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+---
+title: "Parseval's theorem"
+date: 2021-02-22
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
+and 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 FT's definition:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \Inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            \\
+            \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-fourier">Proof.</label>
+We insert the inverse FT into the defintion of the inner product:
+
+$$\begin{aligned}
+    \Inprod{f}{g}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
+    \\
+    &= B^2 \int
+    \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
+    \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
+    \dd{x}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
+    \\
+    &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
+    = \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
+\end{aligned}$$
+
+Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Note that we can equally well do this proof in the opposite direction,
+which yields an equivalent result:
+
+$$\begin{aligned}
+    \inprod{\tilde{f}}{\tilde{g}}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+    \\
+    &= A^2 \int
+    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+    \dd{k}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+    = \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
+\end{aligned}$$
+</div>
+</div>
+
+For this reason, physicists like to define the Fourier transform
+with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+conserves the functions' normalization.
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
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