# Homework Help: Hamiltonian Operator

1. Oct 14, 2007

### UbikPkd

For an atom with one electron and nuclear charge of Z, the Hamiltonian is:

$$H=-~\frac{\nabla^{2}}{2}~- ~\frac{Z}{r}~$$

1) show that the wavefunction:

$$\Psi_{1s}=Ne^{-Zr}$$

is an eigenfunction of the Hamiltonian

2) find the corresponding energy

3) find N, the normalisation constant

In spherical polar coordinates:

$$\nabla^{2}\Psi_{1s}=~\frac{1}{r^{2}}~(~\frac{d}{dr}~[r^{2}~\frac{d\Psi_{1s}}{dr}])$$

by applying H to the wavefunction, i think i've shown that it's an eigenfunction:

1)

$$H\Psi_{1s}=~\frac{-Z^{2}r^{2}}{2}~Ne^{-zr} - ~\frac{Z}{r}~$$

$$H\Psi=n\Psi$$

where n is the eigenvalue, and the bit on the end:

$$- ~\frac{Z}{r}~$$

doesn't matter right, i've still shown it's an eigenvalue?

2)

to find the corresponding energy, don't I need to know N first?

3)

to find N, am I right in thinking:

$$N^{2} \int \Psi* \Psi dx = 1$$

$$N^{2} \int e^{-2zr} dx = 1$$

$$~\frac{-N^{2}}{2z}~e^{-2zr}=1$$

$$N=\sqrt{~\frac{-2}{e^{-2zr}}~}$$

but i think i must have gone wrong somewhere, i mean that doesn't look right. Once i've found N, how do i find the corresponding energy do i just plug N into:

$$H\Psi_{1s}=~\frac{-Z^{2}r^{2}}{2}~Ne^{-zr} - ~\frac{Z}{r}~$$

ie, $$H\Psi_{1s}$$ = the corresponding energy?

i'd appreciate any help, thanks

2. Jan 15, 2009

### cmjrees

You haven't multiplied

$$\frac{Z}{r}$$

right, it should be

$$\frac{Z}{r}~Ne^{-Zr}$$