This is a question that I don't know how to solve, because I keep getting negative kinetic energy.

Suppose there is a quantum mechanical system in d-dimensions that has the following Hamiltonian, where the potential depends only on the radial coordinate:

$$\hat{H}\psi = \left( -\frac{\hbar^2}{2m}\nabla^2-V_0e^{-r/a} \right)\psi=E\psi$$

Use a trial wavefunction $\phi \propto e^{-r/l}$, where $l$ is some constant, to estimate the ground energy $E_{var}=\frac{\left<\phi|\hat{H}|\phi \right>}{\left< \phi|\phi \right>}$ in 1 and 2 dimensions.

My solution:

The potential term of the hamiltonian was fairly easy to integrate, and I had no trouble in either 1 or 2 dimensions. The kinetic term, $\hat{T}=-\frac{\hbar^2}{2m}\nabla^2$, was more difficult. I got it right in 2d, but I could not get it right in 1d. This is what I did:

$$\left<\phi|\hat{T}|\phi \right> = 2\int_0^\infty e^{-r/l}\left(-\frac{\hbar^2}{2m}\nabla^2e^{-r/l}\right)dr = -\frac{\hbar^2}{m} \int_0^\infty e^{-r/l}\left(\frac{d^2}{dr^2}e^{-r/l}\right)dr = -\frac{\hbar^2}{ml^2}\int_0^\infty e^{-2r/l}dr=\frac{\hbar^2}{2ml}(0-1)=-\frac{\hbar^2}{2ml}$$

where I have used $\nabla^2=\frac{d^2}{dr^2}$ in 1 dimension. Now, I realize that this is not the final result for the expected kinetic energy, as I have not yet computed the denominator, and $\phi$ is not normalized. However, the denominator is for sure positive, so I can tell that something is wrong already since I am getting a negative result.


1 Answer 1


In $d=1$, your trial wave function $$\phi(x) = e^{-|x|/\ell} = e^{-x/\ell} \theta(x)+e^{x/\ell} \theta(-x)$$ has a cusp at $x=0.$ Consequently, the first derivative $$\phi^\prime (x) = \frac{1}{\ell}\left[-e^{-x/\ell}\theta(x)+e^{x/\ell}\theta(-x)\right]$$ is discontinuous at $x=0$ with $$\lim\limits_{\epsilon\downarrow 0}\left[\phi^\prime(0+\epsilon)-\phi'(0-\epsilon)\right]=-\frac{2}{\ell}.$$ The second derivative of $\phi$ is thus given by $$\phi^{\prime \prime}(x) = \phi(x)/\ell^2 -2 \delta(x)/ \ell,$$ solving your puzzle. Apart from the missing normalization, the expectation value of the kinetic energy is in fact given by $$\langle \phi |\hat{T} | \phi \rangle = \frac{\hbar^2}{2 m \ell},$$ a positive value, as to be expected for a positive operator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.