Coordinate Representation

The coordinate representation is the description of an abstract quantum state by its amplitudes in a position basis. If $|\psi\rangle$ is a state and $|x\rangle$ is a generalized position eigenket, the one-dimensional wavefunction is

$$\psi(x)=\langle x|\psi\rangle.$$

This equation is the bridge between the abstract formalism and wave mechanics. A wavefunction is not an extra kind of state; it is the position-space representation of the same state.

Required background. State Vectors supplies the basis-expansion language used throughout this page.

Helpful background. The Schrödinger Representation explains the rigorous operator representation behind the generalized-ket notation used formally here.

From Kets To Wavefunctions

In a finite-dimensional Hilbert space, a state can be expanded in an orthonormal basis:

$$|\psi\rangle=\sum_n |n\rangle\langle n|\psi\rangle.$$

The coefficients $\psi_n=\langle n|\psi\rangle$ are the components of the state in that basis. Position representation is the continuous analogue:

$$|\psi\rangle =\int_{-\infty}^{\infty} dx\, |x\rangle\langle x|\psi\rangle =\int_{-\infty}^{\infty} dx\, \psi(x)|x\rangle.$$

The formal completeness relation is

$$\int_{-\infty}^{\infty} dx\, |x\rangle\langle x|=\hat I,$$

with delta normalization

$$\langle x|x'\rangle=\delta(x-x').$$

The kets $|x\rangle$ are not normalizable physical states in $L^2(\mathbb R)$; they are generalized eigenvectors used to represent ordinary states. Delta distributions encode their normalization and completeness.

Representation as a Unitary Map

For a spinless particle on the line with a simple position spectrum and Lebesgue measure, choosing the position representation defines a map

$$\mathcal U_x:\mathcal H\longrightarrow L^2(\mathbb R), \qquad \mathcal U_x|\psi\rangle=\psi(x).$$

This map preserves inner products and is therefore unitary in the generalized-basis sense:

$$\langle\phi|\psi\rangle_{\mathcal H} = \langle\mathcal U_x\phi,\mathcal U_x\psi\rangle_{L^2}.$$

The abstract vector and the function are not two physical states. They are the same state expressed in two mathematical languages. Likewise, two square-integrable functions that differ only on a set of measure zero represent the same element of $L^2$.

Not every quantum state is represented by a single scalar function of position. A finite internal multiplicity gives, for example, $L^2(\mathbb R)\otimes\mathbb C^n$; more general spectra are described by direct integrals. The scalar notation is the simplest case, not the definition of a quantum state.

Generalized basis caution

The formulas

$$\langle x|x'\rangle=\delta(x-x'), \qquad \int dx\,|x\rangle\langle x|=I$$

are distributional identities. They are meaningful when inserted into matrix elements or applied to suitable test states. Expressions such as $\delta(0)$ signal that generalized basis kets are being manipulated as if they were ordinary normalized vectors.

The practical formalism is reliable, but its literal mathematical home involves spectral measures or a rigged Hilbert space.

Inner Products

The abstract inner product becomes an integral over wavefunctions:

$$\langle \phi|\psi\rangle =\int_{-\infty}^{\infty} dx\,\phi^*(x)\psi(x).$$

Normalization of a one-dimensional wavefunction is therefore

$$\langle \psi|\psi\rangle =\int_{-\infty}^{\infty} dx\,|\psi(x)|^2=1.$$

In three dimensions,

$$\psi(\mathbf r)=\langle \mathbf r|\psi\rangle, \qquad \langle \phi|\psi\rangle =\int_{\mathbb R^3} d^3r\,\phi^*(\mathbf r)\psi(\mathbf r).$$

When using curvilinear coordinates, the coordinate measure changes. In spherical coordinates, $d^3r=r^2\sin\theta\,dr\,d\theta\,d\phi$. The measure is part of the representation, not an optional decoration.

More generally, if

$$I=\int dq\,w(q)|q\rangle\langle q|,$$

then the compatible generalized normalization is

$$\langle q|q'\rangle = \frac{\delta(q-q')}{w(q)}.$$

The wavefunction, completeness relation, and inner product must use one consistent weight convention.

Units and Coordinate Measures

The dimensions of a wavefunction follow from its normalization measure. In one Cartesian dimension,

$$\int dx\,|\psi(x)|^2=1$$

implies

$$[\psi]=L^{-1/2}.$$

In three Cartesian dimensions, $[\psi]=L^{-3/2}$. Correspondingly,

$$[\delta(x-x')]=L^{-1}, \qquad [\delta^{(3)}(\mathbf r-\mathbf r')]=L^{-3}.$$

Dimensional checks catch missing Jacobians and incorrect normalization constants.

For a general coordinate label $q$ with measure $d\mu(q)=w(q)\,dq$, completeness and normalization take the form

$$\int dq\,w(q)\,|q\rangle\langle q|=I, \qquad \int dq\,w(q)\,|\psi(q)|^2=1.$$

There are two common conventions under a coordinate change $x=x(q)$:

• keep the wavefunction as a scalar, $\psi_q(q)=\psi_x(x(q))$, and keep the Jacobian in the measure;
• absorb the square root of the Jacobian into a rescaled wavefunction so the new norm uses flat $dq$.

For example, a radial wavefunction $R(r)$ is normalized with $r^2dr$, while the reduced radial function $u(r)=rR(r)$ is normalized with $dr$. These conventions are equivalent only when the differential operator and boundary conditions are transformed consistently.

Worked coordinate change: quadratic map

Let a normalized state be supported on $x>0$ and define $y=x^2$. If $\widetilde\psi(y)=\psi(\sqrt y)$, then

$$1 = \int_0^\infty dx\,|\psi(x)|^2 = \int_0^\infty \frac{dy}{2\sqrt y} |\widetilde\psi(y)|^2.$$

Thus the same component function uses the weighted measure $d\mu(y)=dy/(2\sqrt y)$. If one instead wants flat measure $dy$, define

$$\chi(y) = \frac{\psi(\sqrt y)}{\sqrt{2\sqrt y}}.$$

Then $\int_0^\infty|\chi(y)|^2dy=1$. The position operator becomes multiplication by $\sqrt y$, and its expectation is unchanged:

$$\int_0^\infty dx\,x|\psi(x)|^2 = \int_0^\infty dy\,\sqrt y\,|\chi(y)|^2.$$

The state, measure, operator, and expectation value transform as one package.

Operators In Coordinate Representation

An abstract operator becomes a rule acting on wavefunctions. In one dimension, the position operator acts by multiplication:

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

The momentum operator acts as a derivative on suitable wavefunctions:

$$(\hat p\psi)(x)=-i\hbar\frac{d\psi}{dx}.$$

The canonical commutator is then represented by

$$[\hat x,\hat p]\psi(x)=i\hbar\psi(x),$$

for wavefunctions in a domain where the differentiations and multiplications are meaningful. This domain caveat matters: boundary conditions can change whether a differential expression defines a valid self-adjoint operator.

Matrix, multiplication, differential, and integral-kernel forms are different representations of the same operator only when their domains are transformed along with their actions.

Integral Kernels

Many operators used in wave mechanics admit a function or distribution kernel on a stated test space,

$$A(x,x')=\langle x|\hat A|x'\rangle.$$

Inserting the position completeness relation gives

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

Not every bounded operator on $L^2$ has an ordinary function kernel. Even the identity and differential operators use delta distributions or their derivatives in this notation.

Local multiplication operators have delta-function kernels:

$$\begin{aligned} \langle x|\hat x|x'\rangle &=x\,\delta(x-x'),\\ \langle x|V(\hat x)|x'\rangle &=V(x)\delta(x-x'). \end{aligned}$$

The momentum kernel is distributional:

$$\langle x|\hat p|x'\rangle = -i\hbar\,\partial_x\delta(x-x').$$

Indeed,

$$\int dx'\, \left[-i\hbar\,\partial_x\delta(x-x')\right]\psi(x') = -i\hbar\,\partial_x\psi(x).$$

Nonlocal operators have kernels not proportional to $\delta(x-x')$. The kernel language therefore unifies matrices, differential operators, and genuine integral operators.

Matrix elements become double integrals:

$$\langle\phi|\hat A|\psi\rangle = \int dx\int dx'\, \phi^*(x)A(x,x')\psi(x').$$

For a local differential operator, the delta functions and their derivatives reduce this expression to the familiar single integral, subject to boundary terms and domain conditions.

Hamiltonians As Differential Operators

For a nonrelativistic particle in a real scalar potential $V(x)$, the closed-system Hamiltonian often appears as

$$\hat H =\frac{\hat p^2}{2m}+V(\hat x).$$

In coordinate representation this becomes

$$(\hat H\psi)(x) =-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi(x).$$

This is why wave mechanics is full of differential equations. The abstract energy-eigenvalue equation

$$\hat H|\psi\rangle=E|\psi\rangle$$

becomes the time-independent Schrödinger equation in coordinates:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} +V(x)\psi(x) =E\psi(x).$$

The differential expression must be supplemented by a Hilbert space and boundary conditions before the problem is fully specified.

Time-Dependent Coordinate Wavefunctions

In the Schrödinger picture,

$$\psi(x,t)=\langle x|\psi(t)\rangle.$$

Projecting the abstract equation

$$i\hbar\frac{d}{dt}|\psi(t)\rangle = \hat H|\psi(t)\rangle$$

onto $\langle x|$ gives

$$i\hbar\frac{\partial\psi(x,t)}{\partial t} = (\hat H\psi)(x,t).$$

For a local Hamiltonian this is a partial differential equation. For a nonlocal kernel it is

$$i\hbar\partial_t\psi(x,t) = \int dx'\,H(x,x';t)\psi(x',t).$$

The coordinate representation does not require locality. Local differential Hamiltonians are an important physical subclass.

Momentum Representation Comparison

Position representation is not privileged by the formalism. The same state can be represented by momentum amplitudes

$$\phi(p)=\langle p|\psi\rangle.$$

With the convention used here,

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

The Fourier transform supplies the mathematical change to momentum space, with normalization factors fixed by the declared convention. Physical predictions do not depend on choosing position or momentum representation, but different representations make different questions easier.

General Change of Representation

Let $\{|a\rangle\}$ and $\{|b\rangle\}$ be two generalized or discrete complete bases. Their components are related by the overlap kernel:

$$\psi_b(b) = \langle b|\psi\rangle = \int da\, \langle b|a\rangle\psi_a(a),$$

with sums replacing integrals for discrete labels and the appropriate measure included. The kernel $\langle p|x\rangle$ produces the Fourier transform. A rotation between two finite orthonormal bases produces ordinary unitary matrix multiplication.

Operators transform at the same time. If $\mathcal U$ is the representation map, then

$$\hat A \longmapsto A_{\mathrm{rep}} = \mathcal U\hat A\mathcal U^{-1}.$$

Changing only the state components while leaving the operator formula unchanged generally describes a different calculation. Representation independence comes from transforming states, operators, inner products, and measures consistently.

Multi-Component Wavefunctions

For a particle with an internal orthonormal basis $|s\rangle$, use the product basis $|x,s\rangle$. The coordinate wavefunction is

$$\psi_s(x)=\langle x,s|\psi\rangle.$$

Completeness and normalization become

$$\sum_s\int dx\,|x,s\rangle\langle x,s|=I$$

and

$$\sum_s\int dx\,|\psi_s(x)|^2=1.$$

For spin-$1/2$, $\psi(x)$ is a two-component spinor. Operators can act on position, on the internal index, or on both. A matrix-valued potential $V_{ss'}(x)$ couples components:

$$(\hat V\psi)_s(x) = \sum_{s'}V_{ss'}(x)\psi_{s'}(x).$$

This is the natural coordinate representation for spin-dependent interactions, coupled channels, and multiband models. For a closed-system local potential, the matrix must be Hermitian at each point: $V(x)=V(x)^\dagger$.

Physical Interpretation

The coordinate representation makes spatial questions direct. The probability density for finding the particle near $x$ is $|\psi(x)|^2$, and probabilities come from integrals over regions. Nodes, oscillations, phases, and tails of $\psi(x)$ carry information about confinement, momentum content, currents, tunneling, and boundary behavior.

However, the wavefunction is not a classical wave in space. It is a complex probability amplitude whose squared magnitude gives a probability density through the Born rule. Global phase does not change physical probabilities, while relative phase can affect interference and currents.

The phase of the generalized basis itself is also conventional. Replacing

$$|x\rangle\longmapsto e^{i\chi(x)}|x\rangle$$

changes the component function by the opposite local phase. Differential operator formulas then change as well. This is a passive rephasing of the representation basis; it is mathematically analogous to gauge covariance but is not by itself an electromagnetic gauge transformation.

Boundary of This Page

This page owns the position-space calculation layer: coordinate measures, Jacobians, differential and multiplication operators, kernels, and the domain caveats needed for wave mechanics. The Born interpretation of $|\psi|^2$ belongs to Wavefunctions and Probability Density. Dynamics belongs to the Time-Dependent Schrödinger Equation in Coordinate Space, and detailed domain choices belong to Boundary Conditions. General representation theory and Fourier analysis remain separate mathematical treatments.

Common Mistakes

• Saying $\psi(x)$ is the state rather than the representation of $|\psi\rangle$.
• Treating generalized kets $|x\rangle$ as normalizable vectors.
• Forgetting that a wavefunction has units. In one dimension, normalized $\psi(x)$ has units of length$^{-1/2}$.
• Dropping the coordinate measure in spherical or other curvilinear coordinates.
• Assuming that a differential expression alone defines a Hamiltonian without specifying its domain.
• Confusing position-space probability density $|\psi(x)|^2$ with momentum-space probability density $|\phi(p)|^2$.
• Changing wavefunction coordinates without transforming the measure and operator.
• Assuming every coordinate wavefunction is scalar when internal degrees of freedom are present.
• Reading an arbitrary phase or sign change of basis components as a physical change of state.
• Treating a nonlocal kernel as though it were multiplication by a function.

Where This Is Used

• Wavefunctions and Probability Density interprets $|\psi(x)|^2$ and $|\psi(\mathbf r)|^2$.
• Time-Dependent Schrödinger Equation in Coordinate Space uses coordinate-space Hamiltonians for time evolution.
• State Vectors explains the abstract state language behind this representation.
• Hamiltonians in Coordinate Space develops the differential-operator construction.
• The Schrödinger Representation gives the rigorous representation-theoretic setting.

Exercises

1. Starting from the completeness relation $\int dx\,|x\rangle\langle x|=\hat I$, derive the expansion $|\psi\rangle=\int dx\,\psi(x)|x\rangle$.

Solution

Insert the identity into the state:

$$|\psi\rangle =\hat I|\psi\rangle =\int dx\,|x\rangle\langle x|\psi\rangle.$$

By definition, $\psi(x)=\langle x|\psi\rangle$, so

$$|\psi\rangle=\int dx\,\psi(x)|x\rangle.$$

2. Show that the position-space inner product follows from the same completeness relation.

Solution

Insert the identity between $\langle\phi|$ and $|\psi\rangle$:

$$\langle\phi|\psi\rangle =\int dx\,\langle\phi|x\rangle\langle x|\psi\rangle.$$

Since $\langle x|\psi\rangle=\psi(x)$ and $\langle\phi|x\rangle=\phi^*(x)$, this gives

$$\langle\phi|\psi\rangle =\int dx\,\phi^*(x)\psi(x).$$

3. Derive the action of an operator from its coordinate-space kernel.

Solution

Insert the identity on both sides of $\hat A$:

$$\begin{aligned} \hat A|\psi\rangle &=\int dy\int dx'\, |y\rangle\langle y|\hat A|x'\rangle \langle x'|\psi\rangle. \end{aligned}$$

Project with $\langle x|$ and use

$$A(x,x')=\langle x|\hat A|x'\rangle, \qquad \psi(x')=\langle x'|\psi\rangle.$$

The result is

$$(\hat A\psi)(x) = \int dx'\,A(x,x')\psi(x').$$

4. Show that $u(r)=rR(r)$ converts the radial measure $r^2dr$ to a flat measure.

Solution

Substitute $R(r)=u(r)/r$:

$$\int_0^\infty |R(r)|^2r^2\,dr = \int_0^\infty \left|\frac{u(r)}{r}\right|^2r^2\,dr = \int_0^\infty|u(r)|^2\,dr.$$

The normalization is equivalent, but the radial differential operator and the behavior required at $r=0$ must also be transformed.

5. Write the norm and expectation value of a position-dependent spin-$1/2$ state.

Solution

Write

$$\psi(x) = \begin{pmatrix} \psi_\uparrow(x)\\ \psi_\downarrow(x) \end{pmatrix}.$$

Its norm is

$$\int dx\, \left( |\psi_\uparrow(x)|^2 +|\psi_\downarrow(x)|^2 \right)=1.$$

For a matrix-valued local observable $A(x)$ acting on the spin index,

$$\langle A\rangle = \int dx\,\psi^\dagger(x)A(x)\psi(x).$$

6. Rephase the generalized basis by $|x\rangle'=e^{i\chi(x)}|x\rangle$. Derive the transformed wavefunction and momentum representation.

Solution

The component is

$$\psi'(x)=\langle x|'\psi\rangle=e^{-i\chi(x)}\psi(x).$$

Writing $\psi=e^{i\chi}\psi'$ and transforming the original operator gives

$$\hat p' = e^{-i\chi}(-i\hbar\partial_x)e^{i\chi} = -i\hbar\partial_x+\hbar\chi'(x).$$

The component and operator both change, so expectation values are unchanged.

7. Show that $-i\hbar\partial_x$ is symmetric on an interval for periodic test functions, and identify the boundary term.

Solution

Integration by parts gives

$$\langle\phi,p\psi\rangle-\langle p\phi,\psi\rangle = -i\hbar\,[\phi^*(x)\psi(x)]_a^b.$$

If both functions obey $f(b)=f(a)$, the endpoint products agree and the term vanishes. This demonstrates why the represented differential action is not enough: its domain determines the operator property.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• 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.