The Symplectic Hilbert Space

Given a complex Hilbert space with an inner product and a complex structure , let be the same set considered as a vector space over . Define to be the complexification of :Observe the following structures on :

1. A conjugate that flips the sign of the imaginary part;
2. An inner product , extending the real part of sesquilinearly, makes a Hilbert space:
3. A symplectic form , extending the imaginary part of bilinearly:
4. A skew Hermitian form , extending the imaginary part of sesquilinearly:
5. Extend to linearly, so it becomes a complex structure on :

Proposition

Let , then , and are Lagrangian.

Proof Let us first show that . For any , we can write and , then clearly this forms a decomposition with . Moreover, the only vector that can lie in both and is the zero vector, so the decomposition is unique. In fact, this is the eigenspace decomposition of the operator . For , we have . For all , there holds so and is isotropic. Similarly is also isotropic. To show is maximal, t suffices to show that is nondegenerate on for . Suppose for all , then will vanish on the whole . Since is nondegenerate, , thus is nondegenerate on .

In summary, we have the data

Transpose

Suppose , then is defined as the unique element in such that Moreover, for an operator , the transpose of is defined as where is the adjoint of w.r.t the inner product .

The Symplectic Semigroup

Restricted Symplectic Semigroup

Suppose is a real symplectic Hilbert space, then we write for the restricted symplectic group, that is, a subgroup of such that the operators can be written as for some unitary operator and some Hilbert-Schmidt operator .

Correct Operator & Potapov–Ginzburg Matrix

Given a symplectic Hilbert space , suppose is an (possibly unbounded) operator on . If satisfies where

1. , ;
2. , , ;
3. and are Hilbert-Schmidt,

then is called a correct operator, and the above matrix is called the Potapov–Ginzburg matrix of , denoted as . The collection of correct operators forms a semigroup under the usual composition of operators.

We can use the associated matrix to describe these operators in . Given an operator with the matrix presentation with respect to the decomposition , we have If is invertible, then we can rearrange the above equation to get

Conversely, if is the Potapov–Ginzburg matrix of and is invertible, we can express the block matrix of (w.r.t the decomposition ) using , and :

Proposition

The graph of a correct operator is always Lagrangian. In fact, is Lagrangian if and only if holds.

Proof

Proposition

For some , the followings are equivalent:

1. ;
3. .

Therefore, is a subsemigroup of , and in fact, it is the Shilov boundary of .

Proof

: Suppose , that is , We can express the block matrix of using , and : To show that , it is sufficient to show that is a Hilbert-Schmidt perturbation of the identity. By , we can simplify to Note that is Hilbert-Schmidt, so all of , and are Hilbert-Schmidt. Additionally, for the right-bottom entry, we have Now use the resolvent identity and the fact that is a trace-class, so the right-bottom entry is Hilbert-Schmidt as well.

The Weyl Representation

The Weyl Representation

For some , define the Weyl representation of on the bosonic Fock space : where is some normalization constant, is the second quantization of , and and are defined as follows: for some orthonormal basis of , and , are the matrix elements of and w.r.t this basis.

Note that this definition is valid because both and are Hilbert-Schmidt, so

Proposition

For any , we have the following intertwining property holds:

Proof It is equivalent to show that Since we assume that , there are no shifting terms (i.e., ) To compute the left hand side, we need to compute the following six conjugates for any :

1. ;
2. ;
3. ;
4. ;
5. ;
6. .

First consider . By the BCH formula, we know that Consider the commutator , for any there holds Let for some , then we have where the second last equality holds because is symmetric. Therefore, by linearity, we have which commutes with any creation operator, hence the higher order commutator terms in the BCH formula vanish. So On the other hand, because creation operators commute with each other. 5 and 6 are completely similar, For the second quantization , on each monomial , we have Therefore, , and taking adjoints on the both sides gives . Finally, we can derive the conjugation of :