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.
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 \(0<a<2.5\).
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.