Problem set 21

1. Four charges (two $+q$ and two $-q$) are kept fixed at the four vertices of a square of side $a$ as shown
   
   At the point $P$ which is at a distance $R$ from the centre $( R > > a)$, the potential is proportional to
1. $1/R$
2. $1/R^2$
3. $1/R^3$
4. $1/R^4$
The system forms a quadrupole. The electric field and potential energy variations for various system are as follow:
For monopole $E\propto 1/r^2$ and $V\propto 1/r$
For dipole $E\propto 1/r^3$ and $V\propto 1/r^2$
For quadrupole $E\propto 1/r^4$ and $V\propto 1/r^3$
For octapole $E\propto 1/r^5$ and $V\propto 1/r^4$
Hence, answer is (C).
2. A point charge $q$ of mass $m$ is kept at a distance $d$ below a grounded infinite conducting sheet which lies in the $xy$-plane. For what value of $d$ will the charge remains stationary?
1. $q/4\sqrt{mg\pi\epsilon_0}$
2. $q/\sqrt{mg\pi\epsilon_0}$
3. There is no finite value of $d$
4. $\sqrt{mg\pi\epsilon_0}/q$

When a charge $+q$ is placed below the plate at a distance $d$, then there is an image charge $-q$ above the plate at the same distance $d$. For the charge to remain stationary, weight $mg$ must equal to electrostatic force $\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$. i.e. $$mg=\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$$ $$d=\frac{q}{4\sqrt{mg\pi\epsilon_0}}$$ Hence, answer is (A).
3. The wave function of a state of the hydrogen atom is given by ${\scriptstyle\Psi=\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}}$, where $\psi_{nlm}$ is the normalized eigenfunction of the state with quantum numbers $n,l$ and $m$ in the usual notation. The expectation value of $L_z$ in the state $\Psi$ is
1. $15\hbar/16$
2. $11\hbar/16$
3. $3\hbar/8$
4. $\hbar/8$
Let the normalized wavefunction is given by $$\Psi={\scriptstyle A\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}$$ Applying normalization condition \begin{align*} |\Psi|^2&=|A|^2\left[|\psi_{200}|^2+|2\psi_{211}|^2\right.\\ &\left.+|3\psi_{210}|^2+|\sqrt{2}\psi_{21-1}|^2\right]\\ &=1 \end{align*} we get, $|A|^2\left[1+4+9+2\right]=1$, hence, $A=\frac{1}{4}$. Hence, normalized wavefunction is $$\Psi={\scriptstyle \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}$$ \begin{align*} &< L_z > =\left < \Psi|L_z|\Psi\right > \\ &=\left < \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}\right.\right.\\ &\left.+\sqrt{2}\psi_{21-1}\right]\left|L_z\right|\frac{1}{4}\left[\psi_{200}\right.\\ &\left.\left.+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]\right >\\ &=\frac{1}{16}\left\{\left < \psi_{200}\left|L_z\right|\psi_{200}\right >\right.\\ &+4\left < \psi_{211}\left|L_z\right|\psi_{211}\right >+9\left < \psi_{210}\left|L_z\right|\psi_{210}\right >\\ &\left.+2\left < \psi_{21-1}\left|L_z\right|\psi_{21-1}\right >\right\} \end{align*} But $L_z|\psi_{nlm}> =m\hbar|\psi_{nlm}>$ \begin{align*} < L_z >&=\frac{1}{16}\left\{0+4\hbar+0+2(-1)\hbar\right\}\\ &=\frac{1}{8}\hbar \end{align*} Hence, answer is (D)
4. The energy eigenvalues of a particle in the potential $V(x) =\frac{1}{2}m\omega^2x^2-ax$ are
1. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2}$
2. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega+\frac{a^2}{2m\omega^2}$
3. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{m\omega^2}$
4. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega$
\begin{align*} V(x)&=\frac{1}{2}m\omega^2x^2-ax=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}\right]\\ &=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}+\frac{a^2}{m^2\omega^4}-\frac{a^2}{m^2\omega^4}\right]\\ &=\frac{1}{2}m\omega^2\left[\left(x-\frac{a}{m\omega^2}\right)^2-\left(\frac{a}{m\omega^2}\right)^2\right] \end{align*} Let $\left(x-\frac{a}{m\omega^2}\right)=x'$ \begin{align*} V(x')&=\frac{1}{2}m\omega^2x'^2-\frac{1}{2}m\omega^2\left(\frac{a}{m\omega^2}\right)^2\\ &=\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2} \end{align*} Second term on RHS is a constant. $$H'=-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}$$ \begin{align*} E&=\left < \psi_n|H'|\psi_n\right > \\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\left < \psi_n\left|\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\frac{a^2}{2m\omega^2}\left < \psi_n\left.\right|\psi_n\right >\\ &=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2} \end{align*} Hence, answer is (A).
5. If a particle is represented by the normalized wave function \begin{align*} \psi(x)=\begin{cases}\frac{\sqrt{15}\left(a^2-x^2\right)}{4a^{5/2}}&{\scriptstyle\text{for} -a < x < a}\\ 0 &\text{otherwise} \end{cases} \end{align*} the uncertainty $\Delta p$ in its momentum is
1. $2\hbar/5a$
2. $5\hbar/2a$
3. $\sqrt{10}\hbar/a$
4. $\sqrt{5}\hbar/\sqrt{2}a$
$$\Delta P=\sqrt{<\Delta P^2>-<\Delta P>^2}$$ \begin{align*} & <\Delta P>=\int\psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\psi\:dx\\ &=-\frac{15i\hbar}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial(a^2-x^2)}{\partial x}dx\\ &=\frac{15i\hbar}{a^58}\int_{-a}^{a}(a^2-x^2)x\:dx=0 \end{align*} Since the integrand is an odd function. \begin{align*} & <\Delta P^2>=\int\psi^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\psi\:dx\\ &=-\frac{15\hbar^2}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial^2(a^2-x^2)}{\partial x^2}dx\\ &=\frac{15\hbar^2}{a^58}\int_{-a}^{a}(a^2-x^2)\:dx\\ &=\frac{15\hbar^2}{a^58}\left[a^2x-\frac{x^3}{3}\right]_{-a}^{a}=\frac{5\hbar^2}{2a^2} \end{align*} $$\Delta P=\frac{\sqrt{5}\hbar}{\sqrt{2}a}$$ Hence, answer is (D).