26.4 Complex vector spaces with Hermitian inner product as phase spaces

26.4 Complex vector spaces with Hermitian inner product as phase spaces

Concept links · terms present in this machine draft; source roles are unverified: vector space · complex inner product · Lie algebra · complexification

In many cases of physical interest, the dual phase space $M$ will be a complex vector space with a Hermitian inner product. This will occur for instance when $M$ is a space of complex solutions to a field equation, with examples non-relativistic quantum field theory (see chapter 37) and the theory of a relativistic complex scalar field (see chapter 44.1.2). In such cases, the Bargmann-Fock quantization can be confusing, since it involves complexifying $\mathcal{M},$ which is already a complex vector space. One way to treat this situation is as follows, taking ${\mathcal{M}}_J^{+}=\mathcal{M}$ . In the non-relativistic quantum field theory this gives a consistent Bargmann-Fock quantization of the theory, while in the relativistic case it does not, and in that case a diferent sort of complex structure $J$ is needed, one not related to the complex nature of the field values.

Instead of trying to complexify M, we introduce a conjugate complex vector space $M$ and an antilinear conjugation operation interchanging M and M, with square the identity. In the case of M complex solutions to a field equation, $M$ will be solutions to the complex conjugate equation. Then, Bargmann-Fock quantization proceeds with the decomposition

$$\mathcal{M}\oplus \overline{\mathcal{M}}$$

playing the role of the decomposition

$${\mathcal{M}}_J^{+}\oplus {\mathcal{M}}_J^{-}$$

in our previous discussion.

This determines J: it is the operator that is +$i$ on $\mathcal{M},$ , and −$i$ on $M$. Given a Hermitian inner product $⟨\cdot ,\cdot {⟩}_{\mathcal{M}}$ on $M$, a symplectic structure Ω and indefinite Hermitian product $⟨\cdot ,\cdot ⟩$ on $\mathcal{M}\oplus \overline{\mathcal{M}}$ can be determined as follows, using the relation 26.7

$$⟨u_1,u_2⟩=i\Omega ({\overline{u}}_1,u_2)$$

Writing elements $u_1,u_2\in \mathcal{M}\oplus \overline{\mathcal{M}}$ as

$$u_1=u_1^{+}+u_1^{-},u_2=u_2^{+}+u_2^{-}$$

where $u_1^{+},u_2^{+}\in \mathcal{M}$ and $u_1^{-},u_2^{-}\in \overline{\mathcal{M}}$ , Ω is defined to be the bilinear form such that

$$\Omega (u_1^{+},u_2^{+})=\Omega (u_1^{-},u_2^{-})=0$$

• the Hermitian inner product is recovered on $M$

$$i\Omega (u_1^{-},u_2^{+})=⟨\overline{u_1^{-}},u_2^{+}{⟩}_{\mathcal{M}}$$

• Ω is antisymmetric, so

$$i\Omega (u_1^{+},u_2^{-})=-i\Omega (u_2^{-},u_1^{+})=-⟨\overline{u_2^{-}},u_1^{+}{⟩}_{\mathcal{M}}$$

Basis vectors $z_j$ of M orthonormal with respect to $⟨\cdot ,\cdot {⟩}_{\mathcal{M}}$ satisfy

$$\begin{array}{rl}⟨z_j,z_k⟩& ={\delta }_{jk}\\ ⟨{\overline{z}}_j,{\overline{z}}_k⟩& =-{\delta }_{jk}\\ ⟨z_j,{\overline{z}}_k⟩& =⟨{\overline{z}}_j,z_k⟩=0\end{array}$$

The symplectic form Ω satisfies the usual Poisson bracket relation

$$\Omega (z_j,{\overline{z}}_k)=\{z_j,{\overline{z}}_k\}=i{\delta }_{jk}$$

and one has all the elements needed for the standard Bargmann-Fock quantization. Note that the Hermitian inner product here is indefinite, positive on $\mathcal{M}.$ negative on $\overline{\mathcal{M}}$

To understand better in a basis independent way how quantization works in this case of a complex dual phase space $\mathcal{M}$ , one can use the identification (see section 9.6) of polynomials with symmetric tensor products. In this case polynomials in the $z_j$ get identified with $S^{\ast }(\mathcal{M})$ (since $z_j\in \mathcal{M})$ , while polynomials in the ${\overline{z}}_j$ get identified with $S^{\ast }(\overline{\mathcal{M}})$ (since ${\overline{z}}_j\in \overline{\mathcal{M}})$

We see that the Fock space ${\mathcal{F}}_d$ gets identified with $S^{\ast }(\mathcal{M})$ and using this identification (instead of the one with polynomials) one can ask what operator gives the quantization of an element

$$u=u^{+}+u^{-}$$

where $u^{+}\in \mathcal{M}$ and $u^{-}\in \overline{\mathcal{M}}$ . For basis elements $z_j\in \mathcal{M}$ the operator will be $-i{a}_i^{†}$ , while for ${\overline{z}}_j\in \overline{\mathcal{M}}$ it will be $-i{a}_j$ . We will not enter here into details, which would require more discussion of how to manipulate symmetric tensor products (see for instance chapter 5.4 of [17]). One can show however that the operators ${\Gamma }^{\prime }(u^{+},0),{\Gamma }^{\prime }(u^{-},0)$ defined on symmetrized tensor products by (where $P^{+}$ is the symmetrization operator of section 9.6 and $\widehat{u_j}$ means drop that term).

$${\Gamma }^{\prime }(u^{+},0)P^{+}(u_1\otimes \cdots \otimes u_N)=-i\sqrt{N+1}{P}^{+}(u^{+}\otimes u_1\otimes \cdots \otimes u_N)$$ $${\Gamma }^{\prime }(u^{-},0)P^{+}(u_1\otimes \cdots \otimes u_N)=\frac{-i}{\sqrt{N}}\sum _{j=1}^N⟨\overline{u^{-}},u_j⟩P^{+}(u_1\otimes \cdots \otimes \widehat{u_j}\otimes \cdots \otimes u_N)$$

satisfy the Heisenberg Lie algebra homomorphism relations

$$\begin{array}{rl}[{\Gamma }^{\prime }(u^{-},0),{\Gamma }^{\prime }(u^{+},0)]& ={\Gamma }^{\prime }(0,\Omega (u^{-},u^{+}))\\ & =-i\Omega (u^{-},u^{+})1\\ & =-⟨\overline{u^{-}},u^{+}⟩1\end{array}\text{\hspace{1em}(26.16)}$$

when acting on elements of $S^N(\mathcal{M})$ (which are given by applying the symmetrization operator $P_{+}$ to elements of the N-fold tensor product of $\mathcal{M})$

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