MHB Conjugate Bra Ket Properties for Proving the Schwarz Inequality

[ognik]

Just checking (while trying to prove the Schwarz inequality for $<f|H|g>$, I know $ <f|g>=<g|f>^* $ please confirm/correct :

If $ \psi=f+\lambda g, \:then\: \psi^*=f^*+\lambda^* g^* $

Is $ <f^*|g>=<g^*|f>^* $ and $ <f^*|H|g>=<g^*|H|f>^* $ (H hermitian)?

Is $ <f^*|H|g><g^*|H|f> = - <g^*|H|f><f^*|H|g> $

Thanks

 

[Ackbach]

Just checking (while trying to prove the Schwarz inequality for $<f|H|g>$,

Use the $\LaTeX$ codes

\langle and \rangle

instead of < and > to get better-looking symbols.

I know $ <f|g>=<g|f>^* $ please confirm/correct :

If $ \psi=f+\lambda g, \:\text{then}\: \psi^*=f^*+\lambda^* g^* $

Is $ <f^*|g>=<g^*|f>^* $ and $ <f^*|H|g>=<g^*|H|f>^* $ (H hermitian)?

It's confusing to put the conjugate symbol inside a bra or ket. That's not usually how we do this notation. Remember these rules:
1. The Hermitian conjugate of a bra is the corresponding ket, and vice versa. That is, $|f\rangle^{\dagger}=\langle f|$, and $\langle f|^{\dagger}=|f\rangle$.
2. The Hermitian conjugate of a complex number is its complex conjugate.
3. The Hermitian conjugate of a Hermitian conjugate is the original thing back at you.
4. For a general operator $A$, you have $\langle f|A|g\rangle^*=\langle g|A^{\dagger}|f\rangle$.

So, I would say you have $\langle f|g\rangle^*=\langle g|f\rangle$, and $\langle f|H|g\rangle^*=\langle g|H|f\rangle$, since $H$ is Hermitian.

Is $ <f^*|H|g><g^*|H|f> = - <g^*|H|f><f^*|H|g> $

Thanks

I would say $\langle f|H|g\rangle \, \langle g|H|f\rangle=\langle f|H|g\rangle \langle f|H|g\rangle^*=|\langle f|H|g\rangle|^2$.

 

[ognik]

It's confusing to put the conjugate symbol inside a bra or ket.

Which was what I needed to know to figure out the (too embarrassing to tell) mistake I was making.

The proof fell into place almost too easily after that, thanks ackbach