Problem set 79

1. 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)

2. One of the eigen values of the matrix $\begin{pmatrix}2&3&0\\3&2&0\\0&0&1\end{pmatrix}$ is 5
1. The other two eigenvalues are
1. 0 and 0
2. 1 and 1
3. 1 and -1
4. -1 and -1
2. The normalized eigenvector corresponding to the eigenvalue 5 is
1. $\frac{1}{\sqrt{2}} \begin{pmatrix}0\\-1\\1\end{pmatrix}$
2. $\frac{1}{\sqrt{2}} \begin{pmatrix}-1\\1\\0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\0\\-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\\0\end{pmatrix}$

1. Determinant of matrix = product of eigenvalues.

   Let $\lambda_1$ and $\lambda_2$ be other eigenvalues $$Det=5\times\lambda_1\times\lambda_2=-5$$ $$\lambda_1\times\lambda_2=-1\quad---(1)$$ Trace of matrix=sum of eigenvalues $$5+\lambda_1+\lambda_2=5$$ $$\lambda_1+\lambda_2=0\quad---(2)$$ Solving equation (1) and (2) we get $$\lambda_1=1\quad\lambda_2=-1$$

   Hence, answer is (C)

2. Eigenvalue equation for $\lambda=5$ is $A-\lambda=0$ $$\begin{pmatrix}-3&3&0\\3&-3&0\\0&0&-4\end{pmatrix}\!\begin{pmatrix}x\\y\\z\end{pmatrix}\!=\!\begin{pmatrix}0\\0\\0\end{pmatrix}$$ $$-3x+3y=0$$ $$3x-3y=0$$ $$-4z=0$$ $$\Rightarrow z=0, x=y=c$$ Applying normalization condition $$|c|^2+|c|^2+0=1$$ $$\Rightarrow c=\frac{1}{\sqrt{2}}$$ $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix}$$

   Hence, answer is (d)

3. The powder diffraction pattern of a body centred cubic crystal is recorded by using $Cu K_\alpha$ X-rays of wavelength $1.54 \:A^o$.
1. If the (002) planes diffract at $60^o$, the lattice parameter is
1. $2.67 A^o$
2. $3.08 A^o$
3. $3.56 A^o$
4. $5.34 A^o$
2. Assuming the atomic mass of the constituent atoms to be 50.94 amu, the density of the crystal in units of kg m$^{-3}$ is
1. $3.75 \times 10^3$
2. $4.45 \times 10^3$
3. $5.79 \times 10^3$
4. $8.89 \times 10^3$

Bragg's law is $$2d\sin\theta=n\lambda$$ For first order pattern $n=1$ \begin{align*} d&=\frac{\lambda}{2\sin\theta}\\ &=\frac{1.54}{2\times\sin30}\\ &=1.54 \end{align*} For cubic system $$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ $$a=d\sqrt{h^2+k^2+l^2}=3.08$$

Hence, answer is (B)

For body centered cubic crystal there two atoms per unit cell. $$\rho=\frac{mass~of~two~atoms}{volume~of~cell}$$ 1 atomic mass unit = $1.66053886 \times 10^{-27}$ kilograms $$\rho={\textstyle \frac{2\times50.94\times 1.66053886 \times 10^{-27}}{(3.08\times 10^{-10})^3}}$$ $$\rho=5.79 \times 10^{3}$$

Hence, answer is (C)

Problem set 78

1. The experimentally measured spin $g$ factors of a proton and a neutron indicate that
1. both proton and neutron are elementary point particles
2. both proton and neutron are not elementary point particles
3. while proton is an elementary point particle, neutron is not
4. while neutron is an elementary point pat1icle, proton is not

both proton and neutron are not elementary point particles

Hence, answer is (B)

2. The tank circuit of a Hartley oscillator is shown in the figure. If $M$ is the mutual inductance between the inductors, the oscillation frequency is
   
1. $\frac{1}{2\pi\sqrt{(L_1+L_2+2M)C}}$
2. $\frac{1}{2\pi\sqrt{(L_1+L_2-2M)C}}$
3. $\frac{1}{2\pi\sqrt{(L_1+L_2+M)C}}$
4. $\frac{1}{2\pi\sqrt{(L_1+L_2-M)C}}$

Frequency of Hartley oscillator is given by $$f=\frac{1}{2\pi\sqrt{LC}}$$ where $L=L_1+L_2$ if the coils are assumed to be winded on different cores and $L_1+L_2+2M$ if they are winded on a single core.

Hence, answer is (A)

3. In the given digital logic circuit, $A$ and $B$ form the input. The output $Y$ is
   
1. $Y=\bar A$
2. $Y=A\bar B$
3. $Y=A\oplus B$
4. $Y=\bar B$

\begin{align*} Y&=\overline{\left(\bar A+B\right) \left( A+B\right)}\\ &=\overline{\left(\bar A+B\right)}+ \overline{\left( A+B\right)}\\ &=\overline{\left(\bar A\right)}\bar B+\bar A\bar B\\ &=A\bar B+\bar A\bar B\\ &=\left(A+\bar A\right)\bar B\\ &=\bar B\quad \text{since, }(A+\bar A=1) \end{align*}

Hence, answer is (D)

4. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output for a digital input of 010100, is
1. 1.6 V
2. 2.9 V
3. 3.15 V
4. 5.0 V

For a n-bit DAC $$V_{out}={\textstyle V_R\left(\frac{b_{n-1}}{2^1}+\frac{b_{n-2}}{2^2}+\cdots+\frac{b_{0}}{2^n}\right)}$$ For a 6-bit DAC $$V_{out}=\!\!{\textstyle V_R\!\!\left(\frac{b_{5}}{2^1}\!+\!\frac{b_{4}}{2^2}\!+\!\frac{b_{3}}{2^3}\!+\!\frac{b_{2}}{2^4}\!+\!\frac{b_{1}}{2^5}\!+\!\frac{b_{0}}{2^6}\right)}$$ For 1.0 V output of a digital input of 010100 $$1=\!{\textstyle V_R\!\!\left(\frac{0}{2^1}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}+\frac{0}{2^5}+\frac{0}{2^6}\right)}$$ $$V_R=\frac{16}{5}$$ $$V_{max}\!=\!{\textstyle \frac{16}{5}\!\left(\frac{1}{2^1}\!+\!\frac{1}{2^2}\!+\!\frac{1}{2^3}\!+\!\frac{1}{2^4}+\frac{1}{2^5}\!+\!\frac{1}{2^6}\right)}$$ $$V_{max}=3.15 V$$

Hence, answer is (C)

5. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The values of the resistors are
   
1. $R_1 = 10\: k\Omega$; $R_2 = 1.3 \Omega$
2. $R_1 = 30\: k\Omega$; $R_2 = 1.3 \Omega$
3. $R_1 = 10\: k\Omega$; $R_2 = 1.7 k\Omega$
4. $R_1 = 30\: k\Omega$; $R_2 = 1.7 k\Omega$

Using $$Gain=1+\frac{15k}{R_1}$$ $$R_1=30\:k\Omega$$ Using $$f_c=\frac{1}{2\pi RC}$$ $$R=\frac{1}{2\pi f_cC}=1.7\:k\Omega$$

Hence, answer is (D)