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

It just means that you have chosen to label that particular state as the "ground state" (etc.).In summary, we want to show that \int {f^*}(x)g(x) \cdot dx is an inner product on the set of square-integrable complex functions. This requires that the integrals of both f* and g are finite, which is guaranteed in the set of square-integrable functions. We can use the Schwarz inequality to show that the absolute value of the integral is less than or equal to the square root of the product of the integrals of the absolute values of f and g. It is possible that the absolute value of the integral is greater than the integral of the absolute value of the product,
  • #1
bjnartowt
284
3

Homework Statement


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


Homework 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]


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?
 
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  • #2
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]?
 
  • #3
diazona said:
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]?

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...
 
  • #4
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?
 
  • #5
diazona said:
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?

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?
 
  • #6
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.
 

1. What is an inner product?

An inner product is a mathematical operation that takes two vectors and produces a scalar value. It is similar to the dot product, but it is defined in a more general setting and can be defined for various types of vectors, including functions in Hilbert space.

2. What is a Hilbert space?

A Hilbert space is a mathematical concept that represents a complete vector space. It has many properties that make it useful for studying functions, such as being able to define an inner product and having a notion of distance and convergence.

3. Why must an inner product exist on the set of all functions in Hilbert space?

An inner product is necessary in Hilbert space because it allows us to define the notion of orthogonality and to measure the angle between two vectors. This is important for many applications, including signal processing and quantum mechanics.

4. How is an inner product defined in Hilbert space?

An inner product in Hilbert space is defined as a function that takes two vectors and produces a scalar value. It must satisfy certain properties, such as being linear in the first argument and conjugate symmetric. It can also be defined using the concept of a norm.

5. What are some examples of inner products in Hilbert space?

Some examples of inner products in Hilbert space include the standard inner product in Euclidean space, which is defined as the dot product of two vectors, and the L2 inner product, which is commonly used in signal processing and is defined as the integral of the product of two functions over a given interval.

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