Variational Principle of 3D symmetric harmonic oscillator

[JordanGo]

Homework Statement

Use the following trial function:
$\Psi=e^{-(\alpha)r}$
to estimate the ground state energy of the central potential:
$V(r)=(\frac{1}{2})m(\omega^{2})r^{2}$

The Attempt at a Solution

Normalizing the trial wave function (separating the radial and spherical part):
$\int(A^{2})e^{-2(\alpha)r}r^{2}=1$
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:
$A=\sqrt{\frac{\alpha^{3}}{\pi}}$
Now finding the hamiltonian:
$<H>=<V>+<T>$
I will start with <V>, since this is where I find my issue:
$<V>=\int(A)e^{-(\alpha)r}(\frac{1}{2}m\omega^{2}r^{2})Ae^{-(\alpha)r}$
where the integral is from 0 to infinity.
What I obtain for <V> is:
$<V>=\frac{1}{8\pi}m\omega^{2}$
which of course doesn't have units of energy...
Can someone point out where I went wrong please?

 

[andrien]

<E>=∫ψ*Hψ dv/∫ψ*ψ dv,is the usual way of getting it.why is it bothering you it clearly has the units of energy.
Edit: I have not checked your calculation.

 

[JordanGo]

<V> does not have units of energy...
But thank you for showing me the integration, I did not use dV, which will cause many problems... thanks!