Proving the Potential of Ground State Eigenfunction

[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...

 

[Jhero]

To prove that V(x) = -alpha \hbar^2/mx, we need to use the Schrodinger equation:

H\psi(x) = E\psi(x)

Where H is the Hamiltonian operator, E is the energy, and \psi(x) is the eigenfunction. In this case, the energy is given as E = - \frac{\hbar^2 \alpha^2}{2m} and the eigenfunction is \psi(x) = Nx \exp(-alpha x) for 0 <= x < infinity and zero elsewhere.

Substituting these values into the Schrodinger equation, we get:

H(Nx \exp(-alpha x)) = - \frac{\hbar^2 \alpha^2}{2m}(Nx \exp(-alpha x))

Expanding the Hamiltonian operator, we get:

\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}(Nx \exp(-alpha x)) = - \frac{\hbar^2 \alpha^2}{2m}(Nx \exp(-alpha x))

Simplifying and rearranging, we get:

\frac{d^2}{dx^2}(Nx \exp(-alpha x)) + \frac{2m\alpha^2}{\hbar^2}(Nx \exp(-alpha x)) = 0

Using the fact that N is just a constant, we can simplify further to get:

\frac{d^2}{dx^2}(x \exp(-alpha x)) + \frac{2m\alpha^2}{\hbar^2}(x \exp(-alpha x)) = 0

Now, we can use the definition of the potential energy, V(x) = \frac{\hbar^2}{2m}\frac{d^2}{dx^2}, to get:

V(x)(x \exp(-alpha x)) + \frac{2m\alpha^2}{\hbar^2}(x \exp(-alpha x)) = 0

Simplifying and rearranging, we get:

V(x) = - \frac{2m\alpha^2}{\hbar^2x}

Therefore, we have proven that V(x) = -alpha \hbar^2/mx for 0 <= x < infinity and infinity elsewhere.

To sketch the eigenfunction for the first excited state, we need