Problem set 10

1. The function $f(z)=u(x,y)+iv(x,y)$ is analytic at $z=x+iy$. The value of $\nabla^2u$ at this point is:
1. 0
2. undefined
3. $\pi$
4. $e^{-\pi^2}$
The function $f(z)=u(x,y)+iv(x,y)$ is analytic at $z=x+iy$ means $u(x,y)$ and $v(x,y)$ satisfy Cauchy-Reiman equations namely $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ $$\therefore\nabla^2u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$$ $$\nabla^2u=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)+\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}\right)$$ $$\begin{align*} \nabla^2u&=\frac{\partial }{\partial x}\left(\frac{\partial v}{\partial y}\right)-\frac{\partial }{\partial y}\left(\frac{\partial v}{\partial x}\right)\\ &=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0 \end{align*}$$
2. A head of mass $m$ slides on a smooth rod which is rotating about one end in a vertical plane with uniform angular velocity $\omega$. The Lagrangian of the system is :
1. $L=\frac{1}{2}m\left(\dot r^2+r^2\dot\theta^2\right)-mgr\sin\theta$
2. $L=\frac{1}{2}m\left(r^2\dot\theta^2\right)-mgr\sin\theta$
3. $L=\frac{1}{2}m\left(\dot r^2+\dot\theta^2\right)-mgr\sin\theta$
4. $L=\frac{1}{2}m\left(r^2\dot\theta^2\right)+mgr\sin\theta$
Kinetic energy of head due to sliding on the rod =$\frac{1}{2}m\dot r^2$

Kinetic energy of head due to rotation=$\frac{1}{2}mr^2\dot\theta^2$

Potential energy of the head due to vertical height=$mgr\sin\theta$ $$L=\frac{1}{2}m\left(\dot r^2+r^2\dot\theta^2\right)-mgr\sin\theta$$

3. A partition function of two Bose particles each of which can occupy any of the two energy levels $0$ and $\epsilon$ is
1. $1+e^{-2\epsilon/kT}+2e^{-\epsilon/kT}$
2. $1+e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
3. $2+e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
4. $e^{-2\epsilon/kT}+e^{-\epsilon/kT}$
The partition function for Bose particles is given by $Z=\sum\limits_Re^{-(n_1\epsilon_1+n_2\epsilon_2+\dots)/kT}$, where $n_1$ is number of particles with energy $\epsilon_1$ and so on. Bose particles are indistinguishable particles and any number of particles can be in any state. Hence states of the system can be obtained as follow.
Single particle energy	0	$\epsilon$
	AA
		AA
	A	A
$$\begin{align*} Z&=e^{-(2\times0)/kT}+e^{-(2\times\epsilon)/kT}\\ &+e^{-(1\times0+1\times\epsilon)/kT}\\ &=1+e^{-2\epsilon/kT}+e^{-\epsilon/kT} \end{align*}$$
4. A one dimensional random walker takes steps to left or right with equal probability. The probability that the random walker starting from origin is back to origin after $N$ even number of steps is
   1. $\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}\left(\frac{1}{2}\right)^N$
   2. $\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}$
   3. $2N!\left(\frac{1}{2}\right)^{2N}$
   4. $N!\left(\frac{1}{2}\right)^N$
   The probability $p(m,N)$ that the particle will be at the position $x = ml$ after $N$ steps is given by $${\scriptstyle p(m,N)=\frac{N!}{\left(\frac{N+m}{2}\right)!\left(\frac{N-m}{2}\right)!}p^{\frac{1}{2}\left(N+m\right)}q^{\frac{1}{2}\left(N-m\right)}}$$ Here $p=q=\frac{1}{2}$, the probability of taking steps to right and left, $l$ is step length and $x=ml=0$. Hence $m=0$. Hence $$p(0,N)=\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}\left(\frac{1}{2}\right)^N$$
5. Five electrons (Fermions with spin $1/2\hbar$) are kept in a one-dimensional infinite potential well with width $a$. (Ground state energy of single electron well $=\frac{\hbar^2\pi^2}{2ma^2}$). The first absorption line corresponds to energy:
   1. $\frac{\hbar^2\pi^2}{2ma^2}$
   2. $\frac{5\hbar^2\pi^2}{2ma^2}$
   3. $\frac{7\hbar^2\pi^2}{2ma^2}$
   4. $\frac{11\hbar^2\pi^2}{2ma^2}$
   As electrons are fermions, there can be two electrons with same energy. Hence ground state configuration will be two electrons in $n=1$ state, two electrons in $n=2$ state and one electron in $n=3$ state. The first absorption line corresponds to transition of electron from $n=3$ state to $n=4$ state. Energy of $n=3$ state is $\frac{9\hbar^2\pi^2}{2ma^2}$ and energy of $n=4$ state is $\frac{16\hbar^2\pi^2}{2ma^2}$. Hence first absorption line corresponds to energy $\frac{16\hbar^2\pi^2}{2ma^2}-\frac{9\hbar^2\pi^2}{2ma^2}=\frac{7\hbar^2\pi^2}{2ma^2}$