Problem set 60

1. A beam of unpolarized light in a medium with dielectric constant $\epsilon_1$ is reflected from a plane interface formed with another medium of dielectric constant $\epsilon_2=3\epsilon_1$. The two media have identical magnetic permeability. If the angle of incidence is $60^0$, then the reflected light
1. is plane polarized perpendicular to the plane of incidence
2. is plane polarized parallel to the plane of incidence
3. is circularly polarized
4. has the same polarization as the incident light

This is a simple method (discovered by malus in 1808) of obtaining partially plane polarized light by reflection. The percentage of polarized light is greatest in reflected beam when light beam is incident on the transparent medium with an incident angle equal to the angle of polarization. The vibrations of this plane polarized reflected light are found to be perpendicular to the plane of incidence and therefore, the reflected light is said to be plane polarized in the plane of incidence, when the incident angle is equal to the Brewster angle $\theta_B$, where $\tan\theta_B = n_2/n_1$. $$\begin{align*} \tan\theta_B&=\frac{n_2}{n_1}\\ &=\sqrt{\frac{\epsilon_2}{\epsilon_1}}\\ \tan\theta_B&=\sqrt{3} \end{align*}$$ Hence, $\theta_B=60^0$. As angle of incidence is equal to Brewster angle, reflected light is plane polarized perpendicular to the plane of incidence.

Hence, answer is (A)

2. The partition function of a system of $N$ Ising spins is $Z=\lambda_1^N+\lambda_2^N$, where $\lambda_1$ and $\lambda_2$ are functions of temperature, but are independent of $N$. If $\lambda_1 > \lambda_2$, the free energy per spin in the limit $N\rightarrow\infty$ is
1. $-k_BT\ln{\left(\frac{\lambda_1}{\lambda_2}\right)}$
2. $-k_BT\ln{\lambda_2}$
3. $-k_BT\ln{\left(\lambda_1\lambda_2\right)}$
4. $-k_BT\ln{\lambda_1}$

$$\begin{align*} F&=-kT\ln{Z}\\ &=-kT\ln{\left(\lambda_1^N+\lambda_2^N\right)}\\ F&=-kT\ln{\left(\lambda_1^N\left[1+\left(\frac{\lambda_2}{\lambda_1}\right)^N\right]\right)}\\ \end{align*}$$ For $\lambda_1 > \lambda_2$, $\frac{\lambda_2}{\lambda_1} < 1$, in the limit $N\rightarrow\infty$, $\frac{\lambda_2}{\lambda_1} = 0$ $$F=-kT\ln{\left(\lambda_1^N\right)}=-NkT\ln{\lambda_1^N}$$ $$\frac{F}{N}=-kT\ln{\lambda_1}$$

Hence, answer is (D)

3. The Hamiltonian of a system of $N$ non-interacting spin-1/2 particles is $H=-\mu_0B\sum_iS_i^z$, where $S_i^z=\pm1$ are the components of $i^{th}$ spin along an external magnetic field $B$. At a temperature $T$ such that $e^{\mu_0B/k_BT}=2$, the specific heat per particle is
1. $\frac{16}{25}k_B$
2. $\frac{8}{25}k_B\ln2$
3. $k_B\left(\ln2\right)^2$
4. $\frac{16}{25}k_B\left(\ln2\right)^2$

Because each spin is independent of the others and distinguishable, we can find the partition function for one spin, $Z_1$, and use the relation $Z_N=Z_1^N$ to obtain, the partition function for $N$ spins. $$\begin{align*} Z_1&=\sum\limits_{s=\pm1}e^{-\beta\mu_0 Bs}\\ &=e^{-\beta\mu_0 B}+e^{\beta\mu_0 B}\\ &=2\cosh{\beta\mu_0 B} \end{align*}$$ $$Z_N=\left(2\cosh{\beta\mu_0 B}\right)^N$$ Mean energy $$\begin{align*} E&=-\frac{\partial\ln{Z_N}}{\partial\beta}\\ &=-N\mu B\tanh{\beta\mu_0 B} \end{align*}$$ $$\begin{align*} C_V&=\frac{\partial E}{\partial T}\\ &=\left(\frac{\mu_0 B}{k T}\right)^2Nk\:sech^2{\beta\mu_0 B} \end{align*}$$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(e^{\frac{\mu_0 B}{kT}}+e^{-\frac{\mu_0 B}{kT}}\right)^2}$$ Since, $e^{\mu_0B/k_BT}=2$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(2+\frac{1}{2}\right)^2}$$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{16}{25}$$ $$\frac{C_v}{N}=\left(\ln2\right)^2k\frac{16}{25}$$

Hence, answer is (D)

4. The ground state energy of a particle in the potential $V(x)=g|x|$, estimated using the trial wavefunction $$\psi(x)= \begin{cases} \sqrt{\frac{c}{a^5}}(a^2-x^2),\quad x < |a|\\ 0,\quad x\ge|a| \end{cases}$$ (where $g$ and $c$ are constants) is
1. $\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
2. $\frac{5}{6}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
3. $\frac{3}{4}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
4. $\frac{7}{8}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$

First normalize the wavefunction using normalization condition $\int\limits_{-a}^{a}\psi^*\psi\:dx=1$ $$\frac{c}{a^5}\int\limits_{-a}^{a}(a^2-x^2)^2\:dx=1$$ $$\Rightarrow c=\frac{15}{16}$$ $$\psi(x)=\sqrt{\frac{15}{16a^5}}(a^2-x^2)$$ $$H=-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}+g|x|=T+V$$ $$\begin{align*} < E >& = < \psi |H| \psi > \\ &= < \psi |T| \psi>+ < \psi |V| \psi> \end{align*}$$ $$\begin{align*} < \psi |T| \psi> &={\scriptstyle-\frac{\hbar^2}{2m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2) \frac{d^2(a^2-x^2)}{d x^2}dx}\\ &=\frac{\hbar^2}{m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2)dx\\ &=\frac{5\hbar^2}{4ma^2} \end{align*}$$ $$\begin{align*} < \psi |V| \psi>\! &=\frac{15}{16a^5}g\!\!\int\limits_{-a}^{a}\!(a^2-x^2)^2 |x|dx\\ &=\frac{15}{16a^5}2g\!\!\int\limits_{0}^{a}\!(a^2-x^2)^2 x dx\\ &=\frac{5}{16}ga \end{align*}$$ $$< E > =\frac{5\hbar^2}{4ma^2}+\frac{5}{16}ga$$ $$\frac{d < E > }{da}=0$$ $$\Rightarrow a=2\left(\frac{\hbar^2}{mg}\right)^{1/3}$$ $$< E > =\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$$

Hence, answer is (A)

5. A hydrogen atom is subjected to the perturbation $V_{pert}(\vec r)=\epsilon\cos{\left(2r/a_0\right)}$, where $a_0$ is the Bohr radius. The change in the ground state energy to first order in $\epsilon$ is
1. $\epsilon/4$
2. $\epsilon/2$
3. $-\epsilon/2$
4. $-\epsilon/4$

First order correction is given by $$E_0^{(1)}=\left < \psi_{100}\left|H'\right|\psi_{100}\right >$$ $$\psi_{100}=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$$ $$\begin{align*} E_0^{(1)}&=\frac{\epsilon}{\pi a_0^3}\!\int\limits_0^\infty\! e^{-\frac{2r}{a_0}}\!\cos{\left(\frac{2r}{a_0}\right)}4\pi r^2 dr\\ &={\textstyle\frac{4\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}}\left(\frac{e^{i\frac{2r}{a_0}}+e^{-i\frac{2r}{a_0}} }{2}\right) r^2 dr}\\ &=\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1-i)}r^2dr\\ &+\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1+i)}r^2dr\\ &={\textstyle\frac{2\epsilon}{a_0^3}\left[\frac{2!}{\left(\frac{2}{a_0}(1-i)\right)^3}+\frac{2!}{\left(\frac{2}{a_0}(1+i)\right)^3}\right]}\\ &=\frac{\epsilon}{2}\left[\frac{1}{(1-i)^3}+\frac{1}{(1+i)^3}\right]\\ &={\textstyle\frac{\epsilon}{2}\left[\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1-i}{\sqrt{2}}\right)^3}+\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1+i}{\sqrt{2}}\right)^3}\right]}\\ &=\frac{\epsilon}{4\sqrt{2}}\left[\frac{1}{e^{\frac{-i3\pi}{4}}}+\frac{1}{e^{\frac{i3\pi}{4}}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[e^{\frac{i3\pi}{4}}+e^{\frac{-i3\pi}{4}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\cos{\left(\frac{3\pi}{4}\right)}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\left(-\frac{1}{\sqrt{2}}\right)\right]\\ &=-\frac{\epsilon}{4} \end{align*}$$

Hence, answer is (D)