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+---
+title: "Parseval's theorem"
+firstLetter: "P"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:36:44+01:00
+draft: false
+markup: pandoc
+---
+
+# Parseval's theorem
+
+**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to the
+inner product of their [Fourier transforms](/know/concept/fourier-transform/)
+$\tilde{f}(k)$ and $\tilde{g}(k)$.
+There are two equivalent ways of stating it,
+where $A$, $B$, and $s$ are constants from the Fourier transform's definition:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
+    }
+    \\
+    \boxed{
+        \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+    }
+\end{aligned}$$
+
+For this reason, physicists like to define their Fourier transform
+with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
+conserves the total probability (quantum mechanics) or the total energy
+(optics).
+
+To prove this, we insert the inverse FT into the inner product
+definition:
+
+$$\begin{aligned}
+    \braket{f}{g}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
+    \\
+    &= B^2 \int
+    \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
+    \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
+    \dd{x}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
+    \\
+    &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
+    = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
+\end{aligned}$$
+
+Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Note that we can just as well do it in the opposite direction,
+which yields an equivalent result:
+
+$$\begin{aligned}
+    \braket*{\tilde{f}}{\tilde{g}}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+    \\
+    &= A^2 \int
+    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+    \dd{k}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+    = \frac{2 \pi A^2}{|s|} \braket{f}{g}
+\end{aligned}$$