# How to prove this inequality for the Hamiltonian operator?

I am trying to prove the following:

$$\langle\psi|\hat{H}|\phi\rangle\langle\phi|\hat{H}|\psi\rangle-\langle\psi|\hat{H}|\psi\rangle\langle\phi|\hat{H}|\phi\rangle\leqslant0.$$

I tried some ideas but could reach nowhere. I exploited the fact that $\hat{H}$ is hermitian, and thus the first term in the inequality became $\langle\phi|\hat{H}|\psi\rangle^*\langle\phi|\hat{H}|\psi\rangle=|\langle\phi|\hat{H}|\psi\rangle|^2$ and then by Cauchy Schwartz inequality, $|\langle\phi|\hat{H}|\psi\rangle|^2\leqslant\langle\phi|\phi\rangle\langle\psi|\hat{H}\hat{H}|\psi\rangle$,but I can see this just removes $|\phi\rangle$ from the game.

Another idea was to write $\hat{H}$ in the first term in outer product notation, this gives:$$\langle\psi|\hat{H}|\phi\rangle\langle\phi|\hat{H}|\psi\rangle=\sum_{i,j}E_iE_j\langle\psi|i\rangle\langle i|\phi\rangle\langle\phi|j\rangle\langle j|\psi\rangle$$I tried to work on this to get the inequality, but it got me nowhere.

Any help is appreciated.

• You're on the right track. But why not split it up in another way with Cauchy-Schwartz? Write $H = \sqrt{H} \sqrt{H}$ and give one square root to each of the vectors Psi and Phi. – Luke Jun 20 '17 at 14:59
• You can repeat this proof of the Cauchy–Schwarz inequality inserting $H$ in the middle of all scalar products. – Alexey Sokolik Jun 20 '17 at 15:04

1. 2D Counterexample: Assume $$\langle\psi| \psi \rangle ~=~1~=~\langle\phi| \phi \rangle, \qquad \langle\psi| \phi \rangle~=~0, \qquad \hat{H}~=~| \psi \rangle\langle\psi| -| \phi \rangle\langle\phi|.$$
2. OP's inequality is true for a semi-positive operator $\hat{H}\geq 0$, since then it has a well-defined square root $\sqrt{\hat{H}}$, and it becomes the standard Cauchy–Schwarz inequality.
• Please forgive my ignorance, but that hamiltonian seems to have a negative eigenvalue for $|\phi\rangle$, isn't it supposed to have positive eigenvalues only? – Tofi Jun 20 '17 at 15:49
• @VIP Usually the Hamiltonian is assumed to be bounded from below, for physical reasons. The result however holds only for positive operators, and therefore given an Hamiltonian $H$ bounded from below by $-m$, $m>0$, the result holds for the positive operator $H+m$. Physically, $H+m$ is equivalent to $H$, up to a rescaling of the ground state energy. – yuggib Jun 20 '17 at 16:25