25.5 Normal ordering and the anomaly in finite dimensions

25.5 Normal ordering and the anomaly in finite dimensions

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

For $A\in \mathfrak{u}(d)\subset \mathfrak{sp}(2d,\mathbf{R})$ we have seen that we can construct the Lie algebra version of the metaplectic representation as

$${\tilde{U}}_A^{\prime }=\frac{1}{2}\sum _{j,k}{A}_{jk}(a_j^{†}{a}_k+a_k{a}_j^{†})$$

which gives a representation that extends to sp(2d, $R$), or we can normal order, getting

$$U_A^{\prime }=:{\tilde{U}}_A^{\prime }:=\sum _{j,k}{a}_j^{†}{A}_{jk}{a}_k={\tilde{U}}_A^{\prime }-\frac{1}{2}\sum _{j=1}^d{A}_{jj}1$$

To see that this normal ordered version does not extend to $\mathfrak{sp}(2d,\mathbf{R})$ , observe that basis elements of sp(2d, $R$) that are not in $\mathfrak{u}(d)$ are the linear combinations of $z_j{z}_k$ and ${\overline{z}}_j{\overline{z}}_k$ that correspond to real-valued functions. These are given by

$$\frac{1}{2}\sum _{jk}(B_{jk}{z}_j{z}_k+{\overline{B}}_{jk}{\overline{z}}_j{\overline{z}}_k)$$

for a complex symmetric matrix B with matrix entries $B_{jk}$ . There is no normal ordering ambiguity here, and quantization will give the unitary Lie algebra representation operators

$$-\frac{i}{2}\sum _{jk}(B_{jk}{a}_j^{†}{a}_k^{†}+{\overline{B}}_{jk}{a}_j{a}_k)$$

Exponentiating such operators will give operators which take the state |0⟩ to a distinct state (one not proportional to |0⟩).

Using the canonical commutation relations one can show

$$[a_j^{†}{a}_k^{†},a_l{a}_m]=-a_l{a}_j^{†}{\delta }_{km}-a_l{a}_k^{†}{\delta }_{jm}-a_j^{†}{a}_m{\delta }_{kl}-a_k^{†}{a}_m{\delta }_{jl}$$

and these relations can in turn be used to compute the commutator of two such Lie algebra representation operators, with the result

$$\begin{array}{rl}& [-\frac{i}{2}{\sum }_{jk}(B_{jk}{a}_j^{†}{a}_k^{†}+{\overline{B}}_{jk}{a}_j{a}_k),-\frac{i}{2}{\sum }_{lm}(C_{lm}{a}_l^{†}{a}_m^{†}+{\overline{C}}_{lm}{a}_l{a}_m)]\\ & =\frac{1}{2}{\sum }_{jk}(B\overline{C}-C\overline{B}{)}_{jk}(a_j^{†}{a}_k+a_k{a}_j^{†})={\tilde{U}}_{B\overline{C}-C\overline{B}}^{\prime }\end{array}$$

Note that normal ordering of these operators just shifts them by a constant, in particular

$$U_{B\overline{C}-C\overline{B}}^{\prime }=:{\tilde{U}}_{B\overline{C}-C\overline{B}}^{\prime }:={\tilde{U}}_{B\overline{C}-C\overline{B}}^{\prime }-\frac{1}{2}tr(B\overline{C}-C\overline{B})1\text{\hspace{1em}(25.9)}$$

The normal ordered operators fail to give a Lie algebra homomorphism when extended to sp(2d, $R$), but this failure is just by a constant term. Recall from section 15.3 that even at the classical level, there was an ambiguity of a constant in the choice of a moment map which in principle could lead to an “anomaly”, a situation where the moment map failed to be a Lie algebra homomorphism by a constant term. The situation here is that this potential anomaly is removable, by the shift

$$U_A^{\prime }\to {\tilde{U}}_A^{\prime }=U_A^{\prime }+\frac{1}{2}tr(A)1$$

which gives representation operators that satisfy the Lie algebra homomorphism property. We will see in chapter 39 that for an infinite number of degrees of freedom, the anomaly may not be removable, since the trace of the operator $A$ in that case may be divergent.

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