- #1

JordanGo

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## Homework Statement

Use the following trial function:

[itex] \Psi=e^{-(\alpha)r}[/itex]

to estimate the ground state energy of the central potential:

[itex]V(r)=(\frac{1}{2})m(\omega^{2})r^{2}[/itex]

## The Attempt at a Solution

Normalizing the trial wave function (separating the radial and spherical part):

[itex]\int(A^{2})e^{-2(\alpha)r}r^{2}=1[/itex]

where the integral is from 0 to infinity and A is the normalization constant.

Knowing that the spherical normalization will give 4(pi), then the normalization constant is:

[itex] A=\sqrt{\frac{\alpha^{3}}{\pi}}[/itex]

Now finding the hamiltonian:

[itex] <H>=<V>+<T> [/itex]

I will start with <V>, since this is where I find my issue:

[itex]<V>=\int(A)e^{-(\alpha)r}(\frac{1}{2}m\omega^{2}r^{2})Ae^{-(\alpha)r}[/itex]

where the integral is from 0 to infinity.

What I obtain for <V> is:

[itex] <V>=\frac{1}{8\pi}m\omega^{2}[/itex]

which of course doesn't have units of energy...

Can someone point out where I went wrong please?