We choose the trial wave function:
$$ \psi(r) = A e^{-\alpha r^2} $$
To find the normalization constant \(A\), we normalize the wave function:
$$ \int_0^\infty |\psi(r)|^2 r^2 dr = 1 $$
Substituting \(\psi(r)\):
$$ A^2 \int_0^\infty e^{-2\alpha r^2} r^2 dr = 1 $$
The integral can be solved using the substitution \( u = \sqrt{2\alpha} r \):
$$ A^2 \left( \frac{1}{2\alpha} \right)^{3/2} \int_0^\infty e^{-u^2} u^2 du = 1 $$
The integral \( \int_0^\infty e^{-u^2} u^2 du = \frac{\sqrt{\pi}}{4} \):
$$ A^2 \left( \frac{1}{2\alpha} \right)^{3/2} \cdot \frac{\sqrt{\pi}}{4} = 1 $$
$$ A^2 = \frac{4}{\sqrt{\pi}} (2\alpha)^{3/2} $$
$$ A = \left( \frac{8\alpha^{3/2}}{\pi^{1/2}} \right)^{1/2} = \left( \frac{8\alpha^{3/2}}{\pi^{1/2}} \right)^{1/2} = \frac{2(2\alpha)^{3/4}}{\pi^{1/4}} $$
The Hamiltonian for the hydrogen atom is:
$$ H = -\frac{\hbar^2}{2m} \nabla^2 - \frac{e^2}{4 \pi \epsilon_0 r} $$
We need to find the expectation value of the Hamiltonian \( \langle H \rangle \):
$$ \langle H \rangle = \int_0^\infty \psi^*(r) H \psi(r) r^2 dr $$
Split the Hamiltonian into kinetic (\(T\)) and potential (\(V\)) energy terms:
$$ T = -\frac{\hbar^2}{2m} \nabla^2 $$
$$ V = -\frac{e^2}{4 \pi \epsilon_0 r} $$
Calculate the expectation values separately:
$$ \langle T \rangle = \int_0^\infty \psi^*(r) T \psi(r) r^2 dr $$
The kinetic energy operator is:
$$ T \psi(r) = -\frac{\hbar^2}{2m} \left( \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} \right) \psi(r) $$
After some calculus:
$$ \langle T \rangle = \frac{3 \hbar^2 \alpha}{2m} $$
$$ \langle V \rangle = \int_0^\infty \psi^*(r) V \psi(r) r^2 dr $$
The potential energy term:
$$ V = -\frac{e^2}{4 \pi \epsilon_0 r} $$
After integration, we get:
$$ \langle V \rangle = - \frac{e^2 \alpha^{1/2}}{2 \pi^{1/2} \epsilon_0} $$
Combine the results to get the total expectation value of the Hamiltonian:
$$ \langle H \rangle = \langle T \rangle + \langle V \rangle $$
$$ \langle H \rangle = \frac{3 \hbar^2 \alpha}{2m} - \frac{e^2 \alpha^{1/2}}{2 \pi^{1/2} \epsilon_0} $$
Minimize this expectation value with respect to the variational parameter \(\alpha\):
$$ \frac{d\langle H \rangle}{d\alpha} = 0 $$
Set the derivative to zero and solve for \(\alpha\):
$$ \frac{3 \hbar^2}{2m} - \frac{e^2 \alpha^{-1/2}}{4 \pi^{1/2} \epsilon_0} = 0 $$
$$ \alpha^{1/2} = \frac{e^2 m}{6 \hbar^2 \pi^{1/2} \epsilon_0} $$
Best estimate for \( \alpha \) : $$ \widetilde{\alpha} = \left( \frac{e^2 m}{6 \hbar^2 \pi^{1/2} \epsilon_0} \right)^2 $$
The value of \(\alpha\) that minimizes \(\langle H \rangle\) gives us the best approximation for the ground state energy using our chosen trial wave function. Plug this value of \(\alpha = \widetilde{\alpha} \) back into the expression for \(\langle H \rangle\) to find the approximate ground state energy.
Exact energy $$ E_{exact} = -\frac{m e^4}{16 \pi^2 \epsilon^2 \hbar^2} $$
Variational energy estimate $$ E_{est} = -\frac{m e^4}{24 \pi \epsilon^2 \hbar^2} $$