quantum mechanics - Increasing a potential causes increase in energy levels

Suppose a potential $V(x)$, and suppose a bound particle so the allowed energy levels are discrete. Suppose a second potential $\widetilde{V}(x)$ such that $\widetilde{V}(x) \geq V(x)$ for all $x$ (suppose the potentials are relevant only for some interval). Does this necessarily imply that the eigenvalues of the Hamiltonian with $\widetilde{V}(x)$ will be at least those of the Hamiltonian with $V(x)$? That is, if $E_n$ are the energy levels of the first Hamiltonian and $\widetilde{E}_n$ are the energy levels of the second Hamiltonian, is $E_n\leq \widetilde{E}_n$ for all $n$? How can one show this?

Answer

You can show this by using perturbation theory (only for suitable small changes in the potential).

When you assume, that $\tilde{V}(x) = V(x) + c$ with $c > 0$, then you can write your problem als perturbation: If the unperturbated hamiltonian $\hat{\mathrm H}$ has eigenstates $| \Psi_n \rangle$ with discrete energies, then perturbation-theory states that changing the hamiltonian by a little term $\hat{\mathrm V}_\textrm{perturbation}$ will change the eigenvalues $E_n$ by: $$\Delta E_n = \left\langle \Psi_n \left| \hat{\mathrm V}_\textrm{perturbation}\right| \Psi_n \right\rangle$$ This is valid if you neglect terms of higher Order.

There is one think to watch out regarding perturbation theory: If your energy eigenvalues are degenerated, then the perturbation term has to be diagonal in the subspace that is spanned by the degenerated states.

In our case, $\hat{\mathrm V}_\textrm{perturbation}= c$ is just a multiplication, so: $$\langle \Psi_m | c | \Psi_n\rangle = c \langle \Psi_m | \Psi_n\rangle = c ~\delta_{nm}$$ $\hat{\mathrm V}_\textrm{perturbation}$ is diagonal in any subspace, and we can make use of perturbation theory. You then calculate the energy-shift just by $$\Delta E_n = \langle \Psi_n | c| \Psi_n \rangle = c \langle \Psi_n | \Psi_n \rangle = c >0$$ So if you increase the potential by a constant, then the energy eigenvalues will just shift by that constant.

Edit: One can expand the proof for perturbations that vary with time: Let the change in the potential be $\delta V(x)$ (which now depends on $x$), then you can still calculate the energy-shift by using perturbation theory. In whatever subspace that is formed by degenerate states, you can find a Basis $|\tilde{\Psi}_n \rangle$ for which $\delta V(x)$ is an orthogonal Operator.

In this Basis you then calculate the energy-shift like described above: $$\Delta E_n = \langle \tilde{\Psi}_n | \delta V( \hat{x})| \tilde{\Psi}_n = \int dx |\tilde{\Psi}(x)|^2 \delta V(x) > 0$$ Since $\delta V(x) > 0$. Those are now energyshifts for eigenstates of your "old" hamiltonian. However, those eigenstates are not necessarrily the eigenstates that you started with.