- #1

jdstokes

- 523

- 1

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

[itex](\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[/itex].

Put [itex]\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle[/itex]. Then

[itex] \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0[/itex].

Not quite what I wanted.

Any help would be appreciated.