BRST Symmetry/Indefinite Metric and BRST Cohomology

Indefinite Metric and BRST Cohomology

The standard construction of an indefinite norm 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 norm space is defined by \[ \langle s|s' \rangle \equiv ( s|J |s' ) \ , \] where the angle brackets define the pseudo inner product in the indefinite norm 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 our case the original Hilbert space is identified with the Cartesian product of the fermionic and bosonic Fock spaces. 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 \(Q a^{ \dagger}|0 \rangle=0\ .\)
• The last two 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. Indeed \( \langle s | i \rangle = \langle s |Q|t \rangle = \langle t|Q|s \rangle ^*=0\ .\)

Thus, considering for simplicity the five-dimensional restricted space spanned by the vacuum state and the single-particle states, we see that \( ker \ Q\) is three-dimensional space. Since \(Q\) is nilpotent its kernel contains \(im \ Q\ ,\) the image of \(Q\ ,\) which in the above restricted space is two-dimensional.

Therefore if one selects \( ker \ Q\) as physical subspace of the indefinite norm Fock space one has to face two problems:

• What is the physical meaning of states pseudo-orthogonal to the rest of \( ker \ Q\) such as those in \(im \ Q\ .\)
• Whether the states in \( ker \ Q\) have non-negative norm.

Concerning the first question, the main remark following from the above considerations is that, adding to every state in \( ker \ Q\) arbitrary states in \(im \ Q\) does not change the pseudo-inner products, therefore, from the point of view of the physical interpretation based on the probabilistic interpretation pseudo-inner product, two states in \( ker \ Q\) whose difference belongs to \(im \ Q\) must be considered equivalent \[|s \rangle \sim \ |t \rangle \Longleftrightarrow |s \rangle - \ |t \rangle \in im \ Q \] This \(Q\)-equivalence criterion must be applied to the whole \( ker \ Q\) and hence the linear space of physical states must be identified with \(ker \ Q/im \ Q \ ,\) that is, the linear space of equivalence classes of vectors in \( ker \ Q\ .\)

Coming to the second question, it remains to verify that the pseudo-inner product induces an inner product into \(ker \ Q/im \ Q \) whose completion can thus be identified with the physical Hilbert space \(H_{phys}\ .\) In the present case it is neither difficult, nor immediate, to verify that in the whole B-F Fock space \(ker \ Q \) is the direct sum[1] 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, however trivial, is a Hilbert space since the pseudo-norm of this state is positive.

This proves that \(H_{phys}\) is a Hilbert space.