Time-Independent Perturbation Theory

Hamiltonian: \(H_0 = \frac{p^2}{2\mu} - \frac{e^2}{r} \Rightarrow E_n = -\frac{E_I}{n^2}\).

Relativistic Schr\:odinger Equation: \(\left[\left(p^\mu - \frac{e}{c}A^\mu\right)^2-m^2c^2\right]\Psi=0\) with \(p^\mu = i\hbar\left(c\frac{\partial}{\partial t},-\nabla\right), A^\mu = (\phi,\vec{A})\).

Spin-1/2 charged particles gives the Dirac equation: \(\left[\gamma_\mu\left(p^\mu - \frac{e}{c}A^\mu\right) - mc\right]\psi = 0\). With, \(\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix},\vec{\gamma}=\begin{pmatrix} 0 & \vec{\sigma} \\ -\vec{\sigma} & 0 \end{pmatrix}\). This gives energies of: \(E_{n,j} = m_ec^2[1 + \alpha^2(n-j-1/2+\sqrt{(j+1/2)^2-\alpha^2})^{-2}]^{-1/2} \approx m_e c^2 - \frac{E_I}{n^2} - \frac{m_e c^2\alpha^4}{2n^4}\left(\frac{n}{j+1/2}-\frac{3}{4}\right) + \cdots\). \(j\) is from \(\vec{J} = \vec{L} + \vec{S}\). \(- \frac{m_e c^2\alpha^4}{2n^4}\left(\frac{n}{j+1/2}-\frac{3}{4}\right)\) is the fine structure term.

From the Semi-classical Bohr model, we will see that \(e^-\) in hydrogen atom is weakly relativisitic.

\(\frac{\mu v^2}{r} = \frac{e^2}{r^2},\mu vr = n\hbar\).

For \(n=1\), \(v=\frac{\hbar}{\mu a_0} = \frac{e^2}{\hbar}\), \(a_0 = \frac{\hbar^2}{\mu e^2}\). For \(\frac{v}{c} = \frac{e^2}{\hbar c} = \alpha \approx \frac{1}{137}\).

\(V_{fine} = -\frac{p^4}{8m_e^3c^2} + \frac{1}{2m_e^2 c^2}\frac{1}{r}\frac{dV(r)}{dr}\vec{L}\cdot\vec{S} + \frac{\hbar^2}{8m_e^2c^2}\nabla^2 V(r)\). Numbering each term sequentially \(V_{fine} = V_{mv} + V_{so} + V_{D}\).

• \(V_{mv}\) is due to mass and velocity \(m(v)\).
• \(V_{so}\) is due to spin orbit coupling.
• \(V_D\) is Darwin term - the non-locality term

1. \(V_{mv}\). \(E = c\sqrt{p^2 + m_e^2c^2}\). If \(\frac{v}{c}\ll 1\) then \(E = m_e c^2\sqrt{1 + \frac{p^2}{m_e^2c^2}}\approx m_e c^2\left(1 + \frac{p^2}{2m_e^3c^2} - \frac{p^4}{8m_e^4c^4} + \cdots\right) = m_ec^2 + \frac{p^2}{2m_e^3c^2} - \frac{p^4}{8m_e^3c^2} + \cdots\). \(\frac{\frac{p^4}{8m_e^3c^2}}{\frac{p^2}{2m_e}} = \frac{p^2}{4m_ec^2} = \frac{1}{4}\frac{v^c}{c^2} = \frac{1}{4}\alpha^2\).
2. \(V_{so}\). Consider stationary system \(K\) \((x,y,z)\) for our proton and a moving system \(K'\) \((x',y',z')\) for the electron which has velocity \(\vec{v}\). Then, a magnetic field \(\vec{B}' = -\frac{\vec{v}}{c}\times \vec{E}\), with the electric field being due to the proton and \(\vec{B}'\) is the magnetic field the electron experiences. Now with the magnetic field, we have a magnetic moment of the electron, \(\vec{M}_S = \frac{e}{m_e c}\vec{S}\). So, \(V_{SO} = -\vec{M}_S\cdot\vec{B}' = -\frac{e}{m_ec}\vec{S}\cdot(\vec{v}\times\vec{E})\). Consider \(\vec{F} = e\vec{E} = -\frac{dV}{dr}\frac{\vec{r}}{r}\). For the coulomb interaction, \(\vec{F} = -\frac{e^2}{r^3}\vec{r}\). So, \(-\frac{e^2}{m_e^2c^2r^3}\vec{S}\cdot(\vec{p}\times\vec{r}) = \frac{e^2}{m_e^2c^2 r^3}\vec{L}\cdot\vec{S}\). The factor of two in the original term comes from restricting a degree of freedom. \(\frac{V_{SO}}{\frac{e^2}{r}} \approx \frac{e^2}{m_e c^2 r}\hbar^2\frac{r}{e^2} = \frac{\hbar^2}{m_e^2c^2 r^2} \approx \frac{\hbar^2}{m_e^2c^2a_0^2} = \frac{e^4}{\hbar^2 c^2} =\alpha^2\).
3. For \(\ell = 0\) this term is strongest, i.e. it is relevant to s-states. Because the electron will be bouncing around at small radii, the first derivative averages out but the second derivative becomes ’significant’. \(V_D = \frac{\hbar^2}{8m_e^2 c^2}\nabla^2(\frac{-e^2}{r}) = -\frac{\hbar^2e^2}{8m_e^2c^2}\nabla^2(\frac{1}{r}) = \frac{\pi e^2\hbar^2}{2m_ec^2}\delta(\vec{r})\). \(\langle \Psi|V_D|\Psi\rangle \sim \frac{\pi e^2\hbar^2}{2m_e^2c^2}\int |\Psi|^2\delta(\vec{r})d^3r = \frac{\pi e^2\hbar^2}{2m_e^2c^2}|\Psi(0)|^2\). For the Hydrogen atom, \(\Psi\sim r^\ell \exp\left(\frac{-r}{na_0}\right)Y_\ell^m\). So, this only gives non-zero for \(\ell=0\). From \(\int |\Psi|^2d^3x=1,|\Psi(0)|^2\approx \frac{3}{4\pi a_0^3}\). \(\frac{\langle V_D\rangle}{e^2/r} = \frac{3\pi \hbar^2}{8\pi m_e^2 c^2 a_0^2} = \frac{3}{8}\alpha^2\)

\(V_{hf} = \frac{e}{m_er^3}\vec{L}\cdot\vec{M}_I + \frac{1}{r^3}\left[3(\vec{M}_S\cdot\hat{n})(\vec{M}_I\cdot\hat{n}) - \vec{M}_S\cdot\vec{M}_I\right] + \frac{8\pi}{3}\vec{M}_S\cdot\vec{M}_I\delta(\vec{r})\).

First term is the interaction between the electron’s magnetic field and the proton’s magnetic moment. The second term is the electron-proton dipole-dipole interaction. The third term is Fermi’s contact term.

Our state is then: \(|n\ell m sm_sIm_I\rangle\to|n\ell m;m_sm_I\rangle\).

So, for a 1s state,

1. \(\langle 100;m_s'm_i'|\vec{L}\cot\vec{M}_I|00m_sm_I\rangle=0\)
2. \(\langle 100;m_s'm_i'|\frac{3\gamma_s\gamma_I}{r^3}(\vec{S}\cdot\hat{n})(\vec{I}\cdot\hat{n})-\frac{\gamma_S\gamma_I}{r^3}\vec{S}\cdot\vec{I}|100;m_sm_I\rangle = \langle \psi'|f(r)[3\sum_i S_in_i\sum_jI_jn_j-\sum_{ij}S_iI_j\delta_{ij}]|\psi\rangle = \langle \psi'|f(r)\sum_{ij}S_iI_j(3n_in_j-\delta_{ij})|\psi\rangle = \langle \psi'|f(r)\sum_{ij}S_iI_jY^q_2|\psi\rangle = \langle 100;m_s'm_I'|f(r)\{S_iI_j\}Y_2^q|100m_sm_I\rangle = 0\), \(\ell_1 = 0\) from \(\psi\) and \(\ell_2=2\) from the spherical harmonic. Thus, \(\ell_3\) from \(\psi'\) must be \(\ell_1+\ell_2 = 2\).
3. \(\frac{8\pi}{3}\langle 100;m_s'm_I'|\gamma_S\gamma_I\vec{S}\cdot\vec{I}\delta(\vec{r})|100;m_sm_I\rangle = \frac{8\pi}{3}\gamma_S\gamma_I\langle 100|\delta(\vec{r})|100\rangle\langle m_s'm_I'|\vec{S}\cdot\vec{I}|m_sm_I\rangle = \frac{8\pi}{3}\gamma_S\gamma_I|\Psi_{1s}(0)|^2\langle m_s'm_I'|\vec{S}\cdot\vec{I}|m_sm_I\rangle\). \(|m_sm_i\rangle\to|Fm_F\rangle,\vec{F}=\vec{I}+\vec{S},F=\{0,1\}\).

We know \(\nu_{hf} = 1420405751.768\pm0.001\) Hz.

Beginning of QED. 1.06 GHz. Got Nobel prize in 1955.

See PRL 94, 022001 (in 2005).

For \(n=2\). \(2p_{3/2},2s_{1/2}\). \(1s_{1/2}\to 2p_{3/2}\).