Momentum Representation

Momentum representation is the description of a quantum state by its momentum-space wavefunction. This page treats it as a calculation tool: how to transform, normalize, act with operators, and recognize when momentum space simplifies a problem.

For the conceptual statement that $\psi(x)$ and $\phi(p)$ are two representations of the same abstract state, see Momentum-Space Representation and Position and Momentum Representations. Here the focus is Fourier-space mechanics.

Transform Pair

With the site convention in one dimension,

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

and

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

The formal plane-wave kernel is

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

The complex exponential phase must be dimensionless, so the factor of $\hbar$ is not optional: $px/\hbar$ is dimensionless. The normalization convention is recorded in Fourier Transform Conventions.

For quick transform pairs and the conversion between $k$-space tables and $\phi(p)$, see Fourier Transform Tables for QM.

Inner Products and Probability

Plancherel’s theorem gives

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

Thus a normalized state satisfies

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

and the probability of measuring momentum in an interval $\Delta$ is

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

For two states,

$$\langle \psi\vert\chi\rangle = \int_{-\infty}^{\infty} \phi_\psi(p)^*\phi_\chi(p)\,dp.$$

The theorem behind these identities is Plancherel and Parseval Theorems.

Operator Dictionary

The basic one-dimensional operator actions are:

Abstract operator	Position representation	Momentum representation
$\hat x$	$x\psi(x)$	$i\hbar\,d\phi/dp$
$\hat p$	$-i\hbar\,d\psi/dx$	$p\phi(p)$
$f(\hat p)$	$f(-i\hbar\,d/dx)\psi$	$f(p)\phi(p)$
$V(\hat x)$	$V(x)\psi(x)$	convolution in $p$

The signs follow from the transform convention. For example,

$$\frac{d\phi}{dp} = \frac{1}{\sqrt{2\pi\hbar}} \int \left( -\frac{ix}{\hbar} \right) e^{-ipx/\hbar}\psi(x)\,dx,$$

so

$$\mathcal F_p[x\psi](p) = i\hbar\frac{d\phi}{dp}.$$

Similarly,

$$\mathcal F_p \left[ -i\hbar\frac{d\psi}{dx} \right](p) = p\phi(p),$$

when boundary terms vanish or the identity is interpreted in the appropriate weak sense.

Free Hamiltonian

For a free particle,

$$\hat H = \frac{\hat p^2}{2m}.$$

Momentum representation diagonalizes this Hamiltonian:

$$(\hat H\phi)(p) = \frac{p^2}{2m}\phi(p).$$

The time-dependent Schrödinger equation becomes

$$i\hbar\frac{\partial\phi(p,t)}{\partial t} = \frac{p^2}{2m}\phi(p,t),$$

with solution

$$\phi(p,t) = \exp \left( -\frac{i p^2 t}{2m\hbar} \right) \phi(p,0).$$

The position-space packet is recovered by the inverse Fourier transform:

$$\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar} \exp \left( -\frac{i p^2 t}{2m\hbar} \right) \phi(p,0)\,dp.$$

This is the standard Fourier construction of free-particle wave packets.

Potentials Mix Momenta

A local potential is simple in position space:

$$(V(\hat x)\psi)(x) = V(x)\psi(x).$$

In momentum representation it is generally not multiplication. With the same unitary convention,

$$\mathcal F_p[V\psi](p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \widetilde V(q)\phi(p-q)\,dq,$$

where

$$\widetilde V(q) = \frac{1}{\sqrt{2\pi\hbar}} \int e^{-iqx/\hbar}V(x)\,dx.$$

Thus the Schrödinger equation in momentum space often has the form

$$i\hbar\frac{\partial\phi(p,t)}{\partial t} = \frac{p^2}{2m}\phi(p,t) + \frac{1}{\sqrt{2\pi\hbar}} \int \widetilde V(q)\phi(p-q,t)\,dq.$$

The kinetic term is diagonal, while the potential term couples different momenta. This is the position-momentum tradeoff in its most practical form. The theorem behind the integral is Convolution.

Example: Linear Potential

For a potential $V(x)=Fx$, multiplication by $x$ becomes differentiation in $p$. The Hamiltonian

$$\hat H = \frac{\hat p^2}{2m} +F\hat x$$

acts as

$$(\hat H\phi)(p) = \frac{p^2}{2m}\phi(p) +i\hbar F\frac{d\phi}{dp}.$$

This example is useful because it shows that momentum representation can turn spatial multiplication into differential motion in momentum space. Whether that is simpler depends on the problem and boundary conditions.

Wave Number Versus Momentum

Some references use wave number $k$ rather than momentum $p$:

$$p=\hbar k, \qquad dp=\hbar\,dk.$$

If $a(k)$ is normalized by

$$\int_{-\infty}^{\infty} \lvert a(k)\rvert^2\,dk = 1,$$

then the corresponding momentum-space wavefunction must satisfy

$$\lvert\phi(p)\rvert^2\,dp = \lvert a(k)\rvert^2\,dk.$$

Therefore

$$\phi(p) = \frac{1}{\sqrt{\hbar}} a(p/\hbar)$$

up to the phase convention used for the transform. Treating $a(k)$ and $\phi(p)$ as the same function is a common source of normalization errors.

When Momentum Space Helps

Momentum representation is especially useful for:

• free particles and approximately free asymptotic states;
• translation-invariant Hamiltonians;
• wave packets specified by momentum spread;
• scattering calculations organized by momentum transfer;
• kinetic-energy estimates and Plancherel identities;
• perturbations whose Fourier transform is simpler than their position-space form.

Position representation is often better for hard boundaries, localized measurement questions, and potentials that are simple functions of position. A representation is a computational choice, not a claim about which variables are more real.

Common Mistakes

• Forgetting that $\phi(p)$ is a probability amplitude density, not a probability.
• Dropping the Jacobian when switching between $k$ and $p$.
• Using $\hat x=-i\hbar\,d/dp$ instead of $\hat x=i\hbar\,d/dp$ with the convention on this page.
• Leaving a local potential as $V(p)\phi(p)$ instead of transforming it properly.
• Treating plane waves as normalizable states rather than generalized momentum eigenfunctions.
• Ignoring operator domains when differentiating momentum-space wavefunctions.

Cross-Links

• Momentum-Space Representation
• Position and Momentum Representations
• Fourier Transform
• Inverse Fourier Transform
• Complex Exponentials
• Plancherel and Parseval Theorems
• Convolution
• Wave Packets
• Fourier Transform Tables for QM
• Free Particle

References

• 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.
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.

Exercises

1. Derive the momentum-representation action of $\hat x$.

Solution

Start from

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

Differentiate:

$$\frac{d\phi}{dp} = \frac{1}{\sqrt{2\pi\hbar}} \int \left( -\frac{ix}{\hbar} \right) e^{-ipx/\hbar}\psi(x)\,dx.$$

Therefore

$$i\hbar\frac{d\phi}{dp} = \frac{1}{\sqrt{2\pi\hbar}} \int x e^{-ipx/\hbar}\psi(x)\,dx = \mathcal F_p[x\psi](p).$$

Thus $\hat x$ acts as $i\hbar\,d/dp$ in momentum representation.

2. Solve the free-particle Schrödinger equation in momentum representation.

Solution

The equation is

$$i\hbar\frac{\partial\phi(p,t)}{\partial t} = \frac{p^2}{2m}\phi(p,t).$$

For each fixed $p$, this is an ordinary differential equation in time:

$$\frac{\partial\phi(p,t)}{\partial t} = -\frac{i p^2}{2m\hbar}\phi(p,t).$$

Thus

$$\phi(p,t) = e^{-ip^2t/(2m\hbar)}\phi(p,0).$$

3. Show why a local potential becomes a convolution in momentum space.

Solution

Write

$$V(x) = \frac{1}{\sqrt{2\pi\hbar}} \int \widetilde V(q)e^{iqx/\hbar}\,dq,$$

and

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int \phi(r)e^{irx/\hbar}\,dr.$$

Multiplying and transforming gives a delta constraint $q+r=p$, so

$$\mathcal F_p[V\psi](p) = \frac{1}{\sqrt{2\pi\hbar}} \int \widetilde V(q)\phi(p-q)\,dq.$$

4. Let $\phi(p)$ vanish outside $\Delta=[p_1,p_2]$ and be normalized. What is the probability of measuring momentum in $\Delta$?

Solution

The probability is

$$\int_{p_1}^{p_2} \lvert\phi(p)\rvert^2\,dp.$$

Because $\phi$ is normalized and vanishes outside $\Delta$, this integral is $1$.