I am trying to prove the following:


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.

  • $\begingroup$ 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. $\endgroup$
    – Luke
    Jun 20, 2017 at 14:59
  • $\begingroup$ You can repeat this proof of the Cauchy–Schwarz inequality inserting $H$ in the middle of all scalar products. $\endgroup$ Jun 20, 2017 at 15:04

1 Answer 1

  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.

  • $\begingroup$ Yes, right, the solution I proposed only works for Hamiltonians that are positive. This counter-example shows it is not possible for all Hamiltonians. $\endgroup$
    – Luke
    Jun 20, 2017 at 15:14
  • 1
    $\begingroup$ 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? $\endgroup$
    – Tofi
    Jun 20, 2017 at 15:49
  • 1
    $\begingroup$ A Hermitian operator has real (but not necessarily positive) eigenvalues. $\endgroup$
    – Qmechanic
    Jun 20, 2017 at 16:04
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    $\begingroup$ @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. $\endgroup$
    – yuggib
    Jun 20, 2017 at 16:25

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