# Increasing a potential causes increase in energy levels

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.

1. Let there be given a selfadjoint$^1$ operator $H^{(0)}$ and a (semi)positive operator $V\geq 0$ on a Hilbert space ${\cal H}$. Let the basis of normalized eigenvectors for $H^{(0)}$ be $(|i^{(0)}\rangle)_{i\in I}$ with corresponding eigenvalues $(E^{(0)}_i)_{i\in I}$ ordered such that $$\forall i,j ~\in~I:\quad i~\leq~j\quad\Rightarrow \quad E^{(0)}_i~\leq~E^{(0)}_j.\tag{1}$$ Similarly, let the basis of normalized eigenvectors for $H^{(1)}:=H^{(0)}+V$ be $(|i^{(1)}\rangle)_{i\in I}$ with corresponding ordered eigenvalues $(E^{(1)}_i)_{i\in I}$.

2. Non-degenerate$^2$ perturbation theory yields the following construction: Define a one-parameter family of selfadjoint operators $$H(t)~:=~ H^{(0)}+tV, \qquad t~\in~[0,1].\tag{2}$$ Consider the following initial value problem of coupled 1st order differential equations$^2$ $$\frac{d|i(t)\rangle}{dt}~~=~~\sum_{j\in I\backslash \{i\}} \frac{\langle j(t)| V | i(t) \rangle}{ E_i(t)-E_j(t)}|j(t)\rangle ~~=~~ \sum_{j\in I}A_{ij}(t)|j(t)\rangle , \tag{3}$$ $$A_{ij}(t)~~:=~~\left\{\begin{array}{ccl} \frac{\langle j(t)| V | i(t) \rangle}{ E_i(t)-E_j(t)} &\text{if}& i~\neq~ j \cr 0 &\text{if}& i~=~ j \end{array} \right\}~~=~~-A^{\ast}_{ji}(t), \tag{4}$$ $$\frac{dE_i(t)}{dt}~~=~~\langle i(t)| V | i(t) \rangle~~\geq~~0, \tag{5}$$ $$|i(t\!=\!0)\rangle~~=~~|i^{(0)}\rangle,\qquad E_i(t\!=\!0)~~=~~E^{(0)}_i,\qquad i\in I. \tag{6}$$ We conclude from eq. (5) that the spectrum of $H(t\!=\!1)\equiv H^{(1)}$ is increased compared to the spectrum of $H(t\!=\!0)\equiv H^{(0)}$, i.e. $$\forall i~\in~I:\quad E^{(0)}_i ~\leq~ E^{(1)}_i,\tag{7}$$ as OP wanted to know. Note that the anti-Hermitian property (4) implies that the basis $(|i(t)\rangle)_{i\in I}$ is normalized.

3. In case of degeneracies and level-crossings, the basis $(|i(t)\rangle)_{i\in I}$ is no longer a well-defined/continuous function of $t$. Nevertheless, one may argue that the ordered eigenvalues $E_i(t)$ are still non-decreasing as a function of $t$, i.e. weakly increasing.

4. Alternatively, the increase in the ground state energy can be independently & non-perturbatively deduced from the variational method: $$E_0^{(0)} ~~\leq~~ \langle 0^{(1)}| H^{(0)} | 0^{(1)} \rangle ~~=~~ \langle 0^{(1)}| \left(H^{(1)}\!-\! V \right)| 0^{(1)} \rangle~~\leq~~ E^{(1)}_0. \tag{8}$$

5. A generalization of the above variational method (8), based on the Schur-Horn theorem, leads to the following tower of weaker inequalities: $$\forall j ~\in~I:\quad \sum_{i=0}^j E^{(0)}_i ~\leq~\sum_{i=0}^j E^{(1)}_i.\tag{9}$$ (To prove ineq. (9) work in a basis where $H^{(1)}$ is diagonal, and then apply the Schur-Horn inequality.)

$^1$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ We assume in Section 2 that the spectrum for $H^{(0)}$ is non-degenerate and that no level-crossings occur.

Consider $\tilde{V}(x)=V(x)+\Delta(x)$ where $\Delta(x)>0,\forall x$. Within 1st order perturbation, $E_n=E_n^0+\langle \psi_n|\Delta|\psi_n\rangle=E_n^0+\iint dx_1dx_2\psi_n^*(x_1)\delta(x_1-x_2)\Delta(x_2)\psi_n(x_2)=E_n^0+\int dx|\psi_n(x)|^2\Delta(x)>E_n^0$