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The generators of the Virasoro algebra (actually two copies thereof) appear as constraints in the classical theory of the Polyakov action (after gauge fixing). However, when quantizing only "half" of the constraints are imposed.

I understand it's inconsistent to apply all of the constraints. I also understand this choice has mathematical significance in the sense of picking out highest-weight vectors. Nevertheless, on the surface it seems somewhat arbitrary. What is the physical reason it is "ok" to ignore the other constraints while mandatory to impose these? Also, are there simpler (preferably finite dimensional phase space) examples of quantization where some of the constraints must be dropped?

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Although it is inconsistent to impose $L_n v=0$ for all $n$, it is still possible to impose it for $n\geq -1$. –  Pavel Safronov Dec 10 '11 at 0:33
    
@Dilaton: About your edit: I don't think all TP.SE questions are really research - level . . . –  Dimensio1n0 Aug 19 '13 at 12:22
    
@Dimension10 Qmechanic obvously too ... ;-). However, if you read up the few recent chat discussions concerning a new higher level TP site, one should probably consider including advanced enough graduat stuff too, to obtain a viable site, and consider using research-level as a slightly generalized high-level tag in the same way, auch that for example the questions people have liked on TP are included. This is just a thought ... –  Dilaton Aug 19 '13 at 12:30

2 Answers 2

up vote 8 down vote accepted

You seem to be talking about the "old covariant quantization" in which $L_n$ for positive $n$ and $(L_0-a)$ annihilate physical ket states $|\psi\rangle$, right? It's analogous to the Gupta-Bleuler quantization

http://en.wikipedia.org/wiki/Gupta-Bleuler_quantization

which was a standard procedure used already in electromagnetism. The idea is that the Hermitian conjugate condition guarantees that $L_n$ for a negative $n$ annihilates all the physical bra-vectors $\langle \psi |$. So all the matrix elements $$ \langle \psi | (L_n - a \delta_{n0}) | \phi \rangle =0 $$ vanishes for all pairs of physical states because the matrix inside annihilates either the ket vector or the bra vector. A few more conditions follow.

But I want to say that this treatment has become obsolete in the 1970s when people started to describe gauge theories – and analogously, Virasoro- or diff-symmetric gravitational theories (on the world sheet or in spacetime) – by the modern covariant quantization or BRST quantization.

In that new formalism, you have the $bc$ ghosts, the BRST charge, and physical states are cohomologies of $Q$, the BRST charge. In particular, this also allows you to demand that physical states (representatives) are annihilated by all $L_n$ for both signs of $n$.

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@Squark: that the physical states are highest weight vectors is an oversimplification. It depends on the structure of the matter representation of the Virasoro algebra. Already in the bosonic string for Fock modules at zero momentum (which are always there) this fails. Also for the representations in nonctritical strings,... –  José Figueroa-O'Farrill Dec 10 '11 at 6:05
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@Squark: the "rule" is precisely as in Luboš's answer. You compute BRST cohomology (or if you prefer, relative semi-infinite cohomology) of the Virasoro modules appearing in your conformal field theory. –  José Figueroa-O'Farrill Dec 10 '11 at 12:27
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Dear @Squark, when we're happy to know the BRST formalism, it's a good idea to start with it (there is no central term in it, so no obstruction to impose the annihilation by all generators!), and view the old covariant quantization as a particular convention to choose the representatives from the BRST cohomologies. Because the ghosts-$bc$-free Virasoro generators commute to the central extension, one would need to talk about a complicated representation theory of these "extension" groups and it's not necessarily an insightful approach. –  Luboš Motl Dec 10 '11 at 18:44
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A $Q$-cohomology is the class of all vectors $|\psi\rangle$ that satisfy $Q|\psi\rangle=0$ and that are identified by the equivalence $|\psi\rangle \sim |\psi\rangle + Q|\phi\rangle$ for arbitrary vectors $|\phi\rangle$. In the cohomology, you may find some vectors with no "positive" excitations by the $b,c$-ghosts and a certain proper ghost number. If you take these states and ignore all the $b,c$ ghosts in them, you get states of the old covariant quantization. Also, I wanted to stress that the fact that the Hilbert space is interpreted in a stringy way plays no role: it's still a QFT. –  Luboš Motl Dec 10 '11 at 18:50
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Dear @Squark, right, string theory leads to a world sheet theory with gravity and diff+Weyl (or conformal) symmetry, moduli spaces, etc. But these rules are exactly what you would construct for a QFT/CFT with a dynamical metric and these extra gauge symmetries. It doesn't matter that it's also a "string theory" which is (in spacetime) "more than just a QFT". The world sheet rules are those of a QFT with some gauge symmetries including diff and Weyl, and with the dynamical metric. –  Luboš Motl Dec 11 '11 at 19:01

A modern treatment of this subject can be found in Segreev's book on the Kahler geometry of loop spaces also available online. This line of research started with the seminal work of Bowick and Rajeev: The holomorphic geometry of the closed bosonic string theory and $Diff S^1/S^1$ (Spires) (and independently Kirillov and Yuriev (Please see the reference in Sergeev's book)).

The basic idea is that loop spaces are Kahler manifolds carrying a $Diff S^1$ invariant closed two form (Kahler form). ( These manifolds are quantizeable by selection of a complex structure compatible with the Kahler form and performing Kahler (holomorphic) quantization.)

In the present problem, one seeks to define a $Diff S^1$ equivariant quantizations on them, i.e., to find a representation of $Diff S^1$ invariant functions on these loop spaces as $Diff S^1$ invariant subspace of vectors on a Hilbert space. The invariant vectors are obtained as null vectors of the quantized algebra.

Now, as many other cases in quantum mechanics, one cannot find a faithful representation at the quantum level but only a projective one. This means that the $Diff S^1$ invariant subspace can only be trivial.

This is the main reason that the original solution was found by imposition of half the constraints which is reminiscent of the Kahler quantization itself where one selects holomorphic sections.

The modern solution lies in the fact that a given complex structure on the loop space is not invariant under $Diff S^1$. Thus also a given Kahler quantization is not invariant as well.

The vacuum of the Kahler quantization Hilbert space corresponding to a given complex structure is not invariant under $Diff S^1$. The transformed vacuum is the vacuum corresponding to the Kahler quantization associated with the transformed complex structure.

This leads to the notion of the "Vacuum bundle" which is a line bundle over $Diff S^1/S^1$ comprising all the individual vacua. This bundle has a connection associated with the projective representation of $Diff S^1$, which becomes flat when one takes into account all the fields of the theory (in the case of the bosonic string at $d=26$).

The main result is that the flatness of the connection is equivalent to the unitary equivalence of the quantization Hilbert spaces under the action of $Diff S^1$, which is the required result expressing the equivariance of the quantization.

This treatment has many connections to many contemporary research areas. for example the bundle of complex structures over the loop space is a twistor space as was noticed by Bowick and Rajeev in their original work.

Update:

Sorry for never mentioning it in the original answer (It appears in the references), the loop space under consideration is $\Omega \mathbb{R}^{d-1,1}$ (based loops), (becoming $\Omega \mathbb{R}^{d-1,1}//diff S1$ after the Hamiltonian reduction).

The (unitary) equivalence of the three mentioned quantization methods Goddard Goldstone Rebbi Thorn (GGRT- half of the constraints method), BRST and the Bowick-Rajeev (twistor) can be explained as follows:

GGRT-BRST: The BRST invariance condition of the physical states is equivalent by explicit construction to the GGRT half number of constraints, please see Green Schwarz Witten equation: 3.2.31

BRST - Bowick-Rajeev: The inclusion of the ghost sector of the string is equivalent to tensoring with the anticanonical bundle to the Fock bundle of the bosonic string. This can be understood from the fact that the Ricci tensor of $Diff S1/S1$ is exactly equal to the contribution of the ghost sector to the vacuum bundle curvature (for example by a direct calculation). Thus the half form quantization is equivalent to the introduction of ghosts. This can be further seen in the fact that the physical states have a ghost number of $-\frac{1}{2}$. This is also consistent with the path integral picture, where one must divide by the redundant gauge orbit measure.

In addition to the fact that in this method, one is not required to introduce the ghost fields explicitly, the flatness of the connection establishes the equivalence between the quantizations with respect to different Kahler polarizations.

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There is something I'm missing here. Apparently you're saying LM / diff S1 where M is spacetime is the phase space of the string. However it seems to me the phase space should consist of solutions to the classical equations of motion, rather than slices at constant worldsheet time. Or maybe M is space rather than spacetime? Or a null hypersurface? Also, I still don't understand how it leads to selecting half of the constraints. Is it the case that the complex structure can be chosen invariant under half of the generators? –  Squark Dec 10 '11 at 11:23
    
Or maybe the action of Diff S1 on the loop space is not the trivial action I had in mind but the action of conformal transformations on the worldsheet continuing the loop? Btw, is it unique? Naive intuition suggests we need to assign a "momentum" to the loop in addition to its "position" –  Squark Dec 10 '11 at 11:52
    
I apologize if I'm asking really stupid questions –  Squark Dec 10 '11 at 11:52
    
@ Squark I have posted an answer to your questions in an update to the original answer –  David Bar Moshe Dec 11 '11 at 14:44

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