Problem set 16

1. Consider the digital circuit shown below in which the input $C$ is always high (1).
   
   The truth table for the circuit can be written as

   A	B	Z
   0	0
   0	1
   1	0
   1	1

   The entries in the Z column (vertically) are
1. 1010
2. 0100
3. 1111
4. 1011
AND gate has output high when both inputs are high, OR gate has output high when any one or both inputs high and EXOR gate has high output when any one input is high but not both inputs hight. Hence, answer is (D)
2. Let $p_n(x)$ (where $n = 0,1,2,\dots$) be a polynomial of degree $n$ with real coefficients, defined in the interval $2\leq n\leq 4$. If $\int_2^4p_n(x)p_m(x)dx=\delta_{nm}$, then
1. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(-3-x)$
2. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{3}(3+x)$
3. $p_0(x)=\frac{1}{2}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$
4. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$
For $p_0(x)=\frac{1}{\sqrt{2}}$, we have $$\begin{align*} \int_2^4p_0(x)p_0(x)dx&=\frac{1}{2}\int_2^4dx\\ &=\frac{1}{2}[x]_2^4=1 \end{align*}$$ Hence, $p_0(x)=\frac{1}{\sqrt{2}}$ is correct option. For, $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$, we have $$\begin{align*} &\int_2^4p_1(x)p_1(x)dx\\ &=\frac{3}{2}\int_2^4(3-x)^2dx\\ &=\frac{3}{2}\int_2^4(9-6x+x^2)dx \end{align*}$$ $$\begin{align*} &\int_2^4p_1(x)p_1(x)dx\\ &=\frac{3}{2}\left[9x-3x^2+\frac{x^3}{3}\right]_2^4\\ &=1 \end{align*}$$ $$\begin{align*} &\int_2^4p_0(x)p_1(x)dx\\ &=\sqrt{\frac{3}{4}}\int_2^4(3-x)dx\\ &=\sqrt{\frac{3}{4}}\left[3x-\frac{x^2}{2}\right]_2^4\\ &=0 \end{align*}$$ Hence, orthonormality condition is satified for $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$. Hence, answer is (D)
3. 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.
4. 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 $$\begin{align*} e^{i\omega t(\vec\sigma\cdot\hat n)}&=\\ &I+i\omega t(\vec\sigma\cdot\hat n)\\ &-\frac{(\omega t)^2}{2!}I\\ &-i\frac{(\omega t)^3}{3!}(\vec\sigma\cdot\hat n)\\ &+\frac{(\omega t)^4}{4!}I-\dots \end{align*}$$ $$\begin{align*} e^{i\omega t(\vec\sigma\cdot\hat n)}&=I\left[1-\frac{(\omega t)^2}{2!}+\frac{(\omega t)^4}{4!}-..\right]\\ &+(\vec\sigma\cdot\hat n)\left[\omega t-\frac{(\omega t)^3}{3!}+\dots\right] \end{align*}$$ $$e^{i\omega t(\vec\sigma\cdot\hat n)}=I\cos{\omega t}+\frac{\vec\sigma\cdot\vec B}{B}\sin{\omega t}$$
5. The Hamiltonian of a system with $n$ degrees of freedom is given by $H(q_1, \dots,q_n;p_1,\dots,p_n;t)$, with an explicit dependence on the time $t$ . Which of the following is correct?
1. Different phase trajectories cannot intersect each other.
2. $H$ always represents the total energy of the system and is a constant of the motion.
3. The equations $\dot q_i =\partial H/\partial p_i$, $\dot p_i =-\partial H/\partial q_i$ are not valid since $H$ has explicit time dependence.
4. Any initial volume element in phase space remains unchanged in magnitude under time evolution.
If $H$ depends explicitly on time, then phase trajectories can intersect. A constant Hamiltonian is the total energy if the potential is velocity independent. The equations $\dot q_i =\partial H/\partial p_i$, $\dot p_i =-\partial H/\partial q_i$ are are always valid. The fourth option, Any initial volume element in phase space remains unchanged in magnitude under time evolution, is a Liouville theorem. Hence, answer is (D).