Problem set 62

1. The energy levels of one-dimensional harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ are given by $h\nu\left(n+\frac{1}{2}\right)$ with $n=0,1,2,\dots$. If the potential is changed to $V(x)=\infty$ for $x<0$ and $V(x)=\frac{1}{2}kx^2$ for $x>0$, the energy levels now, will be given by:
1. $h\nu\left(n+\frac{3}{2}\right)$
2. $2h\nu\left(n+\frac{1}{2}\right)$
3. $h\nu\left(n+\frac{1}{2}\right)$, $n$ odd only
4. $h\nu\left(n+\frac{1}{2}\right)$, $n$ even only

The wave function of simple harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ is given by $$\psi_n(x)=N_nH_n\left(\sqrt{\alpha}x\right)e^{-\alpha x^2/2}$$ where $\alpha=\left(\frac{mk}{\hbar^2}\right)^{1/2}$ is constant and $N_n=\left(\frac{\alpha}{\pi}\right)^{1/4}\left(\frac{1}{2^nn!}\right)^{1/2}$ is normalization constant, and $H_n\left(\sqrt{\alpha}x\right)=H_n(z)$ is a Hermite polynomial of degree $n$ given by $$H_0(z)=1$$ $$H_1(z)=2z$$ $$H_2(z)=4z^2-2$$ $$H_3(z)=8z^3-12z$$ The energy eigenvalues are given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{1}{2}h\nu,\frac{3}{2}h\nu,\frac{5}{2}h\nu\dots\\ &\text{for }n=0,1,2,3\dots \end{align*} Now, as $V(x)=\infty$ for $x<0$, the wave functions must vanish at $x=0$. Clearly, one can see that the wavefunction vanish at $x=0$ only for odd values of $n$. The terms with even $n$ do not have node at $x = 0$. Hence, eneegy is given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{3}{2}h\nu,\frac{7}{2}h\nu,\frac{11}{2}h\nu\dots\\ & \text{for }n=1,3,5\dots \end{align*}

Hence, answer is (C)

2. At $t=0$ a one-dimensional harmonic oscillator is in a state given by:$$\Psi(x,0)=\frac{1}{2}u_0(x)+i\frac{\sqrt{3}}{2}u_1(x)$$ where $u_0$ and $u_1$ are first two normalized eigen states. ($\omega$ is natural angular frequency of the oscillator). Then
1. The expectation value of the energy is $\frac{5}{4}\hbar\omega$
2. Energy measurement of this state will always gives its value as $\frac{5}{4}\hbar\omega$
3. The average value of energy is $\hbar\omega$
4. The expectation value of energy is dependent on time

Clearly, one can verify that the wavefunction $\Psi$ is normalized. \begin{align*} & < E > = < \Psi|H|\Psi > \\ &={\scriptstyle\left < \frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\arrowvert H\arrowvert\frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\right > }\\ &=\left < \frac{1}{2}u_0\left|H\right|\frac{1}{2}u_0\right > \\ &+\left < i\frac{\sqrt{3}}{2}u_1\left|H\right|i\frac{\sqrt{3}}{2}u_1\right > \\ &=\frac{1}{4}\left < u_0\left|H\right|u_0\right >\\ & +\!\!\left(\frac{-i\sqrt{3}}{2}\right)\!\!\left(\frac{i\sqrt{3}}{2}\right)\!\!\left < u_1\left|H\right|u_1\right > \\ &=\frac{1}{4}E_0+\frac{3}{4}E_1\\ &=\frac{1}{4}\times\frac{1}{2}\hbar\omega+\frac{3}{4}\times\frac{3}{2}\hbar\omega\\ < E >&=\frac{5}{4}\hbar\omega \end{align*}

Hence, answer is (A)

3. If the $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is $e^{2i\phi}$, then the minimum principal and minimum orbital angular momentum quantum numbers $n$ and $l$ respectively for this eigen function will be:
1. $n=3$, $l=2$
2. $n=2$, $l=1$
3. $n=1$, $l=2$
4. $n=2$, $l=2$

The $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is given by $e^{im\phi}$. Hence, $m=2$. Now, for a given value of $l$, $m$ has values from $-l$ to $+l$. This implies that $l=2$. Also, for a given value of $n$, $l$ takes values from $0$ to $n-1$. Hence, for $l=2$, minimum value of $n$ should be $3$.

Hence, answer is (A)

4. The ground state energy of a particle in an infinite square well is 1eV. If four particles obeying Bose-Einstein statistics are kept in this well, then the ground state energy will be :
1. 30 eV
2. 10 eV
3. 4 eV
4. $\frac{1}{4}$ eV

As particles obey Bose-Einstein statistics, all particles will be the ground state. Hence, energy will be 4 eV.

Hence, answer is (C)

5. Consider a system in contact with a heat and particle reservoir. It may be unoccupied or occupied by one particle with energy 0 and $\epsilon$. The grand partition function will be ($\beta=1/kT$)
1. $Z(\mu,T)=e^{-\epsilon\beta}$
2. $Z(\mu,T)=\left(1+e^{-\epsilon\beta}\right)^{-1}$
3. $Z(\mu,T)=1+e^{-\epsilon\beta}$
4. $Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$

$$Z(\mu,T)=\sum\limits_{i} e^{-\beta n_i(\epsilon_i-\mu)}$$ where $i$ runs over every microstate, $n_i$ number particles occupying the microstate $i$, $\epsilon_i$ is energy of a particles in the $i^{th}$ microstate

Let us write all microstates

Microstate no. $i$	State with enrgy 0	State with enrgy $\epsilon$	energy of microstate $\epsilon_i$	$n_i$
1	0	0	0	0
2	1	0	0	1
3	0	1	$\epsilon$	1

\begin{align*} Z(\mu,T)&= e^{-\beta \times0\times(0-\mu)}+e^{-\beta \times1\times(0-\mu)}\\ &+e^{-\beta\times 1\times(\epsilon-\mu)} \end{align*} $$Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$$

Hence, answer is (D)