# Thread: Prove A=C_dagger C is Hermitian and positive

1. ## Prove A=C_dagger C is Hermitian and positive

An operator A is said to be positive if ${<\psi|A|\psi>}$ $\ge$ 0 for all states | $\psi$> in the space.

PROVE: If C is an arbitrary operator the products $A=C^\dagger C$ and $B=CC^\dagger$ are Hermitian and positive.

Since C is an arbitrary operator then ${C|\psi>} = {|\phi>}$ for some $|\phi>$ in the space.

So $<\psi|C^\dagger C|\psi> = <\phi|\phi>$

Then $A^\dagger = (C^\dagger C)^\dagger$

Which gives $<\psi|(C^\dagger C)^\dagger|\psi> = <\psi|C C^\dagger|\psi> = (|\phi> |(\psi^*) >< (\psi) | <\phi|)^* = <\phi|\phi>$

Can someone verify if I got this correct so far or if I totally botched this up?

Thanks

2. ## Re: Prove A=C_dagger C is Hermitian and positive

An operator $A$ is Hermitian if $A^{\dagger} = A$.

It is easy to check that $(C^{\dagger}C)^{\dagger} = C^{\dagger}(C^{\dagger})^{\dagger} = C^{\dagger}C$

and $(CC^{\dagger})^{\dagger} = (C^{\dagger})^{\dagger}C^{\dagger} = CC^{\dagger}$

Note that $|\psi\rangle^{\dagger} = \langle \psi|$, so that for $A = C^{\dagger}C$:

$\langle \psi|A|\psi\rangle = \langle\psi|C^{\dagger}C|\psi\rangle$, so taking $C|\psi\rangle = |\phi\rangle$

we have $\langle\psi|C^{\dagger}C|\psi\rangle = \langle\phi|\phi\rangle \geq 0$, by the positive-definiteness of the inner product.

A similar observation holds for $CC^{\dagger}$ taking $|\phi\rangle = C^{\dagger}|\psi\rangle$.

(Basically we are regarding $\langle x|y\rangle$ as $|x\rangle^{\dagger}|y\rangle$, which is common for complex inner-product spaces).

3. ## Re: Prove A=C_dagger C is Hermitian and positive

Thank you so much. I was on the right track but messed up a few details.