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Homework Help: Schwartz inequality proof over complex

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider any two vectors, [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex]. 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 [itex]\mathbb{C}[/itex]...

    3. 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
    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 [itex]c[/itex] 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. jcsd
  3. Aug 28, 2011 #2


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

    [tex] 0\leq \langle a|a\rangle + c\langle a|b\rangle +c^2\langle b|b\rangle [/tex]

    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
    [tex] |b\rangle = \langle a|b \rangle|a \rangle + |z\rangle [/tex]
  4. Aug 29, 2011 #3
    Why do you think c must be complex? All the inner products you are using result in real numbers.
  5. Aug 29, 2011 #4


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    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
    [tex] 0 \leq \langle a|a\rangle + 2\Re[c\langle a|b\rangle ] +|c|^2\langle b|b\rangle [/tex]

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

    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
  6. Aug 29, 2011 #5


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    If all else fails, split things into real and imaginary parts.
  7. Aug 29, 2011 #6
    I think I got it, let
    |\psi\rangle = |b\rangle - \frac{ \langle a|b\rangle }{\langle a|a\rangle }|a\rangle
    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}
    and the result clearly follows.

    How does this look?
  8. Aug 29, 2011 #7


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    yep that's looking good to me

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