82
12 Fock Representation
Thus, also λ − 1 ∈ σ(N ) and, since the spectrum of N is non-negative, in order that
this process of lowering the eigenvalues terminates, λ = 0 must be a point of the
spectrum of N and
a j Ψ 0 = 0, ∀ j.
(12.7)
Conversely, if the Fock vacuum Ψ 0 exists, then A H Ψ 0 = P(a
∗
)Ψ 0 , where P(a
∗
)
denotes the polynomial algebra generated by the a
∗ ’s. On such a domain, which
is dense by the irreducibility of A H , N exists as a self-adjoint operator and the
exponential series converges strongly and defines a one-parameter group of unitary
operators, since the monomials of a
∗ applied to Ψ 0 yield eigenstates of N and generate
such a domain.
In the case of a finite number of degrees of freedom, the above argument can be
turned into an analog of von Neumann’s theorem by proving that ∀ j, N j exists as a
self-adjoint operator on D, as a consequence of the regularity condition.
57
As in the finite-dimensional case, all irreducible Fock representations are unitarily
equivalent and one can actually speak of one (irreducible) Fock representation. In
fact, given any two of them, say π 1 , π 2 , with Fock vectors Ψ 01 , Ψ 02 , respectively,
the mapping U defined by
U Ψ 01 = Ψ 02 , U π 1 (A)Ψ 01 = π 2 (A)Ψ 02 , ∀A ∈ A H
and its inverse U
−1 are defined on dense sets since, by irreducibility, Ψ 01 , Ψ 02 are
cyclic vectors. Furthermore, since the matrix elements
(π i (A)Ψ 0i , π i (B)Ψ 0i ), i = 1, 2, ∀A, B ∈ A H
only involve the canonical commutation relations and the Fock condition, (12.6),
they are equal, so that U is unitary.
It is worthwhile to remark that, in an irreducible Fock representation, the zero
eigenvalue of N has multiplicity one. In fact, if Ψ
is orthogonal to Ψ 0 and satisfies
N Ψ
= 0, then a j Ψ
= 0, ∀ j and for any polynomial P(a
∗
) of the a
∗ one has
(Ψ
, P(a
∗
)Ψ 0 ) = (P(a)Ψ
, Ψ 0 ) = 0.
This implies Ψ
= 0, by the cyclicity of Ψ 0 with respect to the polynomial algebra
generated by the a
∗ .
It is clear from the above argument that the characteristic feature of a Fock representation is that the states of its Hilbert space can be described in terms of the
eigenvalues of the N j , all of which exist as self-adjoint operators on D since they are
dominated by N . For this reason, the Fock representation is also called the occupation
number representation.
57 See, e.g. O. Bratteli and D. K. Robinson, Operator Algebras and Quantum Statistical Mechanics,
Vol. 2, Springer 1996, Sect. 5.2.3; see also [SNS 96].
12 Fock Representation
Thus, also λ − 1 ∈ σ(N ) and, since the spectrum of N is non-negative, in order that
this process of lowering the eigenvalues terminates, λ = 0 must be a point of the
spectrum of N and
a j Ψ 0 = 0, ∀ j.
(12.7)
Conversely, if the Fock vacuum Ψ 0 exists, then A H Ψ 0 = P(a
∗
)Ψ 0 , where P(a
∗
)
denotes the polynomial algebra generated by the a
∗ ’s. On such a domain, which
is dense by the irreducibility of A H , N exists as a self-adjoint operator and the
exponential series converges strongly and defines a one-parameter group of unitary
operators, since the monomials of a
∗ applied to Ψ 0 yield eigenstates of N and generate
such a domain.
In the case of a finite number of degrees of freedom, the above argument can be
turned into an analog of von Neumann’s theorem by proving that ∀ j, N j exists as a
self-adjoint operator on D, as a consequence of the regularity condition.
57
As in the finite-dimensional case, all irreducible Fock representations are unitarily
equivalent and one can actually speak of one (irreducible) Fock representation. In
fact, given any two of them, say π 1 , π 2 , with Fock vectors Ψ 01 , Ψ 02 , respectively,
the mapping U defined by
U Ψ 01 = Ψ 02 , U π 1 (A)Ψ 01 = π 2 (A)Ψ 02 , ∀A ∈ A H
and its inverse U
−1 are defined on dense sets since, by irreducibility, Ψ 01 , Ψ 02 are
cyclic vectors. Furthermore, since the matrix elements
(π i (A)Ψ 0i , π i (B)Ψ 0i ), i = 1, 2, ∀A, B ∈ A H
only involve the canonical commutation relations and the Fock condition, (12.6),
they are equal, so that U is unitary.
It is worthwhile to remark that, in an irreducible Fock representation, the zero
eigenvalue of N has multiplicity one. In fact, if Ψ
is orthogonal to Ψ 0 and satisfies
N Ψ
= 0, then a j Ψ
= 0, ∀ j and for any polynomial P(a
∗
) of the a
∗ one has
(Ψ
, P(a
∗
)Ψ 0 ) = (P(a)Ψ
, Ψ 0 ) = 0.
This implies Ψ
= 0, by the cyclicity of Ψ 0 with respect to the polynomial algebra
generated by the a
∗ .
It is clear from the above argument that the characteristic feature of a Fock representation is that the states of its Hilbert space can be described in terms of the
eigenvalues of the N j , all of which exist as self-adjoint operators on D since they are
dominated by N . For this reason, the Fock representation is also called the occupation
number representation.
57 See, e.g. O. Bratteli and D. K. Robinson, Operator Algebras and Quantum Statistical Mechanics,
Vol. 2, Springer 1996, Sect. 5.2.3; see also [SNS 96].