Plancherel and Parseval Theorems

Plancherel and Parseval identities say that Fourier analysis preserves inner products and norms when the transform conventions are chosen consistently. In quantum mechanics, this is the theorem-level reason that a normalized wavefunction stays normalized when written in momentum space.

The slogan is simple:

$$\text{Fourier transformation is a unitary change of representation.}$$

The details matter because the same idea appears in two related settings: discrete Fourier series on a finite interval and Fourier transforms on the full line.

Parseval for Fourier Series

For periodic functions on an interval of length $L$, use the normalized modes

$$e_n(x) = \frac{1}{\sqrt L} \exp \left( \frac{2\pi i n x}{L} \right), \qquad n\in\mathbb Z.$$

If

$$f(x) \sim \sum_{n\in\mathbb Z}c_n e_n(x), \qquad c_n = \int_0^L e_n(x)^*f(x)\,dx,$$

then Parseval’s identity is

$$\int_0^L \lvert f(x)\rvert^2\,dx = \sum_{n\in\mathbb Z} \lvert c_n\rvert^2.$$

More generally, if

$$g(x) \sim \sum_{n\in\mathbb Z}d_n e_n(x),$$

then the inner product is preserved:

$$\int_0^L f(x)^*g(x)\,dx = \sum_{n\in\mathbb Z}c_n^*d_n.$$

This is the infinite-dimensional version of the finite statement that an orthonormal change of basis preserves dot products. The page Periodic Functions and Fourier Series gives the mode expansion and boundary-condition context.

Plancherel for Fourier Transforms

With the wave-mechanics convention

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi(x)\,dx,$$

Plancherel’s theorem says

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx = \int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp.$$

The inner-product version is

$$\int_{-\infty}^{\infty} \psi(x)^*\chi(x)\,dx = \int_{-\infty}^{\infty} \phi_\psi(p)^*\phi_\chi(p)\,dp,$$

where $\phi_\psi$ and $\phi_\chi$ are the Fourier transforms of $\psi$ and $\chi$ with the same convention.

This identity is stronger than a normalization trick. It says the Fourier transform extends to a unitary map on $L^2(\mathbb R)$:

$$\mathcal F: L^2(\mathbb R,dx) \longrightarrow L^2(\mathbb R,dp).$$

The page Fourier Transform states the transform pair, and Inverse Fourier Transform explains reconstruction.

Proof Sketch

For well-behaved functions, such as Schwartz functions, the theorem follows from the delta-kernel identity. Write

$$\phi_\psi(p)^* = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar}\psi(x)^*\,dx,$$

and

$$\phi_\chi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx'/\hbar}\chi(x')\,dx'.$$

Then

$$\begin{aligned} \int \phi_\psi(p)^*\phi_\chi(p)\,dp &= \frac{1}{2\pi\hbar} \int dx \int dx'\, \psi(x)^*\chi(x') \int dp\,e^{ip(x-x')/\hbar} \\ &= \int dx \int dx'\, \psi(x)^*\chi(x')\delta(x-x') \\ &= \int \psi(x)^*\chi(x)\,dx. \end{aligned}$$

For general $L^2$ functions, the theorem is obtained by completing this identity in the $L^2$ norm. This completion step is important: many wavefunctions used in quantum mechanics are not pointwise nice enough for every formal interchange of integrals to be justified directly.

Probability in Momentum Space

If $\psi$ is normalized in position space,

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx = 1,$$

then Plancherel gives

$$\int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp = 1.$$

Thus $\lvert\phi(p)\rvert^2$ is the momentum probability density. For an interval $\Delta$ of momentum values,

$$\mathbb P(p\in\Delta) = \int_\Delta \lvert\phi(p)\rvert^2\,dp.$$

For a finite periodic box, the analogous statement is discrete. If

$$\psi(x) = \sum_{n\in\mathbb Z}c_n e_n(x),$$

and the state is normalized, then

$$\sum_{n\in\mathbb Z} \lvert c_n\rvert^2 = 1.$$

The probability of finding one of a set $S$ of allowed modes is

$$\mathbb P(n\in S) = \sum_{n\in S} \lvert c_n\rvert^2.$$

This is the finite-volume version of momentum-space normalization.

Energy-Like Integral Identities

Plancherel also turns derivative norms into weighted momentum integrals. If $\psi$ is sufficiently regular and its boundary behavior makes integration by parts legitimate, then

$$\mathcal F[\psi'](p) = \frac{ip}{\hbar}\phi(p).$$

Applying Plancherel to $\psi'$ gives

$$\int_{-\infty}^{\infty} \lvert\psi'(x)\rvert^2\,dx = \int_{-\infty}^{\infty} \frac{p^2}{\hbar^2} \lvert\phi(p)\rvert^2\,dp.$$

For a free particle,

$$T = \frac{p^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2},$$

so, under the same domain assumptions,

$$\langle T\rangle = \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \lvert\psi'(x)\rvert^2\,dx = \int_{-\infty}^{\infty} \frac{p^2}{2m} \lvert\phi(p)\rvert^2\,dp.$$

This is why the kinetic energy of a free particle is diagonal in momentum representation. The position-space derivative cost becomes a momentum-space weight $p^2/(2m)$.

For periodic Fourier series, with $k_n=2\pi n/L$,

$$e_n'(x) = ik_n e_n(x).$$

If $f$ is regular enough for termwise differentiation in $L^2$, then

$$\int_0^L \lvert f'(x)\rvert^2\,dx = \sum_{n\in\mathbb Z} k_n^2\lvert c_n\rvert^2.$$

This is the ring analogue of the continuum kinetic-energy identity.

Example: Two Periodic Modes

Let

$$f(x) = a e_0(x)+b e_1(x),$$

where $e_n$ are normalized periodic modes. Parseval gives immediately

$$\int_0^L \lvert f(x)\rvert^2\,dx = \lvert a\rvert^2+\lvert b\rvert^2.$$

Thus $f$ is normalized exactly when

$$\lvert a\rvert^2+\lvert b\rvert^2 = 1.$$

For a particle on a ring, the derivative identity gives

$$\langle T\rangle = \frac{\hbar^2}{2m} \left( k_0^2\lvert a\rvert^2 + k_1^2\lvert b\rvert^2 \right) = \frac{\hbar^2 k_1^2}{2m} \lvert b\rvert^2,$$

because $k_0=0$. No cross term appears because the Fourier modes are orthonormal.

What the Theorems Do Not Say

Plancherel and Parseval are norm and inner-product theorems. They do not say that a Fourier series converges pointwise everywhere, nor that every Fourier integral can be manipulated as an ordinary absolutely convergent integral.

The theorem also does not make plane waves normalizable. A full-line plane wave is a generalized eigenfunction with delta normalization, not an element of $L^2(\mathbb R)$. Normalizable Wave Packets are the physically honest objects when a probability density is required.

Finally, derivative identities require additional regularity. A function can be square-integrable without having a square-integrable derivative. In operator language, the state must lie in the appropriate domain of the momentum or kinetic-energy operator.

Common Mistakes

• Thinking norm preservation means $\lvert\psi(x)\rvert^2=\lvert\phi(p)\rvert^2$ pointwise.
• Mixing Fourier normalization conventions and then applying Parseval with the wrong measure.
• Treating $L^2$ reconstruction as an everywhere-pointwise statement.
• Applying derivative identities to wavefunctions that do not have the needed weak derivative or boundary behavior.
• Using delta-normalized plane waves as though they were normalizable probability densities.
• Forgetting that finite-box sums and continuum momentum integrals have different normalization rules.

Cross-Links

• Fourier Transform
• Inverse Fourier Transform
• Periodic Functions and Fourier Series
• Fourier Transform Conventions
• Position and Momentum Representations
• Wave Packets
• Fourier Transform Table

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

Exercises

1. Let $f(x)=a e_0(x)+b e_2(x)$ in a periodic box of length $L$. Use Parseval to normalize $f$.

Solution

The modes are orthonormal, so

$$\int_0^L \lvert f(x)\rvert^2\,dx = \lvert a\rvert^2+\lvert b\rvert^2.$$

The state is normalized exactly when

$$\lvert a\rvert^2+\lvert b\rvert^2 = 1.$$

2. Starting from the Fourier-transform convention on this page, show formally why Plancherel preserves the inner product.

Solution

For sufficiently well-behaved functions,

$$\begin{aligned} \int \phi_\psi(p)^*\phi_\chi(p)\,dp &= \frac{1}{2\pi\hbar} \int dx \int dx'\, \psi(x)^*\chi(x') \int dp\,e^{ip(x-x')/\hbar} \\ &= \int dx \int dx'\, \psi(x)^*\chi(x')\delta(x-x') \\ &= \int \psi(x)^*\chi(x)\,dx. \end{aligned}$$

The rigorous $L^2$ theorem extends this identity from a dense class of nice functions to all square-integrable functions.

3. Derive the momentum-space expression for the free-particle kinetic energy.

Solution

If integration by parts is justified,

$$\mathcal F[\psi'](p) = \frac{ip}{\hbar}\phi(p).$$

Plancherel gives

$$\int \lvert\psi'(x)\rvert^2\,dx = \int \frac{p^2}{\hbar^2} \lvert\phi(p)\rvert^2\,dp.$$

Therefore

$$\langle T\rangle = \frac{\hbar^2}{2m} \int \lvert\psi'(x)\rvert^2\,dx = \int \frac{p^2}{2m} \lvert\phi(p)\rvert^2\,dp.$$

4. Why is a plane wave not a counterexample to Plancherel’s theorem?

Solution

A full-line plane wave has constant magnitude, so its position-space norm diverges:

$$\int_{-\infty}^{\infty} \lvert e^{ip_0x/\hbar}\rvert^2\,dx = \infty.$$

It is not an element of $L^2(\mathbb R)$. Its momentum-space representation is a delta distribution, also not an ordinary square-integrable function. Plancherel applies to square-integrable states; plane waves belong to the generalized-eigenfunction framework.