0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Take the expectation values of both sides w.r.t. some quantum state. $\endgroup$ – higgsss Mar 14 '18 at 18:11
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.