# Homework Help: Schwartz inequality proof over complex

1. Aug 28, 2011

### jfy4

1. The problem statement, all variables and given/known data
Consider any two vectors, $|a\rangle$ and $|b\rangle$. Prove the Schwartz inequality
$$|\langle a|b \rangle |^2 \leq \langle a|a \rangle \langle b|b \rangle$$

2. Relevant equations
a basic understanding of vector calculus over $\mathbb{C}$...

3. 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
$$0\leq \langle \psi | \psi \rangle = \langle a|a \rangle + c\langle a|b\rangle + c^{\ast}\langle b|a\rangle + |c|^2\langle b|b\rangle$$
which can be written
$$0\leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle$$
So I'm wondering if I can consider this quadratic in $c$ and claim that
$$(2|\langle a|b\rangle |)^2-4\langle a|a\rangle \langle b|b\rangle \leq 0$$

Any help would be appreciated, Thanks.

2. Aug 28, 2011

### 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
$$|b\rangle = \langle a|b \rangle|a \rangle + |z\rangle$$

3. Aug 29, 2011

### Tedjn

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

4. Aug 29, 2011

### 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?

Last edited: Aug 29, 2011
5. Aug 29, 2011

### Hurkyl

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

6. Aug 29, 2011

### jfy4

I think I got it, let
$$|\psi\rangle = |b\rangle - \frac{ \langle a|b\rangle }{\langle a|a\rangle }|a\rangle$$
then
\begin{align} 0\leq \langle \psi|\psi\rangle &= \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) \\ &=\langle b|b\rangle - \frac{\langle a|b\rangle \langle b|a\rangle}{\langle a|a\rangle} \end{align}
and the result clearly follows.

How does this look?

7. Aug 29, 2011

### lanedance

yep that's looking good to me

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

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