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 <math>J </math> with vanishing kernel. The pseudo inner product in the indefinite norm 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 norm space while the round ones define the inner product in the Hilbert space. Furthermore the pseudo-adjoint of an operator <math>O</math> is defined by

<math>O^{ \dagger} J=J O^+ \ .

</math> In our case the original Hilbert space is identified with the Cartesian product of the fermionic and bosonic Fock spaces. 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=[(] \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 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 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>
• The last two relations follow from the 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. Indeed <math> \langle s | i \rangle = \langle s |Q|t \rangle = \langle t|Q|s \rangle ^*=0\ .</math>

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

Therefore if one selects <math> ker \ Q</math> 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 <math> ker \ Q</math> such as those in <math>im \ Q\ .</math>
• Whether the states in <math> ker \ Q</math> have non-negative norm.

Concerning the first question, the main remark following from the above considerations is that, adding to every state in <math> ker \ Q</math> arbitrary states in <math>im \ Q</math> 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 <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> This <math>Q</math>-equivalence criterion must be applied to the whole <math> ker \ Q</math> and hence the linear space of physical states must be identified with <math>ker \ Q/im \ Q \ ,</math> that is, the linear space of ''equivalence classes'' of vectors in <math> ker \ Q\ .</math>

Coming to the second question, it remains to verify that the pseudo-inner product induces an inner product into <math>ker \ Q/im \ Q </math> whose completion can thus be identified with the physical Hilbert space <math>H_{phys}\ .</math> In the present case it is neither difficult, nor immediate, to verify that in the whole B-F Fock space <math>ker \ Q </math> is the direct sumhttp://www.mathreference.com/la-jf,sum.html of <math>im \ Q</math> and the 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, however trivial, is a Hilbert space since the pseudo-norm of this state is positive.

This proves that <math>H_{phys}</math> is a Hilbert space.