Categories: Mathematics, Physics.

Parseval’s theorem

Parseval’s theorem is a relation between the inner product of two functions f(x)f(x) and g(x)g(x), and the inner product of their Fourier transforms f~(k)\tilde{f}(k) and g~(k)\tilde{g}(k). There are two equivalent ways of stating it, where AA, BB, and ss are constants from the FT’s definition:

⟨f(x)∣g(x)⟩=2πB2∣s∣⟨f~(k)∣g~(k)⟩⟨f~(k)∣g~(k)⟩=2πA2∣s∣⟨f(x)∣g(x)⟩\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:

⟨f∣g⟩=∫−∞∞(F^−1{f~(k)})∗ F^−1{g~(k)}dx=B2∫(∫f~∗(k′) eisk′xdk′)(∫g~(k) e−iskxdk)dx=2πB2∬f~∗(k′) g~(k)(12π∫−∞∞eisx(k′−k)dx)dk′dk=2πB2∬f~∗(k′) g~(k) δ(s(k′ ⁣− ⁣k))dk′dk=2πB2∣s∣∫−∞∞f~∗(k) g~(k)dk=2πB2∣s∣⟨f~∣g~⟩\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 δ(k)\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:

⟨f~∣g~⟩=∫−∞∞(F^{f(x)})∗ F^{g(x)}dk=A2∫(∫f∗(x′) e−iskx′dx′)(∫g(x) eiskxdx)dk=2πA2∬f∗(x′) g(x)(12π∫−∞∞eisk(x−x′)dk)dx′dx=2πA2∬f∗(x′) g(x) δ(s(x ⁣− ⁣x′))dx′dx=2πA2∣s∣∫−∞∞f∗(x) g(x)dx=2πA2∣s∣⟨f∣g⟩\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 ⁣= ⁣1/2πA\!=\!B\!=\!1 / \sqrt{2\pi} and ∣s∣ ⁣= ⁣1|s|\!=\!1, because then it nicely preserves the functions’ normalization.

References

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 O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.