delta\psi_n}\) as we can get \(c_m = \frac{\bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} } {E^0_n - E^0_m} \)
note that for degenerate Eigenstates \(E^0_n = E^0_m\) one cannot simply use this

For degenerate Eigenstates \(E^0_n - E^0_m\) we can still solve this if \(\bra{\psi^0_m}\delta\hat{H}\ket{\psi^0_n} = 0\) which is the case when \(\ket{\psi^0_m}\) and \(\ket{\psi^0_n}\) are eigenstates of \(\delta\hat{H}\)
A set of N degenerate eigenstates \(\ket{\psi^0_{kl}}\) of \(\hat{H}_0\) with eigenenergy \(E^0_k\) form a basis for a N-dimensional degenerate subspace
Every state in this subspace is an eigenstate for \(\hat{H}_0\) but not necessarily for \(\delta\hat{H}\). However one can always find M states in this subspace that are eigenstates for \(\delta\hat{H}\) which are the correct states to use in the presence of the pertubation.

For degenerate Eigenstates \(E^0_n - E^0_m\) we can use following trick to solve it:
Find another operator \(\hat{O}\) that  commutes with \(\hat{H}_0\) and \(\delta\hat{H}\).
If you know a set of eigenstates \(\ket{\phi_n}\) of \(\hat{O}\) that all have different eigenvalues than you can use this set as a basis to do pertubation theory because they will also be eigenstates of \(\hat{H}_0\) and \(\delta\hat{H}\).

Q: For Spin-orbit coupling we first calculate the magnetic field experienced by the electron:
[...]
Using this one can find an expression due to the spin-orbit interaction for a perturbation Hamiltonian:
\(\hat{H}_{so} =\) [...]
A: \(\vec{B} = \frac{e}{8\pi \epsilon_0 c^2m_er^3}\) note; the extra factor \(\frac{1}{2}\) is due to the so-called Thomas precession factor; \( -\vec{\mu_S}\cdot\vec{B} = \frac{g_se^2}{16\pi\epsilon_0c^2m^2_er^3}\vec{S}\cdot\vec{B}\)

Commutator relations for \(\vec{S}\) and \(\vec{L}\) :
\([\hat{L}^2,\vec{S}\cdot\vec{L}] = 0\)
\([\hat{S}^2,\vec{S}\cdot\vec{L}] = 0\)
\([\hat{L}_z,\vec{S}\cdot\vec{L}] = i\hbar\hat{S}_x\hat{L}_y - i\hbar\hat{S}_y\hat{L}_x \neq 0\)
\([\hat{S}_z,\vec{S}\cdot\vec{L}] = i\hbar\hat{S}_y\hat{L}_x - i\hbar\hat{S}_x\hat{L}_y \neq 0\)

Q: Total angular operator [...] has commutation relations:
[...]
[...]
A: \(\hat{J}_z = \hat{L}_z + \hat{S}_z\) (e.g. only z-component); \([\hat{J}_z,\vec{S}\cdot\vec{L}] = 0\); \([\hat{J}_z,\hat{H}] = 0,\hat{H} = \hat{H}_0 + \hat{H}_{so}\)

Q: Remembering the states \(\ket{\psi_{n,l,m},m_s}\) which were solutions of the Hydrogen atom we now find that the total angular momentum operator acts on these states as follows
[...] and we can relabel the quantum numbers [...]
A: \(\hat{J}_z\ket{\psi_{n,l,m},m_s} = (m + m_s)\ket{\psi_{n,l,m}, m_s}\); \(m_j = m + m_s\)

If we try to solve the Hydrogen atom considering spin-orbit coupling with a operator that commutes with \(\hat{H} = \hat{H}_0 + \hat{H}_{so}\) we find that:
\(\vec{S}\cdot\vec{L} = \frac{1}{2}(\hat{J}^2 + \hat{S}^2 + \hat{L}^2)\)
But why can't we just use \(\hat{J}_z\)? because the eigenvalues of \(\hat{J}_z\) still contain degeneracies and therefore don't give us the correct basis states.
Acting \(\vec{S}\cdot\vec{L}\) on \(\ket{n,l,s,j,m_j}\) gives:
\(\vec{S}\cdot\vec{L}\ket{n,l,s,j,m_j} = \frac{\hbar^2}{2}(j(j+1)- l(l+1)-s(s+1))\ket{n,l,s,j,m_j}\)
Using this we can find the expectation values for \(\hat{H}_{so}\) and the state \(\ket{n,l,s,j,m_j}\):
\(E^1_{so} =\braket{\hat{H}_{so}b} = \frac{g_sZe^2}{16\pi\epsilon_0c^2m^2_e}\braket{\frac{1}{r^3} }\braket{\vec{S}\cdot\vec{L} }, \braket{\frac{1}{r^3} }=\frac{1}{l(l+\frac{1}{2})(l+1)n^3a^3}\)

Commutation relations for \(\hat{J}^2\):
\([\hat{J}^2,\hat{J}_z] = 0\)
\([\hat{J}^2,\vec{S}\cdot\vec{L}] = 0\)
\([\hat{J}^2,\hat{S}^2] = 0\)
\([\hat{J}^2,\hat{S}^2] = 0\)

Q: Fine structure constant \(\alpha = \) [...]
Unperturbed energies in terms of the fine structure constant
\(E_n =\) [...]
Bohr radius in terms of the fine structure constant
\(a = \) [...]
A: \(\frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac{1}{137.036}\); \(-\frac{1}{2}m_ec^2\alpha^2(\frac{1}{n^2}) = \frac{E_1}{n^2}\); \(\frac{4\pi\epsilon_0\hbar^2}{m_ee^2} = \frac{\hbar}{m_ec}\frac{1}{\alpha}\)

Energy shift of spin-orbit interaction
\(E_{so}^1 =\) \(-\frac{g_s\alpha^2}{4n}E_n[\frac{j(j+1)-l(l+1)-s(s+1)}{l(l+\frac{1}{2})(l+1)}]\)
Notable remarks of the result:
Modification to the order of \(\alpha^2\) therefore \(E_{so}^1\ll E_n\) justifying using perturbation theory in the first place.
Spin-orbit interaction breaks degeneracy between different values of l.
However for \(l=0\) we run into problems. We would expect the interaction to go to zero, however: \(\lim_{l\rightarrow0}E_{so}^1 =-\frac{g_s\alpha^2}{2n}E_n\) Here one would have to use a different perturbation called the Darwin shift which comes from a fully relativistic treatment.

Q: Magnetic moment of the electron and proton
A: \(\vec\mu_p = \frac{g_pe}{2m_p}\vec{S_p}\)
\(\vec\mu = -\frac{g_se}{2m_e}\vec{S}\)
\(|\vec\mu_p|\ll|\vec\mu|\)

The Zeeman effect describes what happens when a atom is placed in a magnetic field.
The perturbation Hamiltonian is \(\delta\hat{H}=-\vec{\mu}\cdot\vec{B}\) Assuming \(\vec{B} = B\hat{z}\) we get \(\delta\hat{H}=\frac{g_seB}{2m_e}\hat{S}_z\)
For eigenstates \(\ket{\frac{1}{2},\pm\frac{1}{2}}\) which are eigenstates of \(\hat{S}_z\) therefore we can already use them:
\(\delta E_+=\) \(\bra{\frac{1}{2},\pm\frac{1}{2} }\delta\hat{H}\ket{\frac{1}{2},\pm\frac{1}{2} }=\frac{g_seB\hbar}{4m_e}\)
\(\delta E_+=\)\(-\frac{g_seB\hbar}{4m_e}\)
This holds if \(l=0\) due to the symmetry of spacial wavefunction however if \(l\neq0\) one has to consider spin-orbit interaction because the Zeeman term no longer commutes with \(H_0\).

Spin-spin coupling \(\vec{S_p}\cdot\vec{S_e}\)is similar to spin-orbit coupling\(\vec{S}\cdot\vec{L}\).
The correct states to use are eigenstates of the total spin: \(\vec{S}=\vec{S_p}+\vec{S_e}\).
As before we use \( \vec{S_p}\cdot\vec{S_e} = \frac{1}{2}(S^2-S_p^2-S_e^2) \).
The possible values are \(S_p^2=S_e^2=\) \(\frac{3}{4}\hbar^2\) and \(S^2=\) \(2\hbar^2\) the triplet state which comes from \(S=1\) or \(S^2=0\) the singlet state which comes from \(S=0\)
Using this we get an engery modification of
\(E_{hf}=\) \(\frac{\mu_0g_pg_se^2\hbar^2}{6\pi m_pm_ea^3}\cdot\bigg\{\frac{1}{4} triplet\\{-\frac{3}{4} singlet}\)
The energy difference between the two states is
\(\Delta E=\) \(5.88\cdot10^{-6}eV\) which corresponds to
\(\nu=\) \(1420mHz\) and \(\lambda=\)\(21cm\)

For time independent perturbation theory we consider a Hamiltonian \(\hat{H}=\hat{H}_0+\hat{H}_1(t)\)
Now we have to distinguish two cases, for \(\hat{H}_1(t)=0\) we can write an arbitrary state in terms of the energy eigenstates \(\{\ket{\phi_n}\}\):
\(\ket{\Psi(t)}=\sum_n c_ne^{-i\omega_nt}\ket{\phi_n}\)
For \(\hat{H}_1(t)\neq0\) we can write an arbitrary state in terms of the energy eigenstates \(\{\ket{\phi_n}\}\) since this is a complet basis:
\(\ket{\Psi(t)}=\sum_n c_n(t)e^{-i\omega_nt}\ket{\phi_n}\)
Plugging this into the S.E. and then taking the inner product with \(\bra{\phi_m}\) gives us:
\(\sum_nc_ne^{-i\omega_nt}\bra{\phi_m}\hat{H}_1(t)\ket{\phi_n}=\sum_ni\hbar\frac{\partial c_n}{\partial t}e^{-i\omega_nt}\braket{\phi_m|\phi_n}=i\hbar\frac{\partial c_n}{\partial t}e^{-i\omega_mt}\)
Back Extra: Note that if a system starts out in an eigenstate of \(\hat{H}_0\), so that \(c_n=1\) for some j and \(c_n = 0, n\neq j\), we know that if \(\hat{H}_{1,mn} =\bra{\phi_m}\hat{H}_1(t)\ket{\phi_n}=0\) then \(c_n = 0, \forall t\)

If we consider an atom that has an electric dipole \(\vec{p}_0=-e\vec{r}\) encounters an electromagnetic wave \(\vec{E}=E_0cos(\omega t)\hat{k} \) we can then write the Hamiltonian: \(\hat{H}_1(t)=\vec{p}_0\cdot\vec{E}=eE_0\hat{\vec{r} }\cdot\hat{k}cos(\omega t) \) 
By just considering two states the Matrix element \(H_{1,ab}=\) \(V_{ab}cos(\omega t)=\braket{\phi_a|eE_0\vec{r}\cdot\hat{k}|\phi_b}\)
Using time-dependent perturbation theory we get that the probability of going from \(\ket{\phi_a}\) to \(\ket{\phi_b}\) is given as \(P_{a\rightarrow b}=|c_b(t)|^2\approx\frac{|V_{ab}|^2}{\hbar^2}(\frac{t}{2})^2sinc^2[(\omega_0-\omega)\frac{t}{2}]\)
Back Extra: Note that this is not an exact result, if \(\omega\) approaches \(\omega_0\) the probability increases and can even go above 1.