qm

1. Wave function of a particle moving in free space is given by, $\psi=e^{ikx}+2e^{-ikx}$. Find the energy of the particle.
one dimensional Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$
For a free particle $V=0$
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$
$\frac{d^2\psi}{dx^2}=-k^2\left(e^{ikx}+2e^{-ikx}\right)=-k^2\psi$
$E=\frac{\hbar^2k^2}{2m}$
2. What are unit and dimensions of wavefunction $\psi$
Wavefunction may be in position space or momentum space. For a 1-dimensional position space wavefunction $\psi(x)$ the normalization condition is $\int\psi^*(x)\psi(x)dx=1$, so $\psi^*(x)\psi(x)$ has units of inverse distance and $\psi(x)$ has units of square root of inverse distance, e.g. $m^{−1/2}$. For a 2-dimensional position space wavefunction $\psi(x,y)$ the normalization condition would be $\iint\psi^*(x,y)\psi(x,y)dxdy=1$, so $\psi^*(x,y)\psi(x,y)$ has units of inverse area and $\psi(x,y )$ has units of square root of inverse area, e.g. $m^{−1}$. Similarly, for 3-dimension $m^{-3/2}$.
In momentum space of 1-dimensional case we have $p^{−1/2}$ i.e. $\frac{1}{\sqrt{kg-m}}$
3. In Young's double slit experiment, if one of the following parameters ($\lambda$, $d$ and $D$) is increased in the same order keeping the other two same, then the fringe width
1. decreases, decreases, increases
2. decreases, increases, increases
3. increases, decreases, increases
4. increases, increases, decreases
Fringe width is the distance between two fringes of the same kind. The formula for fringe width is $\Delta y=\frac{\lambda D}{d}$ If we increase $\lambda$ the fringe width increases. If we increase $d$ the fringe width decreases. If we increase $D$ the fringe width increases. Hence answer is (C)
4. 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)
5. The energy levels of the non-relativistic electron in a hydrogen atom (i.e. in a Coulomb potential $V(r)\propto -1/r$) are given by $E_{nlm}\propto -1/n^2$, where $n$ is the principal quantum number, and the corresponding wave functions are given by $\psi_{nlm}$, where $l$ is the orbital angular momentum quantum number and $m$ is the magnetic quantum number. The spin of the electron is not considered. Which of the following is a correct statement?
1. There are exactly $( 2l+1)$ different wave functions $\psi_{nlm}$, for each $E_{nlm}$.
2. There are $l(l+1)$ different wave functions $\psi_{nlm}$, for each $E_{nlm}$.
3. $E_{nlm}$ does not depend on $l$ and $m$ for the Coulomb potential.
4. There is a unique wave function $\psi_{nlm}$ for each $E_{nlm}$.
(C) $E_{nlm}$ does not depend on $l$ and $m$ for the Coulomb potential.
6. The Hamiltonian of an electron in a constant magnetic field $\vec B$ is given by $H=\mu \vec\sigma\cdot\vec B$ where $\mu$ is a positive constant and $\vec\sigma= (\sigma_1, \sigma_2, \sigma_3 )$ denotes the Pauli matrices. Let $\omega = \mu B/\hbar$ and $I$ be the $2\times2$ unit matrix. Then the operator $e^{iHt/\hbar}$ simplifies to
1. $I\cos{\frac{\omega t}{2}}+\frac{i\vec\sigma\cdot\vec B}{B}\sin{\frac{\omega t}{2}}$
2. $I\cos{\omega t}+\frac{i\vec\sigma\cdot\vec B}{B}\sin{\omega t}$
3. $I\sin{\omega t}+\frac{i\vec\sigma\cdot\vec B}{B}\cos{\omega t}$
4. $I\sin{2\omega t}+\frac{i\vec\sigma\cdot\vec B}{B}\cos{2\omega t}$
$$H=\mu\vec\sigma\cdot\vec B=\mu B\vec\sigma\cdot\hat n$$ where $\hat n$ is unit vecot along $\vec B$. Using $\omega=\mu B/\hbar$, we get $$H=\hbar\omega\vec\sigma\cdot\hat n$$ $$\therefore e^{iHt/\hbar}=e^{i\omega t(\vec\sigma\cdot\hat n)}$$ \begin{align*} e^{i\omega t(\vec\sigma\cdot\hat n)}&=1+i\omega t(\vec\sigma\cdot\hat n)\\ &+\frac{(i\omega t)^2}{2!}(\vec\sigma\cdot\hat n)^2\\ &+\frac{(i\omega t)^3}{3!}(\vec\sigma\cdot\hat n)^3\\ &+\frac{(i\omega t)^4}{4!}(\vec\sigma\cdot\hat n)^4\\ &+\dots\quad (1) \end{align*} But, \begin{align*} (\vec\sigma\cdot\hat n)&=\sigma_xn_x+\sigma_yn_y+\sigma_zn_z\\ &=\begin{pmatrix}0&n_x\\n_x&0\end{pmatrix}+ \begin{pmatrix}0&-n_y\\in_y&0\end{pmatrix}\\ &+\begin{pmatrix}n_z&0\\0&-n_z\end{pmatrix} \end{align*} $$(\vec\sigma\cdot\hat n)=\begin{pmatrix}n_z&n_x-n_y\\n_x+in_y&-n_z\end{pmatrix}$$ It can be verified that $(\vec\sigma\cdot\hat n)^2=I$. Hence, equation (1) becomes \begin{align*} e^{i\omega t(\vec\sigma\cdot\hat n)}&=\\ &I+i\omega t(\vec\sigma\cdot\hat n)\\ &-\frac{(\omega t)^2}{2!}I\\ &-i\frac{(\omega t)^3}{3!}(\vec\sigma\cdot\hat n)\\ &+\frac{(\omega t)^4}{4!}I-\dots \end{align*} \begin{align*} e^{i\omega t(\vec\sigma\cdot\hat n)}&=I\left[1-\frac{(\omega t)^2}{2!}+\frac{(\omega t)^4}{4!}-..\right]\\ &+(\vec\sigma\cdot\hat n)\left[\omega t-\frac{(\omega t)^3}{3!}+\dots\right] \end{align*} $$e^{i\omega t(\vec\sigma\cdot\hat n)}=I\cos{\omega t}+\frac{\vec\sigma\cdot\vec B}{B}\sin{\omega t}$$
7. If the perturbation $H' =ax$, where $a$ is a constant, is added to the infinite square well potential $$ \begin{align*} V(x) &= \begin{cases} 0 & \text{for } 0 \leq x \geq \pi \\ \infty & \text{otherwise} \end{cases} \end{align*} $$ The correction to the ground state energy, to first order in $a$, is
1. $\frac{a\pi}{2}$
2. $a\pi$
3. $\frac{a\pi}{4}$
4. $\frac{a\pi}{\sqrt{2}}$
Ground state wave function is given by $$\psi_0=\sqrt{\frac{2}{L}}\sin{\frac{n\pi x}{L}}=\sqrt{\frac{2}{\pi}}\sin{\frac{n\pi x}{\pi}}$$. Since, $L=\pi$. First order correction to energy is given by \begin{align*} E_0^{(1)}&=<\psi_0|H'|\psi_0>\\ &=\int_0^\pi\psi_0^*\:ax\:\psi_0\:dx\\ &={\scriptstyle \int_0^\pi\sqrt{\frac{2}{\pi}}\sin{\frac{n\pi x}{\pi}}\:ax\:\sqrt{\frac{2}{\pi}}\sin{nx}\:dx}\\ &=\frac{2a}{\pi}\int_0^\pi x\sin^2{nx}\:dx\\ &=\frac{2a}{\pi}\frac{1}{2}\int_0^\pi x(1-\cos{2nx})\:dx\\ &={\scriptstyle \frac{a}{\pi}\left\{\left[\frac{x^2}{2}\right]_0^\pi-\left[x\int \cos{2nx}\:dx\right.\right.}\\ &{\scriptstyle \left.\left.-\int\frac{dx}{dx}\left(\int\cos{2nx}\:dx\right)dx\right]_0^\pi\right\}}\\ &={\scriptstyle \frac{a}{\pi}\left\{\frac{\pi^2}{2}-\left[x\frac{sin{2nx}}{2n}+\frac{\cos{2nx}}{4n^2}\right]_0^\pi\right\}}\\ &=\frac{a}{2\pi} \end{align*}
8. A particle in one dimension moves under the influence of a potential $V(x) = a x^6$, where $a$ is a real constant. For large $n$ the quantized energy level $E_n$ depends on $n$ as:
1. $E_n\sim n^3$
2. $E_n\sim n^{4/3}$
3. $E_n\sim n^{6/5}$
4. $E_n\sim n^{3/2}$
For large value of $n$, the semiclassical approximation is valid and for bound states we may use the Bohr-Sommerfeld quantization condition: $$n\hbar=\oint p\:dq$$ The total energy of system is $E=\frac{p^2}{2m}+ax^6$, hence $$p=\sqrt{2m(E-ax^6)}$$ $$\oint p\:dq=\int_{x_a}^{x_b}\sqrt{2m(E-ax^6)}dx$$ Where $x_a$ and $x_b$ are classical turning points. Turning point is the point at which $E=V(x)$, hence we get, $$E=ax^6$$ Above equation is true for $x=\pm\left(\frac{E}{a}\right)^{1/6}$ $$\oint p\:dq=\int_{-\left(\frac{E}{a}\right)^{1/6}}^{\left(\frac{E}{a}\right)^{1/6}}\sqrt{2m(E-ax^6)}dx$$ Let us scale the integration with substitution $x=\left(\frac{E}{a}\right)^{1/6}$ \begin{align*} &\oint p\:dq\\ &=\sqrt{2m}\left(\frac{E}{a}\right)^{2/3}\int_{-1}^1\sqrt{1-y^6)}dy\\ &=n\hbar \end{align*} Hence, $E_n\sim n^{3/2}$
9. The wave function of a state of the hydrogen atom is given by $\Psi=\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}$, 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)
10. 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).
11. 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).
12. Given $2\times2$ unitary matrix $U$ satisfying $U^\dagger U=UU^\dagger = 1$ with $\det U = e^{i\phi}$, one can construct a unitary matrix $V \:( V^\dagger V = VV^\dagger= 1)$ with $\det V=1$ from it by
1. multiplying $U$ by $e^{i\phi/2}$
2. multiplying any single element of $U$ by $e^{i\phi}$
3. multiplying any row or column of $U$ by $e^{i\phi/2}$
4. multiplying $U$ by $e^{i\phi}$
The general $2\times 2$ unitary matrix is defined by \begin{equation*} U=\begin{pmatrix} a&b\\ -e^{i\phi}b^*&e^{i\phi}a^* \end{pmatrix} \end{equation*} Clearly, as columns or rows of $U$ form an orthonormal set, we have, $|a|^2+|b|^2=1$ and $\det{U}=e^{i\phi}$.
Let us take option (A) i.e. multiplying $U$ by $e^{-i\phi/2}$ \begin{equation*} V=e^{-i\phi}U=\begin{pmatrix} e^{-i\phi/2}a&e^{-i\phi/2}b\\ -e^{i\phi/2}b^*&e^{i\phi/2}a^* \end{pmatrix} \end{equation*} Clearly, we get, $\det{V}=1$.
Hence, option (A) is correct
13. The energies in the ground state and first excited state of a particle of mass $m =\frac{1}{2}$ in a potential $V(x)$ are $-4$ and $-1$, respectively, (in units in which $\hbar = 1$ ). If the corresponding wavefunctions are related by $\psi_1(x)=\psi_0(x) \sinh x$, then the ground state eigenfunction is
1. $\psi_0(x)=\sqrt{sech~ x}$
2. $\psi_0(x)=sech~ x$
3. $\psi_0(x)=sech^2~ x$
4. $\psi_0(x)=sech^3~ x$
Schrodinger equation is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ For $\hbar=1$,$m=m/2$ $$-\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ For ground state $$-\frac{d^2\psi_0}{dx^2}+V\psi_0=-4\psi_0\quad \dots(1)$$ $$-\frac{d^2\psi_1}{dx^2}+V\psi_1=-\psi_1\quad \dots(2)$$ Substituting $\psi_1=\psi_0\sinh{x}$ in equation (2) $${\scriptstyle-\frac{d^2(\psi_0\sinh{x})}{dx^2}+V\psi_0\sinh{x}=-\psi_0\sinh{x}}$$ Solving this we get $$ {\scriptstyle-\frac{d^2\psi_0}{dx^2}\sinh{x}-2\frac{d\psi_0}{dx}\cosh{x}-\psi_0\sinh{x}+V\psi_0=-\psi_0\sinh{x}} $$ Rearranging \begin{align*} &-\frac{d^2\psi_0}{dx^2}-2\coth{x}\frac{d\psi_0}{dx}-\psi_0\sinh{x}\\ &+V\psi_0=-\psi_0\sinh{x} \end{align*} substituting value of $-\frac{d^2\psi_0}{dx^2}+V\psi_0=-4\psi_0$ from equation (1), we get $$\coth{x}\frac{d\psi_0}{dx}=-2\psi_0$$ $$\frac{d\psi_0}{\psi_0}=-2\frac{dx}{\coth{x}}=-2\tanh{x}dx$$ Integrating above equation $$\ln\psi_0=-2\ln{\cosh{x}}=\ln\text{sech}^{2}{x}$$ $$\psi_0=\text{sech}^{2}{x}$$ Hence, answer is (C)
14. The perturbation \begin{align*} H'=\begin{cases} b(a-x),&-a < x < a\\ 0&\text{otherwise} \end{cases} \end{align*} acts on a particle of mass $m$ confined in an infinite square well potential \begin{align*} V(x)=\begin{cases} 0,&-a < x < a\\ \infty&\text{otherwise} \end{cases} \end{align*} The first order correction to the ground-state energy of the particle is
1. $\frac{ba}{2}$
2. $\frac{ba}{\sqrt{2}}$
3. $2ba$
4. $ba$
\begin{align*} &E_0^{(1)}=\int\psi^*H'\psi\:dx\\ &=\frac{1}{a}\int\limits_{-a}^{a}\sin{\frac{\pi x}{2a}}b(a-x)\sin{\frac{\pi x}{2a}}dx\\ &=\frac{b}{a}\int\limits_{-a}^{a}\sin^2{\frac{\pi x}{2a}}\:(a-x)\:dx\\ &=\frac{b}{2a}\int\limits_{-a}^{a}\left(1-\cos{\frac{\pi x}{a}}\right)(a-x)dx\\ &=\frac{b}{2a}\left\{\int\limits_{-a}^{a}(a-x)dx\right.\\ &\left.-\int\limits_{-a}^{a}\cos{\frac{\pi x}{a}}(a-x)dx\right\} \end{align*} Odd functions have integration zero \begin{align*} E_0^{(1)}&=\frac{b}{2a}\left\{a\int\limits_{-a}^{a}dx-a\int\limits_{-a}^{a}\cos{\frac{\pi x}{a}}dx\right\}\\ &=\frac{b}{2a}\left\{2a^2-0\right\}=ba \end{align*} Hence, answer is (D)
15. Let $|0>$ and $|1>$ denote the normalized eigenstates corresponding to the ground and the first excited states of a one-dimensional harmonic oscillator. The uncertainty $\Delta x$ in the state $\frac{1}{\sqrt{2}}\left(|0>+|1>\right)$ is
1. $\Delta x=\sqrt{\hbar/2m\omega}$
2. $\Delta x=\sqrt{\hbar/m\omega}$
3. $\Delta x=\sqrt{2\hbar/m\omega}$
4. $\Delta x=\sqrt{\hbar/4m\omega}$
$$\Delta x=\sqrt{ < x^2 > - < x >^2}$$ Let $u_0=|0 > $, $u_1=|1 > $ and $\psi= |0 > +|1 > =u_0+u_1$. The lowering and raising operators for harmonic oscillator are $$\hat a=\frac{1}{\sqrt{2m\hbar\omega}}\left(m\omega x+ip\right)$$ $$\hat a^\dagger=\frac{1}{\sqrt{2m\hbar\omega}}\left(m\omega x-ip\right)$$ $$x=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat a+\hat a^\dagger\right)$$ Also we have, $\hat a u_n=\sqrt{n}u_{n-1}$, $\hat a^\dagger u_n=\sqrt{n+1}u_{n+1}$. \begin{align*} &< x >=<\psi|x|\psi >\\ &={\scriptstyle\sqrt{\frac{\hbar}{2m\omega}}<\psi|\left(\hat a+\hat a^\dagger\right)|\psi >}\\ &={\scriptstyle\sqrt{\frac{\hbar}{2m\omega}}\frac{1}{2} < (u_0+u_1)|\left(\hat a+\hat a^\dagger\right)|(u_0+u_1) >}\\ &={\scriptstyle\sqrt{\frac{\hbar}{8m\omega}}\left\{ < u_0|\hat a|u_0 > + < u_0|\hat a| u_1 >+ < u_1|\hat a|u_0>\right.}\\ &{\scriptstyle+ < u_1|\hat a|u_1 >+ < u_0|\hat a^\dagger|u_0 > + < u_0|\hat a^\dagger|u_1 >}\\ &{\scriptstyle\left.+ < u_1|\hat a^\dagger|u_0 > + < u_1|\hat a^\dagger|u_1 >\right\}}\\ &={\scriptstyle\sqrt{\frac{\hbar}{8m\omega}}\left\{0+1+0+0+0+0+1+0\right\}}\\ &=\sqrt{\frac{\hbar}{2m\omega}} \end{align*} \begin{align*} &< x^2 > ={\scriptstyle < \psi|x^2|\psi > =\frac{\hbar}{2m\omega} < \psi|\left(\hat a+\hat a^\dagger\right)^2|\psi >} \\ &={\scriptstyle\frac{\hbar}{2m\omega}\frac{1}{2}<(u_0+u_1)|\left(\hat a+\hat a^\dagger\right)^2|(u_0+u_1) > }\\ &={\scriptstyle\frac{\hbar}{4m\omega} < (u_0+u_1)|\left(\hat a\hat a+\hat a^\dagger\hat a^\dagger+\hat a\hat a^\dagger+\hat a^\dagger\hat a\right)|(u_0+u_1) > }\\ &={\scriptstyle\frac{\hbar}{4m\omega}\left\{ < u_0|\hat a\hat a|u_0 > + < u_0|\hat a\hat a|u_1 > + < u_1|\hat a\hat a|u_0 >\right.}\\ &{\scriptstyle + < u_1|\hat a\hat a|u_1 > + < u_0|\hat a^\dagger\hat a^\dagger|u_0 > + < u_0|\hat a^\dagger\hat a^\dagger|u_1 >}\\ &{\scriptstyle + < u_1|\hat a^\dagger\hat a^\dagger|u_0>+ < u_1|\hat a^\dagger\hat a^\dagger|u_1 >+ < u_0|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_0 >}\\ &{\scriptstyle + < u_0|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_1 > + < u_1|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_0 >}\\ &{\scriptstyle\left. + < u_1|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_1 > \right\}} \end{align*} Using identities $\hat a\hat a^\dagger+\hat a^\dagger\hat a=(2n+1)u_n$ and mentioned above, we get $$ < x^2 > =\frac{\hbar}{m\omega}$$ $$\Delta x=\sqrt{\frac{\hbar}{m\omega}-\frac{\hbar}{2m\omega}}=\sqrt{\frac{\hbar}{2m\omega}}$$
16. What would be the ground state energy of the Hamiltonian $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}-\alpha \delta(x)$ if variational principle is used to estimate it with the trial wavefunction $\psi(x)= Ae^{-bx^2}$ with $b$ as the variational parameter? [Hint: ${\scriptstyle \int_{-\infty}^{\infty} x^{2n} e^{-2bx^2}\:dx = (2b)^{-n-\frac{1}{2}}\Gamma\left(n+\frac{1}{2}\right)}$]
1. $-m\alpha^2/2\hbar^2$
2. $-2m\alpha^2/\pi\hbar^2$
3. $-m\alpha^2/\pi\hbar^2$
4. $m\alpha^2/\pi\hbar^2$
According to variational principle ground state energy is given by $$E=\frac{<\psi|H|\psi>}{<\psi|\psi>}$$ \begin{align*} <\psi|\psi>&=|A|^2\int\limits_{-\infty}^{\infty}e^{-2bx^2}\:dx\\ &=|A|^2\sqrt{\frac{\pi}{2b}} \end{align*} Using normalization condition $<\psi|\psi>=1$, we get, $A=\left(\frac{2b}{\pi}\right)^{\frac{1}{4}}$ $$\psi(x)= \left(\frac{2b}{\pi}\right)^{\frac{1}{4}}e^{-bx^2}$$ \begin{align*} & E=<\psi|H|\psi>\\ &=\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:H\:e^{-bx^2}\:dx\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}-\alpha \delta(x)\right)\:e^{-bx^2}\:dx}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{-\frac{\hbar^2}{2m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\frac{d^2}{dx^2}e^{-bx^2}dx\right.}\\ &{\scriptstyle\left.-\int\limits_{-\infty}^{\infty}e^{-bx^2}\alpha \delta(x)\:e^{-bx^2}\:dx\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{-\frac{\hbar^2}{2m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\frac{d}{dx}\left(-2bxe^{-bx^2}\right)dx\right.}\\ &{\scriptstyle\left.-\alpha\int\limits_{-\infty}^{\infty}e^{-2bx^2} \delta(x)\:dx\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\left(e^{-bx^2}-2bxe^{-bx^2}\right)dx-\alpha\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\left[\int\limits_{-\infty}^{\infty}e^{-2bx^2}\:dx-2b\int\limits_{-\infty}^{\infty}xe^{-2bx^2}dx\right]-\alpha\right\}}\\ &=\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\left[\sqrt{\frac{\pi}{2b}}-0\right]-\alpha\right\}\\ &=\frac{\hbar^2b}{m}-\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\alpha \end{align*} $$\frac{\partial E}{\partial b}=\frac{\hbar^2}{m}-\frac{1}{2}\left(\frac{2}{b\pi}\right)^{\frac{1}{2}}\alpha=0$$ gives $$b=\frac{m^2\alpha^2}{2\hbar^4\pi}$$ \begin{align*} E&=\frac{\hbar^2}{m}\frac{m^2\alpha^2}{2\hbar^4\pi}-\left(\frac{2}{\pi}\frac{m^2\alpha^2}{2\hbar^4\pi}\right)^{\frac{1}{2}}\alpha\\ &=-\frac{m\alpha^2}{2\pi\hbar^2} \end{align*}
17. In the scattering of some elementary particles, the scattering cross-section is found to depend on the total energy and the fundamental constants $h$ (Planck’s constant) and $c$ (the speed of light in vacuum). Using dimensional analysis, the dependence of $\sigma$ on these quantities is given by
1. $\sqrt{\frac{hc}{E}}$
2. $\frac{hc}{E^{3/2}}$
3. $\left(\frac{hc}{E}\right)^2$
4. $\frac{hc}{E}$
Scattering cross-section is measured in units of area i.e. $m^2$ $$\frac{hc}{E}=\frac{Js\:m/s}{J}=m$$ $$\left(\frac{hc}{E}\right)^2=m^2$$ Hence, answer is (C)
18. 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.
19. 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)
20. Let $\psi_{nlm}$ denote the eigenstates of a hydrogen atom in the usual notation. The state $\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]$ is an eigenstate of
1. $L^2$ but not of the Hamiltonian or $L_z$
2. the Hamiltonian, but not of $L^2$ or $L_z$
3. the Hamiltonian, $L^2$ and $L_z$
4. $L^2$ and $L_z$, but not of the Hamiltonian
$${\scriptstyle\Psi=\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}$$ As $\psi_{nlm}$ are eigenfunctions of $H$, their linear combination will be eigenfunction of $H$. Now, \begin{align*} L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\ &=l(l+1)\hbar^2\psi_{nlm} \end{align*} \begin{align*} L_z\Psi&{\scriptstyle=L_z\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}\\ &=\frac{1}{5}\left[-3\psi_{211}+\sqrt{5}\psi_{21-1}\right] \end{align*} Hence, $\Psi$ is not eigenfunction of $L_z$. Similarly, one can show that $\Psi$ is not eigenfunction of $L^2$.
Hence, option (B) is correct.
21. The Hamiltonian for a spin-$1/2$ particle at rest is given by $H=E_0(\sigma_z+\alpha\sigma_x)$, where $\sigma_x$ and $\sigma_z$ are Pauli spin matrices and $E_0$ and $\alpha$ are constants. The eigenvalues of this Hamiltonian are
1. $\pm E_0\sqrt{1+\alpha^2}$
2. $\pm E_0\sqrt{1-\alpha^2}$
3. $E_0$ (doubly degenerate)
4. $E_0\left(1\pm\frac{1}{2}\alpha^2\right)$
\begin{align*} H&=E_0(\sigma_z+\alpha\sigma_x)\\ &=E_0\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix}+\alpha\begin{pmatrix}0&1\\1&0\end{pmatrix}\right\}\\ &=E_0\begin{pmatrix}1&\alpha\\\alpha&-1\end{pmatrix} \end{align*} Eigenvalues of $H$ are obtained by solving $\det|H-\lambda I|=0$ $$E_0\begin{vmatrix}1-\lambda&\alpha\\\alpha&-1-\lambda\end{vmatrix}=0$$ $$\lambda=\pm E_0\sqrt{1+\alpha^2}$$ Hence, answer is (A)
22. For a system of independent non-interacting one-dimensional oscillators, the value of the free energy per oscillator, in the limit $T\rightarrow0$, is
1. $\frac{1}{2}\hbar\omega$
2. $\hbar\omega$
3. $\frac{3}{2}\hbar\omega$
4. $0$
(A) $\frac{1}{2}\hbar\omega$
23. In a system consisting of two spin-$\frac{1}{2}$ particles labeled 1 and 2, let $\vec S^{(1)} = \frac{\hbar}{2}\vec\sigma^{(1)}$ and $\vec S^{(2)} = \frac{\hbar}{2}\vec\sigma^{(2)}$ denote the corresponding spin operators. Here $\vec\sigma\equiv (\sigma_x ,\sigma_y ,\sigma_z)$ and $\sigma_x ,\sigma_y ,\sigma_z$ are the three Pauli matrices.
1. In the standard basis the matrices for the operators $ S^{(1)}_x S^{(2)}_y$ and $S^{(1)}_y S^{(2)}_x$ are, respectively,
1. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-1&0\\0&1\end{pmatrix}}$
2. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}i&0\\0&-i\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-i&0\\0&i\end{pmatrix}}$
3. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}$$
4. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&-i\\0&0&i&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}$$
2. These two operators satisfy the relation
1. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=S^{(1)}_z S^{(2)}_z}$
2. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=0}$
3. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=iS^{(1)}_z S^{(2)}_z}$
4. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=0}$
This is a double spin system, represented by $4\times4$ matrix representation. \begin{align*} S^{(1)}_x S^{(2)}_y=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\otimes\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \end{align*} where, $\otimes$ represents tensor product. \begin{align*} &S^{(1)}_x S^{(2)}_y={\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0\begin{pmatrix}0&-i\\i&0\end{pmatrix}&1\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\1\begin{pmatrix}0&-i\\i&0\end{pmatrix}&0\begin{pmatrix}0&-i\\i&0\end{pmatrix}\end{pmatrix}}\\ &= \frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix} \end{align*} \begin{align*} S^{(1)}_y S^{(2)}_x&=\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \otimes\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\\ &=\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix} \end{align*} Hence, answer is (C) \begin{align*} &{\scriptscriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=S^{(1)}_x S^{(2)}_y\:S^{(1)}_y S^{(2)}_x-S^{(1)}_y S^{(2)}_x\:S^{(1)}_x S^{(2)}_y}\\ &{\scriptscriptstyle=\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}\\ &{\scriptscriptstyle-\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}}\\ &={\scriptscriptstyle\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}-\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}}\\ &=0 \end{align*} Hence, answer is (D)
24. If an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bohr radius is approximately
1. 0.68
2. 0.48
3. 0.28
4. 0.91
Ground state wave function is $\psi=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$. Probability that electron is found within Bohr radius is given by \begin{align*} &P(r)=\int\limits_0^{a_0}|\psi|^2\:4\pi r^2 dr\\ & =\frac{4}{a_0^3}\int\limits_0^{a_0}r^2e^{-2r/a_0}\:dr\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2\int e^{-2r/a_0}\:dr\right]_0^{a_0}-\left[\int 2r \left(\int e^{-2r/a_0}\:dr\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\right]_0^{a_0}+a_0\left[\int r e^{-2r/a_0}\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[r\int e^{-2r/a_0}\:dr-\int e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[re^{-2r/a_0}\left(-\frac{a_0}{2}\right)-e^{-2r/a_0}\left(\frac{a_0^2}{4}\right)\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{4}+\frac{a_0^3}{4}\right\}}\\ &=-\left\{-5\times\frac{1}{e^2}+1\right\}=0.32 \end{align*} Hence, probability of finding electron outside Bohr radius=1-0.32=0.68
Hence, answer is (A)
25. A system is known to be in a state described by the wave function $$\psi(\theta,\phi)=\frac{1}{\sqrt{30}}\left[5Y_4^0+5Y_6^0+25Y_6^3\right]$$ where $Y_l^m$ are spherical harmonics. The probability of finding the system in a state with $m=0$ is :
1. Zero
2. $\frac{6}{\sqrt{30}}$
3. $\frac{6}{30}$
4. $\frac{13}{15}$
\begin{align*} |\psi|^2&=\frac{1}{30}\left[25\left|Y_4^0\right|^2+\left|Y_6^0\right|^2\right]\\ &=\frac{26}{30}=\frac{13}{15} \end{align*} Hence, answer is (D)
26. For attractive one-dimensional delta function potential situated at $x=0$, the wave function of the bound state is given by:
1. $\psi(x)=e^{-\alpha x}$
2. $\psi(x)=e^{-\alpha |x|}$
3. $\psi(x)=e^{-\alpha x^2}$
4. $\psi(x)=\sin{\alpha x}$
The attractive delta function is given by $$V(x)=-a\delta(x)$$ The $\delta$-function is infinite at $x=0$ and zero elsewhere. For bound state total energy $E$ of the particle must be less than the potential energy $V(X)$. Hence, $E$ must be negative where $V(x)=0$. The Schrodinger equation at these points will be $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi\quad\text{ or }\frac{d^2\psi}{dx^2}=\alpha^2\psi$$ where, $\alpha=\sqrt{-\frac{2mE}{\hbar^2}}$. The general solution of above equation can be written as $$\psi(x)=Ae^{-\alpha x}+Be^{\alpha x}$$ For $x<0$, the term, $Ae^{-\alpha x}\rightarrow\infty$ as $x\rightarrow-\infty$, and for $x>0$, the term $Ae^{\alpha x}\rightarrow\infty$ as $x\rightarrow\infty$. Hence, we will have to write separate solutions for these two regions. $$\psi_-(x)=Be^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ At $x=0$ these two solutions must be continous. Hence, we get $A=B$. $$\psi_-(x)=Ae^{\alpha x}\quad\text{for} x<0$$ $$\psi_+(x)=Ae^{-\alpha x}\quad\text{for} x>0$$ Thus, $$\psi(x)=Ae^{-\alpha |x|}\quad\text{for all values of } x$$ Hence, answer is (B)
27. The variational method in perturbation theory, when applied to obtain the value of the ground state energy:
1. Always gives exact ground state energy
2. gives energy value lower than the exact ground state energy
3. gives energy value which is sometimes higher than or sometimes lower than the exact ground state energy
4. gives energy value higher than or equal to the exact ground state energy
Variational method gives energy value higher than or equal to the exact ground state energy. Hence, answer is (D).
28. A state of a system with spherically symmetric potential has zero uncertainty in simultaneous measurement of operator $L_x$ and $L_y$. Which of the following statement is true?
1. The state must be $l=0$ state
2. Such a state can never exist as operators $L_x$ and $L_y$ do not commute
3. The state has $l=1$ with $m=0$
4. The state cannot be an eigenstate of $L^2$ operator
\begin{align*} L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\ &=l(l+1)\hbar^2\psi_{nlm} \end{align*} $$\left < L_z \right > =m\hbar, \left < L_z^2 \right > =m^2\hbar^2$$ $$\text{ and } \left < L^2 \right > =l(l+1)\hbar^2$$ The state $\psi_{nlm}$ is symmetric in $x$ and $y$, so \begin{align*} \left < L_x^2 \right >&=\left < L_y^2 \right > \\ &=\frac{\left(\left < L^2 \right >-\left < L_z^2 \right >\right)}{2}\\ &=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2} \end{align*} It can be verified that $\left < L_x \right >=\left < L_y \right >=0$. Using, $\Delta L_i=\sqrt{\left < L_i^2 \right >-\left < L_i \right >^2}$, we have $$\Delta L_x=\Delta L_y=\sqrt{\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}}$$ $$\Delta L_x\Delta L_y=\frac{\left(l(l+1)-m^2\right)\hbar^2}{2}$$ From this relation one can see that uncertainty relation will be zero when $l=0$ and henc, $m=0$.
Hence, answer is (A)
29. The wave function for identical fermions is antisymmetric under particle exchange. Which of the following is a consequence of this property?
1. Heisenberg's uncertainty principle
2. Bohr correspondence principle
3. Bose-Einstein condensation
4. Pauli exclusion principle
(D) Pauli exclusion principle
30. The spin part of two electron wave function is described as a triplet state. The space part of the wave function is given by ($\psi_1$ and $\psi_2$ are two different functions):
1. $\psi_1(r_1)\psi_2(r_2)$
2. $\psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)$
3. $\psi_1(r_1)\psi_2(r_2)+\psi_2(r_1)\psi_1(r_2)$
4. $\psi_1(r_1)\psi_1(r_2)+\psi_2(r_1)\psi_1(r_2)$
Triplet state means $S=s_1+s_2=1$ and $M_s=-1,0,+1$. For $M_s=-1$ we must have $m_{s1}=-\frac{1}{2}$ and $m_{s2}=-\frac{1}{2}$
For $M_s=0$ we must have either $m_{s1}=\frac{1}{2}$ and $m_{s2}=-\frac{1}{2}$ or $m_{s1}=-\frac{1}{2}$ and $m_{s2}=\frac{1}{2}$
For $M_s=+1$ we must have $m_{s1}=+\frac{1}{2}$ and $m_{s2}=+\frac{1}{2}$
Hence, we have three possible spin eigenfunctions $$\phi\left(+\frac{1}{2},+\frac{1}{2}\right)$$ $$\frac{1}{\sqrt{2}}\left[\phi\left(+\frac{1}{2},-\frac{1}{2}\right)+\phi\left(-\frac{1}{2},+\frac{1}{2}\right)\right]$$ $$\phi\left(-\frac{1}{2},-\frac{1}{2}\right)$$ All these functions are symmetric. However, total wavefunction must be antisymmetric. Hence, space part must be antisymmetric.
Hence, answer is (B) i.e. $\psi_1(r_1)\psi_2(r_2)-\psi_2(r_1)\psi_1(r_2)$
31. A transition in which one photon is radiated by the electron in a hydrogen atom when the electron's wave function changes from $\psi_1$ to $\psi_2$ is forbidden if $\psi_1$ and $\psi_2$
1. have opposite parity
2. are both spherically symmetrical
3. are orthogonal to each other
4. are zero at the center of the atomic nucleus
Selection rules:

1.	Principal quantum number	:	$\Delta n=$ anything
2.	Orbital angular momentum	:	$\Delta l = \pm1$
3.	Magnetic quantum number	:	$\Delta m_l = 0, \pm1$
4.	Spin	:	$\Delta s = 0$
5.	Total angular momentum	:	$\Delta j = 0, \pm 1,{\scriptstyle \text{ but} j = 0 \nrightarrow j= 0}$

A. FALSE: It's not related to the selection rules.
B. TRUE: If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. $l = 0 \rightarrow l = 0$ is forbidden as $\Delta l=0$ in this case.
C. FALSE: In any transition, eigenstates should always be mutually orthogonal.
D. FALSE: Eigenstates zero at the center $\Rightarrow l> 0$ could change, for example from 3 to 2. This is allowed.
Hence, answer is (B)
32. A particle moves in one dimension in the potential $V=\frac{1}{2}k(t)x^2$, where $k(t)$ is a time dependent parameter. Then $\frac{d}{dt}\left < V\right >$, the rate of change of the expectation value $\left < V\right >$ of the potential energy, is
1. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{k}{2m}\left < xp+px\right >$
2. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{1}{2m}\left < p^2\right >$
3. $\frac{k}{2m}\left < xp+px\right >$
4. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right >$

$$\frac{d\left <\hat A\right >}{dt}=\left < \frac{\partial \hat A}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\hat H,\hat A\right]\right > $$ \begin{align*} &\frac{d\left < V\right >}{dt}={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\frac{p^2}{2m}+V, V\right]\right >}\\ &={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,V\right]\right > }\\ &={\scriptstyle\left < \frac{\partial \left(\frac{1}{2}k(t)x^2\right)}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,\frac{1}{2}k(t)x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < \left[p^2,x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > }\\ &{\scriptstyle+\frac{ik}{4m\hbar}\left < xp\left[p,x\right]+\left[p,x\right]p+p\left[p,x\right]x+\left[p,x\right]xp\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < -2i\hbar xp-2i\hbar px \right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{k}{2m}\left < xp+ px \right > } \end{align*}

33. 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)

34. Which one of the following relations is true for Pauli matrices $\sigma_x$, $\sigma_y$ and $\sigma_z$
1. $\sigma_x\sigma_y=\sigma_y\sigma_x$
2. $\sigma_x\sigma_y=\sigma_z$
3. $\sigma_x\sigma_y=i\sigma_z$
4. $\sigma_x\sigma_y=-\sigma_y\sigma_x$

Pauli spin matrices are $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$, and $\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ \begin{align*} \sigma_x\sigma_y&=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\ &=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\ &=i\sigma_z \end{align*}

Hence, answer is (C)

35. A particle of mass $m$ is represented by the wavefunction $\psi(x) = Ae^{ikx}$, where $k$ is the wavevector and $A$ is a constant. The magnitude of the probability current density of the particle is
1. $|A|^2\frac{\hbar k}{m}$
2. $|A|^2\frac{\hbar k}{2m}$
3. $|A|^2\frac{\left(\hbar k\right)^2}{m}$
4. $|A|^2\frac{\left(\hbar k\right)^2}{2m}$

$$j=\frac{i\hbar}{2m}\left(\psi\frac{\partial\psi^*}{\partial x}-\psi^*\frac{\partial\psi}{\partial x}\right)$$ $$\psi(x) = Ae^{ikx}$$ $$\psi^*(x) = A^*e^{-ikx}$$ $$\frac{\partial\psi}{\partial x} = ikAe^{ikx}$$ $$\frac{\partial\psi^*}{\partial x} = -ikA^*e^{-ikx}$$ $$\psi^*\frac{\partial\psi}{\partial x} = A^*e^{-ikx} ikAe^{ikx}=ik|A|^2$$ \begin{align*} \psi\frac{\partial\psi^*}{\partial x} &= Ae^{ikx}\left(-ikA^*e^{-ikx}\right)\\ &=-ik|A|^2 \end{align*} \begin{align*} j&=\frac{i\hbar}{2m}\left(-ik|A|^2-ik|A|^2 \right)\\ &=|A|^2\frac{\hbar k}{m} \end{align*}

Hence, answer is (A)

36. The wavefunction of a particle in one-dimension is denoted by $\psi(x)$ in the coordinate representation and $\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$ in the momentum representation. If the action of an operator $\hat T$ on $\psi(x)$ is given by $\hat T\psi(x)=\psi(x+a)$, where $a$ is constant, then $\hat T\phi(p)$ is given by
1. $-\frac{i}{\hbar}ap\phi(p)$
2. $e^{-iap/\hbar}\phi(p)$
3. $e^{+iap/\hbar}\phi(p)$
4. $\left(1+\frac{i}{\hbar}ap\right)\phi(p)$

$$\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$$ \begin{align*} \hat T\phi(p)&=\int \hat T\psi(x) e^{-ipx/\hbar}\:dx\\ &=\int \psi(x+a) e^{-ipx/\hbar}\:dx\\ &=\int \psi(x+a) e^{-ip(x+a)/\hbar}e^{+ipa/\hbar}\:dx\\ &=e^{+ipa/\hbar}\int \psi(x+a) e^{-ip(x+a)/\hbar}\:dx\\ &=e^{+ipa/\hbar}\phi(p) \end{align*}

Hence, answer is (C)

37. A positron is suddenly absorbed by the nucleus of a tritium $(^3_1H)$ atom to turn the latter into a $He^+$ ion. If the electron in the tritium atom was initially in the ground state, the probability that the resulting $He^+$ ion will be in its ground state is
1. 1
2. $\frac{8}{9}$
3. $\frac{128}{243}$
4. $\frac{512}{729}$

Both Tritium and $He^+$ are hydrogenic atoms. The ground state wavefunction of hydrogenic atom is given by $$\psi_{0}=R_{10}Y_{00}$$ where $$R_{10}=2\left(\frac{Z}{a_0}\right)^{3/2}e^{-Zr/a_0}$$ $$Y_{00}=\frac{1}{\sqrt{4\pi}}$$ $$a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}$$ For both Tritium and $He^+$ have reduced mass $\mu$ same. Hence, these two differs in $Z$ values. $$\psi_{0_{He^+}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{2}{a_0}\right)^{3/2}e^{-2r/a_0}$$ $$\psi_{0_{Tritium}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-r/a_0}$$ According to sudden approximation probability of transition is given by $$P=\left|\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\right|^2$$ \begin{align*} &\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\\ &={\textstyle 2^2\frac{1}{4\pi}\left(\frac{1}{a_0}\right)^32^{3/2}\int_0^\infty e^{-3r/a_0}4\pi r^2 dr}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\int_0^\infty r^2 e^{-3r/a_0} dr}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\left(\frac{a_0}{3}\right)^3\int_0^\infty \left(\frac{3}{a_0}r\right)^2 e^{-3r/a_0} d\left(\frac{3}{a_0}r\right)}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\left(\frac{a_0}{3}\right)^32!}\\ &=\frac{2^{9/2}}{27} \end{align*} where we have used formula $\int_0^\infty x^n e^{-x} dx=n!$ $$P=\frac{2^{9}}{27^2}=\frac{512}{729}$$

Hence, answer is (D)

38. The ground state energy of a particle in the potential $V(x)=g|x|$, estimated using the trial wavefunction $$\psi(x)= \begin{cases} \sqrt{\frac{c}{a^5}}(a^2-x^2),\quad x < |a|\\ 0,\quad x\ge|a| \end{cases} $$ (where $g$ and $c$ are constants) is
1. $\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
2. $\frac{5}{6}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
3. $\frac{3}{4}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
4. $\frac{7}{8}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$

First normalize the wavefunction using normalization condition $\int\limits_{-a}^{a}\psi^*\psi\:dx=1$ $$\frac{c}{a^5}\int\limits_{-a}^{a}(a^2-x^2)^2\:dx=1$$ $$\Rightarrow c=\frac{15}{16}$$ $$\psi(x)=\sqrt{\frac{15}{16a^5}}(a^2-x^2)$$ $$H=-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}+g|x|=T+V$$ \begin{align*} < E >& = < \psi |H| \psi > \\ &= < \psi |T| \psi>+ < \psi |V| \psi> \end{align*} \begin{align*} < \psi |T| \psi> &={\scriptstyle-\frac{\hbar^2}{2m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2) \frac{d^2(a^2-x^2)}{d x^2}dx}\\ &=\frac{\hbar^2}{m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2)dx\\ &=\frac{5\hbar^2}{4ma^2} \end{align*} \begin{align*} < \psi |V| \psi>\! &=\frac{15}{16a^5}g\!\!\int\limits_{-a}^{a}\!(a^2-x^2)^2 |x|dx\\ &=\frac{15}{16a^5}2g\!\!\int\limits_{0}^{a}\!(a^2-x^2)^2 x dx\\ &=\frac{5}{16}ga \end{align*} $$ < E > =\frac{5\hbar^2}{4ma^2}+\frac{5}{16}ga$$ $$\frac{d < E > }{da}=0$$ $$\Rightarrow a=2\left(\frac{\hbar^2}{mg}\right)^{1/3}$$ $$ < E > =\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$$

Hence, answer is (A)

39. A hydrogen atom is subjected to the perturbation $V_{pert}(\vec r)=\epsilon\cos{\left(2r/a_0\right)}$, where $a_0$ is the Bohr radius. The change in the ground state energy to first order in $\epsilon$ is
1. $\epsilon/4$
2. $\epsilon/2$
3. $-\epsilon/2$
4. $-\epsilon/4$

First order correction is given by $$E_0^{(1)}=\left < \psi_{100}\left|H'\right|\psi_{100}\right >$$ $$\psi_{100}=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$$ \begin{align*} E_0^{(1)}&=\frac{\epsilon}{\pi a_0^3}\int\limits_0^\infty e^{-2r/a_0}\cos{\left(\frac{2r}{a_0}\right)}4\pi r^2 dr\\ &=\frac{4\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}}\left(\frac{e^{i\frac{2r}{a_0}}+e^{-i\frac{2r}{a_0}} }{2}\right) r^2 dr\\ &=\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1-i)}r^2dr\\ &+\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1+i)}r^2dr\\ &={\textstyle\frac{2\epsilon}{a_0^3}\left[\frac{2!}{\left(\frac{2}{a_0}(1-i)\right)^3}+\frac{2!}{\left(\frac{2}{a_0}(1+i)\right)^3}\right]}\\ &=\frac{\epsilon}{2}\left[\frac{1}{(1-i)^3}+\frac{1}{(1+i)^3}\right]\\ &={\textstyle\frac{\epsilon}{2}\left[\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1-i}{\sqrt{2}}\right)^3}+\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1+i}{\sqrt{2}}\right)^3}\right]}\\ &=\frac{\epsilon}{4\sqrt{2}}\left[\frac{1}{e^{\frac{-i3\pi}{4}}}+\frac{1}{e^{\frac{i3\pi}{4}}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[e^{\frac{i3\pi}{4}}+e^{\frac{-i3\pi}{4}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\cos{\left(\frac{3\pi}{4}\right)}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\left(-\frac{1}{\sqrt{2}}\right)\right]\\ &=-\frac{\epsilon}{4} \end{align*}

Hence, answer is (D)

40. The product of the uncertainties $\left(\Delta L_x\right)\left(\Delta L_y\right)$ for a particle in the state $a|1,1 > +b|1,-1 > $ (where $|l,m > $ denotes an eigenstate of $L^2$ and $L_z$) will be a minimum for
1. $a=\pm ib$
2. $a=0$ and $b=1$
3. $a=\frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}$
4. $a=\pm b$

\begin{align*} L_z |l,m >&=m\hbar|l,m >\\ L^2 |l,m >&=m(m+1)\hbar|l,m >\\ L_+ |l,m >&={\scriptstyle\sqrt{(l-m)(l+m+1)}\hbar|l,m+1 >}\\ L_- |l,m >&={\scriptstyle\sqrt{(l+m)(l-m+1)}\hbar|l,m-1 >} \end{align*} Let $|\psi >=a|1,1 > +b|1,-1 > $ \begin{align*} L_z |\psi >&=a\hbar |1,1 >-b\hbar |1,-1 > \\ L^2 |\psi >&=2a\hbar |1,1 > \\ L_+ |\psi >&=\sqrt{2}b\hbar |1,0 > \\ L_- |\psi >&=\sqrt{2}a\hbar |1,0 > \\ L_+^2 |\psi >&=2b\hbar^2 |1,1 > \\ L_-^2 |\psi >&=2a\hbar^2 |1,-1 > \end{align*} $$\left(\Delta L_x\right)=\sqrt{< L_x^2 > -< L_x >^2 }$$ \begin{align*} &< L_x >=\left <\psi|L_x|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_++L_-|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_+|\psi\right>+\frac{1}{2}\left <\psi|L_-|\psi\right>\\ &=\frac{1}{2}\sqrt{2}b\left <\psi|1,0\right>+\frac{1}{2}\sqrt{2}a\left <\psi|1,0\right>\\ &=0 \end{align*} \begin{align*} &< L_x^2 >=\left <\psi|L_x^2|\psi\right>\\ &=\frac{1}{4}\left <\psi|(L_++L_-)(L_++L_-)|\psi\right>\\ &={\scriptstyle\frac{1}{4}\left <\psi|L_+^2+L_-^2+L_+L_-+L_-L_+|\psi\right>}\\ &={\scriptstyle\frac{1}{4}\left[2a^*b\hbar^2+2ab^*\hbar^2+2\hbar^2|a|^2+2\hbar^2|b|^2\right]}\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+|a|^2+|b|^2\right]\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right] \end{align*} $$\left(\Delta L_x\right)=\sqrt{\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right]}$$ Similarly $$\left(\Delta L_y\right)=\sqrt{\frac{\hbar^2}{2}\left[1-(a^*b+ab^*)\right]}$$ $$\left(\Delta L_x\right)\!\left(\Delta L_y\right)=\!\frac{\hbar^2}{2}\sqrt{1-(a^*b+ab^*)^2}$$ Clearly, for $a=\pm b$, we get,$\left(\Delta L_x\right)\left(\Delta L_y\right)=0$

Hence, answer is (D)

41. A one-dimensional harmonic oscillator is in the state $\psi(x)=\frac{1}{\sqrt{14}}\left[3\psi_0(x)-2\psi_1(x)-\psi_2(x)\right]$, where $\psi_0(x)$, $\psi_1(x)$ and $\psi_2(x)$ are the ground, first and second excited states respectively. The probability of finding the oscillator in the ground state is
1. 0
2. $\frac{3}{\sqrt{14}}$
3. $\frac{9}{14}$
4. 1

Any state of the harmonic oscillator can be expressed in terms of eigen states as follow $$\sum\limits_nc_n\psi_n(x)$$ Probability of finding the system in state $\psi_n(x)$ is given by $|c_n|^2$. Hence The probability of finding the oscillator in the ground state is $\frac{9}{14}$

Hence, answer is (C)

42. The energy levels of one-dimensional harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ are given by $h\nu\left(n+\frac{1}{2}\right)$ with $n=0,1,2,\dots$. If the potential is changed to $V(x)=\infty$ for $x<0$ and $V(x)=\frac{1}{2}kx^2$ for $x>0$, the energy levels now, will be given by:
1. $h\nu\left(n+\frac{3}{2}\right)$
2. $2h\nu\left(n+\frac{1}{2}\right)$
3. $h\nu\left(n+\frac{1}{2}\right)$, $n$ odd only
4. $h\nu\left(n+\frac{1}{2}\right)$, $n$ even only

The wave function of simple harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ is given by $$\psi_n(x)=N_nH_n\left(\sqrt{\alpha}x\right)e^{-\alpha x^2/2}$$ where $\alpha=\left(\frac{mk}{\hbar^2}\right)^{1/2}$ is constant and $N_n=\left(\frac{\alpha}{\pi}\right)^{1/4}\left(\frac{1}{2^nn!}\right)^{1/2}$ is normalization constant, and $H_n\left(\sqrt{\alpha}x\right)=H_n(z)$ is a Hermite polynomial of degree $n$ given by $$H_0(z)=1$$ $$H_1(z)=2z$$ $$H_2(z)=4z^2-2$$ $$H_3(z)=8z^3-12z$$ The energy eigenvalues are given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{1}{2}h\nu,\frac{3}{2}h\nu,\frac{5}{2}h\nu\dots\\ &\text{for }n=0,1,2,3\dots \end{align*} Now, as $V(x)=\infty$ for $x<0$, the wave functions must vanish at $x=0$. Clearly, one can see that the wavefunction vanish at $x=0$ only for odd values of $n$. The terms with even $n$ do not have node at $x = 0$. Hence, eneegy is given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{3}{2}h\nu,\frac{7}{2}h\nu,\frac{11}{2}h\nu\dots\\ & \text{for }n=1,3,5\dots \end{align*}

Hence, answer is (C)

43. At $t=0$ a one-dimensional harmonic oscillator is in a state given by:$$\Psi(x,0)=\frac{1}{2}u_0(x)+i\frac{\sqrt{3}}{2}u_1(x)$$ where $u_0$ and $u_1$ are first two normalized eigen states. ($\omega$ is natural angular frequency of the oscillator). Then
1. The expectation value of the energy is $\frac{5}{4}\hbar\omega$
2. Energy measurement of this state will always gives its value as $\frac{5}{4}\hbar\omega$
3. The average value of energy is $\hbar\omega$
4. The expectation value of energy is dependent on time

Clearly, one can verify that the wavefunction $\Psi$ is normalized. \begin{align*} & < E > = < \Psi|H|\Psi > \\ &={\scriptstyle\left < \frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\arrowvert H\arrowvert\frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\right > }\\ &=\left < \frac{1}{2}u_0\left|H\right|\frac{1}{2}u_0\right > \\ &+\left < i\frac{\sqrt{3}}{2}u_1\left|H\right|i\frac{\sqrt{3}}{2}u_1\right > \\ &=\frac{1}{4}\left < u_0\left|H\right|u_0\right >\\ & +\!\!\left(\frac{-i\sqrt{3}}{2}\right)\!\!\left(\frac{i\sqrt{3}}{2}\right)\!\!\left < u_1\left|H\right|u_1\right > \\ &=\frac{1}{4}E_0+\frac{3}{4}E_1\\ &=\frac{1}{4}\times\frac{1}{2}\hbar\omega+\frac{3}{4}\times\frac{3}{2}\hbar\omega\\ < E >&=\frac{5}{4}\hbar\omega \end{align*}

Hence, answer is (A)

44. If the $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is $e^{2i\phi}$, then the minimum principal and minimum orbital angular momentum quantum numbers $n$ and $l$ respectively for this eigen function will be:
1. $n=3$, $l=2$
2. $n=2$, $l=1$
3. $n=1$, $l=2$
4. $n=2$, $l=2$

The $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is given by $e^{im\phi}$. Hence, $m=2$. Now, for a given value of $l$, $m$ has values from $-l$ to $+l$. This implies that $l=2$. Also, for a given value of $n$, $l$ takes values from $0$ to $n-1$. Hence, for $l=2$, minimum value of $n$ should be $3$.

Hence, answer is (A)

45. In the density matrix representation, the condition for a pure state is:
1. $\hat\rho^2=\hat\rho$
2. $\hat\rho^2=\hat I$
3. $T_r\hat\rho^2=T_r\hat\rho$
4. $T_r\hat\rho=1$

The Density matrix for the pure state $|\psi>$ is given by $$\rho=|\psi><\psi|$$ This density matrix has the following properties:

1. $\hat\rho^2=\hat\rho$ projector
2. $\hat\rho^\dagger=\hat\rho$ hermiticity
3. $T_r\hat\rho=1$ normalization
4. $\hat\rho\geq 0$ positivity

Hence, answer is (A)

46. A particle of mass $m$ is in a potential $V=\frac{1}{2}m\omega^2 x^2$, where $\omega$ is a constant. Let $\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i\hat p}{m\omega}\right)$. In the Heisenberg picture $\frac{d\hat a}{dt}$ is given by
1. $\omega\hat a$
2. $-i\omega\hat a$
3. $\omega\hat a^\dagger$
4. $i\omega\hat a^\dagger$

In Heisenberg picture the time evolution of operator is given by $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]+\frac{\partial\hat A_H(t)}{\partial t}$$ As $\hat A_H(t)$ does not explicitly depend on time $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]$$ For harmonic oscillator $$\hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2 x^2$$ \begin{align*} \frac{d\hat p}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat p\right]\\ &=-m\omega^2\hat x \end{align*} \begin{align*} \frac{d\hat x}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat x\right]\\ &=\frac{\hat p}{m} \end{align*} \begin{align*} \frac{d\hat a}{dt}&=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{d\hat x}{dt}+\frac{i}{m\omega}\frac{d\hat p}{dt}\right)\\ &=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{\hat p}{m}-\frac{i}{m\omega}m\omega^2\hat x\right)\\ &=-i\omega\hat a \end{align*}

Hence, answer is (B)

47. Two different sets of orthogonal basis vectors $\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$ and $\left\{\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix},\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\right\}$ are given for a two-dimensional real vector space. The matrix representation of a linear operator $\hat A$ in these bases are related by a unitary transformation. The unitary matrix may be chosen to be
1. $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
2. $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&0\\1&1\end{pmatrix}$

$$U_{ij}=< \psi_{new_i}|\psi_{old_j} >$$ \begin{align*} U_{11}&=< \psi_{new_1}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{12}&=< \psi_{new_1}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{21}&=< \psi_{new_2}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{22}&=< \psi_{new_2}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=-\frac{1}{\sqrt{2}} \end{align*} $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$

Hence, answer is (C)

48. The wavefunction of a particle in a one-dimensional potential at time $t=0$ is $\psi(x,t=0)=\frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]$ where $\psi_0(x)$ and $\psi_1(x)$ are the ground and the first excited states of the particle with corresponding energies $E_0$ and $E_1$. The wavefunction of the particle at a time $t$ is
1. $\frac{1}{\sqrt{5}}e^{-i(E_0+E_1)t/2\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
2. $\frac{1}{\sqrt{5}}e^{-iE_0t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
3. $\frac{1}{\sqrt{5}}e^{-iE_1t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
4. $\frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]$

$$\hat H\psi_0=E_0\psi_0\quad\hat H\psi_1=E_1\psi_1$$ Time evolution of quantum systems is given by Unitary Transformations, \begin{align*} \left|\psi(x,t)\right > &=\hat U \left|\psi(x,t=0)\right >\\ &=e^{-i\hat Ht/\hbar} \left|\psi(x,t=0)\right >\\ &={\scriptstyle e^{-i\hat Ht/\hbar} \frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]}\\ &={\scriptstyle \frac{1}{\sqrt{5}}\left[2e^{-i\hat Ht/\hbar}\psi_0(x)-e^{-i\hat Ht/\hbar}\psi_1(x)\right]}\\ \end{align*} Using property $f(\hat A)\psi_n=f(a_n)\psi_n$, we get $${\scriptstyle \left|\psi(x,t)\right > =\frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]}$$

Hence, answer is (D)

49. The commutator $[L_x, y]$, where $L_x$ is the $x$-component of the angular momentum operator and $y$ is the $y$-component of the position operator, is equal to
1. 0
2. $i\hbar x$
3. $i\hbar y$
4. $i\hbar z$

$$L_x=yp_z-zp_y$$ \begin{align*} [L_x, y]&=[yp_z-zp_y, y]\\ &=[yp_z, y]-[zp_y, y]\\ &=y[p_z, y]+[y, y]p_z-z[p_y, y]-[z, y]p_y\\ &=i\hbar z \end{align*}

Hence, answer is (D)

50. In hydrogenic states, the probability of finding an electron at $r = 0$ is
1. zero in state $\phi_{1s}(r)$
2. non-zero in state $\phi_{1s}(r)$
3. zero in state $\phi_{2s}(r)$
4. non-zero in state $\phi_{2p}(r)$

non-zero in state $\phi_{1s}(r)$

Hence, answer is (B)

51. An unperturbed two-level system has energy eigenvalues $E_1$ and $E_2$, and eigenfunctions $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ When perturbed, its Hamiltonian is represented by $\begin{pmatrix}E_1&A\\A^*&E_2\end{pmatrix}$
1. The first-order correction to $E_1$ is
1. $4A$
2. $2A$
3. $A$
4. 0
2. The second-order correction to $E_1$ is
1. 0
2. $A$
3. $\frac{A^2}{E_2-E_1}$
4. $\frac{A^2}{E_1-E_2}$
3. The first-order correction to the eigenfunetion $\begin{pmatrix}1\\0\end{pmatrix}$ is
1. $\begin{pmatrix}0\\\frac{A^*}{E_1-E_2}\end{pmatrix}$
2. $\begin{pmatrix}0\\1\end{pmatrix}$
3. $\begin{pmatrix}\frac{A^*}{E_1-E_2}\\0\end{pmatrix}$
4. $\begin{pmatrix}1\\1\end{pmatrix}$

1. The perturbed Hamiltonian is given by $H=H_0+H'$ $$H=\begin{pmatrix}E_1&A\\A^*&E_2\end{pmatrix}$$ As $E_1$ and $E_2$ are eigenvalues $$H_0=\begin{pmatrix}E_1&0\\0&E_2\end{pmatrix}$$ $$H'=\begin{pmatrix}0&A\\A^*&0\end{pmatrix}$$ First order correction to energy $E_1$ is \begin{align*} E_1^{(1)}&=\left < \psi_1|H'|\psi_1 \right > \\ &=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0&A\\A^*&0\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\ &=0 \end{align*}

   Hence, answer is (D)

2. The second order correction to the energy is given by \begin{align*} E_n^{(2)}&=\sum\limits_{m\neq n}\frac{\left|\left < m|H'|n \right > \right|^2}{E_n^{(0)}-E_m^{(0)}}\\ E_1^{(2)}&=\sum\limits_{m\neq 1}\frac{\left|\left < m|H'|1 \right > \right|^2}{E_1^{(0)}-E_m^{(0)}}\\ &=\frac{\left|\left < 2|H'|1 \right > \right|^2}{E_1^{(0)}-E_2^{(0)}}\\ \left < 2|H'|1 \right >&=\begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}0&A\\A^*&0\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}\\ &=A^*\\ E_1^{(2)}&=\frac{A^2}{E_1-E_2} \end{align*}

   Hence, answer is (D)

3. Wave corrected upto first order is given by $$\psi_n^{(1)}=\psi_n^{(0)}+\sum\limits_{m\neq n}\frac{\left|\left < m|H'|n \right > \right|^2}{E_n^{(0)}-E_m^{(0)}}\psi_m^{(0)}$$ $$\psi_1^{(1)}=\psi_1^{(0)}+\frac{\left|\left < 2|H'|1 \right > \right|^2}{E_1^{(0)}-E_2^{(0)}}\psi_2^{(0)}$$ $$\psi_1^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}+\frac{A^*}{E_1-E_2}\begin{pmatrix}0\\1\end{pmatrix}$$ Hence first order correction to the eigenfunction $\begin{pmatrix}1\\0\end{pmatrix}$ is $\begin{pmatrix}0\\\frac{A^*}{E_1-E_2}\end{pmatrix}$

   Hence, answer is (A)

52. The state of a particle of mass $m$ in a one-dimensional rigid box in the interval 0 to $L$ is given by the normalised wavefunction $\psi(x)\!=\!\!\sqrt{\frac{2}{L}}\!\!\left(\frac{3}{5}\sin{\left(\frac{2\pi x}{L}\right)}+\frac{4}{5}\sin{\left(\frac{4\pi x}{L}\right)}\!\right)$. If its energy is measured, the possible outcomes and the average value of energy are, respectively
1. $\frac{h^2}{2mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{73}{50}\frac{h^2}{mL^2}$
2. $\frac{h^2}{8mL^2}$, $\frac{h^2}{2mL^2}$ and $\frac{19}{40}\frac{h^2}{mL^2}$
3. $\frac{h^2}{2mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{19}{10}\frac{h^2}{mL^2}$
4. $\frac{h^2}{8mL^2}$, $\frac{2h^2}{mL^2}$ and $\frac{73}{200}\frac{h^2}{mL^2}$

Eigen function and energy of a particle in a 1D box are given by $$\psi_n(x)=\sqrt{\frac{2}{L}}\sin{\left(\frac{n\pi x}{L}\right)}$$ $$E_n=\frac{n^2h^2}{8mL^2}$$ Hence, $$\psi(x)=c_2\psi_2(x)+c_4\psi_4(x)$$ where, $c_2=\frac{3}{5}$ and $c_4=\frac{4}{5}$ are the expansion coefficients.

Particle may be in its one of the eigenstate, hence possible values of energies are $$E_2=\frac{2^2h^2}{8mL^2}=\frac{h^2}{2mL^2}$$ $$E_4=\frac{4^2h^2}{8mL^2}=\frac{2h^2}{mL^2}$$ Average value of energy is \begin{align*} < E > &= < \psi |H|\psi > \\ &=|c_2|^2E_2+|c_4|^2E_4\\ &=\frac{9}{25}\frac{h^2}{2mL^2}+\frac{16}{25}\frac{2h^2}{mL^2}\\ &=\frac{73}{50}\frac{h^2}{mL^2} \end{align*}

Hence, answer is (A)

53. If $\hat L_x$, $\hat L_y$ and $\hat L_z$ are the components of the angular momentum operator in three dimensions, the commutator $\left[\hat L_x, \hat L_x\hat L_y\hat L_z\right]$ may be simplified to
1. $i\hbar\hat L_x\left(\hat L_z^2-\hat L_y^2\right)$
2. $i\hbar\hat L_z\hat L_y\hat L_x$
3. $i\hbar\hat L_x\left(2\hat L_z^2-\hat L_y^2\right)$
4. 0

\begin{align*} \left[\hat L_x, \hat L_x\hat L_y\hat L_z\right]&= {\scriptstyle \hat L_x \left[\hat L_x, \hat L_y\hat L_z\right]+\left[\hat L_x, \hat L_x\right]\hat L_y\hat L_z}\\ &=\hat L_x \left[\hat L_x, \hat L_y\hat L_z\right]\\ &={\scriptstyle \hat L_x \left\{\hat L_y\left[\hat L_x, \hat L_z\right]+\left[\hat L_x, \hat L_y\right]\hat L_z\right\}}\\ &={\scriptstyle \hat L_x \left\{\hat L_y\left(-i\hbar \hat L_y\right)+\left(i\hbar\hat L_z\right)\hat L_z\right\}}\\ &=i\hbar\hat L_x\left(\hat L_z^2-\hat L_y^2\right) \end{align*}

Hence, answer is (A)

54. The eigenstates corresponding to eigen-values $E_1$ and $E_2$ of a time-independent Hamiltonian are $|1 > $ and $|2 > $ respectively. If at $t=0$, the system is in a state $ |\psi(t=0) > =\sin\theta |1 > +\cos\theta |2 > $ the value of $< \psi(t)|\psi(t) > $ at time $t$ will be
1. 1
2. $(E_1\!\sin^2\theta\!+\!E_2\!\cos^2\theta)/\!\sqrt{E_1^2\!+\!E_2^2}$
3. $e^{iE_1t/\hbar}\sin\theta+e^{iE_2t/\hbar}\cos\theta$
4. $e^{-iE_1t/\hbar}\sin^2\theta+e^{-iE_2t/\hbar}\cos^2\theta$

Time evolution of quantum systems is given by Unitary Transformations, \begin{align*} \left|\psi(t)\right > &=\hat U\left|\psi(t=0)\right > \\ &=e^{-iHt/\hbar}\left\{\sin\theta \left|1\right > +\cos\theta \left|2\right > \right\}\\ &=\sin\theta e^{-iHt/\hbar}\left|1\right > +\cos\theta e^{-iHt/\hbar}\left|2\right > \end{align*} Using property $f(\hat A)\psi_n=f(a_n)\psi_n$ we get \begin{align*} \left|\psi(t)\right > &=\sin\theta e^{-iE_1t/\hbar}\left|1\right > +\cos\theta e^{-iE_2t/\hbar}\left|2\right > \\ \end{align*} Hence, $$\left < \psi(t)\right|\left.\psi(t)\right > =\sin^2\theta+\cos^2\theta=1$$

Hence, answer is (A)

55. Suppose that the Coulomb potential of the hydrogen atom is changed by adding an inverse-square term such that the total potential is $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$, where $g$ is a constant. The energy eigenvalues $E_{nlm}$ in the modified potential
1. depend on $n$ and $l$, but not on $m$
2. depend on $n$ but not on $l$ and $m$
3. depend on $n$ and $m$, but not on $l$
4. depend explicitly on all three quantum numbers $n$, $l$ and $m$

Radial equation for hydrogen atom is $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R=ER}$$ For $V(\vec r)=-\frac{Ze^2}{r}+\frac{g}{r^2}$ equation becomes $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{\hbar^2l(l+1)}{2\mu r^2}R+V(r)R+\frac{g}{r^2}R=ER}$$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2\left(l(l+1)+\frac{2\mu}{\hbar^2}g\right)}{2\mu r^2}\right]R+V(r)R=ER}$$ Let $l_{eff}(l_{eff}+1)=l(l+1)+\frac{2\mu}{\hbar^2}g$ $${\scriptstyle -\frac{\hbar^2}{2\mu}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left[\frac{\hbar^2l_{eff}(l_{eff}+1)}{2\mu r^2}\right]R+V(r)R=ER}$$ $$E\propto\frac{1}{(N+l_{eff}+1)^2}$$

Hence, answer is (A)

56. Given $[x_i,P_j]=i\hbar\delta_{ij}$, $i,j=1,2,3$. $[x_1,P_2^2]$ is:
1. 0
2. $i\hbar P_2$
3. $2x_1$
4. $2P_2$

\begin{align*} [x_1,P_2^2]&=P_2[x_1,P_2]+[x_1,P_2]P_2\\ &=0 \end{align*}

Hence, answer is (A)

57. Which of the following is an eigen state of square of linear momentum operator $P_x^2$?
1. $Ax^2$
2. $A\left(\sin{kx}+\cos{kx}\right)$
3. $Ae^{-\alpha x^2}$
4. $A\sin^2{kx}$

$P_x=-i\hbar\frac{\partial}{\partial x}$. Hnece, $P_x^2=-\hbar^2\frac{\partial^2}{\partial x^2}$.
Clearly, \begin{align*} {\scriptstyle P_x^2\left(A\left(\sin{kx}+\cos{kx}\right)\right)}&={\scriptstyle-\hbar^2\frac{\partial^2}{\partial x^2}\left(A\left(\sin{kx}+\cos{kx}\right)\right)}\\ &=k^2A\left(\sin{kx}+\cos{kx}\right) \end{align*}

Hence, answer is (B)

58. The electron in a hydrogen atom is in a superposition state described by the wavefunction $\psi(\vec r)=A\left[4\psi_{100}(\vec r)-2\psi_{211}(\vec r)+\sqrt{6}\psi_{210}(\vec r)-\sqrt{10}\psi_{21-1}(\vec r)\right]$, $\psi_{nlm}(\vec r)$ normalized wave function. The value of normalization constant, $A$, is:
1. $\frac{1}{3}$
2. $\frac{1}{6}$
3. $6$
4. $36$

Normalization condition is $\int\psi^*(\vec r)\psi(\vec r)dr=1$. As $\psi_{nlm}(\vec r)$ are normalized wave functions, we have, $\int\psi^*_{nlm}(\vec r)\psi_{n'l'm'}(\vec r)dr=\delta_{n'n}\delta_{l'l}\delta_{m'm}$ $$\int\psi^*(\vec r)\psi(\vec r)dr=|A|^2\left(16+4+6+10\right)=1$$ $$A=\frac{1}{6}$$

Hence, answer is (B)

59. Two coherent light sources of intensities I and 9I are used in an interference experiment. The resultant intensity at points where the waves from the two sources with phase difference $\pi$ is :
1. 16I
2. 9I
3. 4I
4. zero

Resultant intensity is given by $$I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$$ $$I_R=I+9I+2\sqrt{9I}\cos\pi$$ $$I_R=10I-6I=4I$$

Hence, answer is (C)

60. Non-relativistic hydrogen atom spectrum is proportional to $-1/n^2$. The degeneracy of $n^{th}$ level is:
1. $n$
2. $2n+1$
3. $n^2$
4. $1/n^2$

The energy levels in the hydrogen atom depend only on the principal quantum number $n$. For given values of $n$ and $l$, the $( 2 l + 1 )$, states with $m_{l}=-l \rightarrow l$ are degenerate. The degree of degeneracy of the energy level $E_n$ is therefore : $\sum _{l=0}^{n-1}(2l+1)=n^{2}$, which is doubled if the spin degeneracy is included.

Hence, answer is (C)

61. Uncertainty relation holds between :
1. Time and space
2. Life time and energy
3. Position and energy
4. Momentum and energy

Life time and energy

Hence, answer is (B)

62. Addition of angular momentum $j_1=1$ and $j_2=\frac{1}{2}$ will result in 6 states, of which the number of linearly independent states with magnetic number $m=-\frac{1}{2}$ is :
1. zero
2. 6
3. 3
4. 2

For $j_1=1$ and $j_2=\frac{1}{2}$, resultant $j=j_1+j_2$ to $|j_1-j_2|$. Hence, $j=\frac{3}{2},\frac{1}{2}$ $$j=\frac{3}{2}\Rightarrow m=\frac{3}{2},\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$$ $$j=\frac{1}{2}\Rightarrow m=\frac{1}{2},-\frac{1}{2}$$ Hence, there are two linearly independent states

Hence, answer is (D)

63. In a scattering event by a spherical symmetric potential, only P-wave scattering occurs. The angular distribution of differential cross-section is proportional to :
1. constant
2. $\cos\theta$
3. $\cos^2{\theta}$
4. $a+\sin\theta$

$\cos^2{\theta}$

Hence, answer is (C)