物理代写|PHYS3034 Quantum mechanics

Phonons in salt. Consider a model of a more complex solid, where there are two kinds of atoms, of masses $m$ and $M$, connected by springs of strength $\kappa_1$ and $\kappa_2$, as in the figure. (This is a cartoon of an ionic solid, like $\mathrm{NaCl}$.)

The unit cell, i.e. the pattern that is repeated, contains two atoms. Let $q_n$ be the deviation from equilibrium position of the $n$th light atom and $Q_n$ be the deviation from equilibrium position of the $n$th heavy atom. Assume periodic boundary conditions. You may wish to rescale the variables $q, Q$ to simplify the dependence on the parameters. (If you prefer, study the special case where $m=M$, but $\kappa_1 \neq \kappa_2$.)
By making the Fourier ansatz
$$
\left(\begin{array}{c}
q_n \
Q_n
\end{array}\right)=\frac{1}{\sqrt{N}} \sum_k e^{\mathrm{i} k n a}\left(\begin{array}{c}
q_k \
Q_k
\end{array}\right)
$$
find the spectrum of normal modes. Introduce creation and annihilation operators and reduce the problem to a collection of decoupled oscillators.

In addition to the acoustic phonons, which have dispersion $\omega_k \sim v_s|k|$ near $k=0$, and represent the quantum avatar of the sound mode, you should find a branch of optical phonons which have $\omega_k \stackrel{k \rightarrow 0}{\rightarrow}$ a finite number. Interpret the optical phonon mode in terms of the motion of the atoms.

Casimir force is regulator-independent. [Bonus problem] Suppose we use a different regulator for the sum in the vacuum energy $\sum_j \hbar \omega_j$. Instead we replace
$$
f(d) \rightsquigarrow \frac{1}{2} \sum_{j=1}^{\infty} \omega_j K\left(\omega_j\right)
$$
where the function $K$ is
$$
K(\omega)=\sum_\alpha c_\alpha \frac{\Lambda_\alpha}{\omega+\Lambda_\alpha} .
$$
We impose two conditions on the parameters $c_\alpha, \Lambda_\alpha$ :

We want the low-frequency answer to but unmodified:
$$
K(\omega) \stackrel{\omega \rightarrow 0}{\rightarrow} 1
$$
this requires $\sum_\alpha c_\alpha=1$.

We want the sum over $j$ to converge; this requires that $K(\omega)$ falls off faster than $\omega^{-2}$. Taylor expanding in the limit $\omega \gg \Lambda_\alpha$, we have
$$
K(\omega) \stackrel{\omega \rightarrow \infty}{\rightarrow} \frac{1}{\omega} \sum_\alpha c_\alpha \Lambda_\alpha-\frac{1}{\omega^2} \sum_\alpha c_\alpha \Lambda_\alpha^2+\cdots .
$$
So we also require $\sum_\alpha c_\alpha \Lambda_\alpha=0$ and $\sum_\alpha c_\alpha \Lambda_\alpha^2=0$.
First, verify the previous claims about $K(\omega)$.
Then compute $f(d)$ and show that with these assumptions, the Casimir force is independent of the parameters $c_\alpha, \Lambda_\alpha$.
[A hint for doing the sum: use the identity
$$
\frac{1}{X}=\int_0^{\infty} d s e^{-s X}
$$
inside the sum to make it a geometric series. To do the remaining integral over $s$, Taylor expand the integrand in the regime of interest.]

Commutation relations of creation operators for general one-particle states. Show that
$$
\mathbf{a}\left(\varphi_1\right) \mathbf{a}^{\dagger}\left(\varphi_2\right)-\zeta \mathbf{a}^{\dagger}\left(\varphi_2\right) \mathbf{a}\left(\varphi_1\right)=\left\langle\varphi_2 \mid \varphi_1\right\rangle,
$$
where these objects are as defined in the lecture notes.

Fermion creation and annihilation algebra.
Consider a single fermion mode c. We showed in lecture that the associated hilbert space is two-dimensional, and is spanned by
$$
|0\rangle, \text { with } \mathbf{c}|0\rangle=0 \text { and }|1\rangle=\mathbf{c}^{\dagger}|0\rangle
$$
(a) Check that the two states are orthogonal.
(b) Show that acting on this Hilbert space it is indeed true that
$$
\mathbf{c}^{\dagger} \mathbf{c}+\mathbf{c c}^{\dagger}=11
$$
as long as $\langle 1 \mid 1\rangle=\langle 0 \mid 0\rangle$.
(c) Check that
$$
[\mathbf{N}, \mathbf{c}]=-\mathbf{c}, \quad\left[\mathbf{N}, \mathbf{c}^{\dagger}\right]=\mathbf{c}^{\dagger}
$$
where $\mathbf{N}=\mathbf{c}^{\dagger} \mathbf{c}$ is the number operator. Notice that this is the same algebra satisfied by bosonic modes.