Schwartz inequality proof over complex

[jfy4]

Homework Statement

Consider any two vectors, |a\rangle and |b\rangle. Prove the Schwartz inequality
<br /> |\langle a|b \rangle |^2 \leq \langle a|a \rangle \langle b|b \rangle<br />

Homework Equations

a basic understanding of vector calculus over \mathbb{C}...

The Attempt at a Solution

I wanted to do this proof almost the same way I do it over \mathbb{R}, except I'm not sure if I can follow through with the normal quadratic part...

I start with |\psi\rangle =|a\rangle + c |b\rangle and using the fact that \langle\psi | \psi \rangle \geq 0 I get
<br /> 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<br />
which can be written
<br /> 0\leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle<br />
So I'm wondering if I can consider this quadratic in c and claim that
<br /> (2|\langle a|b\rangle |)^2-4\langle a|a\rangle \langle b|b\rangle \leq 0<br />

Any help would be appreciated, Thanks.

 

[lanedance]

so for the real case
0\leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle

becomes
0\leq \langle a|a\rangle + c\langle a|b\rangle +c^2\langle b|b\rangle

however for the complex case, that simplification does not occur. so I'm not convinced you can treat the quadratic the same as both the complex parts of c & <a|b> will cause complication - that said I'm not that familiar with this method...

The way I've seen that works for a general inner product space is to write |b> in as a summation of components perpendicular and parallel to |a> and prove it direct form there
|b\rangle = \langle a|b \rangle|a \rangle + |z\rangle

 

[Tedjn]

Why do you think c must be complex? All the inner products you are using result in real numbers.

 

[lanedance]

as the vectors are arbitrary <a|b> may be complex

now unless a constraint is put on c, then I would assume it can also be complex

now it is true that all the terms in the following expression are real, otherwise the inequality would not make sense
0 \leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle

now say we set c to be from the reals, then the inequality becomes
0 \leq \langle a|a\rangle + 2c\Re[\langle a|b\rangle ] +|c|^2\langle b|b\rangle

this is a real quadratic in c, though I'm not sure how the part about Re{<a|b>} could be massaged into the required form?

 

[Hurkyl]

If all else fails, split things into real and imaginary parts.

 

[jfy4]

I think I got it, let
<br /> |\psi\rangle = |b\rangle - \frac{ \langle a|b\rangle }{\langle a|a\rangle }|a\rangle<br />
then
<br /> \begin{align}<br /> 0\leq \langle \psi|\psi\rangle &amp;= \left( \langle b|-\frac{\langle b|a\rangle}{\langle a|a\rangle}\langle a| \right) \left( |b\rangle -\frac{\langle a|b\rangle}{\langle a|a\rangle}|a\rangle \right) \\<br /> &amp;=\langle b|b\rangle - \frac{\langle a|b\rangle \langle b|a\rangle}{\langle a|a\rangle}<br /> \end{align}<br />
and the result clearly follows.

How does this look?

 

[lanedance]

yep that's looking good to me

its worth noting the equality holds when |a> and |b> are parallel