# Groundstate of the Hydrogen Atom

In class, we had an example to use variational methods to find an upperbound on the groundstate of the hydrogen atom, by guessing some wavefunction based on the form of the Hamiltonian \begin{align*} \hat H = \hat T + \hat V &= \frac{\hat p^2}{2m} + V(\hat r)\\ &=-\frac{\hbar^2}{2m} \nabla^2 - \frac{e^2}{4\pi \epsilon_0 \hat r} \end{align*}

My first question is: How do we get from the first line to the second line in the construction of the Hamiltonian? Where does the $\hbar$ come from in the first term? I understand the second term is just the Coloumb repulsion term.

In class, the professor chose a trial wavefunction which is close to the groundstate wavefunction, but sufficiently different for instructive purposes. This wavefunction was

$$\psi(r,a) = \left(\frac{1}{3\pi a^5}\right)^{1/2} r \exp\left(-r/a\right)$$ where $a$ is the optimisation parameter.

We see that the energy of the atom will blow up as $r\rightarrow 0$, according to the potential term in the Hamiltonian. My second confusion, is: Why must the wavefunction we choose have the property that it drops off as $r \rightarrow 0$? The absolute value of the Hamiltonian goes to infinity in this limit, so my intuition tells me that the wavefunction should blow up too? I am trying to see if theres a connection to the expectation value, which has the property that $$<\psi|\hat H| \psi > = \sum_j |c_j|^2 E_j \geq E_0$$

• The constant $a$ would usually be called a "normalization" parameter rather than an "optimization" parameter. – dmckee Jun 28 '18 at 2:19
• I'm not seeing why the energy blows up. $\hat{H}\psi$ is finite everywhere; the potential term does have a $\frac{1}{\hat{r}}$ in it, but that only produces an infinite energy if the wavefunction is nonzero at the origin. Clearly, $\psi(0,a)=0$, so this doesn't happen. – probably_someone Jun 28 '18 at 2:58
• The Hamiltonian is not the energy in quantum mechanics. The absolute value of the Hamiltonian is also not the energy in quantum mechanics. This is because the Hamiltonian is an operator, not a number. It needs an input wavefunction for you to say anything; in particular, the energy of a wavefunction is the expectation value of the Hamiltonian with that wavefunction as an input. – probably_someone Jun 28 '18 at 3:04
• The Hamiltonian is not reliably the energy even in classical mechanics. It is often the energy (and there are specific mathematical tests you can use to tell if that is the case). – dmckee Jun 28 '18 at 15:54

## 1 Answer

Regarding your first question: The second line isn't quite correct. You get there by squaring the momentum operator $\hat{p}^2=(-\mathrm{i}\hbar\nabla)^2=-\hbar^2\nabla^2$. So you are actually missing an $\hbar$ in your expression.

Regarding your second question: Bound states in quantum mechanics are described by vectors in a Hilbertspace. This directly implies that the wavefunction has to be square integrable i.e. $\int\psi^*(r)\psi(r)dr<\infty$. If the wavefunction would diverge for $r\rightarrow 0$ this criterion could not be met. As already explained in the comments the Hamiltonian is not the energy but rather an operator associated with the observable energy. My intuition tells me the opposite of your assumption: When the potential is infinite the wavefunction should vanish, as the probability of a particle being at that location is certainly zero.