# Expected Kinetic Energy in Exponentially decaying potential

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.

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.