Position and Momentum under Canonical Displacements

Position and momentum are canonical generators whose measurement distributions shift predictably under translations and modulations. The precise statements are operator conjugation identities with matching domains, followed by corresponding identities for spectral projections. This page develops those covariance statements on the real line. The construction, maximal domains, and self-adjointness of the Schrödinger operators themselves belong to The Schrödinger Representation.

Required background. The Schrödinger Representation supplies the operators and displacement convention. Projection-Valued Measures supplies spectral event probabilities.

Domain transport by translations and modulations

On $L^2(\mathbb R)$ use

$$Q\psi=x\psi, \qquad P\psi=-i\hbar\psi',$$

with $D(Q)=\{\psi:x\psi\in L^2\}$ and $D(P)=H^1(\mathbb R)$. The prime is a weak derivative. The bounded unitary families act as

$$(T(a)\psi)(x)=\psi(x-a), \qquad (M(b)\psi)(x)=e^{ibx/\hbar}\psi(x).$$

Translations preserve $D(Q)$ because

$$\int x^2|\psi(x-a)|^2dx =\int(y+a)^2|\psi(y)|^2dy,$$

which is finite exactly when $y\psi(y)\in L^2$. The reverse implication follows by translating back. They also preserve $H^1$, since the weak derivative translates with the function.

Modulations preserve $D(Q)$ by their unit modulus. They preserve $D(P)$ by the weak product rule,

$$(M(b)\psi)'=M(b)\bigl(\psi'+ib\psi/\hbar\bigr).$$

Thus all four domain-preservation statements are exact on the line. They are not consequences of unitarity alone: a general unitary need not preserve a specified unbounded-operator domain. Endpoint conditions can spoil the statements, as the interval counterexample shows.

Conjugation moves the conjugate observable

On the appropriate maximal domains, direct calculation gives

$$\begin{aligned} T(a)^*QT(a)&=Q+aI,& T(a)^*PT(a)&=P,\\ M(b)^*PM(b)&=P+bI,& M(b)^*QM(b)&=Q. \end{aligned}$$

These are equalities of self-adjoint operators, including domains. The shifts have the expected units: $a$ is a length, $b$ a momentum. Notice the placement of the adjoint. Reversing the conjugation changes the sign of the shift.

The symmetric Weyl displacement in the chapter convention is

$$W(a,b)=e^{iab/(2\hbar)}T(a)M(b),$$

so

$$(W(a,b)\psi)(x) =e^{ib(x-a/2)/\hbar}\psi(x-a).$$

Its scalar prefactor affects operator composition but cancels from observable conjugation. Consequently $W^*QW=Q+aI$ and $W^*PW=P+bI$. These formulae move a preparation in phase space without choosing an eigenvector basis for either observable.

Spectral probabilities shift even without finite moments

For every Borel set $\Delta\subseteq\mathbb R$, spectral calculus transports the position projection:

$$T(a)^*E_Q(\Delta)T(a)=E_Q(\Delta-a).$$

Thus the translated state $T(a)\psi$ has probability of landing in $\Delta$ equal to the original state’s probability of landing in $\Delta-a$. The analogous statement for modulation is

$$M(b)^*E_P(\Delta)M(b)=E_P(\Delta-b).$$

These bounded-projector identities hold for every normalized Hilbert-space vector, including states with divergent first or second moments. One should use them before discussing means, since a probability distribution exists even when a requested moment does not.

If the relevant first moments are finite, displacement adds $a$ and $b$ to the position and momentum means. If the second moments are finite, the variances remain unchanged. This follows by subtracting the shifted mean from the conjugated operator. A displacement does not squeeze a state; squeezing changes its covariance and is a different canonical transformation.

A displaced packet is a check of both signs

Take any normalized $\phi\in\mathcal S(\mathbb R)$ with zero position and momentum means, and define $\psi=W(a,b)\phi$. Then

$$\langle Q\rangle_\psi=a, \qquad \langle P\rangle_\psi=b.$$

The position density is $|\phi(x-a)|^2$, while the phase $e^{ibx/\hbar}$ produces the positive momentum shift. The remaining constant phase $e^{-iab/(2\hbar)}$ changes neither result. This test is useful when comparing sign conventions across Fourier transforms or displacement-operator definitions.

A unitary change from position to momentum representation is different from this physical displacement. A representation change transforms states and all observables together, preserving every prediction. Applying $W$ to a preparation while keeping the measured observable fixed changes its distribution as above.

Exercise: translate an event instead of a mean

Let a state have probability $r$ of lying in $[u,v]$. What is the probability that $T(a)\psi$ lies in $[u+a,v+a]$? Does the conclusion require $\psi\in D(Q)$?

Solution

It is again $r$, by the spectral projection identity. No membership in $D(Q)$ is required: $E_Q([u,v])$ is bounded on the entire Hilbert space. The domain $D(Q)$ would be needed for the literal vector $Q\psi$ and guarantees a finite second moment, which is stronger than this event-probability statement.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — translations, canonical operators, and spectral calculus.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Chapters 7–8 — Fourier representation and position–momentum operators.