Problem set 17

1. 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 $$\begin{align*} E_0^{(1)}&=<\psi_0|H'|\psi_0>\\ &=\int_0^\pi\psi_0^*\:ax\:\psi_0\:dx\\ &={\scriptstyle \int_0^\pi\sqrt{\frac{2}{\pi}}\sin{\frac{n\pi x}{\pi}}\:ax\:\sqrt{\frac{2}{\pi}}\sin{nx}\:dx}\\ &=\frac{2a}{\pi}\int_0^\pi x\sin^2{nx}\:dx\\ &=\frac{2a}{\pi}\frac{1}{2}\int_0^\pi x(1-\cos{2nx})\:dx\\ &={\scriptstyle \frac{a}{\pi}\left\{\left[\frac{x^2}{2}\right]_0^\pi-\left[x\int \cos{2nx}\:dx\right.\right.}\\ &{\scriptstyle \left.\left.-\int\frac{dx}{dx}\left(\int\cos{2nx}\:dx\right)dx\right]_0^\pi\right\}}\\ &={\scriptstyle \frac{a}{\pi}\left\{\frac{\pi^2}{2}-\left[x\frac{sin{2nx}}{2n}+\frac{\cos{2nx}}{4n^2}\right]_0^\pi\right\}}\\ &=\frac{a}{2\pi} \end{align*}$$
2. Which of the following is an analytic function of the complex variable $z = x + iy$ in the domain $| z |< 2$?
1. $(3+x-iy)^7$
2. $(1+x+iy)^4(7-x-iy)^3$
3. $(1-2x-iy)^4(3-x-iy)^3$
4. $(x+iy-1)^{1/2}$
A function is analytic if it satisfies Cauchy–Riemann condition i.e. a function $f(z)$ is analytic if $\frac{\partial f(z)}{\partial\bar z}=0$ Above functions can be written in terms of $z$ using $z=x+iy$, $\bar z=x-iy$, $x=\frac{z+\bar z}{2}$ and $y=\frac{z-\bar z}{2i}$ $$\begin{align*} &(3+x-iy)^7=(3+\bar z)^7\\ &(1+x+iy)^4(7-x-iy)^4\\ &=(1+z)^4(7-z)^3\\ &(1-2x-iy)^4(3-x-iy)^3\\ &=\left(1-2\frac{z+\bar z}{2}-i\frac{z-\bar z}{2i}\right)^4(3-z)^3\\ &=\left(1-2z-\frac{\bar z}{2}\right)^4(3-z)^3\\ &(x+iy-1)^{1/2}=(z-1)^{1/2} \end{align*}$$ For the functions which contain $\bar z$, $\frac{\partial f(z)}{\partial\bar z}\neq0$. Hence, the functions in the options (A) and (C) are not analytical.
For the functions which are functions of only $z$ have $\frac{\partial f(z)}{\partial\bar z}=0$. Hence, we have option either (B) or (C).
The option (B) i.e. $f(z)=(1+z)^4(7-z)^3$ is a polynomial in $z$ and polynomial in $z$ is analytic everywhere.
For the option (C) i.e. $f(z)=(z-1)^{1/2}$, $f'(z)=\frac{1}{2\sqrt{z-1}}$ and this function is analytic only for $|z|<1$. Our domain is $|z|<2$. Hence, it not analytical over the domain $|z|<2$.
Hence, the correct option is (B).
3. 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, $E_n\sim n^{3/2}$
4. A resistance is measured by passing a current through it and measuring the resulting voltage drop. If the voltmeter and the ammeter have uncertainties of 3% and 4%, respectively, then
1. The uncertainty in the value of the resistance is
1. 7.0%
2. 3.5%
3. 5.0%
4. 12.0%
2. The uncertainty in the computed value of the power dissipated in the resistance is
1. 7%
2. 5%
3. 11%
4. 9%

If $\Delta x$ and $\Delta y$ are uncertainties in measurements of $x$ and $y$, then uncertainty in $z$ i.e. $\Delta z$, for different cases is given by $$\begin{eqnarray} \begin{array}{cc} {\scriptstyle z=x\pm y }&{\scriptstyle\Delta z=\sqrt{(\Delta x)^2+(\Delta y)^2}}\\ {\scriptstyle z=xy} &{\scriptstyle\Delta z=|xy|\sqrt{\left(\frac{\Delta x}{x}\right)^2+\left(\frac{\Delta y}{y}\right)^2}}\\ {\scriptstyle z=\frac{x}{y}} &{\scriptstyle\Delta z=\left|\frac{x}{y}\right|\sqrt{\left(\frac{\Delta x}{x}\right)^2+\left(\frac{\Delta y}{y}\right)^2}}\\ {\scriptstyle z=x^n} &{\scriptstyle\Delta z=\left|n\right|x^{n-1}\Delta x}\\ {\scriptstyle z=f(x,y)} &{\scriptstyle \Delta z=\sqrt{\left(\frac{\partial f}{\partial x}\right)^2(\Delta x)^2+\left(\frac{\partial f}{\partial y}\right)^2(\Delta y)^2}}\\ \end{array} \end{eqnarray}$$ In our case $\Delta V=3\%$ and $\Delta I=4\%$, as $R=\frac{V}{I}$ $$\begin{align*} \Delta R&=\left|\frac{V}{I}\right|\sqrt{\left(\frac{\Delta V}{V}\right)^2+\left(\frac{\Delta I}{I}\right)^2}\\ &=\left|\frac{100}{100}\right|\sqrt{\left(\frac{3}{100}\right)^2+\left(\frac{4}{100}\right)^2}\\ &=\frac{5}{100}\\ \Delta R&=5\% \end{align*}$$ Similarly, using $P=VI$, $\Delta P=5\%$. Hence, anwers are (C) and (B).