View source for Becchi-Rouet-Stora-Tyutin symmetry/Indefinite Metric and BRST Cohomology

<strong> Indefinite Metric and BRST Cohomology </strong>

The standard construction of an [http://en.wikipedia.org/wiki/Krein_space indefinite-metric space]  is based on a [http://en.wikipedia.org/wiki/Hilbert_space Hilbert space]  and on the identification of  a metric [http://en.wikipedia.org/wiki/Hermitian_adjoint Hermitian operator] <math>J </math> with vanishing [http://en.wikipedia.org/wiki/Linear_transformation kernel]. The pseudo [http://en.wikipedia.org/wiki/Inner_product inner product] in the indefinite-metric space
is defined by 
:<math> \langle s|s'  \rangle \equiv ( s|J |s' ) \ ,
</math>
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-[http://en.wikipedia.org/wiki/Hermitian_adjoint adjoint] of an operator <math>O</math> is defined by 
:<math>O^{ \dagger} J=J O^+ \ .
</math>
In the B-F __AUTOLINKER{0|oscillator}oscillator model the total Hilbert space is identified with the [http://en.wikipedia.org/wiki/Cartesian_product Cartesian  product] of the fermionic and bosonic [http://en.wikipedia.org/wiki/Fock_space Fock space]. The  metric Hermitian operator <math>J </math> is identified using the Pseudo-Hermiticity condition for <math>Q</math> which  is equivalent to the Hilbert space relation 
:<math>Q^{ \dagger} J=J Q\ .
</math> 
One can   solve  this relation factorizing <math>J</math> into the product of a bosonic operator <math>j</math> and a fermionic one <math> \mu\ ,</math>
finding 
:<math>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!
</math>
Given <math>J </math> it is easy to verify that 
:<math> \bar AJ=J A  \ , \  \  AJ=J  \bar A  \ , \  \ aJ=J  \bar a  \ , \  \bar aJ=J  a \ ,</math> 
and hence 
:<math> \bar A^+= A^{ \dagger}  \ , \  \ A^+=  \bar A^{ \dagger}  \ , \  \  \bar a^+= a^{ \dagger}  \ , \  \ a^+=  \bar a^{ \dagger} \ .
</math> 
Furthermore one verifies immediately that both <math>H_{B-F} </math> and <math>Q</math> are Pseudo-Hermitian operators.

Further important points are:
  
* The Fock vacuum  <math>|0\rangle</math>  is an [http://en.wikipedia.org/wiki/Eigenvalue eigenvector] of <math>J</math> and has positive pseudo-norm, <math> \langle 0|0  \rangle>0\ .</math> Furthermore <math>Q|0 \rangle=0\ .</math>  
*  Among the [http://en.wikipedia.org/wiki/Quantum_field_theory single-particle states], <math>(A^{ \dagger}+ \bar A^{ \dagger})|0 \rangle/ \sqrt{2}</math> and <math>(a^{ \dagger}+ \bar a^{ \dagger})|0 \rangle/ \sqrt{2}</math> have positive pseudo-norm  while <math>(A^{ \dagger}- \bar A^{ \dagger})|0 \rangle/ \sqrt{2}</math> and <math>(a^{ \dagger}- \bar a^{ \dagger})|0 \rangle/ \sqrt{2}</math> have negative pseudo-norm. 
* Among the single-particle states, <math>Q  \bar A^{ \dagger}|0 \rangle=0</math> and <math>Q a^{ \dagger}|0 \rangle=0\ .</math> These relations follow from the  [http://en.wikipedia.org/wiki/Nilpotent#Nilpotency_in_physics nilpotency] of <math>Q</math> since <math> \bar A^{ \dagger}|0 \rangle=-iQ \bar a^{ \dagger}|0 \rangle</math> and <math>a^{ \dagger}|0 \rangle=iQA^{ \dagger}|0 \rangle\ .</math>   
* In general the states of  <math>im \ Q</math> are pseudo-orthogonal to those of  <math> ker \ Q</math> since <math>Q</math> is pseudo-Hermitian.  

To identify  <math> ker \ Q</math>  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 <math> ker \ Q</math> such as those in <math>im \ Q\ .</math>   
* Showing that the states in <math> ker \ Q</math> have non-negative norm.   

Regarding the first issue, one observes that adding arbitrary states in <math>im \ Q</math> to states in <math> ker \ Q</math> 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 <math> ker \ Q</math> whose difference belongs to <math>im \ Q</math> must be considered equivalent 
:<math>|s \rangle  \sim \ |t \rangle \Longleftrightarrow |s \rangle - \ |t \rangle \in im \ Q
</math>
Hence the linear space of physical states <math>H_{phys}</math> must be identified with <math>ker \ Q/im \ Q \ ,</math> which is the linear space of [http://en.wikipedia.org/wiki/Quotient_space ''equivalence classes''] of vectors in <math> ker \ Q\ .</math>

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

For the B-F model one proves directly that <math>ker \ Q </math> is the [http://www.mathreference.com/la-jf,sum.html ''direct sum''] of <math>im \ Q</math> and the [http://en.wikipedia.org/wiki/Linear_span linear span] of the vacuum vector <math>|0\rangle\ .</math> Therefore  <math>ker \ Q/im \ Q </math> coincides with the equivalence class of the vacuum <math>|0 \rangle</math> which is a Hilbert space since the pseudo-norm of this state is positive.