Proving Cauchy-Schartz inequality in Brak-ket notation

[jdstokes]

[SOLVED] Proving Cauchy-Schartz inequality in Brak-ket notation

$(\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0$.

Put $\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle$. Then

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$.

Not quite what I wanted.

Any help would be appreciated.

 

[kdv]

$(\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0$.

Put $\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle$. Then

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$.

the last line implies

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$
so
$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle \geq \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle$
Isn't this what you wanted to show?

 

[jdstokes]

Thanks for replying.

How do you get $\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$?

 

[jdstokes]

OMG I'm so dumb. No need to answer that. Thanks.