- #1
jfy4
- 649
- 3
Homework Statement
Consider any two vectors, [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex]. Prove the Schwartz inequality
[tex]
|\langle a|b \rangle |^2 \leq \langle a|a \rangle \langle b|b \rangle
[/tex]
Homework Equations
a basic understanding of vector calculus over [itex]\mathbb{C}[/itex]...
The Attempt at a Solution
I wanted to do this proof almost the same way I do it over [itex]\mathbb{R}[/itex], except I'm not sure if I can follow through with the normal quadratic part...
I start with [itex]|\psi\rangle =|a\rangle + c |b\rangle [/itex] and using the fact that [itex]\langle\psi | \psi \rangle \geq 0[/itex] I get
[tex]
0\leq \langle \psi | \psi \rangle = \langle a|a \rangle + c\langle a|b\rangle + c^{\ast}\langle b|a\rangle + |c|^2\langle b|b\rangle
[/tex]
which can be written
[tex]
0\leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle
[/tex]
So I'm wondering if I can consider this quadratic in [itex]c[/itex] and claim that
[tex]
(2|\langle a|b\rangle |)^2-4\langle a|a\rangle \langle b|b\rangle \leq 0
[/tex]
Any help would be appreciated, Thanks.