26.3 Complex structures and quantization

26.3 Complex structures and quantization

Concept links · terms present in this machine draft; source roles are unverified: eigenvalue · Lie algebra · Lie algebra representation · unitary representation · Lie bracket · complexification

Recall that the Heisenberg Lie algebra is the Lie algebra of linear and constant functions on $M$, so can be thought of as

$${\mathfrak{h}}_{2d+1}=\mathcal{M}\oplus \mathbf{R}$$

where the R component is the constant functions. The Lie bracket is the Poisson bracket. Complexifying gives

$${\mathfrak{h}}_{2d+1}\otimes \mathbf{C}=(\mathcal{M}\oplus \mathbf{R})\otimes \mathbf{C}=(\mathcal{M}\otimes \mathbf{C})\oplus \mathbf{C}={\mathcal{M}}_J^{+}\oplus {\mathcal{M}}_J^{-}\oplus \mathbf{C}$$

so elements of ${\mathfrak{h}}_{2d+1}\otimes \mathbf{C}$ can be written as pairs $(u,c)=(u^{+}+u^{-},c)$ where

$$u\in \mathcal{M}\otimes \mathbf{C},u^{+}\in {\mathcal{M}}_J^{+},u^{-}\in {\mathcal{M}}_J^{-},c\in \mathbf{C}$$

This complexified Lie algebra is still a Lie algebra, with the Lie bracket relations

$$[(u_1,c_1),(u_2,c_2)]=(0,\Omega (u_1,u_2))\text{\hspace{1em}(26.12)}$$

and antisymmetric bilinear form Ω on $M$ extended from the real Lie algebra by complex linearity.

For each ${}^{J,}$ we would like to find a quantization that takes elements of ${\mathcal{M}}_J^{+}$ to linear combinations of creation operators, elements of ${\mathcal{M}}_J^{-}$ to linear combinations of annihilation operators. This will give a representation of the complexified Lie algebra

$${\Gamma }_J^{\prime }:(u,c)\in {\mathfrak{h}}_{2d+1}\otimes \mathbf{C}\to {\Gamma }_J^{\prime }(u,c)$$

if it satisfies the Lie algebra homomorphism property

$$[{\Gamma }_J^{\prime }(u_1,c_1),{\Gamma }_J^{\prime }(u_2,c_2)]={\Gamma }_J^{\prime }([(u_1,c_1),(u_2,c_2)])={\Gamma }_J^{\prime }(0,\Omega (u_1,u_2))\text{\hspace{1em}(26.13)}$$

Since we can write

$$(u,c)=(u^{+},0)+(u^{-},0)+(0,c)$$

where $u^{+}\in {\mathcal{M}}_J^{+}$ and $u^{-}\in {\mathcal{M}}_J^{-}$ , we have

$${\Gamma }_J^{\prime }(u,c)={\Gamma }_J^{\prime }(u^{+},0)+{\Gamma }_J^{\prime }(u^{-},0)+{\Gamma }_J^{\prime }(0,c)$$

Note that we only expect ${\Gamma }_J^{\prime }$ to be a unitary representation (with ${\Gamma }_J^{\prime }(u,c)$ skew-adjoint operators) for $(u,c)$ in the real Lie subalgebra ${\mathfrak{h}}_{2d+1}$ (meaning $u^{+}=\overline{u^{-}},c\in \mathbf{R})$ .

For the case of $J=J_0$ , the Lie algebra representation is given on basis elements

$$z_j\in {\mathcal{M}}_{J_0}^{+},{\overline{z}}_j\in {\mathcal{M}}_{J_0}^{-}$$

by

$${\Gamma }_{J_0}^{\prime }(0,1)=-i1,{\Gamma }_{J_0}^{\prime }(z_j,0)=-i{a}_j^{†}=-i{z}_j,{\Gamma }_{J_0}^{\prime }({\overline{z}}_j,0)=-i{a}_j=-i\frac{\partial }{\partial z_j}$$

and is precisely the Bargmann-Fock representation ${\Gamma }_{BF}^{\prime }|$ (see equation 22.5), Note that the operators $a_j$ and $a_j^{†}$ are not skew-adjoint, so ${\Gamma }_{J_0}^{\prime }$ is not unitary on the full Lie algebra ${\mathfrak{h}}_{2d+1}\otimes \mathbf{C}$ , but only on the real subspace ${\mathfrak{h}}_{2d+1}$ of real linear combinations of $q_j,p_j,1$

For more general choices of J we start by taking

$${\Gamma }_J^{\prime }(0,c)=-ic1\text{\hspace{1em}(26.14)}$$

which is chosen so that it commutes with all other operators of the representation, and for c real gives a skew-adjoint transformation and thus a unitary representation. We would like to construct ${\Gamma }_J^{\prime }(u^{+},0)$ as a linear combination of creation operators and ${\Gamma }_J^{\prime }(u^{-},0)$ as a linear combination of annihilation operators. The compatibility condition of equation 26.5 will ensure that the ${\Gamma }_J^{\prime }(u^{+},0)$ will commute, since if $u_1^{+},u_2^{+}\in {\mathcal{M}}_J^{+}$ , by 26.13 we have

$$[{\Gamma }_J^{\prime }(u_1^{+},0),{\Gamma }_J^{\prime }(u_2^{+},0)]={\Gamma }_J^{\prime }(0,\Omega (u_1^{+},u_2^{+}))=-i\Omega (u_1^{+},u_2^{+})1$$

and

$$\Omega (u_1^{+},u_2^{+})=\Omega (J{u}_1^{+},J{u}_2^{+})=\Omega (i{u}_1^{+},i{u}_2^{+})=-\Omega (u_1^{+},u_2^{+})=0$$

The ${\Gamma }_J^{\prime }(u^{-},0)$ will commute with each other by essentially the same argument. To see the necessity of the positivity condition 26.10 on ${}^{J,}$ recall that the annihilation and creation operators satisfy (for $d=1)$ )

$$[a,a^{†}]=1$$

a condition which corresponds to

$$[-ia,-i{a}^{†}]=[{\Gamma }_{J_0}^{\prime }(\overline{z},0),{\Gamma }_{J_0}^{\prime }(z,0)]={\Gamma }_{J_0}^{\prime }(0,\{\overline{z},z\})={\Gamma }_{J_0}^{\prime }(0,-i)=-1$$

Use of the opposite sign for the commutator would correspond to interchanging the role of a and $a^{†}$ , with the state |0⟩ now satisfying $a^{†}|0⟩=0$ and no state $|\psi ⟩$ in the state space satisfying $a|\psi ⟩=0$ . In order to have a state |0⟩ that is annihilated by all annihilation operators and a total number operator with nonnegative eigenvalues (and thus a Hamiltonian with a positive energy spectrum), we need all the commutators $[a_j,a_j^{†}]$ to have the positive sign.

For any choice of a potential basis element $u$ of ${\mathcal{M}}_J^{+}$ , by 26.13 and 26.14 we have,

$$[{\Gamma }_J^{\prime }(\overline{u},0),{\Gamma }_J^{\prime }(u,0)]={\Gamma }_J^{\prime }(0,\Omega (\overline{u},u))=-i\Omega (\overline{u},u)1\text{\hspace{1em}(26.15)}$$

and the positivity condition 26.10 on $J$ will ensure that quantizing such an element by a creation operator will give a representation with non-negative number operator eigenvalues. We have the following general result about the Bargmann-Fock construction for suitable $J:$

Theorem

Given a positive compatible complex structure $J$ on $M$, there is a basis $z_j^{J}of{\mathcal{M}}_J^{+}$ such that a representation of ${\mathfrak{h}}_{2d+1}\otimes \mathbf{C}$ , unitary for the real subalgebra ${\mathfrak{h}}_{2d+1}$ , is given by

$${\Gamma }_J^{\prime }(z_j^J,0)=-i{a}_j^{†},{\Gamma }_J^{\prime }({\overline{z}}_j^J,0)=-i{a}_j,{\Gamma }_J^{\prime }(0,c)=-ic1$$

where $a_j,a_j^{†}$ satisfy the conventional commutation relations, and ${\overline{z}}_j^J$ is the complex conjugate $of{z}_j^J$

Proof. An outline of the construction goes as follows:

1. Define a positive inner product on $M$ by $(u,v{)}_J=\Omega (u,Jv)$ on $M$. By Gram-Schmidt orthonormalization there is a basis of span $\{q_j\}\subset \mathcal{M}$ consisting of $d$ vectors $q_j^J$ satisfying

$$(q_j^J,q_k^J{)}_J={\delta }_{jk}$$

2. The vectors $J{q}_j^J$ will also be orthonormal since

$$(J{q}_j^J,J{q}_k^J{)}_J=\Omega (J{q}_j^J,J^2{q}_j^J)=\Omega (q_j^J,J{q}_j^J)=(q_j^J,q_k^J{)}_J$$

They will be orthogonal to the $q_j^J$ since

$$(J{q}_j^J,q_k^J{)}_J=\Omega (J^2{q}_j^J,q_k^J)=-\Omega (q_j^J,q_k^J)$$

and $\Omega (q_j^J,q_k^J)=0$ since any Poisson brackets of linear combinations of the $q_j$ vanish.

3. Define

$$z_j^J=\frac{1}{\sqrt{2}}(q_j^J-iJ{q}_j^J)$$

The $z_j^J$ give a complex basis of ${\mathcal{M}}_J^{+}$ , their complex conjugates ${\overline{z}}_j^J$ a complex basis of ${\mathcal{M}}_J^{-}$

4. The operators

$${\Gamma }_J^{\prime }(z_j^J,0)=-i{a}_j^{†}=-i{z}_j,{\Gamma }_J^{\prime }({\overline{z}}_j^J,0)=-i{a}_j=-i\frac{\partial }{\partial z_j}$$

satisfy the desired commutation relations and give a unitary representation on linear combinations of the $(z_j^J,0)$ and $({\overline{z}}_j^J,0)$ in the real subalgebra ${\mathfrak{h}}_{2d+1}$

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