Problem set 15

1. For constant uniform electric and magnetic fields $\vec E=\vec E_0$ and $\vec B=\vec B_0$,it is possible to choose a gauge such that the scalar potential $\phi$ and vector potential $\vec A$ are given by
   1. $\phi=0$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$
   2. $\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$
   3. $\phi=-\vec E\cdot \vec r$ and $\vec A=0$
   4. $\phi=0$ and $\vec A=-\vec E t$
   Electric field $\vec E$ and magnetic field $\vec B$ are related to vector and scalar potentials by $$\vec B=\vec\nabla\times\vec A\quad (1)$$ $$\vec E=-\vec\nabla\phi-\frac{\partial\vec A}{\partial t}\quad (2)$$ Substituting values $\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$ in equations (1) and (2), we get $\vec E=\vec E_0$ and $\vec B=\vec B_0$ (Verify this). Hence answer is (B)$\phi=-\vec E\cdot \vec r$ and $\vec A=\frac{1}{2}\left(\vec B_0\times\vec r\right)$.
2. Let $\vec{a}$ and $\vec{b}$ be two distinct three-dimensional vectors. Then the component of $\vec{b}$ that is perpendicular to $\vec{a}$ is given by
1. $\frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}$
2. $\frac{\vec{b}\times(\vec{b}\times\vec{a})}{b^2}$
3. $\frac{(\vec{a}.\vec{b})\vec{b}}{b^2}$
4. $\frac{(\vec{b}.\vec{a})\vec{a}}{a^2}$
$$\begin{align*} \frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}&=\frac{(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a}}{a^2}\\ &=\frac{a^2\vec{b}-ab\cos\theta\vec{a}}{a^2} \end{align*}$$ $$\begin{align*} & \vec{a}\cdot\frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}\\ &=\frac{a^2\vec{a}\cdot\vec{b}-ab\cos\theta\vec{a}\cdot\vec{a}}{a^2}\\ &=\frac{a^3b\cos\theta-a^3b\cos\theta}{a^2}\\ &=0 \end{align*}$$ Hence answer is (A)
3. The wavefunction of a particle is given by $\psi=\left(\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right)$, where $\phi_0$ and $\phi_1$ are the normalized eigenfunctions with energies $E_0$ and $E_1$ corresponding to the ground state and first excited state, respectively. The expectation value of the Hamiltonian in the state $\psi$ is
   1. $\frac{E_0}{2}+E_1$
   2. $\frac{E_0}{2}-E_1$
   3. $\frac{E_0-2E_1}{3}$
   4. $\frac{E_0+2E_1}{3}$
   $H\phi_0=E_0\phi_0$ and $H\phi_1=E_1\phi_1$ $$\begin{align*} \left< H \right>=\frac{\left< \psi |H|\psi \right>}{\left< \psi |\psi \right>} \end{align*}$$ $$\begin{align*} &<\psi|H|\psi>=\\ &\left<\left.\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right|H\left|\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right.\right> \end{align*}$$ $$\begin{align*} &<\psi|H|\psi>=\\ &\left<\left.\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right|\frac{1}{\sqrt{2}}E_0\phi_0+iE_1\phi_1 \right> \end{align*}$$ $$\begin{align*} &<\psi|H|\psi>=\\ &\frac{1}{2}E_0\left<\left.\phi_0\right|\phi_0\right>+E_1(i)(-i) \left<\left.\phi_1\right|\phi_1 \right> \end{align*}$$ Using orthonormality condition $$<\psi|H|\psi>=\frac{1}{2}E_0+E_1$$ $$\begin{align*} <\psi|\psi>&=\\ &\left<\left.\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right|\frac{1}{\sqrt{2}}\phi_0+i\phi_1\right> \end{align*}$$ $$\begin{align*} <\psi|\psi>&=\\ &\frac{1}{2}\left<\left.\phi_0\right|\phi_0\right>+(-i)(i)\left<\left.\phi_1\right|\phi_1\right> \end{align*}$$ $$<\psi|\psi>=\frac{1}{2}+1=\frac{3}{2}$$ $$\begin{align*} \left< H \right> &=\frac{\frac{1}{2}E_0+E_1}{\frac{3}{2}}\\ &=\frac{E_0+2E_1}{3} \end{align*}$$ Hence, answer is (D)
4. Which of the following matrices is an element of the group SU(2)
1. $\begin{pmatrix}1&1\\0&1\end{pmatrix}$
2. $\begin{pmatrix}\frac{1+i}{\sqrt{3}}&\frac{-1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}&\frac{1-i}{\sqrt{3}}\end{pmatrix}$
3. $\begin{pmatrix}2+i&i\\3&1+i\end{pmatrix}$
4. $\begin{pmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}$
The special unitary group of degree $n$, denoted $SU(n)$, is the Lie group of $n×n$ unitary matrices with determinant 1. For $2\times2$ matrix the group is denoted by $SU(2)$. Let $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. As $U$ is unitary, we have $UU^\dagger=U^\dagger U=1$. This gives $$aa^*+bb^*=1\quad(1)$$ $$cc^*+dd^*=1\quad(2)$$ $$ac^*+bd^*=0\quad(3)$$ $$ca^*+db^*=0\quad(4)$$ Also $$det~U=ad-bc=1\quad(5)$$ From (4), we have $$\frac{d}{c}=-\frac{a^*}{b^*}$$ substituting in (5) $$-(aa^*+bb^*)\frac{c}{b^*}=1$$ Using (1) we get $$c=-b^*$$ Similarly, we can show that $$d=a^*$$ Hence, in general the $2\times2$ matrix $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is of $SU(2)$ group if
• Its determinant is 1 and
• It is in the form $U=\begin{pmatrix}a&b\\-b^*&a^*\end{pmatrix}$
These conditions are satisfied by $\begin{pmatrix}\frac{1+i}{\sqrt{3}}&\frac{-1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}&\frac{1-i}{\sqrt{3}}\end{pmatrix}$. Hence answer is (B).
5. The acceleration due to gravity $(g)$ on the surface of Earth is approximately 2.6 times that on the surface of Mars. Given that the radius of Mars is about one half the radius of Earth, the ratio of the escape velocity on Earth to that on Mars is approximately
1. 1.1
2. 1.3
3. 2.3
4. 5.2
Escape velocity is given by $$v=\sqrt{\frac{2GM}{R}}$$ On earth's surface force acting on body is $F=mg$, hence $g=\frac{F}{m}$. However, $F=\frac{GM_Em}{R_E^2}$, hence $g=\frac{\frac{GM_Em}{R_E^2}}{m}$, hence $GM=R_E^2g$ $$v_e=\sqrt{2R_Eg}$$ $$\frac{v_E}{v_M}=\sqrt{\frac{2R_Eg_E}{2R_Mg_M}}=\sqrt{5.2}=2.3$$ Hence, answer is (C)