Problem set 28

1. The solution of the differential equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, with initial condition $x=0$ at $t=0$ is
1. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ \sinh{2t},\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
2. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*}
3. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ 1,\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
4. $x=1-\cos{2t},\quad t\geq 0$
$$\int\frac{dx}{\sqrt{1-x^2}}=2\int dt$$ Let us make substitution $x=\sin\theta$, $dx=\cos\theta$ $$\int d\theta=2t+c_1$$ $$\theta=2t+c_1$$ $$\sin^{-1}x=c_1+2t$$ $$x=c+\sin{2t}$$ At $t=0$, $x=0$, hence, $c=0$ $$x=\sin{2t}$$ However, maximum value of $\sin(2t)$ is 1 $$\sin{2t}=1\quad \text{when } 2t=\frac{\pi}{2}\quad\text{ or }t=\frac{\pi}{4}$$ Hence, \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*} Hence, answer is (B).
2. Given a uniform magnetic field $\vec B=B_0\hat k$ (where $B_0$ is a constant), a possible choice for the magnetic vector potential $\vec A$ is
1. $B_0y\hat i$
2. $-B_0y\hat i$
3. $B_0(x\hat j+y\hat i)$
4. $B_0(x\hat i-y\hat j)$
Using, $\vec B=\vec\nabla \times\vec A$, for options
(A)$\vec B=\vec\nabla \times B_0y\hat i=-B_0\hat k$
(B)$\vec B=\vec\nabla \times (-B_0y\hat i)=B_0\hat k$
(C)$\vec B=\vec\nabla \times (B_0(x\hat j+y\hat i))=0$
(D)$\vec B=\vec\nabla \times (B_0(x\hat i-y\hat j))=2B_0\hat k$
Hence, answer is (B)
3. Consider a charge $Q$ at the origin of 3-dimensional coordinate system. The flux of the electric field through the curved surface of a cone that has a height $h$ and a circular base of radius $R$ is
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{2\epsilon_0}$
3. $\frac{hQ}{R\epsilon_0}$
4. $\frac{QR}{2h\epsilon_0}$
Let us consider a sphere of radius of $h$ (if $h>R$) or of radius $R$ (if $R>h$) with centre at origin. According to Gauss law the flux through entire sphere is $\frac{Q}{\epsilon_0}$. The flux through curved surfaces of cone is half of the flux through sphere i.e. $\frac{Q}{2\epsilon_0}$
Hence, answer is (B)
4. A Hermitian operator $\hat O$ has two normalised eigenstates $|1 > $ and $|2 > $ with eigenvalues $1$ and $2$, respectively. The two states $|u > =\cos\theta|1 > +\sin\theta|2 > $ and $|v > =\cos\phi|1 > +\sin\phi|2 > $ are such that $ < v|\hat O|v > =7/4$ and $ < u|v > =0$. Which of the following are possible values of $\theta$ and $\phi$?
1. $\theta=-\frac{\pi}{6}$ and $\phi=\frac{\pi}{3}$
2. $\theta=\frac{\pi}{6}$ and $\phi=\frac{\pi}{3}$
3. $\theta=-\frac{\pi}{4}$ and $\phi=\frac{\pi}{4}$
4. $\theta=\frac{\pi}{3}$ and $\phi=-\frac{\pi}{6}$
Let $\chi_1=|1 > $ and $\chi_2=|2 > $ \begin{align*} &< v|\hat O|v > =\int v*\hat Ovd\tau\\ &={\scriptstyle\int\left(\cos\phi\chi_1+\sin\phi\chi_2\right)^*\hat O\left(\cos\phi\chi_1+\sin\phi\chi_2\right)d\tau}\\ &={\scriptstyle\cos^2\phi\int\chi_1^*\hat O\chi_1 d\tau +\sin^2\phi\int\chi_2^*\hat O\chi_2d\tau}\\ &{\scriptstyle+\cos\phi\sin\phi\int\chi_1^*\hat O\chi_2d\tau+\cos\phi\sin\phi\int\chi_2^*\hat O\chi_1d\tau}\\ &=\cos^2\phi+2\sin^2\phi\\ &=1+\sin^2\phi \end{align*} $$1+\sin^2\phi=\frac{7}{4}$$ $$\Rightarrow \sin\phi=\frac{\sqrt{3}}{2}\Rightarrow\phi=\frac{\pi}{3}$$ \begin{align*} &< u|v > =\int u^*vd\tau\\ &{\scriptstyle=\int\left(\cos\theta\chi_1+\sin\theta\chi_2\right)^*\left(\cos\phi\chi_1+\sin\phi\chi_2\right)d\tau}\\ &={\scriptstyle\cos\theta\cos\phi\int\chi_1^*\chi_1 d\tau +\sin\theta\sin\phi\int\chi_2^*\chi_2d\tau}\\ &{\scriptstyle+\cos\theta\sin\phi\int\chi_1^*\chi_2d\tau+\sin\theta\sin\phi\int\chi_2^*\chi_1d\tau}\\ &=\cos\theta\cos\phi+\sin\theta\sin\phi\\ &=\cos{(\theta+\phi)}=\cos{(\theta+\frac{\pi}{3})} \end{align*} \begin{align*} &\Rightarrow \cos{(\theta+\frac{\pi}{3})}\\ &=0\Rightarrow\theta+\frac{\pi}{3}\\ &=\frac{\pi}{2}\Rightarrow\theta\\ =\frac{5\pi}{6}\text{ or }-\frac{\pi}{6} \end{align*} Hence, option (A) is correct.
5. The ground state energy of a particle of mass $m$ in the potential $V(x)=V_0\cosh{\left(\frac{x}{L}\right)}$, where $L$ and $V_0$ are constants (and $V_0>>\frac{\hbar^2}{2mL^2}$) is approximately
1. $V_0+\frac{\hbar}{L}\sqrt{\frac{2V_0}{m}}$
2. $V_0+\frac{\hbar}{L}\sqrt{\frac{V_0}{m}}$
3. $V_0+\frac{\hbar}{4L}\sqrt{\frac{V_0}{m}}$
4. $V_0+\frac{\hbar}{2L}\sqrt{\frac{V_0}{m}}$
\begin{align*} V(x)&=V_0\cosh{\left(\frac{x}{L}\right)}\\ &=\frac{V_0}{2}\left(e^{x/L}+ e^{-x/L}\right)\\ &=\frac{V_0}{2}\left\{\left[1+\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right.\\ &\left. +\left[1-\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2-\cdots\right]\right\}\\ &=\frac{V_0}{2}\left\{\left[2+2\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right\}\\ &=V_0+\frac{V_0}{2}\left(\frac{x}{L}\right)^2\\ &=V_0+\frac{1}{2}\frac{V_0}{L^2}x^2\\ &=V_0+\frac{1}{2}kx^2 \end{align*} where $k=\frac{V_0}{L^2}$. Thus, particle is under the action of constant potential $V_0$ and simple harmonic potential $\frac{1}{2}kx^2$. For free particle motion lowest energy is $V_0$ and for harmonic potential lowest energy is $\frac{1}{2}\hbar\omega$. Hence, ground state energy is \begin{align*} E&=V_0+\frac{1}{2}\hbar\omega\\ &=V_0+\frac{1}{2}\hbar\sqrt{\frac{k}{m}}\\ &=V_0+\frac{1}{2}\hbar\sqrt{\frac{V_0}{L^2m}} \end{align*} Hence, answer is (D)