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

[bjnartowt]

Homework Statement

Show that \int {{f^*}(x)g(x) \cdot dx} is an inner product on the set of square-integrable complex functions.

Homework Equations

Schwarz inequality:
\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} }

The Attempt at a Solution

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

...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:
\left\{ {f(x),g(x)} \right\} \in {L^2}

This means the following integrals exist:
\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

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

Schwarz inequality says:
\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} }

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

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

…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:
\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} }

but I'm not sure how to get that. Triangle inequality? Perhaps I am staring too hard, but it doesn't seem so?

 

[diazona]

Are you trying to prove that \int f^*(x)g(x) \mathrm{d}x < \infty, or that \int \lvert f^*(x)g(x)\rvert \mathrm{d}x < \infty?

 

[bjnartowt]

Are you trying to prove that \int f^*(x)g(x) \mathrm{d}x < \infty, or that \int \lvert f^*(x)g(x)\rvert \mathrm{d}x < \infty?

Am trying to prove \int {{f^*}(x)g(x) \cdot dx} < \infty.

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


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

 

[diazona]

OK, that's what I thought... now can you think of circumstances in which \lvert\int f^*(x)g(x)\mathrm{d}x\rvert is convergent but \int f^*(x)g(x)\mathrm{d}x is not?

 

[bjnartowt]

OK, that's what I thought... now can you think of circumstances in which \lvert\int f^*(x)g(x)\mathrm{d}x\rvert is convergent but \int f^*(x)g(x)\mathrm{d}x is not?

Well...gosh, now that I think about it, I can't. It seems sensible, so dare I say:

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

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?

 

[diazona]

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

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 \langle 0 \vert 0 \rangle = 1. 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 \lvert 1\rangle (of course you'd have to change the formulas for the energy levels etc.), or \lvert \psi_0\rangle, or \lvert G\rangle.

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