---
title: Position, Momentum, and Continuous Spectra
module: The Formalism of Quantum Mechanics
moduleNumber: 4
lessonNumber: 4
order: 404
summary: >
  Position and momentum are the observables with no normalizable eigenstates: their
  spectra are continuous, their eigenkets are delta-normalized, and the two are
  Fourier conjugates. We derive the canonical commutator from the momentum operator,
  build the continuous-basis machinery (Dirac deltas replacing Kronecker deltas),
  show the position and momentum wavefunctions are a Fourier-transform pair, and
  compute expectation values in either representation.
topics: [The Formalism of Quantum Mechanics]
sources:
  - book: Shankar
    ref: "Ch. 1 §1.10 Generalization to Infinite Dimensions; Ch. 4 §4.2 The X and P Operators"
  - book: Griffiths & Schroeter
    ref: "Ch. 3; §3.5 Position, §3.6 Momentum Space"
  - book: Cohen-Tannoudji, Diu & Laloë
    ref: "Ch. II §E — Closure relations, the x and p representations"
draft: false
---

Position and momentum are the observables that force the continuous machinery of
the formalism. Neither has a normalizable eigenstate — a particle cannot be at
exactly one point, nor have exactly one momentum, while remaining in $L^2$ — so the
[spectral theorem](/quantum-mechanics/formalism/observables-hermitian-operators-and-eigenvalues)
must be extended to continuous spectra. The eigenkets become delta-normalized, the
sums become integrals, and the two observables turn out to be Fourier conjugates.
Their non-commutation, expressed by the canonical commutator $[\hat x,\hat p]=i\hbar$,
is the algebraic root of the uncertainty principle and the starting point of nearly
every quantization scheme.

## Continuous bases

Discrete orthonormality $\braket{e_i|e_j}=\delta_{ij}$ has no direct continuous
analogue, because a continuum of orthonormal vectors would require an uncountable
sum equal to one. The resolution is to normalize the continuous eigenkets to a
Dirac delta.

> **Definition (Continuous orthonormality and completeness).** A continuous basis
> $\{\ket{\xi}\}$ labeled by a real index $\xi$ satisfies
> $$
> \braket{\xi|\xi'} = \delta(\xi-\xi'), \qquad
> \int \ket{\xi}\bra{\xi}\d\xi = \mathbb{1}.
> $$
> The first is the continuous replacement of $\delta_{ij}$; the second is the
> resolution of the identity with the sum promoted to an integral.

The Dirac delta $\delta(\xi-\xi')$ is not a function but a distribution, defined by
$\int f(\xi')\delta(\xi-\xi')\d\xi'=f(\xi)$. It is the continuous Kronecker delta:
it collapses an integral to the value of the integrand at the matching point,
exactly as $\delta_{ij}$ collapses a sum. Every discrete identity of the formalism
carries over with $\sum_i\to\int\d\xi$ and $\delta_{ij}\to\delta(\xi-\xi')$. A
generic state expands as

$$
\ket{\psi} = \int \ket{\xi}\braket{\xi|\psi}\d\xi = \int \psi(\xi)\,\ket{\xi}\d\xi,
\qquad \psi(\xi) = \braket{\xi|\psi},
$$

with components $\psi(\xi)$ forming a wavefunction rather than a column. The price
of the continuum is that $\ket{\xi}$ is not itself a physical state: its norm is
$\braket{\xi|\xi}=\delta(0)$, which is infinite. Continuous eigenkets are idealized
basis elements, useful as intermediaries, and physical states are always
normalizable superpositions of them.

$$
% caption: A continuous basis labels its eigenkets by a real parameter; the
% orthonormality is a Dirac delta spike at coincident labels, the continuous
% counterpart of the Kronecker delta of a discrete basis.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (0,0) -- (6.2,0) node[right, black!70] {second label};
\draw[black, ->] (0,0) -- (0,3.0) node[above, black!70] {overlap};
% delta spike at coincident labels
\draw[acc, very thick, ->] (3.0,0) -- (3.0,2.6);
\node[acc, anchor=south] at (3.0,2.6) {delta spike};
\node[anchor=north, black!70] at (3.0,-0.08) {labels equal};
\draw[black] (0,0.02) -- (6.0,0.02);
\end{tikzpicture}
$$

## The position operator

The position operator $\hat x$ acts on a wavefunction by multiplication,
$\braket{x|\hat x|\psi}=x\,\psi(x)$, and its eigenkets $\ket{x_0}$ satisfy
$\hat x\ket{x_0}=x_0\ket{x_0}$ with continuous spectrum $x_0\in\mathbb{R}$. In the
position representation the eigenket of $\hat x$ at $x_0$ has wavefunction
$\braket{x|x_0}=\delta(x-x_0)$, a spike at $x_0$, confirming the delta
normalization $\braket{x_0|x_0'}=\delta(x_0-x_0')$. The completeness relation

$$
\int \ket{x}\bra{x}\d x = \mathbb{1}
$$

is the identity inserted throughout wave mechanics to convert abstract kets into
wavefunctions. Position is Hermitian: for real $x$ the multiplication operator
satisfies $\braket{\phi|\hat x\,\psi}=\int\phi^\ast x\psi\d x=\braket{\hat x\phi|\psi}$.

## The momentum operator

Momentum is the generator of spatial translation. The translation operator
$\hat T(a)$ defined by $\hat T(a)\ket{x}=\ket{x+a}$ shifts a wavefunction,
$\braket{x|\hat T(a)|\psi}=\psi(x-a)$. For an infinitesimal shift $\varepsilon$,
Taylor expansion gives

$$
\psi(x-\varepsilon) = \psi(x) - \varepsilon\,\partial_x\psi(x) + O(\varepsilon^2),
$$

so $\hat T(\varepsilon) = \mathbb{1} - \varepsilon\,\partial_x$ to first order.
Writing the generator as a Hermitian operator, $\hat T(\varepsilon)=\mathbb{1}-\tfrac{i}{\hbar}\varepsilon\hat p$,
identifies

$$
\braket{x|\hat p|\psi} = -i\hbar\,\frac{\partial}{\partial x}\,\psi(x),
$$

the momentum operator in the position representation. The factors are fixed by two
requirements: $\hat p$ must be Hermitian (the $-i\hbar$ makes the derivative
self-adjoint on functions vanishing at infinity), and its eigenfunctions must be de
Broglie waves. Solving $\hat p\,u_p(x)=p\,u_p(x)$ gives
$-i\hbar\,u_p'=p\,u_p$, hence

$$
\braket{x|p} = u_p(x) = \frac{1}{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar},
$$

a plane wave of wavelength $2\pi\hbar/p=h/p$, the de Broglie relation recovered.

$$
% caption: Momentum is the generator of translation: an infinitesimal spatial shift
% is the identity minus $\tfrac{i}{\hbar}\varepsilon\hat p$, and exponentiating the
% generator produces a finite displacement of the wavefunction.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (0,0) -- (6.6,0) node[right, black!70] {$x$};
\draw[black, ->] (0,0) -- (0,2.7);
\draw[acc, very thick] plot[domain=0.2:3.2, samples=110]
  (\x, {2.2*exp(-3.0*(\x-1.5)^2)});
\node[acc, anchor=south] at (1.5,2.1) {packet};
\draw[black, very thick, dashed] plot[domain=2.2:5.2, samples=110]
  (\x, {2.2*exp(-3.0*(\x-3.5)^2)});
\node[black!70, anchor=south] at (3.5,2.1) {shifted};
\draw[->, black, thick] (1.5,0.4) -- (3.5,0.4) node[midway, above, black!70] {shift by a};
\end{tikzpicture}
$$

The normalization constant is fixed by requiring delta orthonormality:

$$
\braket{p|p'} = \int\braket{p|x}\braket{x|p'}\d x
= \frac{1}{2\pi\hbar}\int e^{i(p'-p)x/\hbar}\d x = \delta(p-p'),
$$

using the integral representation $\int e^{iky}\d y=2\pi\delta(k)$. Momentum has
continuous spectrum $p\in\mathbb{R}$ and its own completeness relation
$\int\ket{p}\bra{p}\d p=\mathbb{1}$.

## The canonical commutator

The defining algebraic relation of quantum mechanics follows by applying
$[\hat x,\hat p]$ to an arbitrary wavefunction. In the position representation,

$$
\begin{aligned}
\braket{x|[\hat x,\hat p]|\psi}
&= x\bigl(-i\hbar\,\partial_x\psi\bigr) - \bigl(-i\hbar\,\partial_x\bigr)\bigl(x\psi\bigr) \\
&= -i\hbar x\,\partial_x\psi + i\hbar\bigl(\psi + x\,\partial_x\psi\bigr)
= i\hbar\,\psi(x),
\end{aligned}
$$

where the product rule on $\partial_x(x\psi)=\psi+x\partial_x\psi$ leaves only the
term in which the derivative hits $x$. Since this holds for every $\psi$,

$$
\;[\hat x,\hat p] = i\hbar\;
$$

the **canonical commutation relation**. It is the single equation from which the
uncertainty principle, the ladder structure of the oscillator, and the
correspondence with classical Poisson brackets all descend. The classical limit is
explicit: the commutator divided by $i\hbar$ reproduces the Poisson bracket
$\{x,p\}=1$, and Dirac's quantization rule $\{A,B\}\mapsto\tfrac{1}{i\hbar}[\hat A,\hat B]$
elevates every classical canonical pair to operators obeying this relation. In three
dimensions it generalizes to $[\hat x_i,\hat p_j]=i\hbar\,\delta_{ij}$, with
different components of position and momentum commuting, so a particle can have
simultaneously definite $x$-position and $y$-momentum but never definite $x$-position
and $x$-momentum. The trace argument of the
[spectral-theorem lesson](/quantum-mechanics/formalism/observables-hermitian-operators-and-eigenvalues)
shows why: no finite matrices can satisfy $[\hat x,\hat p]=i\hbar$, so the operators
that realize it are necessarily unbounded and act on an infinite-dimensional space.

$$
% caption: The canonical commutator acts on a test function: applying position then
% momentum differs from momentum then position by exactly the term where the
% derivative strikes the multiplying coordinate, leaving $i\hbar$ times the
% function.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0,
  box/.style={draw, minimum width=22mm, minimum height=10mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (psi) at (0,0) {test function};
\node[box] (xp) at (4.2,1.2) {$x$ then $p$};
\node[box] (px) at (4.2,-1.2) {$p$ then $x$};
\node[box, draw=acc, text=acc] (out) at (9.0,0) {a nonzero gap};
\draw[->, black, thick] (psi) -- (xp);
\draw[->, black, thick] (psi) -- (px);
\draw[->, acc, thick] (xp) -- (out) node[midway, above, sloped, black!70] {subtract};
\draw[->, acc, thick] (px) -- (out);
\end{tikzpicture}
$$

## Position and momentum as a Fourier pair

The two representations of a state are connected by the plane-wave overlap. Insert
the momentum completeness into $\psi(x)=\braket{x|\psi}$,

$$
\psi(x) = \int \braket{x|p}\braket{p|\psi}\d p
= \frac{1}{\sqrt{2\pi\hbar}}\int \tilde\psi(p)\,e^{ipx/\hbar}\d p,
$$

and symmetrically, inserting position completeness into $\tilde\psi(p)=\braket{p|\psi}$,

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

The position wavefunction and the momentum wavefunction are a **Fourier-transform
pair**. Localizing one spreads the other: a narrow $\psi(x)$ needs a broad band of
plane waves, so $\tilde\psi(p)$ is wide, and conversely. This is the Fourier
statement that becomes the [uncertainty principle](/quantum-mechanics/formalism/commutators-and-the-generalized-uncertainty-principle)
once the widths are made precise. Both wavefunctions describe the identical abstract
state $\ket{\psi}$; they are its components in two different continuous bases.
Parseval's theorem, $\int\lvert\psi(x)\rvert^2\d x=\int\lvert\tilde\psi(p)\rvert^2\d p$,
expresses that the norm is the same computed in either basis, so
$\lvert\tilde\psi(p)\rvert^2$ is the probability density in momentum.

$$
% caption: Position and momentum wavefunctions are Fourier conjugates: a state
% sharply peaked in position is broad in momentum, and the same state viewed in the
% momentum basis narrows there only by spreading in position.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% narrow in x
\draw[black, ->] (0,0) -- (3.6,0) node[right, black!70] {$x$};
\draw[acc, very thick] plot[domain=0.15:3.4, samples=120]
  (\x, {2.4*exp(-9.0*(\x-1.7)^2)});
\node[acc, anchor=south] at (1.7,2.2) {position};
\draw[black, <->] (1.45,-0.4) -- (1.95,-0.4);
\node[anchor=north, black!70] at (1.7,-0.4) {narrow};
% wide in p
\begin{scope}[xshift=5.4cm]
\draw[black, ->] (0,0) -- (3.6,0) node[right, black!70] {$p$};
\draw[acc, very thick] plot[domain=0.15:3.5, samples=120]
  (\x, {1.5*exp(-0.9*(\x-1.8)^2)});
\node[acc, anchor=south] at (1.8,1.7) {momentum};
\draw[black, <->] (0.85,-0.4) -- (2.75,-0.4);
\node[anchor=north, black!70] at (1.8,-0.4) {broad};
\end{scope}
\end{tikzpicture}
$$

## Expectation values in either representation

An expectation value can be computed in whichever basis is convenient, and the
answer is the same. In the position basis, resolving the identity twice turns
$\braket{\psi|\hat A|\psi}$ into an integral of the wavefunction against the
operator's kernel. For position and momentum the working forms are

$$
\langle\hat x\rangle = \int \psi^\ast(x)\,x\,\psi(x)\d x,
\qquad
\langle\hat p\rangle = \int \psi^\ast(x)\,\Bigl(-i\hbar\,\partial_x\Bigr)\psi(x)\d x,
$$

or, in the momentum basis, with the roles exchanged and $\hat x\to i\hbar\,\partial_p$,

$$
\langle\hat p\rangle = \int \tilde\psi^\ast(p)\,p\,\tilde\psi(p)\d p,
\qquad
\langle\hat x\rangle = \int \tilde\psi^\ast(p)\,\Bigl(i\hbar\,\partial_p\Bigr)\tilde\psi(p)\d p.
$$

Position is diagonal in the position basis and momentum is diagonal in the momentum
basis; each operator is a simple multiplication in its own representation and a
derivative in the other. The symmetry $\hat x\leftrightarrow\hat p$,
$x\leftrightarrow p$, $-i\hbar\partial_x\leftrightarrow +i\hbar\partial_p$ reflects
the Fourier duality and the sign in the canonical commutator.

### The momentum-space Schrödinger equation

The choice of representation extends to the dynamics. The time-dependent Schrödinger
equation $i\hbar\,\partial_t\ket{\psi}=\hat H\ket{\psi}$ becomes, in the momentum
basis with $\hat H=\tfrac{\hat p^2}{2m}+V(\hat x)$,

$$
i\hbar\,\frac{\partial\tilde\psi(p,t)}{\partial t}
= \frac{p^2}{2m}\,\tilde\psi(p,t) + V\!\left(i\hbar\,\frac{\partial}{\partial p}\right)\tilde\psi(p,t).
$$

The kinetic term, a derivative operator in position space, is now a plain
multiplication by $p^2/2m$, while the potential becomes a differential (or, for a
generic $V$, an integral) operator. Momentum space is therefore the natural setting
whenever the kinetic energy dominates and the potential is simple in $p$ — the
[free particle](/quantum-mechanics/wave-mechanics-1d/the-free-particle-and-wave-packet-dynamics),
a uniform force, and scattering problems where the incident and scattered states are
momentum eigenstates. The representations are exchanged by the same Fourier transform
that relates the wavefunctions, so a problem awkward in one basis is often
transparent in the other.

> **Worked example.** A Gaussian state has position wavefunction
> $$\psi(x) = \left(\frac{1}{2\pi\sigma^2}\right)^{1/4} e^{-x^2/4\sigma^2},$$
> normalized so $\int\lvert\psi\rvert^2\d x=1$. Its Fourier transform is again a
> Gaussian,
> $$\tilde\psi(p) = \left(\frac{2\sigma^2}{\pi\hbar^2}\right)^{1/4} e^{-\sigma^2 p^2/\hbar^2},$$
> with width in momentum $\hbar/2\sigma$ set inversely by the position width
> $\sigma$. The expectation values vanish by symmetry,
> $\langle\hat x\rangle=\langle\hat p\rangle=0$, and the spreads are
> $$
> \sigma_x^2 = \langle\hat x^2\rangle = \sigma^2,
> \qquad \sigma_p^2 = \langle\hat p^2\rangle = \frac{\hbar^2}{4\sigma^2}.
> $$
> Their product is $\sigma_x\sigma_p=\hbar/2$, the minimum allowed by the
> uncertainty principle. The Gaussian is the state that saturates the bound, a fact
> derived in general in the next lesson.

The continuous-spectrum apparatus — delta normalization, the Fourier pairing, and
the canonical commutator — is the operator content behind wave mechanics. The next
lesson turns $[\hat x,\hat p]=i\hbar$ into a quantitative limit on simultaneous
knowledge through the
[generalized uncertainty principle](/quantum-mechanics/formalism/commutators-and-the-generalized-uncertainty-principle).

[^shankar-xp]: **Shankar**, _Principles of Quantum Mechanics_ 2nd ed., §1.10 and §4.2 — passage to infinite dimensions, the $X$ and $P$ operators, their delta-normalized eigenkets, and the derivation of $P$ as the generator of translations. Springer, 1994.
[^griffiths-mom]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_ 3rd ed., §3.5–§3.6 — the position and momentum representations, momentum space, and the Fourier relation between $\psi(x)$ and $\tilde\psi(p)$. Cambridge, 2018.