Consider the three ion problem. Let \[\hat{H}=\sum_{ij=1}^3J_{ij}\sigma_i^x\sigma_j^x+B(t)\sum_{i=1}^3\sigma_i^y\]
and use the symmetrized states discussed in the lectures. Let \[J_{ij}=-\frac{\hbar}{2}\sum_{\alpha=1}^N\Omega_i\Omega_jb_i^\alpha
b_j^\alpha\frac{(\delta
k)^2\hbar}{2m}\frac{1}{\omega_\alpha^2-\mu^2}\] by neglecting the
time-dependent terms. For the experiment, assume \(\Omega_i=\Omega_j\). Use the \(b_i^\alpha\)’s derived in class.
a.) There are two important \(J_{ij}\)
values \[\begin{aligned}
&J_{12}=J_{21}=J_{13}=J_{31}=J \\
&J_{13}=J_{31}=J'.
\end{aligned}\] Let \(|J|=2\pi\cdot
800\text{ Hz}\). Find \(J'\)
in the same units under the following assumptions: \[\frac{\omega_{tilt}}{\omega_{com}}=\frac{2.5474}{2.7367}
\ \text{and}\
\frac{\omega_{zig-zag}}{\omega_{com}}=\frac{2.2560}{2.7367}\]
with \(\mu=0.874\cdot\omega_{com}\).
HINT: Compute \(J/J'\) and use the \(J\) value given above after you determine
the sign of \(J\). Your resultant
Hamiltonian should look like \[\hat{H}=J(\sigma_1^x\sigma_2^x+\sigma_2^x\sigma_3^x)+J'\sigma_1^x\sigma_3^x+B(t)(\sigma_1^y+\sigma_2^y+\sigma_3^y)\]
b.) Using the symmetrized states, compute \(H(t)\) in each symmetry sector. Find which
sector the ground state lies in for all \(B\) values.
c.) Assume the adiabatic approximation—the state starts polarized along
the \(y\)-axis for each ion site. At
any later time, we follow the ground state at the given value of \(B\). Pick \(B=-B_0\exp(-t/\tau)\) and \[\begin{aligned}
B_0=& \ 2\pi\cdot10,000\text{ Hz} \\
\tau=& \ 10^{-4}\text{ sec}
\end{aligned}\] Run your simulation out to \(t=0.5\times10^{-3}\text{ s}\). Let \(P(\text{state})\) denote the probability to
be in a given state. Plot \(P(\left|\uparrow\uparrow\uparrow\right\rangle_x)+P(\left|\downarrow\downarrow\downarrow\right\rangle_x),P(\left|\uparrow\downarrow\uparrow\right\rangle_x)+P(\left|\downarrow\uparrow\downarrow\right\rangle_x),\)
and \(P(\left|\uparrow\uparrow\downarrow\right\rangle_x)+P(\left|\downarrow\uparrow\uparrow\right\rangle)+P(\left|\downarrow\downarrow\uparrow\right\rangle)+P(\left|\uparrow\downarrow\downarrow\right\rangle)\)
versus time on the same plot.
d.) Repeat (c) using the sudden approximation. To do this, we start the
state initially polarized along the \(y\)-axis and compute the three overlaps:
\[\begin{aligned}
&\alpha=\left\langle\psi_{gs}\middle|\psi_{pol}\right\rangle \\
&\beta=\left\langle\psi_{1st \
ex}\middle|\psi_{pol}\right\rangle \\
&\gamma=\left\langle\psi_{2nd \
ex}\middle|\psi_{pol}\right\rangle
\end{aligned}\] where \(\left|\psi_{pol}\right\rangle\) is the
state polarized along the \(y\)-axis,
and \(\left|\psi_{gs}\right\rangle,\left|\psi_{1st \
ex}\right\rangle,\) and \(\left|\psi_{2nd \ ex}\right\rangle\) are
the ground state and first two excited states of the system in the
symmetry sector where the ground state lies. Then at any time
later, the sudden approximation says \[\left|\psi(t)\right\rangle_{sudden}=\alpha\left|\psi_{gs}(t)\right\rangle+\beta\left|\psi_{1st}(t)\right\rangle+\gamma\left|\psi_{2nd}(t)\right\rangle.\]
Use this to repeat the same calculation as above and produce the
relevant probability plots.
Assume a spherical well of radius \(R\): \[V(r)=\begin{cases}0 & 0\le r<R \\ \infty
& r>R\end{cases}\] The \(s\)-wave bound states are \(\left|n\right\rangle=\frac{A}{r}\sin{\frac{n\pi
r}{R}}\) and \(A\) is found
through normalization. If an electron is in the ground state and \(R\) suddenly increases to \(R'=R/\nu\) for \(0\le\nu\le1\)...
a.) What is the probability to find the electron in the new ground
state? (plot as a function of \(\nu\))
b.) What is the probability that the electron will be in an excited
state with energy \(n\)?
c.) Sum the probabilities to show that \[\sum_{n'}P_{n\leftarrow 1}=1\] for the
case \(\nu=\frac{1}{2}\).
A tritium nucleus \((Z=1)\)
undergoes beta decay, where a neutron in the nucleus emits an electron
and changes to a proton \((Z=2)\). The
emitted electron receives so much energy that it is not bound to the ion
and flies away almost instantly.
a.) If the electron orbiting the tritium atom is in the ground state,
what is the probability that it lies in the state with principal quantum
number \(n\) in the Helium ion?
b.) Sum up the probability \[\sum_{n=1}^\infty P_{n\leftarrow gs}\]
Note that this sum is not equal to one. Explain why. Note that
you have to do this sum numerically.