Heisenberg Group

The Heisenberg group is the Lie group whose unitary representations encode the canonical commutation relations. It is the group-level home for position translations, momentum translations, Weyl relations, and the central phase that makes quantum phase space noncommutative.

We begin with the group law and its central phase. For the continuity and domain assumptions needed to recover position and momentum operators, see Weyl Form of the Canonical Commutation Relations.

The familiar commutator

$$[X,P]=i\hbar I$$

is the infinitesimal shadow. The Heisenberg group gives the exponentiated structure, where the operators are unitary and domain issues are easier to control.

Helpful background. Lie Algebras explains commutators of generators; Unitary Representations explains their action on states.

Translations and the Weyl relation

Position $X$ and momentum $P$ are unbounded operators. Their formal commutator is meaningful only on suitable common dense domains, as explained in Unbounded Operators. The group-level statement uses unitary operators such as

$$T(q)=\exp\left(-\frac{i}{\hbar}qP\right), \qquad M(p)=\exp\left(\frac{i}{\hbar}pX\right),$$

which represent translations in position and momentum. These unitary operators satisfy exact Weyl relations.

This is why rigorous treatments often state the canonical commutation relations in exponentiated form first and recover $[X,P]=i\hbar I$ infinitesimally.

One-Dimensional Group Law

The one-degree-of-freedom Heisenberg group can be represented by triples

$$(q,p,s)\in\mathbb R^3,$$

where $q$ is a position-translation parameter, $p$ is a momentum-translation parameter, and $s$ is a central phase coordinate with units of action.

A common group law is

$$(q,p,s)(q',p',s') = \left( q+q', p+p', s+s'+\frac12(pq'-qp') \right).$$

The identity is $(0,0,0)$, and the inverse is

$$(q,p,s)^{-1}=(-q,-p,-s).$$

The center consists of elements with $q=p=0$:

$$Z(H)=\{(0,0,s):s\in\mathbb R\}.$$

This central coordinate is the algebraic source of the quantum phase in Weyl relations.

Matrix Model

The same group can be realized by upper triangular matrices:

$$\begin{pmatrix} 1 & p & s+\frac12 pq\\ 0 & 1 & q\\ 0 & 0 & 1 \end{pmatrix}.$$

For example, the upper-right entry of the product is $s+s'+pq/2+p'q'/2+pq'$. Subtracting $(p+p')(q+q')/2$ gives the central coordinate $s+s'+(pq'-qp')/2$, exactly as above. When using this matrix model, choose units for $q$, $p$, and $s$ so the entries are numerical coordinates. This model makes it visible that the group is nonabelian but nilpotent: commutators land in the center, and commutators with central elements vanish.

Lie Algebra

Let $Q$, $P$, and $Z$ denote infinitesimal generators corresponding to $q$, $p$, and $s$. The Heisenberg Lie algebra has the only nonzero bracket

$$[Q,P]=-Z,$$

with

$$[Q,Z]=[P,Z]=0.$$

The minus sign follows from our cocycle: the $q$ generator is the matrix $E_{23}$, the $p$ generator is $E_{12}$, and $[E_{23},E_{12}]=-E_{13}$. Reversing the central coordinate gives the frequently used convention $[Q,P]=Z$.

The unitary representation is $\rho(q,p,s)=e^{is/\hbar}W(q,p)$. Its infinitesimal $q,p,s$ generators are $-i\widehat P/\hbar$, $iX/\hbar$, and $iI/\hbar$, respectively. These abstract translation generators must not be confused with the position and momentum observables, which obey

$$[X,P]=i\hbar I.$$

The factor $i\hbar$ reflects the passage from a real Lie algebra representation by skew-adjoint generators to self-adjoint quantum observables.

Weyl Relations

Define the unitary position translation and momentum translation operators by

$$T(q)=\exp\left(-\frac{i}{\hbar}qP\right), \qquad M(p)=\exp\left(\frac{i}{\hbar}pX\right).$$

They satisfy the Weyl relation

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

This phase is the exponentiated form of the canonical commutator. Expanding to first order in $q$ and $p$ recovers

$$[X,P]=i\hbar I$$

on a suitable common domain.

Weyl Displacement Operators

The symmetric phase-space displacement operator is

$$W(q,p) = \exp\left( \frac{i}{\hbar}(pX-qP) \right).$$

It combines a position translation and a momentum translation. The Baker–Campbell–Hausdorff formula gives

$$W(q,p)W(q',p') = \exp\left( -\frac{i}{2\hbar}(qp'-pq') \right) W(q+q',p+p').$$

The phase contains the standard symplectic form on phase space:

$$\omega((q,p),(q',p'))=qp'-pq'.$$

Thus the real Heisenberg group is a central extension of additive phase-space translations by $\mathbb R$. In the unitary representation, the central coordinate acts as the phase $e^{is/\hbar}$; quotienting $s$ by $2\pi\hbar\mathbb Z$ gives a circle-valued center.

Action on Position and Momentum

The displacement operator translates observables by conjugation:

$$W(q,p)^\dagger X W(q,p)=X+qI,$$

and

$$W(q,p)^\dagger P W(q,p)=P+pI.$$

So $q$ and $p$ really are phase-space translation parameters. The central phase does not change $X$ or $P$ by conjugation, but it is essential for the representation to multiply correctly.

Schrödinger Representation

On $L^2(\mathbb R)$, the standard representation is

$$(X\psi)(x)=x\psi(x), \qquad (P\psi)(x)=-i\hbar\frac{d\psi}{dx}.$$

The unitary translations act as

$$(T(q)\psi)(x)=\psi(x-q),$$

and

$$(M(p)\psi)(x)=e^{ipx/\hbar}\psi(x).$$

Their noncommutation is immediate:

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

while

$$(M(p)T(q)\psi)(x) = e^{ipx/\hbar}\psi(x-q).$$

Hence

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

Stone-von Neumann Theorem

For a finite number of degrees of freedom, the Stone–von Neumann theorem says, roughly, that every irreducible strongly continuous unitary representation of the Weyl relations with the same nonzero central character is unitarily equivalent to the Schrödinger representation.

This is a mathematical reason the position and momentum representations describe the same quantum mechanics for ordinary finite-dimensional phase space. It is not a statement that all representations are identical without hypotheses. The theorem depends on irreducibility, strong continuity, and a fixed central action.

For infinitely many degrees of freedom, as in quantum field theory and thermodynamic limits, inequivalent representations can occur. That is one reason QFT is not merely many copies of the finite-dimensional Stone–von Neumann theorem.

Relation to the Harmonic Oscillator

The harmonic oscillator ladder operators are linear combinations of $X$ and $P$:

$$a = \sqrt{\frac{m\omega}{2\hbar}}X + \frac{i}{\sqrt{2m\hbar\omega}}P,$$ $$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}X - \frac{i}{\sqrt{2m\hbar\omega}}P.$$

Their commutator

$$[a,a^\dagger]=1$$

is another representation of the same Heisenberg algebra. Coherent-state displacement operators are Weyl displacement operators written in oscillator variables.

Common Mistakes

• Treating $[X,P]=i\hbar I$ as an everywhere-defined matrix identity on all of $L^2(\mathbb R)$.
• Forgetting the central phase in products of phase-space translations.
• Confusing ordinary abelian phase-space translations with their nonabelian Heisenberg-group lift.
• Assuming Stone–von Neumann uniqueness holds for infinitely many degrees of freedom.
• Mixing conventions for $W(q,p)$ and then comparing phases without translating signs.
• Calling $X$ and $P$ bounded because their exponentials are unitary.

Exercises

1. Verify the Weyl relation using the Schrödinger representation.

Solution

Using

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

one finds

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

and

$$(M(p)T(q)\psi)(x) = e^{ipx/\hbar}\psi(x-q).$$

Therefore

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

2. Show that the center of the group law consists of elements $(0,0,s)$.

Solution

Let $(q,p,s)$ commute with every $(q',p',s')$. Comparing the central coordinates in the two products gives

$$pq'-qp'=p'q-q'p$$

for all $q'$ and $p'$. This is equivalent to

$$pq'-qp'=-(pq'-qp')$$

for all $q'$ and $p'$, so $pq'-qp'=0$ for all $q',p'$. Choosing $q'=1,p'=0$ gives $p=0$, and choosing $q'=0,p'=1$ gives $q=0$. Thus only $(0,0,s)$ is central.

3. Derive the canonical commutator from the Weyl relation to first order.

Solution

Use

$$T(q)=I-\frac{i}{\hbar}qP+O(q^2), \qquad M(p)=I+\frac{i}{\hbar}pX+O(p^2).$$

Keeping the mixed $pq$ term in

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

gives

$$\frac{pq}{\hbar^2}[P,X] = -\frac{ipq}{\hbar}I.$$

Cancel $pq$ and use $[P,X]=-[X,P]$ to obtain

$$[X,P]=i\hbar I.$$

4. Why does Stone–von Neumann not settle representation questions in quantum field theory?

Solution

The theorem applies to a finite number of canonical degrees of freedom under regularity and irreducibility assumptions. Quantum field theory has infinitely many degrees of freedom, and thermodynamic or continuum limits can produce unitarily inequivalent representations. Therefore the finite-dimensional uniqueness theorem no longer rules out distinct Hilbert-space representations.

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, Academic Press, 1980.