Schrödinger Representation

The Schrödinger representation realizes $n$ canonical position-momentum pairs on

$$\mathcal H=L^2(\mathbb R^n,d^nx).$$

Translations $T(q)$ and modulations $M(p)$ act by

$$(T(q)\psi)(x)=\psi(x-q), \qquad (M(p)\psi)(x)=e^{ip\cdot x/\hbar}\psi(x),$$

and obey

$$T(q)M(p) = e^{-iq\cdot p/\hbar}M(p)T(q).$$

This is a regular irreducible representation of the Weyl relations for finite $n$ and nonzero $\hbar$. Its self-adjoint generators are position by multiplication and momentum by weak differentiation on their maximal domains.

Required background. The Weyl Form of the CCR fixes every sign and defines regularity and irreducibility. The analysis uses $L^2$ equivalence classes, the unitary Fourier transform, weak derivatives, and Schwartz space.

Helpful background. State Vectors distinguishes abstract Hilbert-space vectors from coordinate wavefunctions.

Translations and modulations on L²(Rⁿ)

Vectors in $L^2(\mathbb R^n)$ are equivalence classes of measurable functions equal almost everywhere. The displayed formulas are well defined on those classes because translations preserve null sets and multiplication by a measurable phase preserves almost-everywhere equality.

Both families are unitary. For translations, the change of variables $y=x-q$ gives

$$\|T(q)\psi\|_2^2 = \int_{\mathbb R^n}|\psi(x-q)|^2\,d^nx = \|\psi\|_2^2,$$

and $T(q)^{-1}=T(-q)$. For modulations,

$$|e^{ip\cdot x/\hbar}|=1,$$

so $\|M(p)\psi\|_2=\|\psi\|_2$ and $M(p)^{-1}=M(-p)$.

The additive group laws are

$$T(q)T(q')=T(q+q'), \qquad M(p)M(p')=M(p+p').$$

The cross relation follows directly:

$$\begin{aligned} (T(q)M(p)\psi)(x) &= e^{ip\cdot(x-q)/\hbar}\psi(x-q)\\ &= e^{-iq\cdot p/\hbar}(M(p)T(q)\psi)(x). \end{aligned}$$

Thus the coordinate realization exactly matches the convention fixed on the Weyl page.

Strong continuity and regularity

For every fixed $\psi\in L^2(\mathbb R^n)$,

$$\|T(q)\psi-\psi\|_2\longrightarrow0 \qquad(q\to0).$$

One proves this first for compactly supported continuous functions using uniform continuity, then extends it by density and the unitary norm bound.

Likewise,

$$\|M(p)\psi-\psi\|_2^2 = \int_{\mathbb R^n} |e^{ip\cdot x/\hbar}-1|^2|\psi(x)|^2\,d^nx \longrightarrow0$$

by dominated convergence. Therefore $q\mapsto T(q)$ and $p\mapsto M(p)$ are strongly continuous, and the Schrödinger Weyl system is regular.

Neither family is generally operator-norm continuous. Arbitrarily high-momentum or far-position-localized test vectors can accumulate a nearly maximal phase under an arbitrarily small nonzero parameter change. This is consistent with their unbounded generators.

Position generators and their domains

For each coordinate $j$, the self-adjoint position operator is

$$(Q_j\psi)(x)=x_j\psi(x)$$

on the maximal domain

$$D(Q_j) = \{\psi\in L^2(\mathbb R^n):x_j\psi\in L^2(\mathbb R^n)\}.$$

Its spectral calculus gives

$$M(p) = \exp\!\left( \frac{i}{\hbar}\sum_{j=1}^{n}p_jQ_j \right).$$

For a vector in $D(Q_j)$,

$$Q_j\psi = -i\hbar \left. \frac{d}{ds}M(se_j)\psi \right|_{s=0},$$

where the derivative is a Hilbert-space norm limit. The sign follows from $M(se_j)=e^{isQ_j/\hbar}$.

The PVM of the commuting position tuple acts by indicator multiplication:

$$(E_Q(\Delta)\psi)(x) = \mathbf 1_\Delta(x)\psi(x), \qquad \Delta\in\mathcal B(\mathbb R^n).$$

Momentum generators and Fourier space

For $\psi\in\mathcal S(\mathbb R^n)$, define the Fourier transform by

$$\widehat\psi(k) = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb R^n}e^{-ik\cdot x}\psi(x)\,d^nx.$$

Plancherel’s theorem extends this map uniquely to a unitary operator on all of $L^2(\mathbb R^n)$. For a general $L^2$ equivalence class, the displayed integral is therefore a dense-class definition followed by continuous extension, not a claim of pointwise convergence.

In Fourier space, translation becomes phase multiplication:

$$\widehat{T(q)\psi}(k) = e^{-ik\cdot q}\widehat\psi(k).$$

Stone’s theorem therefore gives

$$\widehat{P_j\psi}(k) = \hbar k_j\widehat\psi(k)$$

on

$$D(P_j) = \{\psi\in L^2(\mathbb R^n):k_j\widehat\psi(k)\in L^2(\mathbb R^n)\}.$$

In position space this is

$$P_j\psi = -i\hbar\,\partial_j\psi,$$

where $\partial_j\psi$ is the weak derivative and belongs to $L^2$. The generator formula is

$$P_j\psi = i\hbar \left. \frac{d}{ds}T(se_j)\psi \right|_{s=0},$$

consistent with $T(se_j)=e^{-isP_j/\hbar}$.

These maximal domains are part of the representation. Writing $P_j=-i\hbar\partial_j$ without the weak-derivative domain specifies only a formal expression.

Schwartz space as a common invariant core

The Schwartz space $\mathcal S(\mathbb R^n)$ consists of smooth functions for which every polynomially weighted derivative decays rapidly. It is dense in $L^2$ and invariant under

$$Q_j, \quad P_j, \quad T(q), \quad M(p),$$

as well as finite products and polynomial combinations of the $Q_j$ and $P_j$. On this common core,

$$[Q_j,P_k]\psi = i\hbar\delta_{jk}\psi, \qquad \psi\in\mathcal S(\mathbb R^n).$$

For example,

$$\begin{aligned} (Q_jP_k-P_kQ_j)\psi &= -i\hbar x_j\partial_k\psi +i\hbar\partial_k(x_j\psi)\\ &= i\hbar\delta_{jk}\psi. \end{aligned}$$

The core makes the calculation legitimate; it does not make $Q_j$ or $P_j$ bounded or everywhere defined.

Irreducibility

The Schrödinger Weyl system is irreducible. A useful bounded-commutant proof has two steps.

1. If a bounded operator $B$ commutes with every modulation $M(p)$, then it belongs to the commutant of the maximal multiplication algebra. Hence $B$ itself is multiplication by some essentially bounded function $g(x)$.

2. If this multiplication operator also commutes with every translation, then

   $$g(x)=g(x-q)$$

   almost everywhere for every $q$. Such an essentially bounded translation-invariant function is constant almost everywhere.

Therefore every bounded operator commuting with all $T(q)$ and $M(p)$ is a scalar multiple of $I$. By the commutant criterion, the representation is irreducible.

The first step imports the standard theorem that the multiplication operators form a maximal abelian von Neumann algebra; the argument is a proof sketch of that step, not a replacement for its measure-theoretic proof.

Regularity and irreducibility have now been established separately. Together with finite $n$ and fixed nonzero $\hbar$, they are the hypotheses that place this representation on the unique side of the Stone–von Neumann theorem.

Generalized position and momentum kets

The symbols $|x\rangle$ and $|p\rangle$ are convenient distributional objects, not elements of $L^2(\mathbb R^n)$. A point-supported delta distribution is not square-integrable, and neither is a plane wave of constant magnitude. Equations such as

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

are representation formulas interpreted through distributions or a rigged Hilbert space.

The actual Hilbert-space spectral projections refer to measurable regions:

$$\Pr_\psi(Q\in\Delta) = \int_\Delta|\psi(x)|^2\,d^nx.$$

No generalized eigenket is needed to define that probability.

Representation is not picture or basis

The Schrödinger representation in this page means a representation of the Weyl or Heisenberg group. It specifies how canonical transformations act on $L^2(\mathbb R^n)$.

The Schrödinger picture is instead a convention for placing time dependence in state vectors rather than observables. One can use the Schrödinger picture in an abstract Hilbert-space representation, and one can discuss the Schrödinger representation in the Heisenberg picture. The terms answer different questions.

Likewise, the momentum-space wavefunction is obtained from the same Schrödinger representation by a unitary Fourier transform. It is not a physically inequivalent representation. Stone–von Neumann uniqueness makes that statement precise for finite canonical systems satisfying its hypotheses.

Scope and limitations

This page owns the realization as a regular irreducible Weyl representation, including its maximal generator domains and common core. Coordinate-wavefunction techniques and Fourier calculations are supporting calculation methods rather than part of this representation theorem’s ownership.

The construction assumes Euclidean configuration space $\mathbb R^n$, finite $n$, and a fixed nonzero central character. Particles on circles or other configuration manifolds have different global groups and boundary data. Quantum fields and infinite systems can have regular representations not unitarily equivalent to this one.

Common pitfalls

Treating $L^2$ vectors as pointwise functions. They are almost-everywhere equivalence classes. Point evaluation is not a continuous functional on general $L^2$.

Omitting generator domains. Multiplication and differentiation formulas do not define self-adjoint operators until their maximal domains or a proven essentially self-adjoint core is stated.

Confusing regularity with irreducibility. Strong continuity establishes regularity. The commutant argument establishes irreducibility.

Turning generalized kets into normalized states. Delta functions and plane waves live outside $L^2$ and require distributional interpretation.

Confusing representation with picture. The Schrödinger representation is about the CCR; the Schrödinger picture is about time dependence.

Exercises

1. Verify unitarity and adjoints

Prove $T(q)^*=T(-q)$ and $M(p)^*=M(-p)$ directly from the inner product.

Solution

A change of variables gives

$$\langle\phi,T(q)\psi\rangle = \int\phi(x)^*\psi(x-q)\,d^nx = \langle T(-q)\phi,\psi\rangle.$$

Complex conjugation of the phase similarly gives

$$\langle\phi,M(p)\psi\rangle = \langle M(-p)\phi,\psi\rangle.$$

Each proposed adjoint is also the inverse, so both operators are unitary.

2. Derive the momentum generator

For $\psi\in\mathcal S(\mathbb R^n)$, compute the strong derivative of $T(se_j)\psi$ at $s=0$.

Solution

Because $T(se_j)\psi(x)=\psi(x-se_j)$,

$$\left. \frac{d}{ds}T(se_j)\psi(x) \right|_{s=0} = -\partial_j\psi(x).$$

Multiplying by $i\hbar$ gives $P_j\psi=-i\hbar\partial_j\psi$.

3. Check the common-core commutator

Calculate $[Q_j,P_k]\psi$ for $\psi\in\mathcal S(\mathbb R^n)$ and explain why the same written expression is not automatically valid for every $L^2$ vector.

Solution

The product rule gives

$$\begin{aligned} [Q_j,P_k]\psi &= -i\hbar x_j\partial_k\psi +i\hbar\partial_k(x_j\psi)\\ &= i\hbar\delta_{jk}\psi. \end{aligned}$$

A general $L^2$ vector need not lie in either generator domain, much less in $D(Q_jP_k)\cap D(P_kQ_j)$. The commutator is therefore a common-domain identity, not an everywhere-defined bounded equation.

4. Strong continuity of modulations

Give the dominated-convergence proof of $\|M(p)\psi-\psi\|_2\to0$.

Solution

Pointwise, $e^{ip\cdot x/\hbar}\to1$ as $p\to0$. Moreover,

$$|e^{ip\cdot x/\hbar}-1|^2|\psi(x)|^2 \leq 4|\psi(x)|^2,$$

and the dominating function is integrable. Dominated convergence sends the squared norm to zero.

5. Complete the commutant argument

Assume $B$ is multiplication by $g\in L^\infty(\mathbb R^n)$ and commutes with every $T(q)$. Show that $g$ is constant almost everywhere.

Solution

Commutation gives, for every test function $\psi$,

$$g(x)\psi(x-q) = g(x-q)\psi(x-q)$$

almost everywhere. Varying $\psi$ yields $g(x)=g(x-q)$ almost everywhere for every $q$. A measurable function invariant under all translations is almost everywhere constant; for instance, convolve $g$ with smooth approximate identities to obtain continuous translation-invariant functions and pass to the almost-everywhere limit.

References

• G. B. Folland, Harmonic Analysis in Phase Space, Princeton University Press, 1989.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014.