Why is the Fourier Transform in QFT Divided by (2pi)?

[captain]

is Fourier analysis in qft just used for going from a position wavefunction to a wavefunction described by the wave vector (k)? also why is the integral divided by [2(pi)]^n where n is the number of dimensions and how do you know when to divide the integral by the 2(pi) factor or not?

 

[olgranpappy]

why is the integral divided by [2(pi)]^n where n is the number of dimensions...

It is a convenient convention. There is no physics there.

 

[olgranpappy]

is Fourier analysis in qft just used for...

Yes. Fourier analysis is always "just used for" the same purpose.

 

[captain]

It is a convenient convention. There is no physics there.

i know n goes up to 4 but i was too lazy to say that.

 

[olgranpappy]

i know n goes up to 4 but i was too lazy to say that.

n is whatever I want it to be...

 

[dextercioby]

Plus that "n" is not necessary natural, think about dimensional regularization of divergent intergrals occurring in loop contributions.

 

[captain]

but why do you need to divide by 2pi to the power of the number of dimensions you are dealing with in the integral? is it because of the whle position and frequency business. then why do you the same for going from position to momentum?

 

[Nesk]

but why do you need to divide by 2pi to the power of the number of dimensions you are dealing with in the integral? is it because of the whle position and frequency business. then why do you the same for going from position to momentum?

It is because the integral leaves an extra factor for which one accounts by dividing by (2pi)^n. It makes the direct and inverse transforms more symmetric, and thus more appealing and easy to remember. Since the mathematics used do not care about what physical quantities are being dealt with, the method is the same regardles of frequency/time or position/momentum transforms.

 

[captain]

It is because the integral leaves an extra factor for which one accounts by dividing by (2pi)^n. It makes the direct and inverse transforms more symmetric, and thus more appealing and easy to remember. Since the mathematics used do not care about what physical quantities are being dealt with, the method is the same regardles of frequency/time or position/momentum transforms.

if you want to go back from from momentum to position after doing a Fourier transform from position to momentum do you still have to divide by (2pi)^n or has that factor already been taken care during the first Fourier transform from position to momentum? does it matter in the order from which variable you go to (like from position to momentum, frequency to momentum, etc.)?

 

[Nesk]

if you want to go back from from momentum to position after doing a Fourier transform from position to momentum do you still have to divide by (2pi)^n or has that factor already been taken care during the first Fourier transform from position to momentum? does it matter in the order from which variable you go to (like from position to momentum, frequency to momentum, etc.)?

The factor must be inlcuded in both direct and inverse Fourier transform.

 

[captain]

how come in math Fourier analysis book it shows to divide by a factor of
(2pi)^(1/2), when in qft it shows to divide by (2pi)^n?

 

[olgranpappy]

It's just convenient to put all the two pi's on one side of the Fourier transform with no square roots.

Math people like the symmetric Fourier transform (with \sqrt{(2 pi)^n} in both transforms) but physicists prefer the non-symmetric version with all the (2 pi)'s in the momentum measure. For example, in one dimension:

Math
[tex]
\tilde f(p) = \int\frac{dx}{\sqrt{2\pi}}f(x)e^{-ipx}
[/tex]
[tex]
f(x)=\int\frac{dp}{\sqrt{2\pi}}\tilde f(p)e^{ipx}
[/tex]

vs. Physics
[tex]
\tilde f(p)=\int dx f(x)e^{-ipx}
[/tex]
[tex]
f(x)=\int \frac{dp}{2\pi}\tilde f(p)e^{ipx}\;.
[/tex]

Again, it is just a convention.

...also, this really has nothing to do in particular with "qft" so I don't know why you keep bring that up...