# 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.