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.

  • $\begingroup$ This will probably get closed as a homework question (we don't do those here) but I'd highly, highly recommend heading over to our chatroom where there's a bunch of people that will be able to help you. Thanks so much, by the way, for showing your work - you'd be surprised how many "give me the solution" questions we get here. Best of luck! $\endgroup$
    – auden
    Feb 19, 2017 at 19:36

3 Answers 3


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.

  • $\begingroup$ Thank you. I am still struggling to arrive at the same integral result as you, but I'm chalking that up to mental exhaustion. As I mentioned, I've spent many hours on this problem - I've been google-searching and integrating almost non-stop for 3 hours trying to solve this this morning. $\endgroup$
    – Izzy
    Feb 19, 2017 at 19:55
  • 5
    $\begingroup$ In that case, it might be good to give it a rest, but don't just take this on faith - don't leave this problem for good until you've been able to replicate the whole chain by yourself, or it's pretty pointless as an exercise. $\endgroup$ Feb 19, 2017 at 20:50

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 :-).

  • 1
    $\begingroup$ OP explicitly sets out to calculate the expectation value of the kinetic energy, and correctly identifies this as $+13.6\:\mathrm{eV}$; this is added to the expectation value of the potential energy, $-27.2\:\mathrm{eV}$, to give the expectation value of the total energy, $-13.6\:\mathrm{eV}$, which coincides with the eigenvalue. $\endgroup$ Feb 19, 2017 at 18:51
  • $\begingroup$ You are correct - I managed to drop the mass somewhere along the way. I just ran through the calculations again including the mass, and wound up with a brand new incorrect answer in units of eV*s^2/m. :S $\endgroup$
    – Izzy
    Feb 19, 2017 at 18:54
  • $\begingroup$ Did you use the Laplacian in spherical coordinates for your calculation? $\endgroup$ Feb 19, 2017 at 18:55
  • $\begingroup$ Yes, of course. I do have one question about that - do I need to apply the LaPlacian to the Jacobian? Doing so earlier was getting me into trouble with non-converging integrals (cos2x/sinx from 0 to pi). $\endgroup$
    – Izzy
    Feb 19, 2017 at 18:58
  • $\begingroup$ The radial function for the ground state that you have is already given in spherical coordinates. So all you need to do is $4\pi \int_0^\infty \mathrm d r \, \psi^*(r) \, \frac{\hbar^2}{2m} \nabla^2_r \psi(r)$ where $\nabla^2_r$ is the first summand in that big Laplacian expression from my answer. The other derivatives would give you zero anyway because the function does not depend on the angles. The factor $4 \pi$ is just the integration of “nothing” over the unit sphere. I am not sure whether there might be a factor $4\pi$ too much, that could come from wrong normalization of my memories. $\endgroup$ Feb 19, 2017 at 19:15

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 $-E\approx 13.6\text{ eV}$


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