Why the 1/Sqrt{2 Pi} in the definition of the Fourier transform?

[AxiomOfChoice]

I've never really thought about this before, but today it hit me: Why do we define the Fourier transform of a function to be

$$\hat f(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{ikx} dx$$

What do we lose if we just define it to be

$$\hat f(k) = \int_{-\infty}^\infty f(x) e^{ikx} dx$$

 

[I like Serena]

I've never really thought about this before, but today it hit me: Why do we define the Fourier transform of a function to be

$$\hat f(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{ikx} dx$$

What do we lose if we just define it to be

$$\hat f(k) = \int_{-\infty}^\infty f(x) e^{ikx} dx$$

The pi-factor is there to insure symmetry between the Fourier transform and its inverse.
Without it, the inverse would need a factor of 1/2pi to compensate.

 

[AlephZero]

Actually, some people do put the whole of the 1/2pi either in the FT, or in the IFT, and nothing in the other one.

If you are reading an "old" book or paper, you need to check which convention they are using.

The bottom line is, you have to "lose" the 1/2pi somewhere, to make IFT[FT(f(x)] = f(x).

 

[AxiomOfChoice]

Right. My quantum instructor actually defined

$$\begin{align*} \mathcal F(f) = \hat f(k) & = \int_{-\infty}^\infty f(x) e^{-ikx}dx,\\ \mathcal{F}^{-1}(\hat f) = f(x) & = \frac{1}{2\pi} \int_{-\infty}^\infty \hat f(k) e^{ikx}dk. \end{align*}$$

And it makes sense to me that we want to have

$$\mathcal{F}^{-1}(\mathcal F(f)) = f.$$

But I don't see why the $2\pi$ is necessary to make that work. Can someone just show me the calculation? I can't seem to find it in any of my books...

 

[HallsofIvy]

Essentially, what you are doing is thinking of the functions $e^{ikx}$ as a (Hamel) basis for the infinite dimensional vector space L₂. Of course, it helps if your basis is "ortho-normal". It is easy to show that the functions are "othogonal":
$$\int_{-\infty}^\infty e^{ikx}e^{-imx}dx= 0$$
for $k\ne m$

You need the
[tex]\frac{1}{\sqrt{2\pi}}[/itex]
to "normalize" them.

 

[disregardthat]

Essentially, what you are doing is thinking of the functions $e^{ikx}$ as a (Hamel) basis for the infinite dimensional vector space L<sub>2[/sup]. Of course, it helps if your basis is "ortho-normal". It is easy to show that the functions are "othogonal":
$$\int_{-\infty}^\infty e^{ikx}e^{-imx}dx= 0$$
for $k\ne m$

You need the
[tex]\frac{1}{\sqrt{2\pi}}[/itex]
to "normalize" them.</sub>

_{Are you sure it is a hamel basis for L^2 space?}

 

[AxiomOfChoice]

I'm familiar with the terms "Hamel basis" and "L^2", but as to whether the set of functions of the form $e^{ikx}$ form a basis, I'm not sure.

My question is this: It seems you're saying that the factor of $1/\sqrt{2\pi}$ is needed to normalize these functions. I just don't see this. Doesn't the normalization condition reduce to

$$1 = (e^{ikx},e^{ikx}) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} e^{ikx} dx = \frac{1}{2\pi} \int_{-\infty}^\infty dx$$

And doesn't that integral fail to exist? I mean, doesn't it blow up? Obviously, I've done something wrong here...

 

[AxiomOfChoice]

Also...am I correct in assuming that, when one is talking about $\left\{ e^{ikx} \right\}$ being a Hamel basis for L^2, we're allowing k to vary over all real numbers? Or do we include complex k as well?

 

[I like Serena]

I'm familiar with the terms "Hamel basis" and "L^2", but as to whether the set of functions of the form $e^{ikx}$ form a basis, I'm not sure.

My question is this: It seems you're saying that the factor of $1/\sqrt{2\pi}$ is needed to normalize these functions. I just don't see this. Doesn't the normalization condition reduce to

$$1 = (e^{ikx},e^{ikx}) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} e^{ikx} dx = \frac{1}{2\pi} \int_{-\infty}^\infty dx$$

And doesn't that integral fail to exist? I mean, doesn't it blow up? Obviously, I've done something wrong here...

Actually the integral reduces to the dirac function.
On the wikipedia page: http://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_transform
it is stated that:

$$\int_{-\infty}^\infty 1 \cdot e^{2\pi i x\xi}\,d\xi = \delta(x)$$

The proof is left out here, but you can see the factor 2pi in the exponent.
Due to the substition rule, 2pi becomes part of the result.

[edit]Note that the integral does not really "blow up", because the function only takes an infinite value on an interval that is infinitesimal small.[/tex]