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(Algebra) Quantum Theory - Cauchy-Schwartz inequality proof

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Given two arbitrary vectors [itex]|\phi_{1}\rangle[/itex] and [itex]|\phi_{2}\rangle[/itex] belonging to the inner product space [itex]\mathcal{H}[/itex], the Cauchy-Schwartz inequality states that:

    [itex]|\langle\phi_{1}|\phi_{2}\rangle|^{2} \leq \langle\phi_{1}|\phi_{1}\rangle \langle\phi_{2}|\phi_{2}\rangle[/itex].

    Consider [itex]|\Psi\rangle = |\phi_{1}\rangle + \lambda|\phi_{2}\rangle[/itex]

    where [itex]\lambda[/itex] is a complex number that can be written as [itex]\lambda = a + ib[/itex].

    a) Write an expression for [itex]\langle\Psi|\Psi\rangle \geq 0[/itex] as a function of [itex]\lambda[/itex] then rewrite as a function of a and b ([itex]f(a,b)[/itex]).

    b) Show that the value of [itex]\lambda[/itex] that minimises [itex]\langle\Psi|\Psi\rangle[/itex] is:

    [itex]\lambda_{min} = -\frac{\langle\phi_{2}|\phi_{1}\rangle}{\langle\phi_{2}|\phi_{2}\rangle}[/itex].

    Hint: Compute the derivatives of [itex]f(a,b)[/itex] wrt a and b, solve these to get [itex]a_{min}[/itex] and [itex]b_{min}[/itex] and then compute [itex]\lambda_{min}[/itex].

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I get [itex]\langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + a(\langle\phi_{1}|\phi_{2}\rangle + \langle\phi_{2}|\phi_{1}\rangle) + ib(\langle\phi_{1}|\phi_{2}\rangle - \langle\phi_{2}\phi_{1}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle = f(a,b)[/itex]. However I can only show:

    [itex]\lambda_{min} = -\frac{\langle\phi_{1}|\phi_{2}\rangle}{\langle\phi_{2}|\phi_{2}\rangle}[/itex],

    by combining Re and Im parts in [itex]f(a,b)[/itex] as follows (and finding the relevant derivatives etc.):

    [itex]\langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + 2a\textrm{Re}(\langle\phi_{1}|\phi_{2}\rangle) + 2b\textrm{Im}(\langle\phi_{1}|\phi_{2}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle[/itex].

    This is the only way I understand how to do it, however, in the solutions for this problem the collection of Re and Im parts is done as follows which I don't understand (in particular the imaginary part):

    [itex]\langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + 2a\textrm{Re}(\langle\phi_{2}|\phi_{1}\rangle) + 2b\textrm{Im}(\langle\phi_{2}|\phi_{1}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle[/itex]

    Thank you
     
  2. jcsd
  3. Apr 13, 2015 #2
    Anyone? Parts 1) and 2) can pretty much be ignored they just provide context for the problem.. I think ultimately it's just a complex number/conjugation question which I haven't understood properly.
     
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