Problem set 25

1. 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)
2. 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)
3. 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 $$ < x^2 > =\frac{\hbar}{m\omega}$$ $$\Delta x=\sqrt{\frac{\hbar}{m\omega}-\frac{\hbar}{2m\omega}}=\sqrt{\frac{\hbar}{2m\omega}}$$
4. 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*}
5. A given quantity of gas is taken from the state $A \rightarrow C$ reversibly, by two paths, $A\rightarrow C$ directly and $A\rightarrow B\rightarrow C$ as shown in the figure below.
   
   
   
   During the process $A\rightarrow C$ the work done by the gas is $100 J$ and the heat absorbed is $150 J$. If during the process $A\rightarrow B\rightarrow C$ the workdone by the gas is $30 J$, the heat absorbed is
1. 20 J
2. 80 J
3. 220 J
4. 280 J
$$\delta Q=dU+\delta W$$ For path $A\rightarrow C$, $\delta W=100 J$ and $\delta Q=150 J$ $$150=du+100$$ $$dU=50 J$$ As, $dU$ is a state function, it does not depend on the path followed. Hence, for the path $A\rightarrow B\rightarrow C$, we have $$\delta Q=50+30=80 J$$ Hence, answer is (B)