Properties of the Laplacian

Show that the Laplacian is a linear, Hermitian, operator.

To show that \(\nabla^2\) is linear, we need only recall that it is the second derivative operator. We know that differentiation is a linear operator. I.e., differentiation of a constant times a function is equal to a constant, times the value obtained by differentiating the function. Similarly, differentiation of a sum of functions is equal to summing the derivatives of the functions. Mathematically, $\( \frac{d^k}{dx^k} \text{constant} \cdot f(x) = \text{constant} \cdot\frac{d^kf}{dx^k} \)\( and \)\( \frac{d^k\left(f(x) + g(x) \right)}{dx^k} = \frac{d^kf(x)}{dx^k} + \frac{d^kg(x)}{dx^k} \)\( So \)\nabla^2\( (corresponding to \)k=2$) is linear.

One can show that \(\nabla^2\) is Hermitian using integration by parts, similar to how we did in the course notes. You can also recognize that the Hermitian property follows directly from Green’s second identity . (In both cases you use the fact that the wavefunction and its derivatives vanish at the end of the interval of integration.)

However, we also know that \(\nabla^2\) is closely related to the momentum operator. $\( \hat{p}^2 = -i \hbar \nabla \cdot -i \hbar \nabla = - \hbar^2 \nabla^2 \)\( In atomic units, then, \)\nabla^2 = - \hat{p}^2$. The following math uses the fact that the momentum operator is Hermitian.

We need to show that $\( \int \Phi(x)^* \nabla^2 \Psi(x) dx = \int \left( \nabla^2 \Phi(x) \right)^* \Psi(x) dx \)\( To this end, we start with the relationship between the Laplacian and the momentum operator, then (repeatedly) invoke the fact the momentum operator is Hermitian. So: \)\( \begin{align} \int \Phi(x)^* \nabla^2 \Psi(x) dx &=\int \Phi(x)^* \left(-\hat{p}^2\right) \Psi(x) dx \\ &=-\int \Phi(x)^* \left(\hat{p}\hat{p} \right) \Psi(x) dx \\ &=-\int \left(\hat{p}\Phi(x)\right)^* \hat{p} \Psi(x) dx \\ &=-\int \left(\hat{p}\hat{p}\Phi(x)\right)^* \Psi(x) dx \\ &=\int \left(-\hat{p}^2\Phi(x)\right)^* \Psi(x) dx \\ &=\int \left(\nabla^2\Phi(x)\right)^* \Psi(x) dx \end{align} \)$

Show that the Laplacian is negative definite.

One can show that \(\nabla^2\) is negative-definite using Green’s first identity or integration by parts. However, as with the first part of this problem, one can also directly invoke the connection to the momentum operator, \(\nabla^2 = - \hat{p}^2\). Since the square of an operator is positive definite, it’s clear that the negative of the square of an operator is negative definite. However, more explicitly: $\( \begin{align} \int \Psi(x)^* \nabla^2 \Psi(x) dx &=\int \Psi(x)^* \left( -\hat{p}^2 \right) \Psi(x) dx \\ &=-\int \Psi(x)^* \hat{p}\hat{p} \Psi(x) dx \\ &=-\int \left(\hat{p}\Psi(x)\right)^* \hat{p} \Psi(x) dx \\ &=-\int \left|\hat{p}\Psi(x)\right|^2 dx \\ &<0 \end{align} \)$