Inverse Fourier Transform

The inverse Fourier transform reconstructs a function from its Fourier components. With the site convention for position and momentum wavefunctions,

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

The forward transform is

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

The signs and normalization factors are paired. Changing one without changing the other breaks reconstruction or norm preservation.

Why the Inverse Matters

The inverse transform is the operational step that turns momentum-space data into a position-space wavefunction. It is used when:

• a free Hamiltonian is diagonal in momentum space;
• a wave packet is specified by its momentum distribution;
• scattering calculations produce amplitudes in momentum variables;
• Green functions are built by transforming back from algebraic momentum-space expressions;
• numerical spectral methods evolve modes and reconstruct functions.

The conceptual page Fourier Transform explains the transform pair. This page focuses on reconstruction and its pitfalls.

Reconstruction Formula

Substitute the forward transform into the inverse:

$$\begin{aligned} \psi(x) &= \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{ipx/\hbar}\phi(p)\,dp\\ &= \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} dp \int_{-\infty}^{\infty} dx'\, e^{ip(x-x')/\hbar}\psi(x'). \end{aligned}$$

The momentum integral is the delta kernel:

$$\frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} e^{ip(x-x')/\hbar}\,dp = \delta(x-x').$$

Therefore

$$\psi(x) = \int_{-\infty}^{\infty} \delta(x-x')\psi(x')\,dx',$$

which returns $\psi(x)$ in the appropriate sense.

For square-integrable wavefunctions, the most robust statement is reconstruction in $L^2$ norm. For smoother functions, stronger pointwise statements may hold. For plane waves and delta functions, the statement is distributional.

The test-function meaning of such statements is explained in Distributions.

Momentum Basis View

In bra-ket notation, the same formula comes from inserting a momentum resolution of identity:

$$\lvert\psi\rangle = \int_{-\infty}^{\infty} \lvert p\rangle\langle p\vert\psi\rangle\,dp.$$

Taking an $x$-representation gives

$$\psi(x) = \int_{-\infty}^{\infty} \langle x\vert p\rangle \phi(p)\,dp.$$

With

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar},$$

this is exactly the inverse Fourier transform.

The notation $\lvert p\rangle$ is formal: momentum eigenkets are generalized eigenvectors, not normalizable Hilbert-space states.

Normalization

With the symmetric convention,

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

Thus if $\phi(p)$ is normalized, the reconstructed $\psi(x)$ is normalized. This is the Plancherel property in the quantum convention.

The canonical theorem page is Plancherel and Parseval Theorems.

The inverse transform is therefore not merely a formal integral. It is a unitary change of representation on the appropriate $L^2$ space.

Example: A Momentum Eigencomponent

Let the momentum-space amplitude be a delta distribution,

$$\phi(p) = \delta(p-p_0).$$

The inverse transform gives

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar}\delta(p-p_0)\,dp = \frac{1}{\sqrt{2\pi\hbar}} e^{ip_0x/\hbar}.$$

This is a plane wave. It is not normalizable on the full line, because the input $\delta(p-p_0)$ is not an ordinary square-integrable function. The example is distributional but extremely useful.

Example: Translating a Packet

Suppose

$$\psi_0(x) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar}\phi_0(p)\,dp.$$

If the momentum-space amplitude is multiplied by a phase,

$$\phi(p) = e^{-ipx_0/\hbar}\phi_0(p),$$

then the inverse transform gives

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ip(x-x_0)/\hbar}\phi_0(p)\,dp = \psi_0(x-x_0).$$

Thus a linear phase in momentum space translates the packet in position space. This is one of the most useful practical checks on Fourier-sign conventions.

Finite Box Versus Continuum

In a periodic box, reconstruction uses a Fourier series:

$$f(x) \sim \sum_{n\in\mathbb Z} c_n e_n(x).$$

On the full line, reconstruction uses the inverse Fourier integral. The large-box limit replaces mode sums by momentum integrals, with spacing

$$\Delta p = \frac{2\pi\hbar}{L}.$$

The finite-volume version is Periodic Functions and Fourier Series. The continuum transform should not be mixed with finite-box normalization without accounting for the sum-to-integral factor.

Common Mistakes

• Reusing the forward-transform sign in the inverse.
• Forgetting the factor of $\hbar$ in the phase $px/\hbar$.
• Mixing a $k$-space inverse with a $p$-space forward transform without the Jacobian $p=\hbar k$.
• Treating reconstruction as pointwise for arbitrary $L^2$ wavefunctions.
• Forgetting that delta and plane-wave examples are distributional.
• Dropping normalization factors and then wondering why probabilities do not match.
• Confusing a finite Fourier series reconstruction with a full-line inverse transform.

Cross-Links

• Fourier Transform
• Plancherel and Parseval Theorems
• Convolution
• Momentum Representation
• Fourier Transform Conventions
• Delta Function
• Distributions
• Position and Momentum Representations
• Wave Packets
• Periodic Functions and Fourier Series
• Fourier Transform Table

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• 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.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. Verify that substituting the forward transform into the inverse gives the delta kernel.

Solution

Substitution gives

$$\psi(x) = \frac{1}{2\pi\hbar} \int dp \int dx'\, e^{ip(x-x')/\hbar}\psi(x').$$

The distributional identity

$$\frac{1}{2\pi\hbar} \int e^{ip(x-x')/\hbar}\,dp = \delta(x-x')$$

then gives

$$\psi(x) = \int \delta(x-x')\psi(x')\,dx' = \psi(x).$$

2. Invert $\phi(p)=\delta(p-p_0)$.

Solution

Use the inverse transform:

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar}\delta(p-p_0)\,dp = \frac{1}{\sqrt{2\pi\hbar}}e^{ip_0x/\hbar}.$$

The result is a delta-normalized plane wave, not a normalizable wave packet.

3. What position-space effect is produced by multiplying $\phi(p)$ by $e^{-ipx_0/\hbar}$?

Solution

The inverse transform becomes

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ip(x-x_0)/\hbar}\phi_0(p)\,dp.$$

This is $\psi_0(x-x_0)$, so the wavefunction is translated by $x_0$ in position space.

4. Why is the inverse transform best understood as norm reconstruction for general square-integrable wavefunctions?

Solution

An arbitrary $L^2$ wavefunction is an equivalence class defined up to changes on sets of measure zero. Pointwise values may be delicate or representative-dependent. The Fourier transform is unitary on $L^2$, so the robust statement is that inverse transformation reconstructs the same Hilbert-space vector in norm. Stronger pointwise claims require additional regularity.