Chapter4problem: Difference between revisions

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'''Solution'''
'''Solution'''
:'''(a)'''
:<math> 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}) </math>
:<math> 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}) </math>


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:<math> <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}}</math>
:<math> <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}}</math>
:<math> = .5\hbar\omega\sqrt{\pi^2/3-2} = .5\hbar\omega(1.136) > .5\hbar\omega </math>
:<math> = .5\hbar\omega\sqrt{\pi^2/3-2} = .5\hbar\omega(1.136) > .5\hbar\omega </math>
We do not need to worry about the discontinuity at <math> \frac{\pm a}{2}</math>. It is true that <math> \frac{d^2 \Psi}{dx^2} </math> has delta functions there, but since <math> \Psi(\frac{\pm a}{2})=0</math> no extra contribution comes from these points.

Revision as of 21:47, 15 April 2010

(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 with the exact energy. Note: This trial wave function has a discontinuous derivative at .

(b) Use 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 <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}}}
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 = .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 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}} . It is true that 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{d^2 \Psi}{dx^2} } has delta functions there, but since 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(\frac{\pm a}{2})=0} no extra contribution comes from these points.