Problem set 32

1. The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be: $H = xp_x- yp_y -\frac{1}{2}x^2 +\frac{1}{2} y^2$ in suitable units. The initial values are given to be $(x(0),y(0)) = (1,1)$ and $(p_x(0),p_y(0) )=(\frac{1}{2}, -\frac{1}{2})$. During the motion, the curves traced out by the particles in the xy-plane and the $p_xp_y$-plane are
1. both straight lines
2. a straight line and a hyperbola respectively
3. a hyperbola and an ellipse, respectively
4. both hyperbolas
$$\dot x=\frac{\partial H}{\partial p_x}=x$$ $$\dot p_x=-\frac{\partial H}{\partial x}=-(p_x-x)=x-p_x$$ $$\dot y=\frac{\partial H}{\partial p_y}=-y$$ $$\dot p_y=-\frac{\partial H}{\partial y}=-(-p_y+y)=p_y-y$$ Solving these equations and applying initial conditions, we get $$\begin{align*} x=e^x\\ y=e^{-x}\\ \end{align*}$$ Or $$x=\frac{1}{y}$$ Which is equation of hyperbola. Similarly, solving equation for $p_x$ and $p_y$, one gets ellipse.
Hence, answer is (C)
2. A static, spherically symmetric charge distribution is given by $\rho(r) = \frac{A}{r} e^{-Kr}$ where $A$ and $K$ are positive constants. The electrostatic potential corresponding to this charge distribution varies with $r$ as
1. $re^{-Kr}$
2. $\frac{1}{r} e^{-Kr}$
3. $\frac{1}{r^2} e^{-Kr}$
4. $\frac{1}{r} (1-e^{-Kr})$
Since $\nabla^2V=-\rho/\epsilon_0$
$\nabla^2V$ must be proportional to $\frac{1}{r} e^{-Kr}$. Taking $\nabla^2=\frac{d^2}{dr^2}+\frac{2}{r}\frac{d}{dr}$, for the option (B) we get $\nabla^2\left(\frac{1}{r} e^{-Kr}\right)=\frac{K^2}{r} e^{-Kr}$
Hence, option (B) is correct.
3. Band-pass and band-reject filters can be implemented by combining a low pass and a high pass filter in series and in parallel, respectively. If the cut-off frequencies of the low pass and high pass filters are $\omega_0^{LP}$ and $\omega_0^{HP}$ , respectively, the condition required to implement the band-pass and band-reject filters are, respectively,
1. $\omega_0^{HP}<\omega_0^{LP}$ and $\omega_0^{HP}<\omega_0^{LP}$
2. $\omega_0^{HP}<\omega_0^{LP}$ and $\omega_0^{HP}>\omega_0^{LP}$
3. $\omega_0^{HP}>\omega_0^{LP}$ and $\omega_0^{HP}<\omega_0^{LP}$
4. $\omega_0^{HP}>\omega_0^{LP}$ and $\omega_0^{HP}>\omega_0^{LP}$
One can easily verify that the correct option is (B)
4. Non-interacting bosons undergo Bose-Einstein Condensation (BEC) when trapped in a three-dimensional isotropic simple harmonic potential. For BEC to occur, the chemical potential must be equal to
1. $\hbar\omega/2$
2. $\hbar\omega$
3. $3\hbar\omega/2$
4. $0$
For a three-dimensional isotropic simple harmonic potential, the lowest energy is $3\hbar\omega/2$ called as zero-point energy. Hence, option (C) is correct.
5. Linearly independent solution of the differential equation $$\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=0$$ are:
1. $e^{-x}$, $e^{-2x}$
2. $e^{-x}$, $e^{2x}$
3. $e^{-2x}$, $e^{x}$
4. $e^{2x}$, $e^{x}$
Auxiliary equation is $$D^2+3D+2=0$$ $$(D+1)(D+2)=0$$ Hence, roots are $-1$ and $-2$

solutions are $e^{-x}$, $e^{-2x}$

Hence, answer is (A)