Creation and Annihilation as Operators

Creation and annihilation operators have different analytic behavior for bosons and fermions. Fermionic operators associated with a fixed one-particle vector are bounded. Bosonic operators are closed and unbounded, and their commutation relations initially belong to a common invariant core. This page establishes the domains and norm estimates needed to turn the familiar occupation formulas into operator statements.

Required background. Fock space construction supplies the Hilbert direct sum and mode factorization. Creation and annihilation operators gives the occupation-number interpretation.

Creation operators on the finite-particle core

Let $\mathfrak h$ be a complex Hilbert space, with inner product linear in its second argument. On the algebraic finite-particle space $\mathcal F_{\mathrm{fin}}^\pm$, define

$$a^*(f)\psi_n=\sqrt{n+1}\,P^\pm_{n+1}(f\otimes\psi_n).$$

Here $P^\pm_{n+1}$ is the orthogonal symmetrization or antisymmetrization projection. Annihilation is the adjoint matrix action between adjacent sectors. Thus $a^*(f)$ is linear in $f$, while $a(f)$ is antilinear. Initially these operators preserve the finite-particle space, on which all their finite products are defined.

For bosons,

$$[a(f),a^*(g)]=\langle f,g\rangle I, \qquad [a(f),a(g)]=0.$$

For fermions the corresponding identities use anticommutators. These statements follow by inserting or contracting a tensor slot and collecting its permutation signs; the linked occupation formulas exhibit that calculation. What remains to establish here is how the operators extend beyond finite sectors.

Exact bosonic domains involve the occupied mode

Fix $f\ne0$, put $e=f/\|f\|$, and factor bosonic Fock space as

$$\mathcal F_+(\mathfrak h) \simeq\mathcal F_+(\mathbb Ce)\otimes \mathcal F_+(e^\perp).$$

A vector has an expansion $\psi=\sum_{n\geq0}|n\rangle\otimes\eta_n$ with $\sum_n\|\eta_n\|^2<\infty$. The closures of the finite-particle operators act as weighted shifts:

$$a(f)\psi=\|f\|\sum_{n\geq1}\sqrt n\,|n-1\rangle\otimes\eta_n,$$ $$a^*(f)\psi=\|f\|\sum_{n\geq0}\sqrt{n+1}\,|n+1\rangle\otimes\eta_n.$$

Both domains are exactly

$$D(a(f))=D(a^*(f)) =\left\{\psi:\sum_n n\|\eta_n\|^2<\infty\right\} =D(N_e^{1/2}),$$

where $N_e=d\Gamma(|e\rangle\langle e|)$. Weighted-shift convergence proves closedness. Testing against vectors supported in one sector identifies the Hilbert-space adjoints, so the star in $a^*(f)$ denotes the actual adjoint of the closed annihilation operator.

On this common domain,

$$\|a(f)\psi\|^2=\|f\|^2\|N_e^{1/2}\psi\|^2,$$ $$\|a^*(f)\psi\|^2 =\|a(f)\psi\|^2+\|f\|^2\|\psi\|^2.$$

Since $N_e\leq N$ as quadratic forms,

$$\|a(f)\psi\|\leq\|f\|\|N^{1/2}\psi\|, \qquad \|a^*(f)\psi\|\leq\|f\|\|(N+I)^{1/2}\psi\|.$$

Thus $D(N^{1/2})$ is a useful uniform domain for every $f$. It need not be the maximal domain for one fixed mode. Arbitrarily large occupation in orthogonal modes does not contribute to $N_e$.

Total-particle cutoffs approximate a vector in the graph norm of either fixed-mode operator. Consequently $\mathcal F_{\mathrm{fin}}^+$ is a core, including when $\mathfrak h$ is infinite-dimensional. For $f=0$ both operators are simply the everywhere-defined zero operator.

Fermionic boundedness and its exact constant

The CAR on finite-particle vectors give

$$\|a(f)\psi\|^2+\|a^*(f)\psi\|^2 =\|f\|^2\|\psi\|^2.$$

Both operators therefore extend uniquely to bounded operators on all fermionic Fock space, with norms at most $\|f\|$. Acting with $a^*(f)$ on the vacuum proves its norm is exactly $\|f\|$; adjoint equality gives the same result for annihilation.

The bounded extensions satisfy the CAR everywhere. The bound is independent of the number of available modes. Fermionic Fock space can still carry unbounded Hamiltonians and an unbounded total number operator when the one-particle space is infinite-dimensional. Boundedness of each $a(f)$ does not bound the infinite sum of occupation operators.

Fields and the limits of a formal commutator

For bosons define the symmetric field on the finite-particle core by

$$\phi(f)=\frac{a(f)+a^*(f)}{\sqrt2}.$$

A vector initially supported in sectors at most $m$ satisfies an estimate of the form

$$\|\phi(f)^r\psi\| \leq(\sqrt2\|f\|)^r \sqrt{\frac{(m+r)!}{m!}}\;\|\psi\|.$$

The corresponding power series with coefficients $t^r/r!$ converges for every finite $t$. These are analytic vectors. Nelson’s analytic-vector theorem therefore makes $\phi(f)$ essentially self-adjoint on this core. Its self-adjoint closure generates bounded Weyl unitaries; Weyl relations explain the domain-independent formulation.

The equation $[a(f),a^*(f)]=\|f\|^2I$ does not mean both products exist on $D(N_e^{1/2})$. In the mode decomposition their common product domain is $D(N_e)$, which is smaller. Likewise, a common core identity for two arbitrary unbounded operators does not by itself imply strong commutation of their spectral measures. Exponentiation here is justified by the specific Fock field construction.

Exercises

A maximal domain larger than the number domain. Choose orthogonal unit vectors $e,g$ and a normalized vector with all particles in mode $g$, having $n$-particle amplitudes proportional to $1/n$ for $n\geq1$. Compare its domains.

Solution

Its squared norm involves $\sum n^{-2}<\infty$, but its expected total number involves $\sum n^{-1}=\infty$. It is outside $D(N^{1/2})$. Nevertheless $N_e\psi=0$, so it belongs to both fixed-mode domains, with $a(e)\psi=0$ and $\|a^*(e)\psi\|=\|\psi\|$. This example separates total occupation control from the exact domain of a specified operator.

Products require more decay. In one bosonic mode take amplitudes $c_n=Cn^{-5/4}$ for $n\geq1$. Which of $a$, $a^*$, and $a^*a$ act on this vector?

Solution

The sum $\sum n|c_n|^2=\sum n^{-3/2}$ converges, so both first-order operators act. But $\sum n^2|c_n|^2=\sum n^{-1/2}$ diverges. Since $a^*a=N$ with domain $D(N)$, the product does not act on the vector. Subtracting two undefined products is not a way to extend the commutator.

References

• J. Dereziński, “Introduction to representations of the canonical commutation and anticommutation relations,” in Large Coulomb Systems, Lecture Notes in Physics 695, Springer, 2006, pp. 63–143, doi:10.1007/3-540-32579-4_3, author text.