Problem set 51

1. The trace of $3\times3$ matrix is 2. Two of its eigenvalues are 1 and 2. The third eigenvalue is
1. -1
2. 0
3. 1
4. 2

The trace of a matrix is the sum of the eigenvalues. Hence, $1+2+x=2$, hence $x=-1$

Hence, answer is (A)

2. A particle is moving in an inverse square force field. If the total energy of the particle is positive, the trajectory of the particle is
1. circular
2. elliptical
3. parabolic
4. hyperbolic

Orbit	Eccentricity $\epsilon$	Energy $E$
Circular	$\epsilon=0$	$E < 0$
Parabolic	$\epsilon=1$	$E=0$
Elliptical	$0 < \epsilon < 1$	$E < 0$
Hyperbolic	$\epsilon > 1$	$E\ge0$

Hence, answer is (D)

3. In an electromagnetic field, which one of the following remains invariant under Lorentz transformation?
1. $\vec E\times\vec B$
2. $E^2-c^2B^2$
3. $B^2$
4. $E^2$

The Lorentz transformation equations of electric and magnetic field are given by

$E_x'=E_x$	$B_x'=B_x$
${\scriptstyle E_y'=\gamma\left(E_y-vB_z\right)}$	${\scriptstyle B_y'=\gamma\left(B_y+\frac{v}{c^2}E_z\right)}$
${\scriptstyle E_z'=\gamma\left(E_z+vB_y\right)}$	${\scriptstyle B_z'=\gamma\left(B_z-\frac{v}{c^2}E_y\right)}$

$$\begin{align*} &E^{2'}-c^2B^{2'}= E_x^{2'}+E_y^{2'}+E_z^{2'}\\ &-c^2\left(B_x^{2'}+B_y^{2'}+B_z^{2'}\right)\\ &={\textstyle E_x^2+\left(\gamma\left(E_y-vB_z\right)\right)^2}\\ &{\textstyle +\left(\gamma\left(E_z+vB_y\right)\right)^2}\\ &-c^2{\scriptstyle \left(B_x^2+\left(\gamma\left(B_y+\frac{v}{c^2}E_z\right) \right)^2+\left(\gamma\left(B_z-\frac{v}{c^2}E_y\right)\right)^2 \right)}\\ &=E^2-c^2B^2 \end{align*}$$

Hence, answer is (B)

4. A sphere of radius $R$ has uniform volume charge density. The electric potential at a point $r(r < R)$ is
1. due to the charge inside a sphere of radius $r$ only
2. due to the entire charge of the sphere
3. due to the charge of the spherical shell of inner and outer radii $r$ and $R$, only
4. independent of $r$

For region $r < R$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ $$E4\pi r^2=\frac{\frac{4}{3}\pi r^3\rho}{\epsilon_0}$$ $$E=\frac{\rho}{3\epsilon_0}r$$ $$\begin{align*} V&=-\int\limits_0^r\frac{\rho}{3\epsilon_0}r\:dr\\ &=-\frac{\rho}{6\epsilon_0}r^2 \end{align*}$$

Hence, answer is (A)

5. A free particle is moving in $+x$ direction with a linear momentum $p$. The wavefunction of the particle normalized in a length $L$ is
1. $\frac{1}{\sqrt{L}}\sin{\frac{p}{\hbar}x}$
2. $\frac{1}{\sqrt{L}}\cos{\frac{p}{\hbar}x}$
3. $\frac{1}{\sqrt{L}}e^{-i\frac{p}{\hbar}x}$
4. $\frac{1}{\sqrt{L}}e^{i\frac{p}{\hbar}x}$

For a free particle potential $V=0$, hence, for a one dimensional motion, Schrodinger equation is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$$ $$i.e. \frac{d^2\psi}{dx^2}=-\frac{2mE}{\hbar^2}\psi$$ $$or \frac{d^2\psi}{dx^2}=-k^2\psi$$ where, $k^2=\frac{2mE}{\hbar^2}$. The possible solutions of the above equation are $$\begin{align*} \psi(x)=A \begin{cases} \sin{kx}=\sin{\sqrt{\frac{2mE}{\hbar^2}}x}\\ \cos{kx}=\cos{\sqrt{\frac{2mE}{\hbar^2}}x}\\ e^{\pm ikx}=e^{\pm i\sqrt{\frac{2mE}{\hbar^2}}x} \end{cases} \end{align*}$$ where $A$ is normalization constant. For a particle moving in $+x$ direction, the wavefunction should satisfy momentum eigenvalue equation $$\hat p_x\psi(x)=-i\hbar\frac{d\psi}{dx}=p\psi(x)$$ It can be easily verified that both $\sin{kx}$ and $\cos{kx}$ are not eigenfunctions of $\hat p_x$. Hence, possible solutions are $$\psi(x)=Ae^{\pm ikx}$$ Out of these, for $k > 0$, $e^{ikx}$ represents particle moving from left to right ($+x$ direction) and $e^{-ikx}$ represents particle moving from right to left ($-x$ direction). Hence, possible solution is $$\psi(x)=Ae^{ikx}$$ Using normalization condition, we have $$|A|^2\int\limits_{0}^{L}dx=1$$ Hence, $A=\frac{1}{\sqrt{L}}$. $$\begin{align*} \psi(x)&=\frac{1}{\sqrt{L}}e^{ikx}\\ &=\frac{1}{\sqrt{L}}e^{i\sqrt{\frac{2mE}{\hbar^2}}x}\\ &=\frac{1}{\sqrt{L}}e^{i\frac{p}{\hbar}x} \end{align*}$$

Hence, answer is (D)