24.3 Implications of the choice of z, z

24.3 Implications of the choice of z, z

Concept links · terms present in this machine draft; source roles are unverified: eigenvalue · U(1) · Lie bracket · complexification

The definition of annihilation and creation operators requires making a specific choice, in our case

$$\overline{z}=\frac{1}{\sqrt{2}}(q+ip),z=\frac{1}{\sqrt{2}}(q-ip)$$

for complexified coordinates on phase space, which after quantization becomes the choice

$$a=\frac{1}{\sqrt{2}}(Q+iP),a^{†}=\frac{1}{\sqrt{2}}(Q-iP)$$

Besides the complexification of coordinates on phase space $M$, the choice of z introduces a new piece of structure into the problem. In chapter 26 we’ll examine other possible consistent such choices, here will just point out the various diferent ways in which this extra structure appears.

• The Schr¨odinger representation of the Heisenberg group comes with no particular distinguished state. The unitarily equivalent Bargmann-Fock representation does come with a distinguished state, the constant function 1. It has zero eigenvalue for the number operator $N=a^{†}a$ so can be thought of as the state with zero “quanta”, or the “vacuum” state and can be written |0⟩. Such a constant function could also be characterized (up to scalar multiplication), as the state that satisfies the condition

$$a|0⟩=0$$

• The choice of coordinates $z$ and $\overline{z}\mathrm{gives}$ a distinguished choice of Hamiltonian function, $h=z\overline{z}$ . After quantization this corresponds to a distinguished choice of Hamiltonian operator

$$H=\frac{1}{2}(a^{†}a+a{a}^{†})=a^{†}a+\frac{1}{2}=N+\frac{1}{2}$$

With this choice the distinguished state |0⟩ will be an eigenstate of H with eigenvalue $\frac{1}{2}$ .

• The choice of the coordinate $z$ gives a decomposition

$$\mathcal{M}\otimes \mathbf{C}=\mathbf{C}\oplus \mathbf{C}\text{\hspace{1em}(24.10)}$$

where the first subspace $C$ has basis vector $z,$ the second subspace has basis vector $z$.

• The decomposition 24.10 picks out a subgroup $U(1)\subset SL(2,\mathbf{R})$ , those symplectic transformations that preserve the decomposition. In terms of the coordinates $z,\overline{z},$ the Lie bracket relations 16.15 giving the action of $\mathfrak{sl}(2,\mathbf{R})$ on $\mathcal{M}$ become

$$\{z\overline{z},z\}=-iz,\{z\overline{z},\overline{z}\}=i\overline{z}$$ $$\left\{\frac{z^2}{2},z\right\}=0,\left\{\frac{z^2}{2},\overline{z}\right\}=iz$$ $$\left\{\frac{{\overline{z}}^2}{2},z\right\}=-i\overline{z},\left\{\frac{{\overline{z}}^2}{2},\overline{z}\right\}=0$$

The only basis element of sl(2, $R$) does not mix the $z$ and $\overline{z}$ coordinates is $z\overline{z}.$ . We saw (see equation 24.1) that upon exponentiation this basis element gives the subgroup of SL(2, $R$) of matrices of the form

$$\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$$

• Quantization of polynomials in $z$, $z$ involves an operator ordering ambiguity since a and $a^{†}$ do not commute. This can be resolved by the following specific choice, one that depends on the choice of $z$ and $\overline{z}:$

Definition

Normal ordered product

Given any product P of the a and $a^{†}$ operators, the normal ordered product of $P_z$ , written $:P:$ is given by re-ordering the product so that all factors $a^{†}$ are on the $left$ , all factors a on the right, for example

$$:a^2{a}^{†}a(a^{†}{)}^3:=(a^{†}{)}^4{a}^3$$

For the case of the Hamiltonian $H_{;}$ , the normal ordered version

$$:H:=:\frac{1}{2}(a{a}^{†}+a^{†}a):=a^{†}a$$

could be chosen. This has the advantage that it acts trivially on |0⟩ and has integer rather than half-integer eigenvalues on $\mathcal{F}.$ . Upon exponentiation one gets a representation of $U(1)$ with no sign ambiguity and thus no need to invoke a double covering. The disadvantage is that $:H:$ gives a representation of u(1) that does not extend to a representation of sl(2, $R$).

Previous section: 24.2 Intertwining operators in terms of a and a ^ · Next section: 24.4 SU(1, 1) and Bogoliubov transformations · 中文版本

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原书 PDF · 印刷页 266、267、268、269、270、271、272、273、274、275

来源版本：2025-10-20

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