Categories: Mathematics, Physics.

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 \(\tilde{f}(k)\) 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. Note that we can just as well do it in the opposite direction, which yields an equivalent result:

\[\begin{aligned} \braket*{\tilde{f}}{\tilde{g}} &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k} \\ &= A^2 \int \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big) \Big( \int g(x) \exp(i s k x) \dd{x} \Big) \dd{k} \\ &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x} \\ &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x} \\ &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x} = \frac{2 \pi A^2}{|s|} \braket{f}{g} \end{aligned}\]

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