Problem set 22

1. Given the usual canonical commutation relations, the commutator $[A,B]$ of $A=i(xp_y-yp_z)$ and $B=i(yp_z+zp_y)$ is
1. $\hbar(xp_z-p_xz)$
2. $-\hbar(xp_z-p_xz)$
3. $\hbar(xp_z+p_xz)$
4. $-\hbar(xp_z+p_xz)$
Poisson bracket of two functions $f(q_i,p_i)$ and $g(q_i,p_i)$ is defined by \begin{align*} &\left\{f,g\right\}=\sum\limits_{i=1}^{N}\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\\ &=\frac{\partial f}{\partial x}\frac{\partial g}{\partial p_x}-\frac{\partial f}{\partial p_x}\frac{\partial g}{\partial q_x}+\frac{\partial f}{\partial q_y}\frac{\partial g}{\partial p_y}\\ &-\frac{\partial f}{\partial p_y}\frac{\partial g}{\partial q_y}+\frac{\partial f}{\partial q_z}\frac{\partial g}{\partial p_z}-\frac{\partial f}{\partial p_z}\frac{\partial g}{\partial q_z} \end{align*} $A=i(xp_y-yp_x)$ and $B=(yp_z+zp_y)$ \begin{align*} \left\{A,B\right\}&=ip_y\cdot 0-(-iy)\cdot 0\\ &+\:(-ip_x)\cdot z-ix\cdot p_z\\ &+\:0\cdot y-0\cdot p_y\\ &=-i(p_xz+xp_z) \end{align*} However, commutator bracket, $[q,p]=i\hbar\left\{q,p\right\}$ $$[\hat A,\hat B]=i\hbar\left\{A,B\right\}=\hbar(p_xz+xp_z)$$ Hence, option (C) is correct.
2. The entropy of a system, $S$, is related to the accessible phase space volume $\Gamma$ by $S = k\ln \Gamma(E, N,V)$ where $E$, $N$ and $V$ are the energy, number of particles and volume respectively. From this one can conclude that $\Gamma$
1. does not change during evolution to equilibrium
2. oscillates during evolution to equilibrium
3. is a maximum at equilibrium
4. is a minimum at equilibrium
Answer is (C) i.e. $\Gamma$ is a maximum at equilibrium. This is because, when system attains equilibrium its entropy and hence number of accessible states becomes maximum (equilibrium state corresponds to the most disordered state).
3. Let $\Delta W$ be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about $\Delta W$ is correct?
1. $\Delta W$ is a perfect differential if the process is isothermal
2. $\Delta W$ is a perfect differential if the process is adiabatic
3. $\Delta W$ is always a perfect differential
4. $\Delta W$ cannot be a perfect differential
There are two types of functions viz. State functions and path functions. In case of state functions, the change in the state function depends only on the initial and final states of the system and not on the path from the initial to the final state (e.g. internal energy $U(S,V)$, enthalpy $H(S,P)$, Gibb's free energy $G(T,P)$ Helmholtz free energy $F(T,V)$ etc. are state functions). In case of path functions, the change in the path functions depends on the path followed from the initial to the final state (e.g..$dQ$, $dW$ etc. are path functions). State functions are exact differential while path functions are inexact differential.
According to first law of thermodynamics $$dU=dQ+dW$$ Here $dU$ is exact, but, $dQ$ and $dW$ are inexact. However, for adiabatic process $dQ=0$. Hence, we have $$dU=dW$$. As $dU$ is always exact, for adiabatic process, $dW$ is exact.
Hence answer (B) is correct.
4. Consider a system of three spins $S_1$, $S_2$ and $S_3$ each of which can take values $+1$ and $-1$. The energy of the system is given by $E = -J\left[ S_1 S_2 + S_2 S_3 + S_3 S_1\right]$, where $J$ is a positive constant. The minimum energy and the corresponding number of spin configurations are, respectively,
1. $J$ and 1
2. $-3J$ and 1
3. $-3J$ and 2
4. $-6J$ and 2
Clearly, for $S_1= S_2 = S_3=1$ or $S_1= S_2 = S_3=-1$, we get minimum energy $E=-3J$. Hence, there are 2 configurations. Hence, answer (C) is correct.
5. The minimum energy of a collection of 6 non-interacting electrons of spin-$\frac{1}{2}$ and mass $m$ placed in a one dimensional infinite square well potential of width $L$ is
1. $14\pi^2\hbar^2/mL^2$
2. $91\pi^2\hbar^2/mL^2$
3. $7\pi^2\hbar^2/mL^2$
4. $3\pi^2\hbar^2/mL^2$
For a particle in one dimensional infinite square well potential of width $L$, energy is given by $$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\quad n=1,2,\dots$$ In each level there can be two electrons with up and down spin. Hence, total energy of the system is \begin{align*} E_{total}&=2\frac{1^2\pi^2\hbar^2}{2mL^2}+2\frac{2^2\pi^2\hbar^2}{2mL^2}\\ &+2\frac{3^2\pi^2\hbar^2}{2mL^2}\\ &=\frac{14\pi^2\hbar^2}{mL^2} \end{align*} Hence, answer is (A)