# Excited state

1. Nov 14, 2006

### stunner5000pt

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

$$E = - \frac{\hbar^2 \alpha^2}{2m}$$

and its eigenfunction is given by

$$\psi(x) = Nx \exp(-alpha x)$$ if 0 <= x < infinity
zero elsewhere

Prove that $$V(x) = -alpha \hbar^2/mx$$ 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...

#### Attached Files:

File size:
7.3 KB
Views:
51
• ###### 2.JPG
File size:
8.4 KB
Views:
62
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?