Let $\{ \ket{n} \}$ be an orthonormal eigenbasis for $Q$ so that $Q\ket{n}=q_n\ket{n}$. In this basis then
$$
\bra{m}e^{iQ}\ket{n} = \bra{m}e^{iq_n}\ket{n} = e^{iq_n}\delta_{mn} \equiv u_{mn}
$$
This matrix $u_{mn}$ is orthogonal since
$$
\sum_l u^*_{lm}u_{ln} = \sum_l (e^{iq_m}\delta_{lm})^*e^{iq_n}\delta_{ln} = \sum_l e^{i(q_n-q_m)} \delta_{lm}\delta_{ln} = e^{i(q_n-q_m)} \delta_{mn} = \delta_{mn}
$$
Any operator whose matrix representation in an orthonormal basis is unitary is also unitary.
Q2 - G6.4
$\hat{Q}$ is not strictly an operator but rather is a function that takes in a pair of operators $\hat{A}$, $\hat{B}$ and returns an operator according to the rule:
$$
\hat{Q}(\hat{A},\hat{B}) = \sum_{bc} Q_{bc} \hat{A}^b \hat{B}^c
$$
We are being asked to check that this function $\hat{Q}$ has the property that
if we take the operator $\hat{Q}(\hat{x},\hat{p})$ returned by plugging in $\hat{x}$ and $\hat{p}$ into $\hat{Q}$ and sandwich it with $\hat{T}^{\dagger}$ and $\hat{T}$, i.e. $\hat{T}^{\dagger} \hat{Q}(\hat{x},\hat{p}) \hat{T}$,
we get the operator $\hat{Q}(\hat{x}+a,\hat{p})$ we would have gotten by plugging in $\hat{x}+\hat{a}$ and $p$ into $\hat{Q}$, where
$\hat{a}$ is the identity operator scaled by a factor $a \in \R$, i.e. for any $\ket{\psi}$ we get $\hat{a}\ket{\psi}=a\ket{\psi}$
This is just application of the product rule $\left( \frac{d}{dx} \left( f g \right) \right)(x) = \left(\frac{d}{dx}f\right)(x) g(x) + f(x) \left(\frac{d}{dx}g\right)(x)$.
b),c)
We can rewrite Griffiths equation 6.14 as
$$
\left( -\frac{d^2}{dz^2} - 2iy \frac{d}{dz} + v(z) \right) u_{ny} = \left(\epsilon_{ny} - y^2\right) u_{ny}
$$
where
$ \alpha \in \R $,
which I've introduced to vary the strength of the potential
In other words we're expressing the position $x$ in units of $a$, the "crystal momentum" $q$ in units of $\frac{1}{a}$, and the energy $E_{nq}$ in units of $E_o$.
We will solve for the eigenfunctions $u_{ny}$ approximately by "discretizing" the z-coordinate to only take values $\frac{m}{M}$, where $m = 0,1,2,\ldots,M-1$.
The functions $f$ on this discretized space now take discrete numbers $m$ as their arguments instead of a continuous variable $z$, i.e. $f(z) \to f_m \sim f(\frac{m}{M})$.
The eigenvalue equation then becomes
$$
- \frac{\left( u_{m+1} - 2 u_m + u_{m-1} \right)}{\frac{1}{M^2}} + 2 i y \frac{ \left( u_{m+1} - u_{m-1} \right) }{2 \frac{1}{M}} + v_m u_m = \left( \epsilon_{ny} - y^2 \right) u_m
$$
or
$$
\sum_{m'} h_{m'm} u_{m'} = \left( \epsilon_{ny} - y^2 \right) u_m
$$
where
$$
h_{m'm} = - M^2 \left( \delta_{m'(m+1)} - 2 \delta_{m'm} + \delta_{m'(m-1)} \right) + iyM \left( \delta_{m'(m+1)} - \delta_{m'(m-1)} \right) + v_m \delta_{m'm}
$$
Note that for every index pair $m',m$we should really be writing $(m'\bmod M),(m \bmod M)$ to appropriately handle the boundaries, where $a \bmod b$ is the remainder of $ a \div b $
Here's the code I used to calculate the hamiltonian:
import numpy as np
alpha = 1.
M = 5
y = 1.
grid = 1. * np.arange(M) / M # discrete grid of z coordinates
def potential(z):
return -alpha if np.abs(z-.5) < .25 else 0.
d = np.identity(M) # create the matrix delta_{m'm}
d_plus = np.roll(d,1,0) # shift elements of d to create delta_{m'(m+1)}
d_minus = d_plus.transpose()
v = np.diag(map(potential,grid)) # create v_m \delta_{m'm}
# construct the hamiltonian
h = -M**2 * (d_plus - 2 * d + d_minus) + 1j * y * M * ( d_plus - d_minus ) + v
and here's the hamiltonian I get after setting $M=5,\alpha=y=1$:
50, 0i
-25, -5i
0, 0i
0, 0i
-25, 5i
-25, 5i
50, 0i
-25, -5i
0, 0i
0, 0i
0, 0i
-25, 5i
49, 0i
-25, -5i
0, 0i
0, 0i
0, 0i
-25, 5i
49, 0i
-25, -5i
-25, -5i
0, 0i
0, 0i
-25, 5i
50, 0i
Note that it doesn't really matter whether we call $m$ the column and $m'$ the row or vice versa.
Check that transposing the matrix just transforms $y \to -y$.
The pair, as you saw last week, are gauranteed to be degenerate by virtue of the potential's reflection symmetry.
Any numerical software package worth its salt will have an eigenvalue solver.
Using the numpy.linalg.eigvals function we find the eigenvalues and eigenvectors of the $h$ matrix.
The resulting band structure diagram is:
The full code used to generate the plot is shown below:
from matplotlib import pyplot as plt
import numpy as np
alpha = 20.
M = 100
grid = 1. * np.arange(M) / M # discrete grid of z coordinates
def potential(z):
return -alpha if np.abs(z-.5) < .25 else 0.
d = np.identity(M) # create the matrix delta_{m'm}
d_plus = np.roll(d,1,0) # shift elements of d to create delta_{m'(m+1)}
d_minus = d_plus.transpose()
v = np.diag(map(potential,grid)) # create v_m \delta_{m'm}
h_root = -M**2 * (d_plus - 2 * d + d_minus) + v
for y in np.linspace(0,np.pi,100):
# construct the hamiltonian for current value of y
h = h_root + 1j * y * M * ( d_plus - d_minus )
# find the eigenvalues of h,
# add on the y^2 to get the true energy,
# and select lowest 2 eigenvalues
e1,e2 = sorted(np.linalg.eigvals(h)+y**2)[:N]
# plot for positive y and negative y
plt.scatter([y,y],[e1,e2],s=1)
# plot for negative y
plt.scatter([-y,-y],[e1,e2],s=1)
plt.title('band structure for finite square well')
plt.xlabel(r'crystal momentum $q$ (units of $\frac{1}{a}$)',fontsize=15)
plt.ylabel(r'energy $E_{nq}$ (units of $\frac{\hbar^2}{2ma^2}$)',fontsize=15)
plt.show()
Q4 - G6.7
a)
The key here is that for commuting operators $A$ and $B$, it holds that
$$
e^{A+B} = e^A e^B = e^B e^A
$$
Now $\hat{p}_1$ and $\hat{p}_2$ commute since $\frac{\partial}{\partial x_1} \frac{\partial}{\partial x_2} = \frac{\partial}{\partial x_2}\frac{\partial}{\partial x_1}$.
So for any $\psi(x_1,x_2)$ we have that
\begin{align}
& \left( e^{\frac{a}{i\hbar}\hat{P}} \psi \right) (x_1,x_2) = \\ &
\left( e^{\frac{a}{i\hbar}\left( \hat{p}_1 + \hat{p}_2 \right) } \psi \right) (x_1,x_2) = \\ &
\left( \left( e^{\frac{a}{i\hbar}\hat{p}_1}e^{\frac{a}{i\hbar}\hat{p}_2} \right) \psi \right) (x_1,x_2) = \\ &
\left( e^{\frac{a}{i\hbar}\hat{p}_1} \left( e^{\frac{a}{i\hbar}\hat{p}_2} \psi \right)\right) (x_1,x_2) = \\ &
\left( e^{\frac{a}{i\hbar}\hat{p}_2} \psi \right)(x_1+a,x_2) = \\ &
\psi (x_1+a,x_2+a) = \\ &
\left( \hat{T}_a \psi \right) (x_1,x_2)
\end{align}
b)
From the discussion in Griffiths section 6.2 we find that to show that an observable $\hat{A}$ is conserved it is enough to show that the hamiltonian commutes with $e^{ia\hat{A}}$ for any $a \in \R$ (assume we've scaled $\hat{A}$ by whatever constant necessary to make it unitless).
To this end, we first recognize that $-\hbar^2\frac{\partial^2}{\partial x_i^2} = \hat{p}^2_i$ and see that
\begin{align}
& \hat{T}^{\dagger}(a) \hat{H} \hat{T}(a) \\ = &
\hat{T}^{\dagger}(a) \left( \frac{\hat{p}^2_1}{2m_1} + \frac{\hat{p}^2_2}{2m_2} + V(|x_1 - x_2|) \right) \hat{T}(a) \\ = &
\hat{T}^{\dagger}(a) \frac{\hat{p}^2_1}{2m_1} \hat{T}(a) + \hat{T}^{\dagger}(a) \frac{\hat{p}^2_2}{2m_2} \hat{T}(a) + \hat{T}^{\dagger}(a) V(|x_1 - x_2|) \hat{T}(a) \\ = &
\hat{T}^{\dagger}(a) \frac{\hat{p}^2_1}{2m_1} \hat{T}(a) + \hat{T}^{\dagger}(a) \frac{\hat{p}^2_2}{2m_2} \hat{T}(a) + V\left(|(x_1-a) - (x_2-a)|\right) \\ = &
\hat{T}^{\dagger}(a) \frac{\hat{p}^2_1}{2m_1} \hat{T}(a) + \hat{T}^{\dagger}(a) \frac{\hat{p}^2_2}{2m_2} \hat{T}(a) + V\left(|x_1 - x_2|\right) \\ = &
\hat{t}^{\dagger}_2(a) \hat{t}^{\dagger}_1(a) \frac{\hat{p}^2_1}{2m_2} \hat{t}_1(a) \hat{t}_2(a) + \hat{t}^{\dagger}_1(a) \hat{t}^{\dagger}_2(a) \frac{\hat{p}^2_2}{2m_2} \hat{t}_2(a) \hat{t}_1(a) + V\left(|x_1 - x_2|\right) \\ = &
\hat{t}^{\dagger}_2(a) \frac{\hat{p}^2_1}{2m_2} \hat{t}_2(a) + \hat{t}^{\dagger}_1(a) \frac{\hat{p}^2_2}{2m_2} \hat{t}_1(a) + V\left(|x_1 - x_2|\right) \\ = &
\frac{\hat{p}^2_1}{2m_2} + \frac{\hat{p}^2_2}{2m_2} + V\left(|x_1 - x_2|\right) \\ = &
\hat{H}
\end{align}
where $\hat{t}_j(a) \equiv e^{\frac{a \hat{p}_j}{i\hbar}}$.
All of the simplifying steps are just applications of Griffiths equation 6.8, which you proved earlier.
Q5 - G7.2
a)
We know the spectrum $E'_n$ of a harmonic oscillator with spring constant $k+\epsilon$, they are:
$$
E'_n = \hbar \omega' ( n + \frac{1}{2} )
$$
where
$$
\omega' = \sqrt{\frac{k+\epsilon}{m}}
$$
we can rewrite the energies in terms of the energy levels of the unperturbed oscillator $E_n$:
$$
E'_n = \sqrt{1+\frac{\epsilon}{k}} E_n \approx \left( 1 + \frac{1}{2} \frac{\epsilon}{k} - \frac{1}{4} \left(\frac{\epsilon}{k}\right)^2 \right) E_n
$$
b)
The perturbation is the change in the hamiltonian, so that the new hamiltonian is the sum $H_o + H'$ of the old hamiltonian $H_o$ plus the perturbation $H'$.
So since our potential energy went from $\frac{1}{2}kx^2 \to \frac{1}{2}\left( k + \epsilon \right) x^2$, the perturbation is:
$$
H' = \frac{1}{2}\left( k + \epsilon \right) x^2 - \frac{1}{2} k x^2 = \frac{1}{2} \epsilon x^2 = \frac{\epsilon}{k} V_o(x)
$$
Where $V_o(x)$ is the potential energy of the unperturbed system.
The first order correction to the energy levels is given by:
$$
ev{H'}_n =
\ev{ \frac{\epsilon}{k} \hat{V}_o } =
\frac{\epsilon}{k} \ev{ \hat{V}_o } =
\frac{\epsilon}{k} \frac{1}{2} E_n
$$
To first order this agrees exactly with the result in part a).
Q5 - G7.3
a)
We're experts at this problem by now.
The bosonic ground state is $\psi_1(x_1)\psi_1(x_2)\ket{ 0 0 } \equiv \Psi_o(x_1,x_2)$ and the first excited state is $\frac{1}{\sqrt{2}} \left( \psi_1(x_1) \psi_2(x_2) + \psi_1(x_2) \psi_2(x_1) \right) \ket{ 0 0 }\equiv \Psi_1(x_1,x_2)$, where
The $\psi_n(x) = \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right)$ are the single particle-in-box states and
$\ket{ S = 0, M_S = 0}$ is the spin state $\ket{s_1 = 0, m_{s_1} = 0}\ket{s_2 = 0, m_{s_2} = 0}$, which is the only way to combine to spin zero states.
