PHYS 5002: Homework 7

Perturbed two-level system

Consider a general two-level system with a Hamiltonian: \[\hat{H}_0=\begin{pmatrix}E_0^0 & 0 \\ 0 & E_1^0\end{pmatrix}, \ \hat{V}=\begin{pmatrix}V_{00} & V_{01} \\ V_{10} & V_{11}\end{pmatrix}\] where \(E_0^0\le E_1^0\) and \(V_{01}=V_{10}\).
 
a.) Find the exact eigenvalues of \(\hat{H}=\hat{H}_0+\hat{V}\)
 
b.) Expand both energies to order \(V^2\).
 
c.) No Crossing Theorem: As \(V\) increases, one might expect the lower level to become higher than the higher level. Show this cannot occur if \(V_{01}\ne0\).
 
d.) Specialize to the degenerate case \(E_0^0=E_1^0\). Can you expand the energies for small \(V\)? To what order in \(V\) are the corrections to \(E_0\)?
 
e.) Now consider also \(V_{00}=V_{11}=0\). Compare the exact result to an incorrect application of Rayleigh-Schrödinger perturbation theory to first order. Compare to Wigner-Brillouin perturbation theory at second order.

More degenerate perturbation theory

In the case where \(\hat{P}_k\hat{V}\hat{P}_k=\hat{V}_k=0\), the degeneracy is not lifted at all to first order in degenerate perturbation theory. Then, all \(E_{k,n_k}^{(1)}=0\) and \(\left|k,n_k\right\rangle_\parallel^{(1)}\) are undetermined.
 
a.) Show that second order perturbation theory gives another eigenvalue problem for \(\left|k,n_k\right\rangle_\parallel^{(2)}\).
 
b.) Show the equation for the energy shift is \(\det(V_k-E_k^{(2)})=0\) with \[V_{ij}^k=\sum_{k',n_{k'}}\frac{\left\langle ki\middle|\hat{V}\middle|k'n_{k'}\right\rangle\left\langle k'n_k'\middle|\hat{V}\middle|kj\right\rangle}{E_k^0-E_{k'}^0}\] where \(\left|\cdot\right\rangle\) refers to the unperturbed kets. Here \(k\) denotes the degenerate subspace and i runs through all of its members, while \(k'\) is an energy eigenstate that is not degenerate with the \(k\) states, but could be degenerate itself.This is known as the matrix of second order matrix elements. Note that it takes a moment to really follow how to calculate this.
 
c.) Consider the Hamiltonian: \[\hat{H}_0=\begin{pmatrix}E_0^0 & 0 & 0 \\ 0 & E_0^0 & 0 \\ 0 & 0 & E_1^0\end{pmatrix}, \ \hat{V}=\begin{pmatrix}0 & 0 & a \\ 0 & 0 & b \\ a^* & b^* & 0\end{pmatrix}\] Use perturbation theory to compute all three energies to \(O(V^2)\) and determine \(\left|k,n_k\right\rangle_\parallel^{(2)}\) for the \(2\times 2\) degenerate space.
 
d.) Compute the energies of the Hamiltonian in (c) exactly. Compare the energies to the perturbative expansion by plotting on the same plot. Choose \(E_0^0=0\) and \(E_1^0=1\) and \(a=b\) with \(0<a<5\) and \(a=\frac{1}{2}b\) with \(0<a<2.5\).

Hyperfine splitting

We want to consider the effect of nuclear spins on the atomic energy levels. In particular, we will work with hydrogen taking into account that the nucleus is a proton with spin one half and has a magnetic moment.
 
Let \(\hat{\mathbf{S}}\) be the electron spin operator and \(\hat{\mathbf{I}}\) be the proton spin operator. The magnetic moment of the proton is \[\mathbf{m}_p=g_p\frac{e\hbar}{2Mc}\mathbf{I}\] where \(M\) is the mass of the proton and \(g_p=5.56\) and of the electron is \[\mathbf{m}_e=-g_e\frac{e\hbar}{2mc}\mathbf{S}\] where \(m\) is the mass of the electron and \(g_e=2\).
 
If we start analyzing classically, we imagine the proton moment provides a magnetic field \[\mathbf{B}=\nabla\times\mathbf{A}, \ \text{with}\quad\mathbf{A}=-\mathbf{m}_p\times\nabla \left (\frac{1}{r}\right )\] The electron experiences an interaction with this field. \[\hat{H}'=-\mathbf{m}_p\cdot\mathbf{B}=\mathbf{m}_e\cdot\mathbf{m}_p\nabla^2\left (\frac{1}{r}\right )+(\mathbf{m}_e\cdot\nabla)(\mathbf{m}_p\cdot\nabla)\left (\frac{1}{r}\right )\] or \[\hat{H}'=\frac{e^2g_eg_p\hbar^2}{4nMc^2}\left\{-\mathbf{S}\cdot\mathbf{I} \ \nabla\left (\frac{1}{r^2}\right )+\mathbf{S}\cdot\nabla \ \mathbf{I}\cdot\nabla \ \left (\frac{1}{r}\right )\right \}.\] We will treat this to first order as a perturbation of the \(l=0\) levels of hydrogen (ignoring fine structure). For \(l=0\) wavefunctions, the spatial wavefunction is symmetric, so \[\int d\Omega \ \mathbf{S}\cdot\nabla \ \mathbf{I}\cdot\nabla \ \left (\frac{1}{r}\right )=\frac{\mathbf{S}\cdot\mathbf{I}}{3}\nabla^2\left (\frac{1}{r}\right )\int d\Omega\] So we write \[\hat{H}'=-\frac{2}{3}\frac{e^2g_eg_p\hbar^2}{4mMc^2}\nabla^2\left (\frac{1}{r}\right ) \ \mathbf{S}\cdot\mathbf{I}.\] But, \(\nabla^2\left (\frac{1}{r}\right )=-4\pi\delta(\mathbf{r})\) so \[\hat{H}'=\frac{4\pi}{3}\frac{e^2g_eg_p\hbar^2}{2mMc^2}\delta(\mathbf{r})\mathbf{S}\cdot\mathbf{I}.\] In the absence of the perturbation, the \(l=0\) states are fourfold degenerate (electron spins up and down and proton spins up and down). Work on the ground state \(n=1\). The perturbation will split the levels into two different levels. Compute the energy separation of these levels to lowest order in \(\hat{H}'\).
 
Express your answer analytically in terms of \(e,g_e,g_p,m,M,c,\text{etc.}\) Numerically determine the wavelength (in cm) of a photon with \(\Delta E\). This photon is very important in radio astronomy. Note that it is very easy to make a mistake in your numerical calculation, so work carefully and check things frequently. Your final answer will be on the order of a foot.