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)