Question #203738

Use the variational principle to obtain an upper limit to ground state energy of a particle in one dimensional box.


1
Expert's answer
2021-06-07T09:34:48-0400

=<H>=<ψ/H/ψ>=ψ(Hψ)dz= <H>=<\psi/H/\psi>=\int_{-\infty}^{\infty}\psi*(H\psi)dz

H=eBxmSy=ehˉBym[10 0i]/ψ>=[sinθcosθ]H=\frac{eB_x}{m}Sy=\frac{e\bar{h}By}{m}[_1^0 \space _0^{-i}] /\psi>=[^{cos\theta}_{sin\theta}]

So, <ψ/=[cosθsinθ]<\psi/=[cos\theta sin\theta]

=<H>=<ψ/H/ψ/>=<H>=<\psi/H/\psi/>

=[cosθsinθ][10 0i][sinθcosθ]ehBym=[cos \theta sin \theta][_1^0 \space_0^{-i}][_{sin\theta}^{cos \theta}] \frac{ehBy}{m}

=[icosθsinθ+sinθcosθ]ehBym=0=[-icos\theta sin\theta+sin \theta cos \theta]\frac{ehBy}{m}=0

<H>=<ψ/μ/ψ>=0\therefore <H>= <\psi/\mu/\psi>=0


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