Fourier transforms
Hilbert space structure
\begin{gather*}
\int \dd{\omega} e^{i\omega(t_1 - t_2)} = 2 \pi \delta(t_1 - t_2) \\
\int_0^L \dd{x} e^{2 \pi i (n-m) x / L} = L\delta_{nm}
\end{gather*}
Derivation
The first identity we verify by integrating against test function $f(t_1,t_2)$ for $\omega \in [-\Omega, \Omega]$\begin{align*}
\int \dd{\Delta t}\dd{\omega} f(t_2 + \Delta t,t_2) e^{i \omega \Delta t}
&= \int \dd{\Delta t} f(t_2 + \Delta t, t_2)
\eval{\frac{e^{i \omega \Delta t}}{i \Delta t}}_{\omega = -\Omega}^{\Omega} \\
&= \int \dd{\Delta t} f(t_2 + \Delta t, t_2) \frac{e^{i \Omega \Delta t}- e^{-i\Omega \Delta t}}{i \Delta t} \\
&= 2 \pi f(t_2, t_2)
\end{align*}
where the last equality follows from evaluating each term via the residue
theorem, closing the contour in the upper and lower half-plane
respectively. This requires $f(t_2+\Delta t, t_2)$ to grow more slowly
than $e^{\Omega \abs{\Im \Delta t}}$ over the arc at infinity, meaning $f$
is band-limited: for generic test functions $f$, we therefore take $\Omega
\to \infty$.The second identity can be shown by direct integration, for now assuming $n,m$ may be real.
\begin{align*} \int_0^L \dd{x} e^{2 \pi i (n-m) x / L} = \frac{L}{2 \pi
i (n-m)}\qty[e^{2 \pi i (n-m)} - 1] = L\delta_{nm}\end{align*}
For the last equality, we assume $n,m$ are integers. When $n \neq m$,
$n-m$ is an integer and the expression vanishes while when $n - m = 0$ by
taking the regular limit $n \to m$.Space-time sign convention
\begin{align*}
f(\vb x, t) &= \int \dd[d]{\vb k}\dd{\omega} f(\vb k, \omega)e^{i \vb k \cdot \vb x - i \omega t} \\
f(\vb k, \omega) &= \int \frac{\dd[d]{\vb x}\dd{t}}{(2\pi)^{d+1}} f(\vb x, t)e^{-i \vb k \cdot \vb x + i \omega t} \\
\end{align*}
Explanation
The relative sign between $\vb k \cdot \vb x$ and $\omega$ is chosen to be "compatible" with the Schrodinger equation $i \pdv{f}{t} = -\pdv[2]{f}{x}$, yielding the de Broglie dispersion $\omega = k^2$. Ultimately the sign convention comes from the sign convention chosen for the Schrodinger equation.Normalization convention
\begin{align*}
f(t) &= \int \dd{\omega} f(\omega) e^{-i \omega t} \\
f(\omega) &= \int \frac{\dd{t}}{2 \pi} f(t)e^{i \omega t} \\
\end{align*}
Sokhotski–Plemelj theorem
\begin{align*}
\int_0^\infty \dd{t} e^{-i \omega t} = \pi \delta(\omega) - \, \PV \frac{i}{\omega}
\end{align*}
Derivation
Identical to derivation of the delta function identity $\int \dd{t} e^{i \omega t} = 2 \pi \delta(\omega)$ except here the lower boundary term at $t = 0$ gives the integral against $1/\omega$. The indented contour used to evaluate the integral implies the Cauchy principal value. This can be brought to the usual form of Sokhotsi–Plemelj by explicitly parameterizing the shifted contour with $\omega \to \omega - i \epsilon$ where formally\[
\int_0^\infty \dd{t} e^{-i(\omega - i \epsilon)t} = \frac{-i}{\omega - i \epsilon}
\]
Translation symmetries
If $f(x,y) = f(x+a, y+a)$ for any $a$,
\begin{align*}
f(k_x,k_y) &\propto \delta(k_x+k_y) \\
f(x,y) &= f(x-y)
\end{align*}
Derivation
\begin{align*}
f(x,y) &= \int \dd{k_x} \dd{k_y} f(k_x, k_x) e^{i k_x x + i k_y y} \\
&= \int \dd{k_x} \dd{k_y} e^{ia(k_x + k_y)} f(k_x, k_y) e^{i k_x x + i k_y y}
\end{align*}
To be true for any $a$ we must have $f(k_x, k_y) = \delta(k_x + k_y) f_1(k_x, k_y)$ therefore
\begin{align*}
f(x,y) = \int \dd{k_x} f_1(k_x, -k_x) e^{i k_x(x-y)} \\
\end{align*}
where the latter is the (inverse) Fourier transform of a function $f(x,y) = f(x-y)$.
Real-valued functions
If $f(x)$ is real,
\[ f(k) = f^*(-k) \]
Derivation
As $f(x)$ is real, $f^*(x) = f(x)$ which implies\[\int \dd{k} f(k) e^{ikx} = \int \dd{k} f^*(k) e^{-ikx}\]
Fourier series
If $f$ is $L$-periodic $f(x) = f(x+L)$, then
\[f(k) = \sum_{n=-\infty}^\infty c_k\delta(k-2\pi n / L)\]
where $c_k$ are the Fourier coefficients. They satisfy
\[ c_n = \frac{1}{L} \int_0^L \dd{x} f(x) e^{-2\pi i m x / L} \]
Derivation
\begin{align*} f(x) &= \int \dd{k} f(k) e^{ikx} \\ &= \int \dd{k}
e^{ikL} f(k) e^{ikx} \end{align*}
For $(e^{ikL} - 1)f(k) = 0$ then $f(k)$ must be supported on $k =
\frac{2\pi n}{L}$ for integer $n$. To retrieve the Fourier coefficients,
note\begin{align*}
f(x) &= \int \dd{k} \sum_n c_n \delta(x - 2 \pi n / L) e^{ikx} \\
&= \sum_n c_n e^{2\pi i n x / L}
\end{align*}
and one can integrate against $e^{-2\pi i m x / L}$ using the Hilbert
space inner product.Dirac comb
\[ \sum_m e^{2 \pi i m x / a} = a\sum_m \delta(x - ma)\]
Derivation
\begin{align*}
\sum_m e^{2 \pi i m x / a}
&= \sum_{m=0}^\infty e^{2\pi i m x /a}
+ \sum_{m=1}^\infty e^{-2\pi i m x / a} \\
&= \frac{1}{1 - e^{2\pi i (x + i0)/a}} - \frac{1}{1- e^{2\pi i (x-i0)/a}}
\end{align*}
where we use $x \pm i0$ as bookkeeping notation for the contour
deformation to analytically continue the sums. The poles are located where
$x = na$ for integer $n$. Therefore, integration against a test function
$f(x)$ is a sum of two contours, one from left-to-right in the upper half
plane and one from right-to-left in the lower half plane. Their sum
becomes a clockwise contour $\mathcal{C}$ which encircles all poles, which
can be evaluated with residue theorem, yielding\begin{align*}
\int_{\mathcal C} \dd{x} \frac{f(x)}{1 - e^{2\pi i x/a}} = a\sum_{m} f(ma)
\end{align*}
which is the definition of the delta function.Lattice transform
If $f(x) = a \sum_{n} f_n\delta(x-na)$, then
\[ f_n = \int_{-\pi/a}^{\pi/a} \dd{k} f(k) e^{ikx}\]
where $k \in [-\pi/a, \pi/a]$ is referred to as the (first) Brillouin zone.
Derivation
We directly compute\begin{align*}
f(k) &= \int \frac{\dd{x}}{2\pi} \sum_n f_n a
\delta(x - na) e^{-ikx} \\
&= \frac{a}{2\pi} \sum_n f_n e^{-ikna}
\end{align*}
indicating that $f_n$ are formally the Fourier coefficients of $f(k)$. The
Brillouin zone is therefore just the dual of a real-space periodic domain.