The zeroth-order eigenstates are the $\ket{n,l,j,m_j}$ states.
These states diagonalize the perturbation matrix $W'(n,j)_{(l',m_j')(l,m_j)}$, where $l,l' = j\pm1/2$ and $m_j,m_j' = -j, -j+1,\ldots,j-1,j$, and the eigenvalues are given by Griffiths equation 7.79:
$$
E_Z^1 = \mu_B B_{\text{ext}} g_J(j,l) m_j
$$
where
$$
g_J(j,l) = (2j+1)/(2l+1)
$$
For the $n=2$ states we have two energy levels: one for $j=1/2$ and another for $j=3/2$.
The $j=1/2$ level has four states:
$l=0, m_j = \pm 1/2$ and
$l=1, m_j = \pm 1/2$ and
The $j=3/2$ level has four states as well: $l = 1/2, m_j = \pm 3/2, \pm 1/2$.
Here is the plot I compute:
and here is the code used to generate the plot:
import numpy as np
from matplotlib import pyplot as plt
mub = 5.788E-5 / 27.2 # mu_B B_ext in hartrees / tesla
alpha = 1. / 137.036 # fine structure constant
mubmax = .4 # max magnetic field
e = {1:0,3:alpha**2 / 32.} # zero field energy levels (within additive constant)
e1 = e[1]
e3 = e[3]
# plot labels for zero field energy levels
plt.text(
-.1*mubmax,
e1,
r'$-\frac{1}{8} - \alpha^2 \frac{5}{128}$',
horizontalalignment='right',
verticalalignment='center'
)
plt.text(
-.1*mubmax,
e3,
r'$-\frac{1}{8} - \alpha^2 \frac{1}{128}$',
horizontalalignment='right',
verticalalignment='center'
)
for j in (1,3):
for mj in range(j+1):
for l in {1:(0,1),3:(1,)}[j]:
# calculate g-factor
gj = (j+1.)/(2.*l+1)
# convert mj label to actual m_j value
MJ = mj - j / 2.
# compute zero field and max field energies
y_coords = [e[j],e[j]+mub*gj*.5*MJ*mubmax]
# plot energy as function of field
plt.plot([0,mubmax],y_coords,color='black')
# plot label of quantum state
plt.text(
1.1*mubmax,
y_coords[1],
'$j=\\frac{%d}{2},\\,m_j=%s\\frac{%d}{2},\\,l=%d$' %
(j,'-' if 2*mj-j < 1 else '',abs(2*mj-j),l),
horizontalalignment='left',verticalalignment='center'
)
plt.title('weak field Zeeman effect for $n=2$')
plt.xlabel(r'$B_{\mathregular{ext}}$ (T)')
plt.yticks([]) # turn off y labels
plt.show()
Q2 - G7.44
Plugging in $s=0$ we get:
$$
\frac{1}{n^2}\ev{1} - a \ev{r^{-1}} + \frac{0}{4}\left( \cdots \right) = 0
$$
after noting that $\ev{1} = 1$ and doing some rearranging we find
$$
\ev{r^{-1}} = \frac{1}{an^2}
$$
as you proved last week.
Plugging in $s=1$ we get:
$$
\frac{2}{n^2}\ev{r} - 3a\ev{1} + \frac{1}{4}\left[ (2l + 1)^2 - 1\right] a^2 \ev{r^{-1}}
$$
from which we can solve for $\ev{r}$:
$$
\ev{r}_{nl} = \frac{a}{2} \left[ 3n^2 - l ( l + 1 ) \right]
$$
the rest follow similarly.
Q3 - G7.45
a)
The perturbation to this non degenerate level is given by the expectation value $\ev{H'_S}$.
Since the $\ket{ n, l, m_l }$ states have definite parity, and the perturbation has odd parity, the expectation value will vanish (i.e. it involves that integral of an overall odd function over all space).
b)
The diagonal elements of the perturbation matrix in the $\ket{n=2,l,m_l}$ degenerate subspace vanish by the same argument described in part a).
For the off-diagonal elements we begin by noting that the action of the perturbation on a state with definite $L_z = \hbar m_l$ will produce be another state with $L_z = \hbar m_l$.
This is a result of the perturbation commuting with $L_z$.
See Q8 from week 2 if the connection is unclear to you.
Therefore the perturbation can only "couple" states with the same $m_l$, i.e. the matrix elements $W_{(l',m'_l)(l,m_l)}$ vanish unless $m'_l = m_l.$
Since the $l=1,\,m_l=\pm1$ are the only $n=2$ states with $m_l=\pm1$, they constitute "good states" for degenerate pertubation theory, they couple to no other degenerate states.
Their first correction is thus given by their diagonal matrix elements, which, as discussed, necessarily vanishes.
The only non-zero elements of the perturbation matrix are then the two that connect the states $m_l = 0,/, l=0,1$.
Since the perturbation is hermitean it suffices to compute one and the other can be obtained by complex conjugation.
Consider then $W_{(l=1,m_l=0),(l=0,m_l=0)}\equiv w_{10}$:
general approach
Rearrange the recursion relation for the legendre polynomials $P_l(x)\equiv P^0_l(x)$:
$$
(l+1)P_{l+1}(x) = (2l+1)xP_l(x) - l P_{l-1}(x)
$$
1
to the form
$$
xP_l(x) = \frac{1}{2l+1} \left( (l+1)P_{l+1}(x) + l P_{l-1}(x) \right)
$$
Applying this relation for $l=0$ and keeping only terms with $l=0,1$ we find
$$
x P_0(x) \to P_1(x)
$$
or, equivalently,
$$
\cos \theta Y^0_0(\theta,\phi) =
\frac{1}{\sqrt{4\pi}} \cos \theta P_0(\cos \theta) \to
\frac{1}{\sqrt{4\pi}} P_1(\cos \theta) =
\frac{1}{\sqrt{3}} Y^0_1(\theta,\phi)
$$
fast approach
Observe from a table of spherical harmonics that
$$
\cos \theta Y^0_0(\theta,\phi) = \frac{1}{\sqrt{3}} Y^0_1(\theta,\phi)
$$
The matrix element $w_{10}$ is thus
$$
eE_{\text{ext}} \frac{1}{\sqrt{3}} \bra{n=2,\,l=1}r\ket{n=2,\,l=0} =
-3 eE_{\text{ext}} a_o
$$
We computed $\bra{n=2,\,l=1}r\ket{n=2,\,l=0}$ by observing that:
$$
R_{20}(z) = 2\sqrt{3}\frac{1-z/2}{z} R_{21}(z)
$$
Where $z\equiv r/a_o$.
Therefore
$$
r R_{20}(z) = a_o z R_{20}(z) = a_o 2\sqrt{3}\left( 1 - z / 2 \right) R_{21}(z)
$$
We find then
$$
\bra{n=2,\,l=1}r\ket{n=2,\,l=0} = a_o 2\sqrt{3} \left( 1 - \frac{ 1 }{2} \ev{\frac{r}{a_o}}_{n=2,\,l=1} \right)
$$
but from an earlier problem we already found that
$$
\ev{r}_{nl} = \frac{a_o}{2} \left[ 3n^2 - l(l+1) \right]
$$
So that finally
$$
\bra{n=2,\,l=1}r\ket{n=2,\,l=0} = -3 \sqrt{3} a_o
$$
The $2 \times 2$ perturbation matrix between the $m_l = 0$ states is:
$$
w_{10}
\begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
$$
we are pretty sick of this matrix by this point. The eigenvectors are
$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ \pm 1 \end{bmatrix}
$$
with eigenvalues $\pm w_{10}$.
The $E_2$ level thus splits into three sublevels with shifts $0,\pm w_{10}$.
c)
The "good" wavefunctions are $\psi_{21\pm1}$ and $\frac{1}{\sqrt{2}}\left( \psi_{210} \pm \psi_{200} \right)$.
The $x$ and $y$-components vanish for all four states. To see this, write
$$
x = r \sin \theta \cos \phi = r \sin \theta \frac{1}{2} \left( e^{i \phi} + e^{-i \phi} \right)
$$
and similarly
$$
y = r \sin \theta \sin \phi = r \sin \theta \frac{1}{2i} \left( e^{i \phi} - e^{-i \phi} \right)
$$
Since all the good wavefunctions are states of definite $m_l$, the $\phi$-integral part of the expectation value will vanish.
The expectation value for the $z$-component of the dipole moment is this same matrix elements we computed for part b), just scaled by $-\frac{1}{E_{\text{ext}}}$:
The kinetic energy of a gaussian wavefunction was computing in Griffiths equation 8.5.
To compute the potential energy we first compute:
$$
2 \int_0^{\infty} dx x^4 e^{-2bx^4} =
2 \int_0^{\infty} \frac{d(\sqrt{2b}x)}{\sqrt{2b}}\sqrt{2b}^{-4} \left(\sqrt{2b}x\right)^4 e^{-2bx^2} =
2 (2b)^{-5/2} \int_0^{\infty} dz z^4 e^{-z^2} =
2 (2b)^{-5/2} \sqrt{\pi} \frac{3!!}{8} =
\frac{ 3 \sqrt{\pi} (2b)^{-5/2} }{4}
$$
Mixing the constants back in we find:
$$
\ev{V} = \alpha \frac{\sqrt{2b}}{\pi} \frac{ 3 \sqrt{\pi} (2b)^{-5/2} }{4} =
\alpha \frac{ 3 }{4}(2b)^{-2}
$$
Recalling that
$$
\ev{T} = \frac{\hbar^2}{4m} 2b
$$
Minimizing with respect to $2b$ we find:
$$
\frac{\partial}{\partial 2b}\ev{T+V} = \frac{\hbar^2}{4m} - \alpha \frac{3}{2} (2b)^{-3} = 0
$$
Leads to minimum energy width parameter of
$$
2b = \sqrt[3]{6 \frac{\alpha m}{\hbar^2}} \equiv \beta
$$
which gives a upper bound on the ground state energy of
$$
E_{\text{min}} = \frac{\hbar^2}{4m} \beta + \alpha \frac{3}{4} \frac{1}{\beta^2}
$$
Q4: 8.3
The kinetic energy is given in Griffiths equation 8.13:
$$
\ev{T} = \frac{6 \hbar^2}{ma^2}
$$
Inspecting equation 8.10 we find the modulus square of the wavefunction at the origin is
$$
| \psi(0;a) |^2 = A^2 a^2 / 4 = 3 / a
$$
giving an expectation of the potential energy of
$$
\ev{V} = - A^2 \alpha 3 / a = - 36 \alpha / a^4
$$
For a grand total of
$$
\ev{H} = 6 \frac{\hbar^2}{m} a^{-2} - 36 \alpha a^{-4}
$$
This function is minimized at
$$
a = \sqrt{\frac{12 \alpha m}{\hbar^2}}
$$
or
$$
a^{-2} = \frac{\hbar^2}{12 \alpha m}
$$
For an upper bound of
$$
E_{\text{min}} = \frac{1}{4} \frac{\hbar^4}{\alpha m^2}
$$
You're free to work the algebra for this one, but recognize that this trial function is proportional to the first excited state for the harmonic oscillator.
If we match the parameter $b$ to the coefficient in exponential term in the known solution, i.e. $b \to \frac{1}{2} \frac{m\omega}{\hbar}$ we will get
$$
E_{\text{min}} = \frac{3}{2} \hbar \omega
$$
footnotes
1
It should not be surprising that multiplying a legendre polynomial gives us a linear combination of legendre polynomials.
They are defined as the eigenfunctions to the hermitean differential operator $(1-x^2)\frac{d^2}{dx^2} - 2x \frac{d}{dx}$ (with eigenvalues $l(l+1)$), and therefore constitute a basis of functions (on the interval $[-1,1]$).
