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An inner product must exist on the set of all functions in Hilbert space

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]\int {{f^*}(x)g(x) \cdot dx} [/itex] is an inner product on the set of square-integrable complex functions.

    2. Relevant equations
    Schwarz inequality:
    [tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]

    3. The attempt at a solution

    Rephrase what we want to prove: see if:
    [itex]\int {{f^*}(x)g(x) \cdot dx} < \infty [/itex]

    ...is true.

    Since we are considering the set of square-integrable-functions: “f” and “g” are in the set of square-integrable functions: they are elements of Hilbert space:
    [tex]\left\{ {f(x),g(x)} \right\} \in {L^2}[/tex]

    This means the following integrals exist:
    [tex]\int_{ - \infty }^{ + \infty } {{f^*}(x)f(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {f(x)} \right|}^2} \cdot dx} < \infty {\rm{ }}\int_{ - \infty }^{ + \infty } {{g^*}(x)g(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {g(x)} \right|}^2} \cdot dx} < \infty [/tex]

    In turn: this gaurantees:
    [tex]\sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } < \infty [/tex]

    Schwarz inequality says:
    [tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]

    Together: the previous two equations require:
    [tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \infty [/tex]

    Another inequality is that:
    [tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \int {\left| {{f^*}(x)g(x)} \right| \cdot dx} [/tex]

    …which actually ruins what we’re trying to prove. Gah. Well, it doesn't counter-prove it, but I've obviously used the wrong inequality. Somehow it must be that:
    [tex]\int {\left| {{f^*}(x)g(x)} \right| \cdot dx} \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]

    but i'm not sure how to get that. Triangle inequality? Perhaps I am staring too hard, but it doesn't seem so?
  2. jcsd
  3. Aug 25, 2010 #2


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    Homework Helper

    Are you trying to prove that [itex]\int f^*(x)g(x) \mathrm{d}x < \infty[/itex], or that [itex]\int \lvert f^*(x)g(x)\rvert \mathrm{d}x < \infty[/itex]?
  4. Aug 25, 2010 #3
    Am trying to prove [itex]\int {{f^*}(x)g(x) \cdot dx} < \infty [/itex].

    I think we're valid up to the step:
    [tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \infty [/tex]

    If you think of something, let me know. I'm off quantum, and practicing electro-magnetism problems for Physics GRE...
  5. Aug 25, 2010 #4


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    OK, that's what I thought... now can you think of circumstances in which [itex]\lvert\int f^*(x)g(x)\mathrm{d}x\rvert[/itex] is convergent but [itex]\int f^*(x)g(x)\mathrm{d}x[/itex] is not?
  6. Aug 25, 2010 #5
    Well....gosh, now that I think about it, I can't. It seems sensible, so dare I say:

    [itex]\lvert\int f^*(x)g(x)\mathrm{d}x\rvert>\int f^*(x)g(x)\mathrm{d}x[/itex]

    Oh yeah, the absolute-value is that of just a number, so that's as true as |-6| > -6, and > only becomes >= if the number is 0.

    Duh. Thanks so much. ...err.. I think. I think it's obvious. I'm way too cautious after a summer's research of "oops"-es coming from an inattentiveness to subtleties.

    Now let's have some fun by just haphazardly going way off topic : )

    Do null kets demand that case, where <generic bra|0> = 0, where |0> is the null ket? Like in vacuum-states?
  7. Aug 25, 2010 #6


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    That's it, pretty much :wink: If you want something more like a proper justification, start by expressing the value of the integral in complex polar form, re.

    By the way, it has to be |A| ≥ A, in case A is a positive number.

    Now about those "null kets": the inner product doesn't necessarily have to be zero. For instance, the normalized ground state of a harmonic oscillator satisfies [itex]\langle 0 \vert 0 \rangle = 1[/itex]. Something similar applies for a vacuum state in QFT; it has a nonzero norm. But then again, I don't think it really makes sense to call it a "null ket" since there's nothing "null" about it. Remember that the quantum number that appears in the ket is really just a label for the state, and the label is kind of arbitrary. The ground state of a SHO could just as well have been labeled [itex]\lvert 1\rangle[/itex] (of course you'd have to change the formulas for the energy levels etc.), or [itex]\lvert \psi_0\rangle[/itex], or [itex]\lvert G\rangle[/itex].

    Anyway, the point is that just because you write a particular ket as [itex]\lvert 0 \rangle[/itex], doesn't mean the associated function is actually equal to zero.
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