# Anticommutator expression

I need to show that this expression is contradictory. The is no more information is given for $\hat{b}$. $$\hat{b}^{\dagger}\hat{b}+\hat{b}\hat{b}^{\dagger}=-I$$

• Take the expectation values of both sides w.r.t. some quantum state. – higgsss Mar 14 '18 at 18:11

The "hat" on your b's suggests Hilbert space operators, so I'm going to use that assumption. This implies that starting from a state $|\psi\rangle$, both $|\psi^\prime\rangle \equiv\hat{b}|\psi\rangle$ and $|\psi^{\prime\prime}\rangle \equiv\hat{b}^\dagger|\psi\rangle$ are valid states in your Hilbert space, as well. This means that taking $\langle\psi|\dots|\psi\rangle$ on both sides gives:

$\langle\psi^\prime| \psi^\prime\rangle + \langle\psi^{\prime\prime}| \psi^{\prime\prime}\rangle=-\langle\psi|\psi\rangle$

Since all your states need to have well-defined norm, this equation implies that the sum of two positive numbers is a negative one, which is clearly impossible.