At level one of the closed string states have the generic form $\gamma_{\mu\nu}\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ and are massless. It makes sense to decompose $\gamma_{\mu\nu}$ into a symmetric traceless, antisymmetric, and trace part, with the dilaton associated with the trace. That the dilaton is then a scalar seems obvious. However, we realise that the state $\eta_{\mu\nu}\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ is unphysical since the $L_{+1}$ condition requires the polarisation to be transverse, whereas here we have $\eta_{\mu\nu}p^\mu=p_\nu\neq0$. So we introduce some symmetric tensor $C_{\mu\nu}=p_\mu\overline p_\nu+\overline p_\mu p_\nu$, where $\overline p$ satisfies $\overline p\cdot p=1$ and $\overline p\cdot\overline p=0$. Then we can write down a modified state $(\eta_{\mu\nu}-C_{\mu\nu})\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ which is physical. I want to see explicitly that a state of this form is a scalar, since naively it has more than one degree of freedom. I think that this comes down to the fact all but one of the degrees of freedom are spurious, and I've shown that this state is indeed orthogonal to all physical states at levels zero and one of the closed string, except itself. Does this quality of being 'nearly spurious' confirm that we really do have a scalar? Thanks in advance.


The answer to your question can be understood directly by removing $n^{\mu}$ altogether, and seeing explicitly that the only remaining quantum number is $k^{\mu}$, implying immediately that the dilaton vertex operator is a scalar.

Primarily, note that a fixed-picture dilaton vertex operator, $\hat{\mathscr{V}}$, of momentum $k^{\mu}$ that satisfies the Virasoro constraints is easy to write down (as you mention) if we choose a vector with components $n^{\mu}$ such that $k\cdot n=1$ and $n^2=0$. Up to normalisation (which is $g_c/\sqrt{D-2}$ with $D$ the number of non-compact dimensions and $g_c$ the closed string coupling, as determined by unitarity or factorisation), $$ \boxed{\hat{\mathscr{V}}=c\tilde{c}\,\eta^{\perp}_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}e^{ik\cdot x},\qquad \eta^{\perp}_{\mu\nu}=\eta_{\mu\nu}-n_{\mu}k_{\nu}-k_{\mu}n_{\nu}\,} $$ where we have set $\alpha'=2$, and transversality, $\eta^{\perp}_{\mu\nu}k^{\mu}=0$, is manifest. (All vertex operators I write are to be considered normal-ordered in the $z,\bar{z}$ frame.) $\hat{\mathscr{V}}$ is (as you have checked) a conformal primary of weight $(h,\tilde{h})=(0,0)$ when $k^2=0$. That is, $$ \tilde{L}_n\cdot \hat{\mathscr{V}}=L_n\cdot \hat{\mathscr{V}}=0,\qquad {\rm for}\qquad n\geq0, $$ with $L_n,\tilde{L}_n$ the standard Virasoro generators of the full matter plus ghost energy-momentum tensor. (It is also true that $b_n\cdot \hat{\mathscr{V}}=\tilde{b}_n\cdot \hat{\mathscr{V}}=0$ for $n\geq0$, where $b_n,\tilde{b}_n$ are anti-ghost modes, but this will not play a dominant role here.) Incidentally, I'm using the operator/state correspondence here, according to which: $$ \alpha_{-1}|1\rangle\leftrightarrow i\partial x(z),\qquad \tilde{\alpha}_{-1}|1\rangle\leftrightarrow i\bar{\partial} x(\bar{z}),\qquad c_1|1\rangle\leftrightarrow c(z),\qquad \tilde{c}_1|1\rangle\leftrightarrow \tilde{c}(z) $$ $$ |0;k\rangle\equiv e^{ik\cdot x_0}|1\rangle\leftrightarrow e^{ik\cdot x(z,\bar{z})} $$

Your question is how to see that $\hat{\mathscr{V}}$ is a scalar. And it is a good question: clearly, a spacetime scalar (from the spacetime viewpoint) should only depend on $k^{\mu}$ and nothing else (and different scalars are distinguished by their mass, and there may in general be a degeneracy for fixed mass, compactification data may also enter, etc., which will also enter in the expression for the mass and the string coupling, $g_c$, but this is irrelevant for the dilaton). So why is $\hat{\mathscr{V}}$ a scalar, given we introduced a vector $n^{\mu}$ which seemingly might introduce new parameters into the state (it doesn't) while breaking manifest spacetime covariance (which it does as it singles out a distinguished direction in spacetime)? We can answer this in a simple way by restoring manifest covariance (i.e. removing the apparent $n^{\mu}$-dependence), as follows.

Recall that the physical state conditions for any vertex operator $\hat{\mathscr{V}}$ are that it lives in the cohomology of the BRST charge, $Q_B$. The crucial point is that not all states in the cohomology class of the dilaton are conformal primaries, i.e. not all of them are invariant under conformal transformations, but are perfectly acceptable physical states nevertheless as they are annihilated by $Q_B$ (and cannot be written as $Q_B\cdot \mathscr{W}$ for some $\mathscr{W}$). To bring this point into plain view, note that we can hop from one state in the dilaton cohomology class to another (i.e. from one description of the same physical state to another) by adding BRST exact terms, $$ \hat{\mathscr{V}}\rightarrow \hat{\mathscr{V}}'=\hat{\mathscr{V}}+Q_B\cdot \mathscr{W}, $$ so if the dilaton is a scalar there must be a choice of $\mathscr{W}$ for which the $n^{\mu}$ dependence drops out, leaving only the dependence on $k^{\mu}$. Of course, if $\mathscr{V}$is physical so will $\hat{\mathscr{V}}'$ be physical. The choice that does the job is to take: $$ \mathscr{W}=\big(c\,n_{\mu}\partial x^{\mu}-\tilde{c}\,n_{\mu}\bar{\partial} x^{\mu}\big)e^{ik\cdot x}. $$ Computing $Q_B\cdot \mathscr{W}$ (which I leave as an exercise, see Polchinksi's book for the explicit expression for $Q_B$) leads to the new vertex operator, \begin{equation}\label{eq:V'=V+exact} \begin{aligned} \boxed{\hat{\mathscr{V}}'=\Big(c\tilde{c}\,\eta_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}-\tfrac{1}{2}c\partial^2c+\tfrac{1}{2}\tilde{c}\bar{\partial}^2\tilde{c}\Big)e^{ik\cdot x}\,} \end{aligned} \end{equation} Clearly, from the above $\hat{\mathscr{V}}'$ and $\hat{\mathscr{V}}$ describe the same physical state. Covariance is restored (we have completely removed the ugly dependence on $n^{\mu}$) and most importantly the state is manifestly completely specified by the choice of momentum: $k^{\mu}$. So this is a scalar, given that there is no other quantum number characterising the state.

Your question is therefore answered, but I should mention a drawback. Although covariance is restored, the vertex operator $\hat{\mathscr{V}}'$ is not a conformal primary (it is not annihilated by $L_n,\tilde{L}_n$ for $n>0$), and so is not conformally invariant. This is clearly no problem as BRST invariance is a more general concept; we don't need conformal invariance, although it does help to have it for many computations.

Finally, I will mention the analogue of $\hat{\mathscr{V}}$ and $\hat{\mathscr{V}}'$ in the integrated picture, that I denote by $\mathscr{V}$ and $\mathscr{V}'$ respectively. Because this involves some additional technology that would take a few more paragraphs to explain (given it doesn't exist in standard textbooks), and because the answer to your question has already emerged, I will simply state the results*: $$ \boxed{\mathscr{V} = \int d^2z \,\eta^{\perp}_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}e^{ik\cdot x}\,} $$ for the conformal primary, and: $$ \boxed{\mathscr{V}' = \int d^2z\,\Big(\eta_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}-R_{z\bar{z}}\Big)e^{ik\cdot x}\,} $$ for the covariant (non-primary) integrated picture vertex operator. Again, you see that $\eta_{\mu\nu}$ appears and not $\eta_{\mu\nu}^{\perp}$ in $\mathscr{V}'$, so the $n^{\mu}$ dependence again drops out. Here $R_{z\bar{z}}$ is the worldsheet Ricci tensor. Care is needed in interpreting these expressions, and the "coordinates" $z,\bar{z}$ are not really worldsheet coordinates but are best interpreted as frame coordinates, see Polchinski for all the glory details where everything I have covered is explained in much greater detail. I will leave it at this.

*(This elaborates on @Nogueira's comment about it not being possible to write down covariant physical states without ghosts -- the point I'm highlighting here is that in the integrated picture one can instead write down covariant vertex operators without ghost BRST technology.)

  • $\begingroup$ When you do that, avoiding ghost excitations, you are going to find a covariant polarization for the state plus constraints and redundancies coming from $R_{z\bar{z}}$. This constraint and redundancies will return if you do the what you said. This does not prove your point that there is a covariant manner to obtain the physical state without introducing ghost. You are just hidden this fact by working with integrated vertex operators. $\endgroup$ – Nogueira Mar 17 '18 at 15:46
  • $\begingroup$ my comment refers to 3rd boxed eqn.. In the Polyakov formalism (which is what people refer to when they say ocq), there are no ghosts and one integrates over vertex operators. I don’t know how to construct a covariant state without ghosts, so i agree on this point with your last comment. $\endgroup$ – Wakabaloola Mar 17 '18 at 16:07
  • $\begingroup$ If you correct the part that mistake my statement I erase my downvote, in the same way as a ghost state erases the longitudinal and time-like polarizations of a covariant physical states ;) $\endgroup$ – Nogueira Mar 17 '18 at 16:25
  • $\begingroup$ Ok, so you should agree that you mistake my statement? I agree with you that there is covariant integrated vertex operators in OCQ, but this just hidden the fact that there is no way to solve the constraints and the redundancies to obtain the physical states in a covariant manner without introducing ghost states. $\endgroup$ – Nogueira Mar 17 '18 at 16:28
  • 1
    $\begingroup$ yes that’s correct (if that is what you meant when you wrote it), and the boundary terms you are referring to in moduli space are associated to a global obstruction associated to the Euler number, not complex structure moduli, although this statement is coordinate dependent. also the map integrated to fixed vertex operators is indeed many to one. $\endgroup$ – Wakabaloola Mar 17 '18 at 16:31

Actually there is two things here. The constraint $L_{+1}|\psi\rangle=0$ and the gauge redundancy $|\psi\rangle \cong |\psi\rangle +L_{-1}|\xi\rangle$. For $N=1$, massless particles, with state $|\psi\rangle=e_{\mu\nu}\alpha_{-1}^{\mu}\tilde{\alpha}_{-1}^{\nu}|0;k\rangle$ we have as constraints

$$ k^2=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k^{\mu}e_{\mu\nu}=k^{\nu}e_{\mu\nu}=0 $$

and as gauge redundancy

$$ e_{\mu\nu}\cong e_{\mu\nu}+a_{\mu}k_{\nu}+k_{\mu}b_{\nu} $$

where the $a_{\mu}$ and $b_{\mu}$ parametrizes the redundancy under the constraint $a\cdot k=b\cdot k=0$.

In order to obtain the spectra from it we should solve the constraint and fix the redundancy. There is no way to do that in a covariant manner without the introduction of ghost states (BRST) (I'm not talking about vertex operators, I'm talking about physical states). The reason for that is the fact that time-like components and longitudinal components of the polarization tensor $e_{\mu\nu}$ is not physical, and there is no way to single out this component in a covariant manner. One way to preserve covariance is the introduction of ghost states to erase this components rather than single out them.

Picking the $e_{\mu\nu}=0$ for $\mu=0,1$ and $\nu=0,1$ in the frame where the momentum is $(k^{\mu})=(E,E,0,...,0)$ both solve the constraint as fixes the gauge redundancy. This left a $SO(D-2)$ matrix $e_{ij}$, for $i,j=2,...,D-1$. The scalar particle, the dilation, will be the trace of this matrix.

Note that the dilaton is not the trace of the space-time matrix $e_{\mu\nu}$, but only the trace of the transverse part of the $e_{\mu\nu}$. The time-like and longitudinal polarization are fixed by the constraints and gauged away by the gauge redundancy. This should be expected by the fact that massless particle little group is $SO(D-2)$ and not $SO(D)$, so the state should be at the representation of $SO(D-2)$.

If you pick the state $e_{\mu\nu}=\eta_{\mu\nu}$, this will not satisfy the constraint. A way to solve it is by subtracting the longitudinal and time-like components, in this frame this will be $e_{\mu\mu}=0$ for $\mu=0,1$, and the other components intact. The gauge redundancy in this frame will be adding

$$ e_{\mu0}\cong e_{\mu0}+a_{\mu}E $$

$$ e_{\mu1}\cong e_{\mu1}+a_{\mu}E $$

$$ e_{0\mu}\cong e_{0\mu}+b_{\mu}E $$

$$ e_{1\mu}\cong e_{1\mu}+b_{\mu}E $$

We can fix this components to zero. (Actually this components are already zero, so we just need to avoid the gauge transformations, i.e $a=b=0$). This will give at the end a matrix $e_{ij}=\delta_{ij}$ for $i,j=2,...,D-1$ and zero otherwise. This is your dilaton state.

$$ |\psi\rangle=\phi\delta_{i,j}\alpha_{-1}^{i}\alpha_{-1}^{j}|0;k\rangle $$

just one component $\phi$, as expected from a scalar particle.

  • $\begingroup$ On what are you basing the claim `.. we should solve the constraint and fix the redundancy. There is no way to do that in a covariant manner without the introduction of ghost states (BRST)' ? $\endgroup$ – Wakabaloola Mar 17 '18 at 8:52
  • $\begingroup$ @Wakabaloola the constraints and redundancies on the polarization tensor $e_{\mu\nu} $ contains the four-vector $k^{\mu}$. In order to solve this explicity we should pick a frame for $k^{\mu} $ and this breaks covariance. This appears in order to anihilates the negative and zero norms states. Introduction of ghost from the BRST formalism can resolve that since they can cancel the contribution of negative and zero norm states in a covariant gauge. $\endgroup$ – Nogueira Mar 17 '18 at 14:40
  • $\begingroup$ In your answer you writte down the vertex operator of a state that solve both the constraint and the redundancy in a covariant manner by the coust of introducing ghost modes, the $\partial^2c$ term. $\endgroup$ – Nogueira Mar 17 '18 at 14:44
  • $\begingroup$ the point is that you don’t need the ghosts as you can see in what I called $\mathscr{V}’$ $\endgroup$ – Wakabaloola Mar 17 '18 at 14:50
  • $\begingroup$ @Wakabaloola The fact that the integrated vertex operator can be written without the ghost fields does not imply that the state created by this vertex operator does not have ghost excitation. The map between states and vertex operators are less straightforward for integrated vertex operators. The more clear way to do that is to work with unintegrated vertex operators. So you claim is not right, since in unintegrated form the vertex operator does have a ghost excitation $\partial^2c$ $\endgroup$ – Nogueira Mar 17 '18 at 15:24

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