Problem set 74

1. The wavefunction of a particle in a one-dimensional potential at time $t=0$ is $\psi(x,t=0)=\frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]$ where $\psi_0(x)$ and $\psi_1(x)$ are the ground and the first excited states of the particle with corresponding energies $E_0$ and $E_1$. The wavefunction of the particle at a time $t$ is
1. $\frac{1}{\sqrt{5}}e^{-i(E_0+E_1)t/2\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
2. $\frac{1}{\sqrt{5}}e^{-iE_0t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
3. $\frac{1}{\sqrt{5}}e^{-iE_1t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
4. ${\scriptstyle \frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]}$

$$\hat H\psi_0=E_0\psi_0\quad\hat H\psi_1=E_1\psi_1$$ Time evolution of quantum systems is given by Unitary Transformations, \begin{align*} \left|\psi(x,t)\right > &=\hat U \left|\psi(x,t=0)\right >\\ &=e^{-i\hat Ht/\hbar} \left|\psi(x,t=0)\right >\\ &={\scriptstyle e^{-i\hat Ht/\hbar} \frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]}\\ &={\scriptstyle \frac{1}{\sqrt{5}}\left[2e^{-i\hat Ht/\hbar}\psi_0(x)-e^{-i\hat Ht/\hbar}\psi_1(x)\right]}\\ \end{align*} Using property $f(\hat A)\psi_n=f(a_n)\psi_n$, we get $${\scriptstyle \left|\psi(x,t)\right > =\frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]}$$

Hence, answer is (D)

2. The commutator $[L_x, y]$, where $L_x$ is the $x$-component of the angular momentum operator and $y$ is the $y$-component of the position operator, is equal to
1. 0
2. $i\hbar x$
3. $i\hbar y$
4. $i\hbar z$

$$L_x=yp_z-zp_y$$ \begin{align*} [L_x, y]&=[yp_z-zp_y, y]\\ &=[yp_z, y]-[zp_y, y]\\ &={\scriptstyle y[p_z, y]+[y, y]p_z-z[p_y, y]-[z, y]p_y}\\ &=i\hbar z \end{align*}

Hence, answer is (D)

3. In hydrogenic states, the probability of finding an electron at $r = 0$ is
1. zero in state $\phi_{1s}(r)$
2. non-zero in state $\phi_{1s}(r)$
3. zero in state $\phi_{2s}(r)$
4. non-zero in state $\phi_{2p}(r)$

non-zero in state $\phi_{1s}(r)$

Hence, answer is (B)

4. Each of the two isolated vessels, $A$ and $B$ of fixed volumes, contains $N$ molecules of a perfect monatomic gas at a pressure $P$. The temperatures of $A$ and $B$ are $T_1$ and $T_2$, respectively. The two vessels are brought into thermal contact. At equilibrium, the change in entropy is
1. $\frac{3}{2}Nk_B\ln{\left[\frac{T_1^2+T_2^2}{4T_1T_2}\right]}$
2. $Nk_B\ln{\left(\frac{T_2}{T_1}\right)}$
3. $\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]}$
4. $2Nk_B$

Energy of $N$ monatomic gas molecules is given by $$E=\frac{3}{2}Nk_BT$$ $$E_A=\frac{3}{2}Nk_BT_1$$ $$E_B=\frac{3}{2}Nk_BT_2$$ $$C_A=C_B=C=\frac{3}{2}Nk_B$$ When two bodies are in thermal contact, final temperature is given by $$T_f=\frac{C_AT_1+C_BT_2}{C_A+C_B}$$ $$T_f=\frac{T_1+T_2}{2}$$ Change in entropy of body $A$ is \begin{align*} \Delta S_A&=\int_{T_1}^{T_f}\frac{dQ}{T}\\ &=\int_{T_1}^{T_f}\frac{CdT}{T}\\ &=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_1}} \end{align*} $$\Delta S_B=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_2}}$$ Total change in entropy is $$\Delta S=\Delta S_A+\Delta S_B$$ \begin{align*} \Delta S&=\frac{3}{2}Nk_B\left(\ln{\frac{T_f}{T_1}}+\ln{\frac{T_f}{T_2}} \right)\\ &=\frac{3}{2}Nk_B\ln{\left[\frac{T_f^2}{T_1T_2}\right]} \\ &=\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]} \\ \end{align*}

Hence, answer is (C)

5. The mean internal energy of a one-dimensional classical harmonic oscillator in equilibrium with a heat bath of temperature $T$ is
1. $\frac{1}{2}k_BT$
2. $k_BT$
3. $\frac{3}{2}k_BT$
4. $3k_BT$

Energy of harmonic oscillator is given by $$E=\frac{p_x^2}{2m}+\frac{1}{2}kx^2$$ According to equipartition theorem each quadratic degrees of freedom has energy $\frac{1}{2}k_BT$. Energy is quadratic in $p_x$ and also in $x$. Hence, $$E=\frac{1}{2}k_BT+\frac{1}{2}k_BT=k_BT$$

Hence, answer is (B)