Calculating the (expected) kinetic energy of an electron in the ground state of a Coulomb potential? [closed]

Question

I've been struggling with this all week to no avail.

I'm asked to calculate the expectation value of kinetic energy for an electron in the ground state of a Coulomb potential. I know that it ought to be $ 13.6 \, \mathrm{eV}$, but I am having a difficult time arriving there.

In general, the expectation value of, say, $Q$, is

$$\langle Q \rangle = \int \psi^* \hat Q \psi \, \mathrm dV $$

over all space. In the case of kinetic energy, $\hat Q$ would be equal to

$$ \frac{-\hbar^2}{2m} \nabla^2 $$

and, in the case of a ground-state electron, we would have

$$ \psi = \sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}} \exp(- r / a) $$

with $a$ being the Bohr radius.

However, for the life of me, I cannot get this integral to work. For a while, I was continually coming up with either 0 or a non-converging integral, until I stumbled on some piece of information (that I can't find convincing proof of, either in my textbook or on the internet) that the square angular momentum (that is, $(\mathrm d^2/ \mathrm d \theta^2 + \mathrm d^2/ \mathrm d \phi^2) \theta\psi$) is equal to $l(l+1)$ - in my case, 0, since $l = 0$ in the ground state $(1,0,0)$. This simplified things and gave me an integral I could get to converge. However, it seems to converge to $$ \frac{\hbar^2}{a^2} $$ which not only has the wrong units of (energy time per length)^2 but also has the wrong value.

Please help. This homework problem has taken an embarrassingly long time and a lot of scratch paper to do already.

Answer 1

The factorization into a radial part plus the angular momentum operator is true, but you don't really need it; instead, you can simply use the Laplacian in spherical coordinates, $$ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r} +\frac{1}{r^2\sin(\theta)}\frac{\partial}{\partial \theta}\sin(\theta)\frac{\partial}{\partial \theta}+ \frac{1}{r^2\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2}, $$ and note that the angular derivatives vanish with your spherically-symmetric wavefunction.

From here you just need to calculate the action of the kinetic energy operator on the ground state, \begin{align} \hat{Q}\psi &= \frac{-\hbar^2}{2m}\nabla^2\psi \\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\nabla^2 \exp(- r / a) \\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2\frac{\partial}{\partial r} \exp(- r / a)\right] \\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\frac{\partial}{\partial r}\left[r^2\frac{-1}{a} \exp(- r / a)\right] \\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\frac{1}{r^2}\left[2r\frac{-1}{a} \exp(- r / a)+r^2\frac{1}{a^2} \exp(- r / a)\right] \\& = \frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\left[\frac{-2}{ar} \exp(- r / a)+\frac{1}{a^2} \exp(- r / a)\right], \end{align} and then integrate against the wavefunction itself: \begin{align} \langle\psi|\hat{Q}|\psi\rangle &=\int \psi(\mathbf r)^* \, \hat{Q}\psi(\mathbf r)\mathrm d\mathbf r \\& = \int_0^{\infty} \psi(r)^* \hat{Q}\psi(r) 4\pi r^2\mathrm dr \\& = \int_0^{\infty} \left(\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\exp(- r / a)\right) \\& \qquad\times\left(\frac{-\hbar^2}{2m}\sqrt{\frac{1}{4 \pi}} \frac{2}{a^{3/2}}\left[\frac{-2}{ar} \exp(- r / a)+\frac{1}{a^2} \exp(- r / a)\right]\right) 4\pi r^2\mathrm dr \\& = \frac{-\hbar^2}{2m} \frac{4}{a^3}\int_0^{\infty} \left(\frac{-2r}{a} e^{- 2r / a}+\frac{r^2}{a^2} e^{-2 r / a}\right) \mathrm dr \\& = \frac{-\hbar^2}{2m} \frac{4}{a^3} \left(\frac{-a}{4} \right) \\& = +\frac{\hbar^2}{2ma^2} \\& = 13.6\:\mathrm{eV}, \end{align} as required.

Answer 2

You want the energy of the ground state, the whole energy. Therefore you need to compute the expectation value of your Hamiltonian, not only of the kinetic operator. This will give you more terms and also gives you a chance to get the electric charge to enter.

In your current computation, I think you have somehow dropped the mass $m$ from the final result. It should be still there, right?

Then to address the angular momentum: It is correct, you can switch to spherical coordinates and have the solution $$ \psi(r, \theta, \psi) = A(r) B(\theta) C(\psi) \,, $$ where $A$, $B$, and $C$ are just function of one variable. The thing $BC(\theta, \psi)$ are the spherical harmonics. They are the eigenfunctions of the angular part of the Laplacian operator (the one with second derivatives of the angles) with eigenvalues $l(l+1)$. Since $l = 0$ in the ground state, you have no angular part and just a radial function.

Then in spherical coordinates, you have to be careful to use the correct $\nabla^2$. Look at this awesome listing. There you will find that the Laplacian on a function $f(r, \theta, \psi)$ is given by $$ {1 \over r^2}{\partial \over \partial r}\!\left(r^2 {\partial f \over \partial r}\right) \!+\!{1 \over r^2\!\sin\theta}{\partial \over \partial \theta}\!\left(\sin\theta {\partial f \over \partial \theta}\right) \!+\!{1 \over r^2\!\sin^2\theta}{\partial^2 f \over \partial \varphi^2} \,. $$

You need to care only about the radial part. That is what you need to use as your kinetic energy (together with $\hbar^2/2m$).

I hope this will give you a way to solve the problem. Quantum mechanics will always take an insane amount of scratch paper. In all the theoretical physics lectures that I have taken, the homework problems would always be really long calculations. And even if you already solved it, it would still require a couple of pages to write everything down. The whole solution to the hydrogen atom is a long calculation and I think I would need at least 10 pages to get somewhere. So don't worry about the paper needed :-).

Answer 3

Emilio Pisanty has explained how to work out the answer by calculating the integral. You can get around doing that (but that should not come at the expense of learning how to compute the integral), by using a result from first order perturbation theory. It can be shown that the change in the eigenvalue of an operator due to adding a perturbation is, to first order, the expectation value of that perturbation in the original unperturbed eigenstate.

This means that you can calculate the expectation value of the kinetic energy by multiplying the kinetic term in the Hamiltonian by $1+\epsilon$, the exact energy eigenvalue is trivial to work out, the the expectation value of the kinetic energy can be extracted by expanding this in powers of $\epsilon$, it is is the coefficient of $\epsilon$ in this expansion. The ground state energy can be written as:

$$E = -\frac{m e^4}{2\hbar^2}$$

If we multiply the kinetic term $\frac{p^2}{2m}$ in the Hamiltonian by $1+\epsilon$, this amounts to dividing the mass by $1+\epsilon$, which to first order in $\epsilon$ is the same as multiplying by $1-\epsilon$. Therefore, the coefficient of $\epsilon$ is $-E\approx 13.6\text{ eV}$