$2\pi$ and Feynman Rules I notice a $2\pi$ term in the $\delta$-function when trying to construct an amplitude using the Feynman Rules. The $2\pi$ also appears as an integration measure to enforce normalisation in the phase space, what is the origin of this $2\pi$ term? Where does it come from? 
 A: I suppose you talk about the standard $2\pi$ that appears in the rules for Fourier transform. The factor of $2\pi$ or $1/2\pi$ or two factors of $1/\sqrt{2\pi}$ have to appear "somewhere" in the Fourier transform rules because this is what the mathematics implies. At any rate, if this is your question, it is a mathematical question and you may learn it in some basic enough crash courses on calculus, perhaps even at

https://en.wikipedia.org/wiki/Fourier_transform

The number $k=2\pi$ is the minimum positive number for which $\exp(i k x) = 1$ for $x=1$. That's because $\exp(2\pi i)=1$ – Gauss' and Feynman's most favorite identity in all of mathematics. A schoolkid knows this constant $2\pi$ that appears in the exponent as the circumference of the unit circle. That's why $2\pi$ is the natural spacing in the space of frequencies and the spacing is needed for a translation between "a sum over angular frequencies or wave numbers" to an integral over the same variables. (This conversion is easy if you think about the Riemann integral $\int f(x)dx=\lim \sum f(x)\Delta x$ etc.: the $\Delta x$ factors are the spacings.) That's how the $2\pi$ appears.
Equivalently, if you want the $\delta$-function, the $2\pi$ appears in the identity
$$ \int_{-\infty}^\infty \exp(ikx) dx = 2\pi \delta(k) $$
To prove that, you may regulate the integral to the interval $(-L/2,+L/2)$. To make the exponential function periodic on that interval, $k$ has to be a multiple of $2\pi/L$, and the integral will only be zero if $k=0$. So the integral is $L$ times the "Kronecker delta" imposing $k=0$ and because the spacing between allowed $k$ is $2\pi/L$, the integral i.e. $L$ times the Kronecker delta may be converted to $2\pi/L$ times $L$ times the delta-function which is $2\pi$ times the delta-function because the $L$ factors cancel.
Various multiples of $\pi$ appear at tons of other places in quantum mechanics and quantum field theory and physics in general for many closely related reasons. For example, the evaluation of Feynman's diagrams leads to the volume of a 4-sphere or related integrals and the volumes of unit $n$-spheres are rational multiplies of a power of $\pi$, too. (That's, for example, why it's natural to include $1/4\pi$ from the surface of the sphere in the Coulomb's law. The initial example with $2\pi$ is a special example as well because the aforementioned unit circle is a 1-dimensional sphere, too.) There's just way too much maths here to prepare one for all occurrences of $\pi$ in physics. This constant appears pretty much everywhere in physics and one can't really do physics without knowing many of these mathematical identities involving $\pi$.
A: When beginning by calculate transition amplitudes in position space, and taking the Fourier transform of these amplitudes, to get the transition amplitude in momentum space, you get terms (for instance in a $2 \to 2$ interaction) in $\int d^4v e ^{-i(p_1+p_2-p_3-p_4)v}$, and this is equals to $(2\pi)^4 \delta^4(p_1+p_2-p_3-p_4)$
An example of such an amplitude in position space is (for instance in a $\phi^4$ theory):
$A (x_1,x_2,x_3,x_4) = \int d^4w ~d^4z ~\delta(x_1-w)~\delta(x_2-w) [D(w-z)]^2~\delta(z-x_3)\delta(z-x_4)$
Here $D$ is the propagator in position space.
When taking the Fourier transform $A (p_1,p_2,p_3,p_4) = \int dx_1 dx_2 dx_3dx_4 A (x_1,x_2,x_3,x_4) e^{i (p_1x_1 +p_2x_2+p_3x_3+p_4x_4)}$, 
you get a term $e ^{i(w (p_1+p_2) - z(p_3+p_4))}$, which you may write $e ^{i w (p_1+p_2 -p_3-p_4)} e ^{i(w - z) (p_3+p_4)}$, the first term gives you the $(2\pi)^4 \delta^4(p_1+p_2-p_3-p_4)$ term, while the second term gives the Fourier transform of $[D(x)]^2$, at the point $p_3+p_4= p_1+p_2$, which is  a convolution $\int dp \tilde D(p) \tilde D(p_1+p_2 -p)$, where $\tilde D(p)=\frac{1}{p^2}$
A: So the integral is $L$ times the "Kronecker delta" imposing $k=0$ and because the spacing between allowed $k$ is $2\pi/L$, the integral i.e. $L$ times the Kronecker delta may be converted to $2\pi/L$ times $L$ times the delta-function which is $2\pi$ times the delta-function because the $L$ factors cancel.
