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.

The representation-theoretic owner proving regularity and irreducibility of the full Weyl system on $L^2(\mathbb R^n)$ is The Schrödinger Representation.

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

Choose the site 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

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

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.

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.

Boundary with Physical Interpretation

The unitary transform alone says that two functions represent the same abstract vector. Quantum postulates additionally interpret $\lvert\psi(x)\rvert^2dx$ and $\lvert\phi(p)\rvert^2dp$ as probabilities, identify $X$ and $P$ with laboratory quantities, and specify dynamics.

For the physical momentum-space workflow, including probabilities, free dynamics, mixed states, and scattering uses, see Momentum-Space Representation.

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 site Fourier convention, prove

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

Solution

By definition,

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

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

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

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

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

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.

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.