- #1

stunner5000pt

- 1,461

- 2

A particle in a potnetial V(x) has a definite energy

[tex] E = - \frac{\hbar^2 \alpha^2}{2m} [/tex]

and its eigenfunction is given by

[tex] \psi(x) = Nx \exp(-alpha x) [/tex] if 0 <= x < infinity

zero elsewhere

Prove that [tex] V(x) = -alpha \hbar^2/mx [/tex] 0 <= x < infinity

infinity elsewhere

Given that the eigenfunction describes the ground state of the aprticle in teh potnetial V(x) roughly sketch the the eigenfunction whifchcdescribes the first excited state

ground state ook like sketch 1

no zeroes

i think the first excited state will look like sketch

it has to have 1 zero...

[tex] E = - \frac{\hbar^2 \alpha^2}{2m} [/tex]

and its eigenfunction is given by

[tex] \psi(x) = Nx \exp(-alpha x) [/tex] if 0 <= x < infinity

zero elsewhere

Prove that [tex] V(x) = -alpha \hbar^2/mx [/tex] 0 <= x < infinity

infinity elsewhere

Given that the eigenfunction describes the ground state of the aprticle in teh potnetial V(x) roughly sketch the the eigenfunction whifchcdescribes the first excited state

ground state ook like sketch 1

no zeroes

i think the first excited state will look like sketch

it has to have 1 zero...