(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An electron in a one dimensional crystal is bound by:

[tex]U(x) = \frac{-\overline{h}^{2}x^{2}}{mL^{2}\left(L^{2}-x^{2}\right)}[/tex]

for

[tex]\left|x\right| < L[/tex]

and

[tex]x = infinity[/tex]

for

[tex]\left|x\right| \geq L[/tex]

Show that a stationary state for the electron in the potential well

[tex]\psi(x) = A\left(1-\frac{x^{2}}{L^{2}}\right)[/tex]

satisfies the Schrodinger's Equation

and find E

2. Relevant equations

[tex]\frac{-\overline{h}^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+U(x)\psi = E\psi[/tex]

3. The attempt at a solution

from Schrodinger's:

[tex]\frac{d^{2}\psi}{dx^{2}} = \frac{-2m}{\overline{h}^{2}}\left(\frac{\overline{h}^{2}x^{2}}{mL^{2}\left(L^{2}-x^{2}\right)}+E\right)\psi[/tex]

and from the guess solution:

[tex]\frac{d^{2}\psi}{dx^{2}} = \frac{-2A}{L^{2}} = \frac{-2}{\left(L^{2}-x^{2}\right)}\psi[/tex]

and so equating [tex]\frac{d^{2}\psi}{dx^{2}}[/tex],

I deduced that it satisfies the Schrodinger's equation but only when E = 0 and x = L.

Am I right?

I am also concerned because the potential, U, is negative 'inside' the well...

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# Schrodinger's Equation in 1D

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