Bogoliubov Implementability

A Bogoliubov transformation always acts on the classical phase space and hence on the abstract canonical algebra. It need not act by a unitary on a chosen Fock space. For bosons, Shale’s theorem says that the obstruction is global: the antilinear part must be Hilbert–Schmidt. Modewise mixing can be finite even when unitary implementability fails; the squared mixing must be summable over all modes.

Required background. Flat-Spacetime Quantum Field Theory supplies mode Bogoliubov transformations and particle number, while Variational and Symplectic Geometry defines the bounded symplectic transformations whose quantum implementation is at issue.

Helpful background. Partial Differential Equations and Green Functions provides the solution-space maps induced by linear evolution, and Causality and Global Structure establishes the Cauchy evolution setting in which two splittings can be compared.

For immediate use, well-posed linear Cauchy evolution maps one solution space symplectically to another. Comparing complex structures before and after that map produces the real-linear transformation whose antilinear part is tested below.

Linear and antilinear parts

Fix a real symplectic Hilbert space $(\mathcal V,\sigma)$, a positive compatible complex structure $J$, and its bosonic Fock representation. Let $S:\mathcal V\to\mathcal V$ be bounded, invertible, real-linear, and symplectic. Relative to multiplication by $i=J$, decompose it as

$$S_+=\frac12(S-JSJ), \qquad S_-=\frac12(S+JSJ).$$

The first part commutes with $J$ and is complex-linear; the second anticommutes with $J$ and is complex-antilinear. In mode notation these are the $\alpha$ and $\beta$ parts. Moreover,

$$[S,J]=2S_-J,$$

so $S_-$ is Hilbert–Schmidt exactly when $[S,J]$ is. For any orthonormal basis $(e_n)$ of the one-particle space,

$$\lVert S_-\rVert_{\mathrm{HS}}^2 =\sum_n\lVert S_-e_n\rVert^2.$$

The value is basis independent. Boundedness and symplecticity are hypotheses of the theorem; a formally defined, unbounded mode map lies outside this statement.

The bosonic Shale theorem

Theorem (Shale). Under the hypotheses above, there exists a unitary $U_S$ on the symmetric Fock space such that

$$U_S W(v)U_S^*=W(Sv)$$

for every $v\in\mathcal V$ if and only if $S_-$ is Hilbert–Schmidt. When it exists in the irreducible Fock representation, the implementer is unique up to a phase. Equivalently, the automorphism of the Weyl algebra is always defined, but it is spatial in this representation precisely under the Hilbert–Schmidt condition. This is the modern form of Shale 1962, Theorem 4.1, pp. 157–158.

If $J'=SJS^{-1}$ is a second positive complex structure and the corresponding one-particle norms are boundedly comparable, the same criterion may be expressed as a Hilbert–Schmidt condition on $J-J'$. It is a statement about global unitary equivalence of two pure Fock representations. It does not say that their abstract CCR algebras differ, nor does it say that local restrictions of physically admissible representations must be inequivalent.

For a diagonal bosonic transformation,

$$b_n=\alpha_n a_n+\beta_n a_n^*, \qquad \lvert\alpha_n\rvert^2-\lvert\beta_n\rvert^2=1,$$

the criterion reduces to $\sum_n\lvert\beta_n\rvert^2<\infty$. In an implementable transformation the old-vacuum expectation of the new total number, or the new-vacuum expectation of the old total number with the corresponding orientation, is

$$\sum_n\langle b_n^*b_n\rangle=\sum_n\lvert\beta_n\rvert^2.$$

Thus finite total particle production and Hilbert–Schmidt implementability coincide in this pure diagonal setting. A finite answer for each fixed $n$ is a strictly weaker statement.

Why necessity and sufficiency use the same sum

For finitely many squeezed modes, the canonical relations can be implemented by a product of squeeze unitaries. In a diagonal real basis write $\alpha_n=\cosh r_n$ and $\beta_n=\sinh r_n$. The vector annihilated by all $b_n$ has the formal product form

$$\Omega_S\propto \prod_n\exp\!\left[-\frac12\tanh r_n\,(a_n^*)^2\right]\Omega.$$

For a finite cutoff $N$, its normalization contains

$$\prod_{n=1}^{N}(\cosh r_n)^{-1/2}.$$

The infinite product has a nonzero limit exactly when $\sum_n\log(\cosh r_n)$ converges. For small $r_n$, $\log(\cosh r_n)\sim r_n^2/2$ and $\sinh^2r_n\sim r_n^2$, so this is precisely square summability of $\beta$. More explicitly, for normalized cutoffs $M\gt N$ the overlap is $\langle\Omega_N,\Omega_M\rangle=\prod_{n=N+1}^{M}(\cosh r_n)^{-1/2}$. If the sum diverges, this tends to zero as $M\to\infty$ for every fixed $N$, so the sequence is not Cauchy. Adjacent cutoffs need not become orthogonal. The transformed Gaussian vacuum has no vector representative in the original Fock space; this establishes the obstruction without assuming that every Fock vector has finite number expectation.

The general proof replaces the mode product by the polar and singular-value analysis of the antilinear operator. For sufficiency, finite-rank approximations to $S_-$ have finite-mode implementers; the Hilbert–Schmidt estimate makes their squeezed vacua and their action on the dense Weyl orbit Cauchy, after phases are chosen consistently. The limit is a unitary implementer. Conversely, an implementer supplies a normalizable transformed vacuum; the determinant or singular-value estimate forces the antilinear singular values to be square summable. The two directions therefore test the same Hilbert–Schmidt norm, not merely pointwise decay.

Finite-mode success and infinite-mode failure

Take

$$\beta_n=\frac1{\sqrt n}, \qquad \alpha_n=\sqrt{1+\frac1n}.$$

Every mode satisfies the bosonic normalization, and each finite truncation has a perfectly good squeeze unitary. Nevertheless,

$$\sum_{n=1}^{\infty}\lvert\beta_n\rvert^2 =\sum_{n=1}^{\infty}\frac1n=\infty.$$

There is no unitary on the original full Fock space implementing the infinite transformation. Saying “the Bogoliubov coefficients are finite” therefore misses the theorem’s essential global hypothesis. A continuum calculation has the analogous requirement that the integral-kernel $\beta$ be Hilbert–Schmidt; informal expressions such as an infinite spatial volume or a delta function squared signal that the global question needs a regulator-independent formulation.

Failure of this criterion is not a breakdown of the quantum theory. The transformed quasifree state still defines a GNS representation of the same Weyl algebra. What fails is one proposed identification of that representation with the original Fock Hilbert space by a single global unitary.

Fermions: the Shale–Stinespring criterion

The fermionic theorem has a parallel conclusion but a different classical group and normalization. Fix a CAR polarization $P$ on the one-particle space. A one-particle unitary, or an orthogonal map in the self-dual formulation, is implementable in the polarized fermionic Fock representation exactly when its off-diagonal block

$$P S(\mathbf1-P), \qquad (\mathbf1-P)SP$$

are Hilbert–Schmidt; equivalently, $[P,S]$ is Hilbert–Schmidt. These two off-diagonal blocks are not generally adjoints of one another, even when $S$ is unitary. This is the Shale–Stinespring condition associated with the Clifford algebra, as developed in Shale and Stinespring 1964, pp. 365–381.

For fermionic pair-mixing blocks the normalization is $\lvert\alpha_n\rvert^2+\lvert\beta_n\rvert^2=1$, not the bosonic difference. Pauli exclusion changes the squeeze construction, but infinitely many non-square-summable off-diagonal amplitudes still obstruct implementation in the chosen Fock representation. Calling the bosonic result “Shale–Stinespring” conceals this distinction; here Shale condition refers to CCR and Shale–Stinespring condition to CAR.

Exercise: locate the threshold

For independent bosonic modes let $\beta_n=(n+1)^{-p}$ with $p>0$ and $\alpha_n=\sqrt{1+\beta_n^2}$. For which values of $p$ is the transformation unitarily implementable? What happens at the borderline even though every mode is finite?

Solution

The canonical normalization holds mode by mode for every positive value of $p$. Shale’s condition asks a different question: the squared antilinear amplitudes form the series $\sum_n(n+1)^{-2p}$. The standard comparison test says this converges exactly when $2p>1$, so implementation holds precisely for $p>1/2$. At the borderline $p=1/2$ the harmonic series diverges. Every finite truncation is implemented and every individual coefficient is finite, but the sequence of truncated squeezed vacua has no nonzero Fock-space limit. This is the promised separation between modewise control and global Hilbert–Schmidt control.

References

• Shale, David, and W. Forrest Stinespring. “States of the Clifford Algebra.” Annals of Mathematics 80, no. 2 (1964): 365–381. DOI 10.2307/1970397.
• Shale, David. “Linear Symmetries of Free Boson Fields.” Transactions of the American Mathematical Society 103, no. 1 (1962): 149–167. DOI 10.1090/S0002-9947-1962-0137504-6.