Fourier transform and momenta flow in Feynman diagram Imagine that it is necessary to evaluate some (general) Feynman diagram in a coordinate representation (simply $x$-representation) and the only $x$-dependent part is given by some function $F(x)$. Then, having to re-express it in the momentum representation (simply $p$-representation), one has to perform the Fourier transform, i.e. to calculate $F(p)=\int\mathrm{d}^{4}x\,F(x)e^{-i p\cdot x}$. It is of course assumed that in this particular case there is no choice other than start in the $x$-representation and then move to the $p$-representation. The question is the following. It has been pointed out to me that the minus sign in the exponential corresponds to the convention that the momentum $p$ in the diagram is incoming into the vertex. Is it really true? Because I have never realized it and took the Fourier transform in my calculations to be of the form $F(p)=\int\mathrm{d}^{4}x\,F(x)e^{+i p\cdot x}$ (i.e. with the plus sign) although I have always considered the momentum to be incoming. Therefore - is there really a connection between the sign in the exponential and a convention of the momenta flow? If so, is my convention with the minus sign correct?
 A: Short answer 
you can choose whatever convention you want as long as you use it consistently. 
What does it mean consistently? It means that, chosen a convention, you have to use it to compute ALL the quantities of your theory as well as ALL the solution of the differential equation you may encounter (like the ones you solve to compute propagators).
Long answer
Take a two-point function, where on the external legs the momentum is the same(because of momenta conservation) and we call it $p^\mu$.
Let us consider the following cases:
First case
The convention is the following :


*

*All momenta are incoming with + sign.

*The exponential of the Fourier Transform is with - sign for all incoming momenta.


The two point function in the coordinate space is a function $K(x,y)$ and assuming translation invariance, we can Fourier transform it as
$$
K(x,y)= K(|x-y|)=\int d^4 p \, d^4 q e^{-ipx}e^{-iqy} \, K(p,q) \delta^{(4)}(p+q) = \int d^4 p e^{-i p (x-y)}\,K(p)
$$
where $p$ is the momentum flowing from $x$ and $q$ the momenta flowing from $y$. Both are incoming momenta by convention and then they come with a minus sign in the exponential.
What about the case you cited?
Second case
The convention is the following :


*

*All momenta are incoming with + sign.

*The exponential of the Fourier Transform is with + sign for all incoming momenta.


The two point function then can be trasformed as
$$
K(x,y)= K(|x-y|)=\int d^4 p \, d^4 q e^{+ipx}e^{+iqy} \, K(p,q) \delta^{(4)}(p+q) = \int d^4 p e^{i p (x-y)}\,K'(p)
$$
Then, you can transform $p \rightarrow -p$ and you get
$$
K(x,y)= \int d^4 p e^{-i p (x-y)}\,K'(-p)
$$
and if $K'=K$, you see that the result may be different wrt the previous cases! What is happening?!
In general, $K' \neq K$ and this is because in deriving $K(x,y)$ you solved the differential equation
$$ 
-\int d^4 \,y\, \partial^2 \delta^{(4)}(x-y) K(y,z) = \delta^{(4)}(x-z)
$$
by Fourier transforming and you have to use the same convention also here!
In the first case, $K(p)$ is obtained by transforming with a minus sign in the exponential whereas in the second case $K'(p)$ is obtained using the plus sign. For scalar theory there is no problem since K(p) is an even function(the operator to invert is $\partial^2$ and this cancel any sign difference), so maybe this is why your results have been always correct. 
The lesson to be learnt is then : you can choose whatever convention you want as long as you use it consistently. 
