# Becchi-Rouet-Stora-Tyutin symmetry/Indefinite Metric and BRST Cohomology

Indefinite Metric and BRST Cohomology

The standard construction of an indefinite-metric space is based on a Hilbert space and on the identification of a metric Hermitian operator $$J$$ with vanishing kernel. The pseudo inner product in the indefinite-metric space is defined by $\langle s|s' \rangle \equiv ( s|J |s' ) \ ,$ where the angle brackets define the pseudo inner product in the indefinite-metric space while the round ones define the inner product in the Hilbert space. Furthermore the pseudo-adjoint of an operator $$O$$ is defined by $O^{ \dagger} J=J O^+ \ .$ In the B-F oscillator model the total Hilbert space is identified with the Cartesian product of the fermionic and bosonic Fock space. The metric Hermitian operator $$J$$ is identified using the Pseudo-Hermiticity condition for $$Q$$ which is equivalent to the Hilbert space relation $Q^{ \dagger} J=J Q\ .$ One can solve this relation factorizing $$J$$ into the product of a bosonic operator $$j$$ and a fermionic one $$\mu$$, finding $J = \mu j=[( \bar a^{ \dagger}-a^{ \dagger})(a- \bar a)+1] \sum_{n=0}^ \infty( \bar A^{ \dagger}-A^{ \dagger})^n(A- \bar A)^n/n!$ Given $$J$$ it is easy to verify that $\bar AJ=J A \ , \ \ AJ=J \bar A \ , \ \ aJ=J \bar a \ , \ \bar aJ=J a \ ,$ and hence $\bar A^+= A^{ \dagger} \ , \ \ A^+= \bar A^{ \dagger} \ , \ \ \bar a^+= a^{ \dagger} \ , \ \ a^+= \bar a^{ \dagger} \ .$ Furthermore one verifies immediately that both $$H_{B-F}$$ and $$Q$$ are Pseudo-Hermitian operators.

Further important points are:

• The Fock vacuum $$|0\rangle$$ is an eigenvector of $$J$$ and has positive pseudo-norm, $$\langle 0|0 \rangle>0$$. Furthermore $$Q|0 \rangle=0$$.
• Among the single-particle states, $$(A^{ \dagger}+ \bar A^{ \dagger})|0 \rangle/ \sqrt{2}$$ and $$(a^{ \dagger}+ \bar a^{ \dagger})|0 \rangle/ \sqrt{2}$$ have positive pseudo-norm while $$(A^{ \dagger}- \bar A^{ \dagger})|0 \rangle/ \sqrt{2}$$ and $$(a^{ \dagger}- \bar a^{ \dagger})|0 \rangle/ \sqrt{2}$$ have negative pseudo-norm.
• Among the single-particle states, $$Q \bar A^{ \dagger}|0 \rangle=0$$ and $$Q a^{ \dagger}|0 \rangle=0$$. These relations follow from the nilpotency of $$Q$$ since $$\bar A^{ \dagger}|0 \rangle=-iQ \bar a^{ \dagger}|0 \rangle$$ and $$a^{ \dagger}|0 \rangle=iQA^{ \dagger}|0 \rangle$$.
• In general the states of $$im \ Q$$ are pseudo-orthogonal to those of $$ker \ Q$$ since $$Q$$ is pseudo-Hermitian.

To identify $$ker \ Q$$ with the physical invariant subspace of the indefinite-metric Fock space one has to address two issues:

• Understanding the physical meaning of states pseudo-orthogonal to the rest of $$ker \ Q$$ such as those in $$im \ Q$$.
• Showing that the states in $$ker \ Q$$ have non-negative norm.

Regarding the first issue, one observes that adding arbitrary states in $$im \ Q$$ to states in $$ker \ Q$$ does not change their pseudo-inner products. Therefore, from the point of view of the physical interpretation based on the probabilistic interpretation of the pseudo-inner product, two states in $$ker \ Q$$ whose difference belongs to $$im \ Q$$ must be considered equivalent $|s \rangle \sim \ |t \rangle \Longleftrightarrow |s \rangle - \ |t \rangle \in im \ Q$ Hence the linear space of physical states $$H_{phys}$$ must be identified with $$ker \ Q/im \ Q$$, which is the linear space of equivalence classes of vectors in $$ker \ Q$$.

As for the second question, if the inner product induced on $$ker \ Q/im \ Q$$ by the pseudo-inner product on the original space is definite positive, $$H_{phys}$$ is a Hilbert space. This has to be investigated on a case by case basis.

For the B-F model one proves directly that $$ker \ Q$$ is the direct sum of $$im \ Q$$ and the linear span of the vacuum vector $$|0\rangle$$. Therefore $$ker \ Q/im \ Q$$ coincides with the equivalence class of the vacuum $$|0 \rangle$$ which is a Hilbert space since the pseudo-norm of this state is positive.