Chapter4problem

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

(a) Using the wave function $\psi ={\begin{cases}A*cos({\frac {\pi *x}{a}})&{\frac {-a}{2}}<x<{\frac {a}{2}}\\0&otherwise\end{cases}}$

obtain a bound on the ground state energy of the one-dimensional harmonic oscillator. Compare $<H>_{min}$ with the exact energy. Note: This trial wave function has a discontinuous derivative at ${\frac {\pm a}{2}}$.

(b) Use $\Psi =B*sin({\frac {\pi *x}{a}})$ on the interval (-a,a) to obtain a bound on the first excited state. Compare to the exact answer.

Solution

(a)

$1=\int |\Psi |^{2}dx=\int \limits _{-a/2}^{a/2}cos^{2}({\frac {\pi *x}{a}})\,dx=|A|^{2}*{\frac {a}{2}}\Rightarrow A={\sqrt {(}}{\frac {2}{a}})$

$<T>=-{\frac {\hbar ^{2}}{2m}}\int \Psi {\frac {d^{2}\Psi }{dx^{2}}}={\frac {\hbar ^{2}}{2m}}({\frac {\pi }{a}})^{2}\int \Psi ^{2}dx={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$

$<V>=.5m\omega ^{2}\int x^{2}\Psi ^{2}dx=.5m\omega ^{2}{\frac {2}{a}}\int \limits _{-a/2}^{a/2}x^{2}cos^{2}({\frac {\pi x}{a}}dx$ $={\frac {m\omega ^{2}}{a}}({\frac {a}{\pi }})^{2}\int \limits _{-\pi /2}^{\pi /2}y^{2}cos^{2}(y)dy$ $={\frac {m\omega ^{2}a^{2}}{\pi ^{3}}}[{\frac {y^{3}}{6}}+({\frac {y^{2}}{4}}-{\frac {1}{8}})sin(2y)+{\frac {ycos(2y)}{4}}]_{-a/2}^{a/2}$ $={\frac {m\omega ^{2}a^{2}}{4\pi ^{2}}}({\frac {\pi ^{2}}{6}}-1)$

$<H>={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}+{\frac {m\omega ^{2}a^{2}}{4\pi ^{2}}}({\frac {\pi ^{2}}{6}}-1)$

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

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

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

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

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, $<\Psi |\Psi _{gs}>=0$. $<H>\geq E_{fe}$ where $E_{fe}$ is the energy of the first excited state.

$1=\int |\Psi |^{2}dx=|B|^{2}\int _{-a}^{a}sin^{2}({\frac {\pi x}{a}})dx=|B|^{2}a\Rightarrow B={\frac {1}{{\sqrt {(}}a)}}$

$<T>={\frac {-\hbar ^{2}}{2m}}\int \Psi {\frac {d^{2}\Psi }{dx^{2}}}dx={\frac {\hbar ^{2}}{2m}}{\frac {\pi ^{2}}{a^{2}}}\int \Psi ^{2}dx={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}$ $<V>=.5m\omega ^{2}\int x^{2}\Psi ^{2}dx=.5m\omega ^{2}a^{-1}\int _{-a}^{a}x^{2}sin^{2}({\frac {\pi x}{a}})dx={\frac {m\omega ^{2}}{2a}}{\frac {a^{3}}{\pi ^{3}}}\int _{\pi }^{\pi }y^{2}sin^{2}(y)dy$ $={\frac {m\omega ^{2}a^{2}}{2ma^{2}}}+{\frac {m\omega ^{2}a^{2}}{4\pi ^{2}}}({\frac {2\pi ^{2}}{3}}-1)$ ;${\frac {\partial <H>}{\partial a}}={\frac {-\pi ^{2}\hbar ^{2}}{ma^{3}}}+{\frac {m\omega ^{2}a}{2\pi ^{2}}}({\frac {2\pi ^{2}}{3}}-1)=0$

$\rightarrow a=\pi {\sqrt {(}}{\frac {\hbar }{m\omega }})({\frac {2}{2\pi ^{2}/3-1}})^{1/4}$

$<H_{min}>={\frac {\pi ^{2}\hbar ^{2}}{2m\pi ^{2}}}{\frac {m\omega }{\hbar }}{\sqrt {(}}{\frac {2\pi ^{2}/3-1}{2}})+{\frac {m\omega ^{2}}{4\pi ^{2}}}({\frac {2\pi ^{2}}{3}}-1)\pi ^{2}{\frac {\hbar }{m\omega }}{\sqrt {(}}{\frac {2}{2\pi ^{2}/3-1}})$

$=.5\hbar \omega {\sqrt {(}}{\frac {4\pi ^{2}}{3}}-2)=.5\hbar \omega (3.341)>{\frac {3}{2}}\hbar \omega$