Say we have two Hamiltonians $\hat{H}_1$ and $\hat{H}_2$ that differ only in their potential energies and $$V_2(x) > V_1(x)$$ for all $x$. Is the energy of the ground state of system 2 necessarily larger than that of system 1?
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Yes, because $$E_1~=~\langle \psi_1 |\hat{H}_1 |\psi_1\rangle~\stackrel{\begin{array}{c}\text{def. of }|\psi_1\rangle\text{ being}\cr \text{ground state for }\hat{H}_1\end{array}}{\leq}~\langle \psi_2 |\hat{H}_1 |\psi_2\rangle~\stackrel{\begin{array}{c}\hat{H}_2-\hat{H}_1\geq 0\cr \text{semipos. op.}\end{array}}{\leq}~\langle \psi_2 |\hat{H}_2 |\psi_2\rangle~=~E_2, $$ where $|\psi_i\rangle$ denote a ground state for system $i\in\{1,2\}.$
answered Oct 24, 2018 at 13:01
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$\begingroup$ This argument doesn't show that $E_2 > E_1$ though. $\endgroup$ Commented Oct 25, 2018 at 11:21
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$\begingroup$ It seems to me that your argument does prove that $E_2 > E_1$, since $H_2 - H_1$ is strictly positive. $\endgroup$ Commented Oct 25, 2018 at 17:03