Chapter4problem

(Problem submitted by team 9, based on problem 7.11 of Griffiths)

(a) Using the wave function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi= \begin{cases} A*cos(\frac{\pi*x}{a}) & \frac{-a}{2}<x<\frac{a}{2} \\ 0 & otherwise \end{cases}}

obtain a bound on the ground state energy of the one-dimensional harmonic oscillator. Compare Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <H>_{min} } with the exact energy. Note: This trial wave function has a discontinuous derivative at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pm a}{2}} .

(b) Use Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi = B*sin(\frac{\pi*x}{a}) } on the interval (-a,a) to obtain a bound on the first excited state. Compare to the exact answer.

Solution

(a)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1= \int |\Psi|^2 dx = \int\limits_{-a/2}^{a/2} cos^2(\frac{\pi*x}{a})\, dx = |A|^2*\frac{a}{2} \Rightarrow A=\sqrt(\frac{2}{a}) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <T> = -\frac{\hbar^2}{2m} \int \Psi \frac{d^2 \Psi}{dx^2} = \frac{\hbar^2}{2m} (\frac{\pi}{a})^2\int\Psi^2dx = \frac{\pi^2\hbar^2}{2ma^2} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <V> = .5m\omega^2\int x^2\Psi^2dx = .5m\omega^2\frac{2}{a}\int\limits_{-a/2}^{a/2}x^2cos^2(\frac{\pi x}{a}dx } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{m\omega^2}{a}(\frac{a}{\pi})^2\int\limits_{-\pi/2}^{\pi/2}y^2cos^2(y)dy } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{m\omega^2a^2}{\pi^3}[\frac{y^3}{6}+(\frac{y^2}{4}-\frac{1}{8})sin(2y)+\frac{ycos(2y)}{4}]_{-a/2}^{a/2} } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{m\omega^2a^2}{4\pi^2}(\frac{\pi^2}{6}-1) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <H> = \frac{\pi^2 \hbar^2}{2ma^2} + \frac{m\omega^2a^2}{4\pi^2}(\frac{\pi^2}{6}-1)}

${\frac {\partial <H>}{\partial a}}=-{\frac {\pi ^{2}\hbar ^{2}}{ma^{3}}}+{\frac {m\omega ^{2}a}{2\pi ^{2}}}({\frac {\pi ^{2}}{6}}-1)=0$

$\Rightarrow a=\pi {\sqrt {\frac {\hbar }{m\omega }}}({\frac {2}{\pi ^{2}/6-1}})^{1/4}$

$<H>_{min}={\frac {\pi ^{2}\hbar ^{2}}{2m\pi ^{2}}}{\frac {m\omega }{\hbar }}{\sqrt {\frac {\pi ^{2}/6-1}{2}}}+{\frac {m\omega ^{2}}{4\pi ^{2}}}(\pi ^{2}/6-1)\pi ^{2}{\frac {\hbar }{m\omega }}{\sqrt {\frac {2}{\pi ^{2}/6-1}}}$

$=.5\hbar \omega {\sqrt {\pi ^{2}/3-2}}=.5\hbar \omega (1.136)>.5\hbar \omega$

We do not need to worry about the discontinuity at ${\frac {\pm a}{2}}$. It is true that ${\frac {d^{2}\Psi }{dx^{2}}}$ has delta functions there, but since $\Psi ({\frac {\pm a}{2}})=0$ no extra contribution comes from these points.

b

Because this trial function is odd, it is orthogonal to the ground state. So, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\Psi|\Psi_{gs}>=0} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <H> \ge E_{fe} } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{fe} } is the energy of the first excited state.