# Physical interpretation of Parseval's theorem

I have read that Parseval's theorem, relating the norm of a function $f$ and the norm of its Fourier transform $g(k)$:

$$\int |f(x)|^2 dx=\int|g(k)|^2 dk$$

has the simple physical interpretation of "conservation of energy". I just don't see this, so can you suggest me a way to think about it?

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it’s not quite right--- the conservation of energy assumes each Fourier mode is oscillating separately, so that the energy is either a sum over modes or a sum over positions, and this is a consequence of Parseval's theorem. Proving Parseval's theorem is best using the abstract idea that the integral is the "length" of the function considered as a vector, and the length doesn't depend on your choice of orthonormal basis. – Ron Maimon Apr 8 '12 at 23:20

## 2 Answers

Think of a vector $\mathbf{V}$. As seen in a coordinate system $S$ with basis vectors $\hat{e}_i$, it can be written $$\mathbf{V} = \sum_i V_i \hat{e}_i$$ where $V_i$ are the components of $\mathbf{V}$ in $S$. As seen from another coordinate system $S'$ with basis vectors $\hat{e}_i'$, it has a representation $$\mathbf{V} = \sum_i V_i' \hat{e}_i'.$$ Obviously the length of the vector is independent of the coordinate system used to represent it. In other words, we must have $$\sum_i V_i^2 = \sum_i (V_i')^2$$

Proceeding with this analogy, for a function $f(x)$ one can have a position space representation in $\delta$-function basis as $$f(x) = \int f(x') \delta(x-x') dx'$$ where the "component" of $f(x)$ along the "basis vector" $\delta(x-x')$ is $f(x')$ and we sum (integrate since $x$ is a continuous variable) over all the possible "axes".

One can look at the same function in Fourier-space representation as $$f(x) = \int g(k) e^{-i k x} dk$$ where $e^{-ikx}$ are the "basis vectors" and $g(k)$ are the "components" of $f(x)$ along these basis vectors.

You would then agree that $$\int |f(x)|^2 dx = \int |g(k)|^2 dk$$

So Parseval's theorem is just the restatement of the invariance of the length of a "vector" independent of the representation used.

• If $|f(x)|^2$ is proportional to the energy, then Parseval's theorem is a statement of the conservation of the energy as seen in the real-space domain or the Fourier-space domain

• If $f(x)$ is a quantum-mechanical wavefunction, $|f(x)|^2$ is proportional to the probability density. Parseval's theorem is then a statement of the conservation of the probability as seen in the position-space representation or the momentum-space representation.

See also Parseval's identity

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The question asked for an understanding of Parseval's theorem as a statement of conservation of energy, not just a proof of the theorem. – kleingordon Apr 8 '12 at 20:04

In short - Energy in time domain is equal to energy in frequency domain.

EDIT: Think of f(x) as a signal, then Parseval's theorem says that total energy of this signal over time equals to energy of this signal over frequencies.

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