Position and Momentum Representations

Position and momentum wavefunctions are two coordinate representations of one abstract Hilbert-space vector. On the full line, the unitary map between them is a Fourier transform with momentum variable $p$ and phase $px/\hbar$.

The mathematical dictionary is:

• position space realizes position by multiplication and momentum by differentiation;
• momentum space realizes momentum by multiplication and position by differentiation;
• the Fourier map carries vectors, operators, domains, and inner products together;
• generalized position and momentum kets are distributional kernels, not normalizable Hilbert-space vectors.

The physical interpretation of the amplitudes belongs to Wavefunctions as Representations and Momentum-Space Representation. This page develops the transform and operator mathematics.

For the regularity and irreducibility of the Weyl operators on $L^2(\mathbb R^n)$, see The Schrödinger Representation.

Helpful background. Review Fourier transforms to convert derivatives into multiplication and track normalization.

Abstract Representation Maps

Let $\mathcal H$ be the abstract state Hilbert space. A position representation is a unitary map

$$U_x:\mathcal H\longrightarrow L^2(\mathbb R,dx),$$

and the coordinate function of $\lvert\psi\rangle$ is

$$\psi(x) = (U_x\psi)(x) = \langle x\vert\psi\rangle.$$

Similarly, a momentum representation is a unitary map

$$U_p:\mathcal H\longrightarrow L^2(\mathbb R,dp),$$

with

$$\phi(p) = (U_p\psi)(p) = \langle p\vert\psi\rangle.$$

Unitarity means that either coordinate description preserves the abstract inner product:

$$\begin{aligned} \langle\chi\vert\psi\rangle &= \int_{\mathbb R} \chi(x)^*\psi(x)\,dx\\ &= \int_{\mathbb R} \widetilde\chi(p)^*\phi(p)\,dp. \end{aligned}$$

The functions are not two states. They are the images of the same vector under two unitary coordinate maps.

Generalized Position and Momentum Kernels

The symbols $\lvert x\rangle$ and $\lvert p\rangle$ are generalized eigenvectors satisfying formal relations

$$\begin{aligned} X\lvert x\rangle &=x\lvert x\rangle,\\ P\lvert p\rangle &=p\lvert p\rangle. \end{aligned}$$

On the full line they are delta normalized:

$$\begin{aligned} \langle x\vert x'\rangle &=\delta(x-x'),\\ \langle p\vert p'\rangle &=\delta(p-p'). \end{aligned}$$

Their formal resolutions of the identity are

$$\begin{aligned} \int_{\mathbb R} \lvert x\rangle\langle x\rvert\,dx &=I,\\ \int_{\mathbb R} \lvert p\rangle\langle p\rvert\,dp &=I. \end{aligned}$$

These formulas are distributional statements about kernels or spectral measures. Neither generalized ket belongs to $L^2(\mathbb R)$. See Generalized Eigenvectors for the rigged-space interpretation.

Position–Momentum Overlap

Use the Fourier convention

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

Complex conjugation gives

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

The normalization is fixed by the distributional identity

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

The overlap kernel has dimensions

$$[\langle x\vert p\rangle] = (\text{action})^{-1/2},$$

consistent with position and momentum delta normalizations.

Fourier Transform Pair

Insert the position resolution of the identity:

$$\begin{aligned} \phi(p) &= \langle p\vert\psi\rangle\\ &= \int_{\mathbb R} \langle p\vert x\rangle \langle x\vert\psi\rangle\,dx. \end{aligned}$$

Therefore

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

The inverse transform is

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

On test functions these are ordinary integrals. Plancherel’s theorem extends the transform uniquely to a unitary map on all of $L^2(\mathbb R)$, where the integrals may need to be interpreted as $L^2$ limits.

The $p$-space transform differs from the wave-number transform by

$$p=\hbar k.$$

Changing between $p$ and $k$ changes both the measure and normalization factor. See Fourier Transform for the full convention table.

Plancherel and Representation Invariance

Unitarity gives

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

More generally,

$$\int_{\mathbb R} \chi(x)^*\psi(x)\,dx = \int_{\mathbb R} \widetilde\chi(p)^*\phi(p)\,dp.$$

Norms, inner products, orthogonality, and matrix elements are therefore representation invariant when every object is transformed consistently. This is the Hilbert-space content of Plancherel and Parseval.

Operators under a Change of Representation

If $A$ is an abstract operator, its coordinate representatives are

$$\begin{aligned} A_x &=U_xAU_x^{-1},\\ A_p &=U_pAU_p^{-1}. \end{aligned}$$

The Fourier transform

$$\mathcal F_\hbar=U_pU_x^{-1}$$

relates them:

$$A_p = \mathcal F_\hbar A_x \mathcal F_\hbar^{-1}.$$

This is unitary equivalence, not a change in the physical operator. Spectra and matrix elements are unchanged, although the coordinate action can become more or less local.

For an integral-kernel operator in position space,

$$(A_x\psi)(x) = \int_{\mathbb R} A(x,x')\psi(x')\,dx',$$

the transformed kernel is obtained by Fourier transforming both variables with the corresponding signs.

Position and Momentum Operators

On the full line, the standard position-space actions are

$$\begin{aligned} (X\psi)(x) &=x\psi(x),\\ (P\psi)(x) &=-i\hbar\frac{d\psi}{dx}(x). \end{aligned}$$

The natural self-adjoint domain of $X$ is

$$\mathcal D(X) = \left\{ \psi\in L^2(\mathbb R): x\psi\in L^2(\mathbb R) \right\}.$$

The standard self-adjoint momentum domain on the line is the Sobolev space

$$\mathcal D(P)=H^1(\mathbb R),$$

whose elements have a weak derivative in $L^2$. The derivative formula does not act on every $L^2$ equivalence class.

Fourier transformation interchanges multiplication and differentiation:

$$\begin{aligned} (P\phi)(p) &=p\phi(p),\\ (X\phi)(p) &=i\hbar\frac{d\phi}{dp}(p), \end{aligned}$$

on their transformed domains. The domain of multiplication by $p$ is

$$\mathcal D(P) = \left\{ \phi\in L^2(\mathbb R): p\phi\in L^2(\mathbb R) \right\},$$

while $X$ in momentum space has the corresponding $H^1$ domain.

Deriving the Operator Dictionary

For a Schwartz function $\psi$, integration by parts gives

$$\begin{aligned} \mathcal F_\hbar \left[ -i\hbar\frac{d\psi}{dx} \right](p) &= \frac{-i\hbar}{\sqrt{2\pi\hbar}} \times \int_{\mathbb R} e^{-ipx/\hbar}\psi'(x)\,dx \\ &\quad=p\phi(p). \end{aligned}$$

The boundary term vanishes because Schwartz functions and all their derivatives decay rapidly. Differentiating the transform with respect to $p$ gives

$$\begin{aligned} i\hbar\frac{d\phi}{dp} &= \frac{1}{\sqrt{2\pi\hbar}} \int_{\mathbb R} x e^{-ipx/\hbar}\psi(x)\,dx\\ &= \mathcal F_\hbar[x\psi](p). \end{aligned}$$

These calculations establish the formulas first on a common dense test domain. Self-adjoint operators are then obtained by closure with their proper domains.

Canonical Commutator and Domain

On the Schwartz space $\mathcal S(\mathbb R)$,

$$\begin{aligned} [X,P]\psi &= x(-i\hbar\psi') +i\hbar(x\psi)'\\ &= i\hbar\psi. \end{aligned}$$

Thus

$$[X,P]=i\hbar I$$

on this common invariant domain. The equality is not an unrestricted identity between products defined on every $L^2$ vector. For unbounded operators, $XP$ and $PX$ each have domain conditions.

The same computation in momentum space uses $X=i\hbar\,d/dp$ and $P=p$. See Canonical Commutation Relations for the physical role and Weyl form.

Multiplication and Convolution

A position-dependent function $V(x)$ acts locally in position space:

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

Define its momentum transform by

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

when the transform exists in the required sense. In momentum space, multiplication becomes convolution:

$$(\widetilde{V\psi})(p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{\mathbb R} \widetilde V(p-p')\phi(p')\,dp'.$$

Conversely, functions of momentum are diagonal in momentum space and generally nonlocal in position space. This is why free-particle kinetic energy is simple in momentum representation while a local potential is simple in position representation. For a general function $f$, the notation $f(-i\hbar\,d/dx)$ means the spectral functional calculus of $P$, not necessarily a convergent series of derivatives. The convolution theorem and its normalization are developed in Convolution.

Free Evolution in Momentum Space

For a particle of mass $m>0$ on the full line, the free Hamiltonian $H=P^2/(2m)$ acts by multiplication in momentum space. On its operator domain, the Schrödinger equation becomes

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

At each momentum this is an ordinary differential equation in time, with solution

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

The phase multiplier defines unitary evolution for every $L^2$ initial state; a strong time derivative additionally requires $p^2\phi(p,0)\in L^2$. Its modulus is one, so the momentum distribution does not change. Position-space spreading comes from the different phases accumulated by different momentum components:

$$\psi(x,t) =\frac{1}{\sqrt{2\pi\hbar}} \int_{\mathbb R} e^{ipx/\hbar-ip^2t/(2m\hbar)}\phi(p,0)\,dp.$$

As with the inverse transform itself, this integral may require an $L^2$ interpretation. Free Particle develops the physical model and wave-packet dynamics. Momentum space is useful here because evolution is diagonal; position space can be more useful when boundaries or localized measurements determine the question.

A Linear Potential Translates the Momentum Distribution

Let $V(x)=Fx$ with constant real slope $F$, so that the mechanical force is $-F$. For $H=P^2/(2m)+FX$, the operator dictionary gives, initially on Schwartz wavefunctions,

$$i\hbar\partial_t\phi =\frac{p^2}{2m}\phi+i\hbar F\partial_p\phi,$$

or

$$\partial_t\phi-F\partial_p\phi =-\frac{ip^2}{2m\hbar}\phi.$$

Along a characteristic $p(t)=p(0)-Ft$, the right-hand side changes only the phase. Hence

$$\lvert\phi(p,t)\rvert^2 =\lvert\phi(p+Ft,0)\rvert^2.$$

The profile moves toward lower momentum when $F>0$, consistently with the force $-F$. A change of integration variable preserves normalization and, when the first absolute momentum moment is finite, gives

$$\langle P\rangle_t=\langle P\rangle_0-Ft.$$

This example also checks the sign of $X=i\hbar\partial_p$. Reversing that sign would accelerate the packet against the force. A linear potential has a distributional Fourier transform; the derivative expression above avoids treating it as an ordinary integrable convolution kernel.

Translations

The unitary generated by momentum,

$$T(a)=e^{-iaP/\hbar},$$

acts in position space as

$$(T(a)\psi)(x)=\psi(x-a).$$

In momentum space it is multiplication by a phase:

$$(T(a)\phi)(p) = e^{-iap/\hbar}\phi(p).$$

Similarly,

$$B(b)=e^{ibX/\hbar}$$

multiplies a position wavefunction by $e^{ibx/\hbar}$ and translates the momentum wavefunction:

$$(B(b)\phi)(p)=\phi(p-b).$$

These identities are the representation-level form of the Heisenberg–Weyl translation structure.

Worked Example: Gaussian Transform

Take the normalized position-space Gaussian

$$\psi(x) = \left( \frac{\alpha}{\pi} \right)^{1/4} e^{-\alpha x^2/2}, \qquad \alpha>0.$$

Using the Gaussian integral with a linear term gives

$$\phi(p) = \left( \frac{1}{ \pi\alpha\hbar^2 } \right)^{1/4} e^{-p^2/(2\alpha\hbar^2)}.$$

Both norms equal one. The variances are

$$\begin{aligned} (\Delta x)^2 &=\frac{1}{2\alpha},\\ (\Delta p)^2 &=\frac{\alpha\hbar^2}{2}, \end{aligned}$$

so

$$\Delta x\,\Delta p=\frac{\hbar}{2}.$$

The reciprocal widths are a Fourier property. The interpretation as a sharp uncertainty bound requires the quantum variance definitions developed in Core Formalism.

Basis Phases and Convention Dependence

The generalized basis is not fixed by eigenvalue equations alone. Rephasing

$$\lvert p\rangle \longmapsto e^{i\chi(p)}\lvert p\rangle$$

changes the momentum wavefunction by

$$\phi(p) \longmapsto e^{-i\chi(p)}\phi(p).$$

Operator representatives transform with the same basis change, leaving matrix elements invariant. A $p$-dependent phase can add a connection-like term to derivative operators, so one may not rephase the basis while leaving all operator formulas untouched.

Fourier conventions also differ in sign and normalization. A correct calculation uses one transform pair consistently and checks that the momentum derivative and translation phases match it.

Higher Dimensions

For $\mathbf x,\mathbf p\in\mathbb R^d$,

$$\phi(\mathbf p) = \frac{1}{(2\pi\hbar)^{d/2}} \int_{\mathbb R^d} e^{-i\mathbf p\cdot\mathbf x/\hbar} \psi(\mathbf x)\,d^dx.$$

The operator dictionary becomes

$$\begin{aligned} \mathbf X &\longleftrightarrow i\hbar\nabla_{\mathbf p},\\ \mathbf P &\longleftrightarrow \mathbf p \end{aligned}$$

in momentum space, with the reverse multiplication–derivative assignment in position space.

Curvilinear coordinates require the correct measure and may change the differential expression representing a self-adjoint momentum component. One should not transplant the Cartesian formula without checking the inner product and domain.

Finite Intervals and Periodic Boxes

On a periodic interval of length $L$, momentum is discrete:

$$p_n=\frac{2\pi\hbar n}{L}, \qquad n\in\mathbb Z.$$

The normalized modes are

$$u_n(x) = \frac{1}{\sqrt L} e^{i2\pi nx/L}.$$

The position–momentum transform becomes a Fourier series rather than a Fourier integral.

Boundary conditions are essential. Periodic and quasiperiodic endpoint conditions give self-adjoint realizations of the first derivative with different momentum spectra. Dirichlet conditions are standard for second-order Hamiltonians but do not by themselves make the first-derivative momentum operator self-adjoint. The differential expression $-i\hbar\,d/dx$ is not an operator until its domain is specified.

Numerical Fourier Representations

For a uniform grid with $N$ points and spacing $\Delta x$, a standard discrete Fourier transform gives momentum spacing

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

The represented momentum range is finite and periodic. Reliable use requires:

• a declared ordering of positive and negative frequencies;
• transform normalization consistent with the discrete inner products;
• quadrature factors $\Delta x$ and $\Delta p$;
• enough position range to suppress wraparound;
• enough resolution to suppress aliasing;
• consistent phase origins when grids do not begin at zero;
• convergence checks under both box enlargement and grid refinement.

An FFT computes a periodic discrete transform, not the continuum transform without approximation. See Fast Fourier Transform for implementation details.

Common Mistakes

• Treating $\lvert x\rangle$ or $\lvert p\rangle$ as normalizable vectors.
• Calling position and momentum wavefunctions two different states.
• Mixing $p$ and wave number $k$ without transforming the measure.
• Combining a forward transform from one convention with an inverse transform from another.
• Forgetting the factor of $\hbar$ in phases and differential operators.
• Transforming a state but not the operator acting on it.
• Applying derivative formulas outside their operator domains.
• Dropping boundary terms without a decay or boundary-condition argument.
• Assuming a local multiplication operator remains local after Fourier transformation.
• Treating the finite-grid FFT as an exact continuum Fourier transform.
• Ignoring basis-dependent phase conventions.

Exercises

1. Let $\psi\in\mathcal S(\mathbb R)$. Starting from the Fourier convention, prove

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

   By differentiating the same transform with respect to $p$, also show that $\mathcal F_\hbar[x\psi](p)=i\hbar\phi'(p)$.

Solution

By definition,

$$\mathcal F_\hbar[-i\hbar\psi'](p) = \frac{-i\hbar}{\sqrt{2\pi\hbar}} \times \int_{\mathbb R} e^{-ipx/\hbar}\psi'(x)\,dx.$$

Integrating by parts and using the rapid decay of $\psi$ gives

$$\begin{aligned} \mathcal F_\hbar[-i\hbar\psi'](p) &= \frac{i\hbar}{\sqrt{2\pi\hbar}} \times \int_{\mathbb R} \left( -\frac{ip}{\hbar} \right) e^{-ipx/\hbar}\psi(x)\,dx \\ &\quad= p\phi(p). \end{aligned}$$

Differentiation under the integral, justified by rapid decay, gives

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

Multiplying by $i\hbar$ proves the position-operator formula with its positive sign.

2. Fourier transform the normalized Gaussian

   $$\psi(x) = \left( \frac{\alpha}{\pi} \right)^{1/4} e^{-\alpha x^2/2}$$

   and verify its momentum-space norm.

Solution

Use

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

Substitution into the transform gives

$$\phi(p) = \left( \frac{1}{ \pi\alpha\hbar^2 } \right)^{1/4} e^{-p^2/(2\alpha\hbar^2)}.$$

Then

$$\begin{aligned} \int_{\mathbb R} \lvert\phi(p)\rvert^2\,dp &= \frac{1}{\sqrt{\pi\alpha\hbar^2}} \int_{\mathbb R} e^{-p^2/(\alpha\hbar^2)}\,dp\\ &=1. \end{aligned}$$

3. Show that $T(a)=e^{-iaP/\hbar}$ translates a Schwartz wavefunction:

   $$(T(a)\psi)(x)=\psi(x-a).$$

   Verify the same statement in momentum representation.

Solution

In momentum representation, $P$ is multiplication by $p$, so

$$(T(a)\phi)(p) = e^{-iap/\hbar}\phi(p).$$

Taking the inverse Fourier transform gives

$$\begin{aligned} (T(a)\psi)(x) &= \frac{1}{\sqrt{2\pi\hbar}} \times \int_{\mathbb R} e^{ipx/\hbar} e^{-iap/\hbar}\phi(p)\,dp \\ &= \frac{1}{\sqrt{2\pi\hbar}} \times \int_{\mathbb R} e^{ip(x-a)/\hbar}\phi(p)\,dp \\ &= \psi(x-a). \end{aligned}$$

This proof applies through the unitary Fourier transform and does not assume that a general Schwartz function equals its Taylor series.

4. On the periodic interval $[0,L]$, let

   $$u_n(x) = \frac{1}{\sqrt L} e^{i2\pi nx/L}.$$

   Verify orthonormality and find the eigenvalue of $-i\hbar\,d/dx$ on $u_n$.

Solution

For integers $m,n$,

$$\begin{aligned} \langle u_m\vert u_n\rangle &= \frac{1}{L} \int_0^L e^{i2\pi(n-m)x/L}\,dx\\ &= \delta_{mn}. \end{aligned}$$

Differentiation gives

$$\begin{aligned} -i\hbar\frac{du_n}{dx} &= -i\hbar \left( \frac{i2\pi n}{L} \right)u_n\\ &= \frac{2\pi\hbar n}{L}u_n. \end{aligned}$$

Thus

$$p_n=\frac{2\pi\hbar n}{L}.$$

The result depends on the periodic domain; changing the endpoint condition changes the allowed spectrum.

5. Solve the free-particle Schrödinger equation in momentum space. If $\phi(p,0)$ is normalized and vanishes outside an interval $\Delta$, what is the probability of measuring momentum in $\Delta$ at time $t$?

Solution

For each fixed $p$, the equation is

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

Integrating gives $\phi(p,t)=e^{-ip^2t/(2m\hbar)}\phi(p,0)$. The phase does not change the modulus or support, so

$$\mathbb P_t(P\in\Delta) =\int_\Delta\lvert\phi(p,t)\rvert^2\,dp =\int_\Delta\lvert\phi(p,0)\rvert^2\,dp=1.$$

This leaves room for spreading in position space: momentum probabilities alone do not specify the relative phases that enter the inverse transform.

6. For $V,\psi\in\mathcal S(\mathbb R)$, derive the convolution formula for $\mathcal F_\hbar[V\psi]$ with its normalization factor.

Solution

Insert the inverse transform of $\psi$ into the transform of $V\psi$. Rapid decay makes the resulting double integral absolutely integrable, so its order can be exchanged:

$$\begin{aligned} \mathcal F_\hbar[V\psi](p) &=\frac{1}{2\pi\hbar}\int_{\mathbb R}dp'\,\phi(p') \int_{\mathbb R}dx\,e^{-i(p-p')x/\hbar}V(x)\\ &=\frac{1}{\sqrt{2\pi\hbar}} \int_{\mathbb R}\widetilde V(p-p')\phi(p')\,dp'. \end{aligned}$$

The first equality contains two transform prefactors. Recognizing the inner integral as $\sqrt{2\pi\hbar}\,\widetilde V(p-p')$ leaves the single prefactor in the result. For singular or growing potentials, the ordinary-integral argument must be replaced by an appropriate distribution or operator-domain treatment.

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 2009.
• 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.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.