Categories: Mathematics, Physics.

# Parseval’s theorem

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 $\tilde{f}(k)$ 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}

We insert the inverse FT into the definition 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') \: e^{i s k' x} \dd{k'} \Big) \Big( \int \tilde{g}(k) \: e^{- i s k x} \dd{k} \Big) \dd{x} \\ &= 2 \pi B^2 \iint \tilde{f}^*(k') \: \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty e^{i s x (k' - k)} \dd{x} \Big) \dd{k'} \dd{k} \\ &= 2 \pi B^2 \iint \tilde{f}^*(k') \: \tilde{g}(k) \: \delta\big(s (k' \!-\! k)\big) \dd{k'} \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. 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') \: e^{- i s k x'} \dd{x'} \Big) \Big( \int g(x) \: e^{i s k x} \dd{x} \Big) \dd{k} \\ &= 2 \pi A^2 \iint f^*(x') \: g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty e^{i s k (x - x')} \dd{k} \Big) \dd{x'} \dd{x} \\ &= 2 \pi A^2 \iint f^*(x') \: g(x) \: \delta\big(s (x \!-\! x')\big) \dd{x'} \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}

For this reason, physicists like to define the Fourier transform with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely preserves the functions’ normalization.

## References

1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.