In class, we derived an identity for the hydrogen energy
eigenfunction in a form reminiscent of the position wavefunction: \[\left|\psi_{nl}\right\rangle_{norm}=(-i)^{n-l-1}\left(\frac{na_0}{2}\right)^{n-1}\sqrt{\frac{(2n-1)!(n-l-1)!}{(n+l)!}}\frac{1}{\hat{r}^l}P_n^l(\hat{r}_x,\hat{r}_y,\hat{r}_z)\left(\frac{2\hat{r}}{na_0}\right)^lL_{n-l-1}^{2l+1}\left(\frac{2\hat{r}}{na_0}\right)\left|\phi_n\right\rangle\]
a.) We want to look at \[\left|\psi_{21}\right\rangle=\frac{a_0}{\hat{r}}P_n^1(\hat{r}_x,\hat{r}_y,\hat{r}_z)\left(\frac{\hat{r}}{a_0}\right)L_0^3\left(\frac{\hat{r}}{a_0}\right)\left|\phi_2\right\rangle\]
Recall \(L_0^3=1\) and choose \(P_n^1(\hat{r}_x,\hat{r}_y,\hat{r}_z)=\sqrt{\frac{3}{4\pi}}\hat{r}_\alpha\)
which is a \(p\)-wave state. So, \[\left|\psi_{21}\right\rangle_{norm}=\sqrt{\frac{3}{4\pi}}\hat{r}_\alpha\left|\phi_2\right\rangle\]
for \(\alpha=x,y,\) or \(z\).
Compute \(\left\langle
p_x,p_y,p_z\middle|\psi_{21}\right\rangle_{norm}=\psi_{21}(p_x,p_y,p_z)\)
recalling that \[\hat{p}_\alpha\left|\phi_2\right\rangle=\frac{i\hbar}{2a_0}\frac{\hat{r}_\alpha}{\hat{r}}\left|\phi_2\right\rangle\]
Use this to rewrite \(\left\langle\mathbf{p}\middle|\hat{r}_x\middle|\phi_2\right\rangle=\sqrt{\frac{3}{4\pi}}\frac{2a_0}{i\hbar}\left\langle\mathbf{p}\middle|\hat{r}\hat{p}_\alpha\middle|\phi_2\right\rangle\).
Then, use the commutator \([\hat{p}_\alpha,\hat{r}]=-i\hbar\frac{\hat{r}_\alpha}{\hat{r}}\)
to show that \[\psi_{21}(\mathbf{p})=\sqrt{\frac{3}{4\pi}}\frac{2a_0p_\alpha}{i\hbar}\left\langle\mathbf{p}\middle|\hat{r}\middle|\phi_2\right\rangle+\sqrt{\frac{3}{4\pi}}2a_0\left\langle\mathbf{p}\middle|\frac{\hat{r}_\alpha}{\hat{r}}\middle|\phi_2\right\rangle\]
Now, one last trick is needed: Write \(\hat{r}=\sum_{\alpha}\frac{\hat{r}_\alpha\hat{r}_\alpha}{\hat{r}}\)
and replace \(\frac{\hat{r}_\alpha}{\hat{r}}\left|\phi_2\right\rangle=\frac{i\hbar}{2a_0}\hat{p}_\alpha\left|\phi_2\right\rangle\)
etc. Note further that you should calculate \(\left\langle\mathbf{p}\middle|\hat{r}\middle|\phi_2\right\rangle\)
separately and substitute the result into the final answer. You
will end up needing to compute \(\left\langle\mathbf{p}\middle|\phi_2\right\rangle\),
which can be done similar to how we did this in Lecture
11. Finish the problem to compute \(\psi_{21}(\mathbf{p})\). You do not need to
evaluate \(\left\langle
0_p\middle|\phi_2\right\rangle\).
b.) Now consider \(\left|\psi_{20}\right\rangle=-ia_0\sqrt{\frac{3}{4\pi}}L_1^1\left(\frac{\hat{r}}{a_0}\right)\left|\phi_2\right\rangle\)
where \(L_1^1\left(\frac{\hat{r}}{a_0}\right)=2-\frac{\hat{r}}{a_0}\)
so \[\psi_{20}(\mathbf{p})=-i\sqrt{\frac{3}{4\pi}}a_0\left\langle\mathbf{p}\middle|\left
(2-\frac{\hat{r}}{a_0}\right )\middle|\phi_2\right\rangle.\]
Calculate \(\psi_{20}(\mathbf{p})\)
using the same techniques as above.
Using \(\hat{H}=\hat{H}_0+\hat{V}\)
where \(V_{nm}=\left\langle
n\middle|\hat{V}\middle|m\right\rangle\) we have the
non-degenerate perturbation theory through fourth order is \[E_n=E_n^0+\text{ first order correction~} +
\text{~second order correction~} + \text{~third order correction~} +
\text{~fourth order correction}\] where \[\text{first order correction}=V_{nn}\]
\[\text{second order correction}=\sum_{m\ne
n}\frac{V_{mn}}{E_n^0-E_m^0}\] \[\text{third order correction}=\sum_{m\ne
n}\sum_{m\ne
n'}\frac{V_{nm}V_{mm'}V_{m'n}}{(E_n^0-E_m^0)(E_n^0-E_{m'}^0)}-V_{nn}\sum_{m\ne
n}\frac{|V_{nm}|^2}{(E_n^0-E_m^0)^2}\] and last but not least,
\[\begin{gathered}
\text{fourth order correction}=\sum_{m=\ne n}\sum_{m'\ne
n}\sum_{m''\ne
n}\frac{V_{nm}V_{mm'}V_{m'm''}V_{m''n}}{(E_n^0-E_m^0)(E_{n}^0-E_{m'}^0)(E_{n}^0-E_{m''}^0)}-\sum_{m\ne
n}\sum_{m'=n}\frac{|V_{nm}|^2|V_{nm'}|^2}{(E_n^0-E_m^0)(E_n^0-E_{m'}^0)}
\\ -V_{nn}\sum_{m\ne n}\sum_{m'\ne
n}\frac{V_{nm}V_{mm'}V_{m'n}}{(E_n^0-E_m^0)(E_n^0-E_{m'}^0)}\left(\frac{1}{E_n^0-E_m^0}+\frac{1}{E_n^0-E_{m'}^0}\right)+V_{nn}^2\sum_{m\ne
n}\frac{|V_{nm}|^2}{(E_n^0-E_m^0)^3}
\end{gathered}\] Now, take \(\hat{H_0}=\frac{\hat{p}^2}{2m}+\frac{1}{2}k\hat{x}^2\)
and \(\hat{V}=\frac{1}{2}\Delta
k\hat{x}^2\)
a.) Compute \(E_n\) exactly, and Taylor
expand in a series in \(\Delta k\)
through fourth order in \(\Delta
k\).
b.) Compute \(E_n\) through fourth
order in \(\hat{V}\) using the above
formula for Rayleigh-Schrödinger perturbation theory.
c.) Compute \(E_n\) to second in
Wigner-Brillouin perturbation theory for the ground state
only.
d.) Plot \(E(\text{second order RS}),
E(\text{fourth order RS}), E(\text{second order WB}),\) and the
Taylor series through fourth order for \(0\le\frac{\Delta k}{k}\le3\) (for the
ground state only). Comment on the quality of the different
approximations.
Consider the isotropic simple harmonic oscillator: \[\hat{H}=\frac{\hat{p}_x^2+\hat{p}_y^2+\hat{p}_z^2}{2m}+\frac{1}{2}k(\hat{x}^2+\hat{y}^2+\hat{z}^2)\]
a.) Using the factorization method in Cartesian coordinates, find the
energy eigenvalues.
b.) Rewrite \[\hat{H}=\frac{\hat{p}_r^2}{2m}+\frac{\hat{L}^2}{2m\hat{r}^2}+\frac{1}{2}k\hat{r}^2\]
and find the eigenvalues using the factorization method in spherical
coordinates. Recall \[\hat{H}_l=\frac{\hat{p}_r^2}{2m}+\frac{\hbar^2l(l+1)}{2m\hat{r}^2}+\frac{1}{2}k\hat{r}^2\]
Verify they are the same as those found in (a).
HINT: Use \[\hat{B}_r^\dagger(l)=\frac{1}{\sqrt{2m}}\left(\hat{p}_r+\left[m\omega\alpha\hat{r}+\frac{\hbar\beta}{\hat{r}}\right]\right)\]
with both \(\alpha\) and \(\beta\) both nonzero.
c.) Plot the energy levels for discrete values of \(l\).
a.) Find the ground state wavefunction (use either Cartesian or
spherical coordinates).
b.) Add a perturbation \(\hat{V}=\frac{1}{2}\Delta k\hat{x}^2\). The
ground state is non-degenerate. Compute \(E_{gs}\) through second order in
Rayleigh-Schrödinger perturbation theory.
c.) Compute the energy in the Cartesian basis and compare to the
perturbative energy.