We can construct an energy eigenbasis $\{ \Psi_n \otimes \ket{S M_s} \}$ of distinguishable states the for hamiltonian defined in equation 5.38 by taking the tensor product of wavefunctions $\Psi_n(\bm{x_1},\bm{x_2})$ with states $\ket{S M_S}$ of definite total spin angular momentum $S^2$ where $\bm{S} = \bm{s_1}+\bm{s_2}$.
Since the hamiltonian commutes with the permutation operator $P$ defined by $(P\Psi)(\bm{x_1},\bm{x_2}):=\Psi(\bm{x_2},\bm{x_1})$, the wavefunctions $\Psi_n(\bm{x_1},\bm{x_2})$ can be written to have definite permutation symmetry so that
$$
\Psi_n(\bm{x_1},\bm{x_2})=\sigma(n)\Psi_n(\bm{x_2},\bm{x_1})
$$
where $\sigma(n) \in \{+1,-1\}$.
We also note that the spin states $\ket{S M_S}$ are also of definite permutation symmetry in that
$$
\ip{(m_s)_1,(m_s)_2}{S M_S}=\tilde{\sigma}(S) \ip{(m_s)_2,(m_s)_1}{S M_S}
$$
where $m_s = \pm 1$ and again $\tilde{\sigma}(S) \in \{+1,-1\}$.
In particular $\tilde{\sigma}(1) = +1$ (triplet state is symmetric) and $\tilde{\sigma}(0) = -1$ (singlet state is antisymmetric).
Exchanging the two particles is equivalent to applying the permutation operator to both the spatial and spin parts of the state. The result on a basis vector is then
$$
\Psi_n \otimes \ket{S M_s} \to \sigma(n) \tilde{\sigma}(S) \Psi_n \otimes \ket{S M_s}
$$
The combinations $n,S$ where $\sigma(n) \tilde{\sigma}(S)$ is positive (negative) comprise a basis for the bosonic (fermionic) subspace of the full space of distinguishable particles.
Q2 - G5.23
From Griffiths equation 5.57
$$
P = \frac{(3\pi^2)^{2/3}\hbar^2}{5m}\rho^{5/3}
$$
we find
$$
\frac{dP}{dV} = \frac{(3\pi^2)^{2/3}\hbar^2}{5m} \left( \frac{5}{3} \rho^{2/3} \frac{d \rho}{d V} \right) = \frac{5}{3} \frac{(3\pi^2)^{2/3}\hbar^2}{5m} \rho^{2/3} \left( -\frac{\rho}{V} \right) = - \frac{1}{V} \frac{5}{3} \frac{(3\pi^2)^{2/3}\hbar^2}{5m} \rho^{5/3} = - \frac{5}{3} \frac{1}{V} P
$$
So that $B \equiv - V \frac{dP}{dV} = \frac{5}{3} P$.
If a material has a mass density $\mu$, and the weight of each atom is $M$, and each atom contributes one electron, then the electron density is $\frac{\mu}{ M }$. Plugging this in and substituting in the known values for copper we get a degeneracy pressure of:
$$
P_{cu} =
\frac{(3\pi^2)^{2/3}\hbar^2}{5m} \left( \frac{\mu}{ M } \right)^{5/3} =
\frac{(3\pi^2)^{2/3}(\hbar c)^2}{5mc^2} \left( \frac{\mu}{M} \right)^{5/3} \approx
\frac{(3\pi^2)^{2/3}(1240 \text{ev nm})^2}{5\cdot 511 \text{keV}} \left( 9 \text{g cm}^{-3} \frac{6\cdot 10^{23}}{60\text{g}} \right)^{5/3}
$$
doing the calculation really sloppily by hand got me $2.4 \times 10^{14} \text{N/m}^2$. I suspect I made a mistake. If anyone knows the answer for sure let me know and I'll update the page.
Q3 - G5.24
a)
Using the trig relation $\sin \left( a - b \right) = \sin a \cos b - \sin b \cos a$ we find that
$$
\sin \left( k x \right) + e^{-iqa} \sin \left( k \left( a - x \right) \right) = \sin \left( k x \right) + e^{-iqa} \left( \sin ka \cos kx - \sin kx \cos ka \right)
$$
so that, ignoring a normalization factor
$$
A = e^{iqa} - \cos kx
$$
and
$$
B = \sin ka
$$
These constants can easily be seen to satisy Griffiths equation 5.70, as required.
b)
We just observe here that we can satisfy equations 5.68 and 5.69 at $k=\frac{\pi}{a}$ by setting $B=0$ and $q=\frac{\pi}{a}$.
Then $\psi(x) \propto \sin \left(\frac{\pi}{a}x\right)$.
An electron in this state avoids the delta spikes (probability at $x=na$ goes to zero for all $n$).
Q4 - G5.27
Consider the parity operator (in one dimension) $R$ which operates on wavefunctions according to the rule $\left( R \psi \right)(x) = \psi(-x)$.
If we have an endlessly repeating potential $V(x) = \sum_{n=-\infty}^{+\infty} v(x+an)$ where $v(x)$ is symmetric, i.e. $v(-x) = v(x)$, then $R$ commutes with $V$:
\begin{align}
& \left( \left( R \hat{V} \right) \psi \right)(x) \\ = &
\left( R \left( \hat{V} \psi \right) \right)(x) \\ = &
\left( \hat{V} \psi \right)(-x) \\ = &
V(-x) \psi(-x) \\ = &
\sum_{n}v(-x+na) \psi(-x) \\ = &
\sum_{n}v(-(x-na)) \psi(-x) \\ = &
\sum_{n}v(x-na) \psi(-x) \\ = &
\sum_{n}v(x+(-n)a) \psi(-x) \\ (!) = &
\sum_{n}v(x+na) \psi(-x) \\ = &
V(x) \psi(-x) \\ = &
V(x) \left(R\psi\right)(-x) \\ = &
\left( \hat{V} \left(R\psi \right) \right)(-x) \\ = &
\left( \left( \hat{V} R \right) \psi \right)(-x)
\end{align}
The line marked with a (!) exploits the fact that summing $n$ from $-N$ to $+N$ is the same as summing from $+N$ to $-N$ (here $N\to\infty$).
Since the parity operator also commutes with kinetic energy (you should check it if you don't buy this), this means that the parity operator commutes with the hamiltonian.
This means that if we operator $R$ on an energy eigenstate $\psi_q$ with energy $E_q$, we're gauranteed to get another eigenstate with the same energy:
$$
H \left( R \psi_q \right) =
\left( H R \right) \psi_q =
\left( R H \right) \psi_q =
R \left( H \psi_q \right) =
R \left( E_q \psi_q \right) =
E_q \left( R \psi_q \right)
$$
If the state $R\psi_q$ is not a multiple of $\psi_q$ then this implies that the energy level $E_q$ is degenerate. Let's see how $R$ operates on our Bloch states $\psi_q(x)$:
$$
(R\psi_q)(x+a) =
\psi_q(-x-a) =
e^{iq(-a)}\psi_q(-x) =
e^{i(-q)a}\psi_q(-x) =
e^{i(-q)a}\left(R \psi_q\right)(-x)
$$
So that $R\psi_q$ has acts differently under translation than $\psi_q$, specifically like a Bloch wavefunction with "crystal momentum" $-q$ instead of $+q$,…unless of course $e^{i(-q)a}=e^{i(+q)a}$!
Since $q$ values separated by integer multiples of $\frac{2\pi}{a}$ are equivalent, we only need to check in the range $[0,2\pi)$.
Since the delta potential is symmetric, this above applies and we find that only for the values $q=0,\frac{\pi}{a}$ does this hold, so that every energy level except for $E_o$ and $E_{\pi/a}$ is degenerate.
Q5 - G5.28
Using Python and the technique suggested by Griffiths, I got the following plot:
here is the code I used to generate the plot:
from matplotlib import pyplot as plt
import numpy as np
trial_energies = np.linspace(0,30,30000)
betas = (.1,1.,10.,100.)
colors = ('blue','green','red','black')
for index, beta in enumerate(betas):
momenta = []
energies = []
for trial_energy in trial_energies:
k = np.sqrt(2*trial_energy)
test = np.cos(k) + beta / k * np.sin(k)
if test**2 < +1:
energies.append(trial_energy)
momenta.append(np.arccos(test))
plt.scatter(
momenta,
energies,
s=1,
color=colors[index%len(colors)],
label='$\\beta$='+str(beta)
)
plt.title('band structure for delta spike train')
plt.xlabel(r'momentum $q$ (in units of $a^{-1}$)',fontsize=15)
plt.ylabel(r'energy (in units of $\frac{\hbar^2}{2m a^2}$)',fontsize=15)
plt.legend()
plt.show()
Notice how the band structure looks like the energy-momentum relationship for a free particle when the potential is weak, and how the bands become narrow (i.e. gaps become large) when the potential is strong.
In fact, why don't we take a look at a wider range of the spectrum for a strong potential, say $\beta=100$ in our units:
The energy bands are so narrow so that the system might be better described as having energy levels similar to our localized systems (a harmonic oscillator, a particle in a single box, an electron orbiting a single nucleus).
We might also observe the level spacing.
Read off the (approximate) energies of the different "levels".
What series do they form?
What system that we've already studied has a similar spectrum?
Can you put together a picture of what the electrons are doing as the delta spikes become more and more repulsive (i.e. $\beta \to \infty$)?
What happens to the coefficients $A$ and $B$ in the solution (equation 5.66)?
Q6 - G5.32
a)
The key observation is that the energies of the spin up and spin down particles can be added together to determine the total energy.
The condition that the spin up and down particles separately occupy their lowest energy state requires the particles to fill up from the lowest energy state upwards.
We then add the energies from two octants, one for the spin up of radius $k^+_F$ and one for the spin down or radius $k^-_F$. The energy density in $k$-space for each octant is one half of what it was when $k$-space contained both spin states, so the total energy $E'_{tot}$ can be written as:
$$
E'_{tot} = E_{tot}(k^+_F) + E_{tot}(k^-_F)
$$
Where the unprimed $E_{tot}(k_F)$ is Griffiths equation 5.56 expressed as function of $k_F$.
$k^{\pm}_F$ however no longer has the same relationship to $N_{\pm}$ that $k_F$ did to $N$.
$k_F$ is defined as the radius of the octant necessary to fit $N$ states.
However we can only fit half as many states in each individual octant as we can when the octant contains both spin states.
So we will need to double the volume of the octant, or increase $k^{\pm}_F$ by a factor $\sqrt[3]{2}$, i.e.
$$
k^{\pm}_F(N_{\pm}) = \sqrt[3]{2} k_F(N_{\pm})
$$
where $k_F(N)$ is defined in Griffiths 5.52.
The rest of part a) is to plug and chug to find $E'_{tot}$ as a function of $N_+$ and $N_-$.
b)
From part a) you will have found that
$$
E'_{tot} \propto N_+^{5/3} + N_-^{5/3}
$$
defining:
$$
\bar{N} \equiv \frac{N_+ + N_-}{2}
$$
and
$$
\Delta{N} \equiv \frac{N_+ - N_-}{2}
$$
substituting these new terms in we get
$$
E'_{tot} \propto (\bar{N}+\Delta N)^{5/3} + (\bar{N}-\Delta N)^{5/3}
$$
or
$$
E'_{tot} \propto \underbrace{(1+x)^{5/3}}_\text{T+} + \underbrace{(1-x)^{5/3}}_\text{T-}
$$
where $x \equiv \frac{\Delta N}{\bar{N}}$.
Our setup suggests a taylor expansion, which is precisely what we'll do.
Note that the first term (T+) is a reflection of the second (T-), which implies that their even order terms are equal but their odd order terms are opposite, so their sum will cancel.
The zeroth order for both T+ and T- is 1, which combine to give a zeroth order contribution of 2.
The next (non-vaninshing) term is the second order term.
As mentioned it is only necessary to compute the expansion coefficient $\frac{1}{2}\frac{d^2}{d x^2}\Bigr|_{x=0}(1+x)^{5/3}$ of one piece, so we will just calculate it for T+:
$$
\frac{d^2}{d\epsilon^2} (1 + \epsilon)^{5/3} = \frac{5}{3} \times \frac{2}{3} (1 + \epsilon)^{-1/3}
$$
which when we evaluate at $x = 0$ becomes $\frac{10}{9}$. We find then that:
$$
E'_{tot} \propto 1 + \frac{5}{9}x^2
$$
the rest, like in part a), is plugging and chugging.
Q7 - G5.37
a)
For a free particle we know that momentum eigenstates $\phi_p(x)\propto e^{i\frac{px}{\hbar}}$ are energy eigenstates.
They also happen to already be in Bloch form:
$$
\phi_p(x+a) =
e^{i\frac{p(x+a)}{\hbar}} =
e^{i\frac{p}{\hbar}a} e^{i\frac{px}{\hbar}} =
e^{i\frac{p}{\hbar}a} \phi_p(x)
$$
so it is a state of "crystal momentum" $q=\frac{p}{\hbar}$.
The energy of the state is the familiar $E = \frac{p^2}{2m} = \frac{\hbar^2}{2m}q^2$ so we get
$$
\frac{dE}{dq} = \frac{\hbar^2}{m}q = \frac{\hbar^2}{m} \sqrt{\frac{2mE}{\hbar^2}} = \hbar \sqrt{\frac{2E}{m}}
$$
b)
Defining $y\equiv qa$ and using the definitions in Griffiths 5.72 we can rewrite equation 5.71 as
$$
\cos y = \cos z + \beta\frac{\sin z}{z}
$$
Taking the differential of both sides yields:
$$
- \sin y dy = - \sin z dz + \beta \left( \frac{\cos z}{z} - \frac{\sin z}{z^2} \right) dz
$$
or
$$
\frac{dy}{dz} = \frac{ \sin z \left( 1 + \frac{\beta}{z^2} \right) - \cos z \frac{\beta}{z}}{\sin y}
$$
Recognizing that $\sin y = \sqrt{1 - \cos^2 y}$ we get
$$
\frac{dy}{dz} = \frac{ \sin z \left( 1 + \frac{\beta}{z^2} \right) - \cos z \frac{\beta}{z}}{\sqrt{ 1 - \left(\cos z + \beta\frac{\sin z}{z}\right)^2}}
$$
Q8 - white dwarf
a)
We would like to find the radius that minimizes $E+E_{grav}$.
This is accomplished in the usual way by taking a derivative and setting to zero.
b)
I could have made a mistake but I get something like $2.2 \times 10^7\text{m}$ of radius, or 44,000km in diameter.
Neptune appears to be the closest in size.
