07.03.2007, 08:06
elSalvadore schrieb:aza du dummbatz! jetzt bekommst du unnuetzes zeugs!
For two arbitrary hermitian operators A: H ’ H and B: H ’ H, and any element x of H such that A B x and B A x are both defined (so that in particular, A x and B x are also defined), then
\langle B A x | x \rangle = \langle A x | B x \rangle = \langle B x | A x \rangle^{*}
In an inner product space the Cauchy-Schwarz inequality holds.
\left|\langle B x | A x \rangle\right |^2 \leq \|A x \|^2 \|B x \|^2