Fourier transform standard practice for physics

Question

I'm very confused about the definition of the inverse Fourier transform between time and frequency spaces. In many places, including Wikipedia, the Fourier transform pairs are defined by:

\begin{align} f(t)=&\int_{-\infty}^\infty\tilde{f}(\omega)e^{i\omega t}\frac{d\omega}{2\pi} \\ \tilde{f}(\omega)=&\int_{-\infty}^\infty f(t)e^{-i\omega t}dt \, . \end{align}

In other sources, such as "Many-body quantum theory in condensed matter physics" by Bruus and Flensberg, or these lecture notes, they are instead defined by:

\begin{align} f(t)=&\int_{-\infty}^{\infty}\tilde{f}(\omega)e^{-i\omega t}\frac{d\omega}{2\pi} \\ \tilde{f}(\omega)=&\int_{-\infty}^\infty f(t)e^{i\omega t}dt \, . \end{align}

Why are there these two different definitions? I thought maybe they mean the same thing, but this would only seem to be the case for $\tilde{f}(\omega)=\tilde{f}(-\omega)$ and I don't see why this should be true in general.

Indeed, I originally noticed this discrepancy when I tried to calculate the impedance of a capacitor using the second definition above, and obtained: $$Z=-\frac{1}{i\omega C} \, ,$$ and when I looked up the solution, found that it is generally accepted to be: $$Z=\frac{1}{i\omega C} \, ,$$ which is what I obtain using the second definition above. I don't understand how these two functions could have the same physical interpretation. That's why I asked here instead of the math stackexchange, because I'm sure they can be defined either way formally, but it seems like if we want to think physically about time and frequency, one must be correct.

Can anybody tell me which one is correct, or explain why they are equivalent?

Answer 1

There are many possible choices regarding the overall scaling coefficients as well as the scaling coefficient converting time and frequency. It is possible to summarize these conventions succinctly using two numbers $a$ and $b$. I use the same notation as used in the Mathematica Fourier Transform function.

We define the Fourier Transform:

$$ \mathcal{FT}_{a,b}[f(t)](\omega) = \sqrt{\frac {|b|}{(2\pi)^{1-a}}}\int_{-\infty}^{+\infty} e^{+i b \omega t} f(t) dt $$

And the inverse Fourier Transform

$$ \mathcal{FT}_{a,b}^{-1}[\tilde{f}(\omega)](t) = \sqrt{\frac{|b|}{(2\pi)^{1+a}}}\int_{-\infty}^{+\infty} e^{-i b \omega t} \tilde{f}(\omega) d\omega $$

Let $$ \tilde{f}_{a,b}(\omega) = \mathcal{FT}_{a,b}[f(t)](\omega) $$ $$ \check{f}_{a,b}(t) = \mathcal{FT}_{a,b}^{-1}[\tilde{f}_{a,b}(\omega)](t) $$

It can be shown via the Fourier inversion theorem that for the classes of functions we care about in physics $\check{f}_{a,b}(t) = f(t)$ for any $a$ and $b$. That is, for these definitions of the Fourier Transform and Inverse Fourier transform the two operations are inverses of eachother.

It's turns out that in the engineering and scientific literature there are many conventions that people choose depending mostly on what they are used to.

The first convention in the OP is $(a,b) = (1,-1)$ which is commonly used in physics, about as commonly as $(a,b) = (1,+1)$ which is the second convention you have shown.

In addition you will also see conventions where $(a,b) = (0,\pm1)$ where the factor of $2\pi$ is split evenly between the transform and inverse transform showing up with a square root.

Furthermore, usually in math or signal processing you will come across the $(a,b) = (0,\pm 2\pi)$ convention in which there is NO prefactor of $2\pi$ on either the transform or the inverse transform but now instead of angular frequency $\omega$ represents a cyclic frequency and a $2\pi$ appears in all of the exponentials.

All of these different conventions have advantages and disadvantages which may make one choice of convention more attractive than another depending on the application. The main point is that in any problem, whichever convention is chosen should be kept the same throughout the whole problem.

To get back to the OP's main question now. In the language set up in this answer the OP is basically asking if it matters whether $b=+1$ or $b=-1$. The short answer is that it does not matter. Either way works and converts the original signal as a function of time into a function of frequency. The difference has to do with how we interpret positive and negative frequencies. Consider $$ f^1(t) = e^{+i\omega_0 t} $$ $$ f^2(t) = e^{-i \omega_0 t} $$

The phasor for the first function rotates counterclockwise in phase space whereas the second rotates clockwise in phasespace.

If we choose the $b=-1$ convention then $\tilde{f}^1_{1,-1}(\omega)$ will have a nonzero contribution at $+\omega_0$ whereas $\tilde{f}^2_{1,-1}(\omega)$ will have a nonzero contribution at $-\omega_0$. We might say $f^1$ is a positive frequency signal while $f^2$ is negative.

However, if we choose $b=+1$ then everything reverses. $\tilde{f}^1_{1,+1}(\omega)$ will have a nonzero contribution at $-\omega_0$ while $\tilde{f}^2_{1,+1}(\omega)$ will have a contribution at $+\omega_0$. now $f^1$ is negative frequency and $f^2$ is positive frequency!

Thuse we see that both $b=+1$ and $b=-1$ give answer that we can interpret as frequencies with the only difference between the two being what we call positive and negative frequencies. As a note I personally prefer $(a,b)=(1,+1)$ because it makes the formula for the Fourier transform (which I use more often than the inverse transform) as simple as possible. No prefactor and no minus sign in the exponent.

edit: As you have pointed out sometimes these signs can have a substantial effect on some physical quantity such as reversing the sign (inverting the phase) of the complex impedance of a capacitor. Unfortunately this is something we just have to deal with and try to be consistent with our own conventions and those used by the references we consult. Of course you will find both conventions give the same answer for a real measurable quantity such as $V(t)$ across the resistor.

Answer 2

I think a better way to memorize this is to remember the definition with no normalization constants, that is, $$ \hat f(\mathbf k) =\int_{\mathbb R^n} f(\mathbf x)e^{-i2\pi \mathbf k\cdot \mathbf x} \mathrm d\mathbf x,\\ f(\mathbf x) =\int_{\mathbb R^n} \hat f(\mathbf k)e^{i2\pi \mathbf x\cdot \mathbf k} \mathrm d\mathbf k. $$ From this, it is easy to show that with arbitrary constants $A$ and $B$, we can equivalently define $$ \hat f(\mathbf k) =A^n \int_{\mathbb R^n} f(\mathbf x)e^{-i2\pi AB\mathbf k\cdot \mathbf x} \mathrm d\mathbf x,\\ f(\mathbf x) =B^n \int_{\mathbb R^n} \hat f(\mathbf k)e^{i2\pi AB\mathbf x\cdot \mathbf k} \mathrm d\mathbf k. $$

Answer 3

I really think that this is a very good question. Both, conventions are correct and here I attempt to explain why. So, the Fourier transform is derived from the basic Fourier series. For aperiodic signals we can not define a legitimate real value of periodicity and go for the transform by allowing the period to extend to infinity.

The Fourier series is defined as $$f(x)=\sum_{0}^{+\infty}\left\{A_k\cos{\frac{2\pi kx}{x_0}}+B_k\sin{\frac{2\pi kx}{x_0}}\right\}\\ A_0=\frac{1}{x_0}\int_{x_0}f(\tau)\mathrm{d}\tau, \quad A_k=\frac{2}{x_0}\int_{x_0}f(\tau)\cos{\frac{2\pi k\tau}{x_0}}\mathrm{d}\tau, \quad B_k=\frac{2}{x_0}\int_{x_0}f(\tau)\sin{\frac{2\pi k\tau}{x_0}}\mathrm{d}\tau$$ Converting sine and cosine of the first equation into exponential forms, we get $$f(x)=\sum_{k=0}^{\infty}\left\{\left(\frac{A_k-iB_k}{2}e^{i2\pi kx/x_0}\right)+\left(\frac{A_k+iB_k}{2}e^{-i2\pi kx/x_0}\right)\right\}=\sum_{k=0}^{\infty}\left\{\left(C_k e^{i2\pi kx/x_0}\right)+\left(D_k e^{-i2\pi kx/x_0}\right)\right\}$$

Now, it is understandable that $C_{-k}=D_k$ So, I can write $$f(x)=\sum_{k=-\infty}^{+\infty}C_k e^{i2\pi kx/x_0}=\sum_{k=-\infty}^{+\infty}D_k e^{-i2\pi kx/x_0}$$ So, in my opinion that is where the road divergence starts from. When I started to derive it from the first ($C_k$) convention, I ended up with $$F(\omega)=\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}\mathrm{d}x$$ And when I used the second convention($D_k$), I ended up deriving $$F(\omega)=\int_{-\infty}^{+\infty}f(x)e^{i\omega x}\mathrm{d}x$$

Answer 4

I like this summary of the range of definitions on John D Cook's blog:

https://www.johndcook.com/blog/fourier-theorems/

Answer 5

For Bruus and Flensberg which deals with quantum theory, I will try to describe the motivation for the sign convention in the exponent:

For position-momentum Fourier transform, it follows from the de Broglie's relation. $$ -i\hbar \frac{d}{d x} \leftrightarrow p = \hbar k $$

Therefore, the positional Fourier transform should be such that its action on the derivative is $\frac{d}{d x} \to ik$ $$ \tilde{f}(k) = \int e^{-ikx} f(x) dx $$

For time-energy Fourier transform, the Schrodinger equation and Planck's relation give a natural choice of sign. $$ i\hbar \frac{d}{dt} \leftrightarrow E = \hbar \omega $$

Therefore, the temporal Fourier transform should be such that its action on the derivative is $\frac{d}{d t} \to -i\omega$ $$ \tilde{f}(\omega) = \int e^{i\omega t} f(t) dx $$

By adopting these conventions, it becomes easier to interpret our formulas consistently. The convention for where to put the $2\pi$ factor is a separate issue.