1. Introduction
It is valuable to look for various modifications of Einstein's general relativity since quantum theory of general relativity is known to be nonrenormalizable in the conventional perturbative approach. Recently, f(R) gravity has attracted much attention for several reasons, especially from the viewpoint of cosmology [1–5]. It is also of interest that f(R) gravity belongs to the class of higher-derivative gravity but is free from ghosts and consequently unitary, although it is not renormalizable.
Thus far there have been many studies of f(R) gravity from both classical and quantum aspects, but as far as we know there have been no studies performing its manifestly covariant quantization in the de Donder gauge condition (or harmonic gauge condition) for general coordinate invariance, investigating its global symmetries or verifying the physical states on the basis of the BRST formalism [6]. Performing such manifestly covariant quantization in an explicit way is one of aims of this article.
Another motivation behind this article is to understand the BRST formalism of the higher-derivative gravity that was recently called quadratic gravity. The quadratic gravity is known to be renormalizable but is not unitary owing to the presence of massive ghosts [7], and its Lagrangian takes a symbolic form ${{ \mathcal L }}_{QG}=\sqrt{-g}\left(\frac{1}{2{\kappa }^{2}}R+\alpha {R}^{2}+\beta {C}_{\mu \nu \rho \sigma }^{2}\right)$. In the manifestly covariant quantization procedure of quadratic gravity, it is mentioned in [8–10] that the equal-time commutation relation (ETCR) between the metric and its time derivative identically vanishes, $[{g}_{\mu \nu },{\dot{g}}_{\rho \sigma }^{{\prime} }]=0$,1 whereas it is shown in [11, 12] that in general relativity this ETCR is nonvanishing. Of course, this vanishing ETCR does not imply that the gravitational modes become trivial in quadratic gravity since dynamical degrees of freedom associated with the graviton and the other modes are involved in auxiliary fields such as the Nakanishi–Lautrup field and the Stuckelberg field. Nevertheless, this vanishing ETCR not only enables us to easily carry out complicated calculations of the ETCRs relevant to the metric field but might also exhibit an interesting feature of the quadratic gravity, such as a sort of triviality of the gravitational sector at high energies where the higher-derivative terms are more dominant than the Einstein–Hilbert term. Furthermore, the ETCR, $[{g}_{\mu \nu },{\dot{g}}_{\rho \sigma }^{{\prime} }]=0$ means that the canonical formalism of the quadratic gravity is not smoothly connected with that of general relativity in the suitable limit α, β → 0.
In this article, we would like to verify whether the vanishing ETCR is valid or not in a simpler theory where the conformal tensor-squared term ${C}_{\mu \nu \rho \sigma }^{2}$ is dropped and an R2 term is replaced with the f(R) term in quadratic gravity, that is, the Lagrangian is of the form ${{ \mathcal L }}_{0}=\sqrt{-g}\left(\frac{1}{2{\kappa }^{2}}R+f(R)\right),$ since the BRST quantization of the ${C}_{\mu \nu \rho \sigma }^{2}$ term requires us to treat a rather complicated Stuckelberg symmetry [8–10]. We will see that the ETCR, $[{g}_{\mu \nu },{\dot{g}}_{\rho \sigma }^{{\prime} }]$ in f(R) gravity is nonvanishing and has a more complicated expression than that of general relativity. This observation might imply that the ${C}_{\mu \nu \rho \sigma }^{2}$ term may play a role for making the ETCR vanishing in quadratic gravity. In fact, in conformal gravity described by the Lagrangian ${{ \mathcal L }}_{{C}^{2}}=\sqrt{-g}\,\beta {C}_{\mu \nu \rho \sigma }^{2}$ it is already known that the ETCR $[{g}_{\mu \nu },{\dot{g}}_{\rho \sigma }^{{\prime} }]$ has a rather simpler expression compared with that of general relativity [13]. On this we would refer to related articles [14–21].
We close this section with an overview of this article. In section 2 , we review a relation between f(R) gravity and scalar–tensor gravity. In section 3 , we construct a BRST-invariant action, derive field equations and argue its global symmetries. In section 4 , starting with the canonical commutation relations, we calculate various ETCRs where special attention is paid to the ETCR between the metric and its time derivative. In section 5 , we clarify the physical content by imposing the Kugo–Ojima condition [6] and show that the physical S-matrix is unitary in the theory being considered. The final section is devoted to the conclusion. Two appendices are included for technical details. In Appendix A , two different derivations of the field equation for the bμ field are given, and in Appendix B we present two different derivations of $[{b}_{\mu },{\dot{b}}_{\nu }^{{\prime} }]$.
2. Relation between f(R) gravity and scalar–tensor gravity
We wish to consider a model of f(R) gravity defined by the Lagrangian density
$\begin{eqnarray}{{ \mathcal L }}_{0}=\sqrt{-g}\left[\frac{1}{2{\kappa }^{2}}R+f(R)\right],\end{eqnarray}$
where κ is defined as ${\kappa }^{2}=8\pi G=\frac{1}{{M}_{{\rm{P}}{\rm{l}}}^{2}}$ through the Newton constant G and the reduced Planck mass MPl.2 Since general relativity nicely describes many gravitational and cosmological phenomena at low energies, we have extracted the Einstein–Hilbert term from the f(R) term. This definition is slightly different from that of conventional f(R) gravity where the f(R) term includes the Einstein–Hilbert term.Since the Lagrangian (1 ) possesses higher-derivative terms in the f(R) term it is necessary to transform it into a first-order formalism in the canonical quantization procedure. To this end let us first impose a constraint χ = R by introducing a Lagrange multiplier field φ1 ) and (2 ) are equivalent at a quantum level since in (2 ) the path integral over φ produces the delta functional δ(χ − R) and then integrating over χ exactly leads to (1 ).
$\begin{eqnarray}{{ \mathcal L }}_{\phi -\chi }=\sqrt{-g}\left[\frac{1}{2}R+\phi (R-\chi )+f(\chi )\right].\end{eqnarray}$
The two Lagrangians (Next, let us take the variation with respect to χ in (2 ), from which we have the field equation 3 ) into the Lagrangian (2 ), we can obtain4 ) is equivalent to (1 ) at least classically since we have made use of the field equation for χ. Note that χ in (4 ) is implicitly defined in terms of φ through equation (3 ). When the one-variable function $\phi ={f}^{{\prime} }(\chi )$ is not only continuous and differentiable at χ = R but also f″(R) ≠ 0, the inverse function theorem says that $\phi ={f}^{{\prime} }(\chi )$ can be uniquely solved in the neighborhood of χ = R and be expressed by3 3 ) and (5 ) into (4 ), we arrive at our classical Lagrangian which consists of the Einstein–Hilbert term, the non-minimal coupling term and a potential 7 ). Compared with conventional scalar–tensor gravity [23], the Lagrangian (7 ) contains the Einstein–Hilbert term but does not include the kinetic term for the scalar field $-\sqrt{-g}\,\frac{1}{\phi }\,{g}^{\mu \nu }{\partial }_{\mu }\phi {\partial }_{\nu }\phi $ (recall that φ has a mass dimension 2), although it is implicitly provided by the non-minimal coupling term $\sqrt{-g}\phi R$, as will be seen later.
$\begin{eqnarray}\phi ={f}^{{\prime} }(\chi ),\end{eqnarray}$
where ${f}^{{\prime} }(\chi )\equiv \frac{\rm{d}\,f}{\,\rm{d}\chi }$. Inserting ( $\begin{eqnarray}{{ \mathcal L }}_{\chi }=\sqrt{-g}\left[\left(\frac{1}{2}+{f}^{{\prime} }(\chi )\right)R-{f}^{{\prime} }(\chi )\chi +f(\chi )\right].\end{eqnarray}$
It is obvious that the Lagrangian ( $\begin{eqnarray}\chi =F(\phi ),\end{eqnarray}$
where F is the inverse function of ${f}^{{\prime} }(\chi )$, that is, $\begin{eqnarray}F({f}^{{\prime} }(\chi ))=\chi ,\quad {f}^{{\prime} }(F(\phi ))=\phi .\end{eqnarray}$
Then, substituting equations ( $\begin{eqnarray}{{ \mathcal L }}_{c}=\sqrt{-g}\left[\frac{1}{2}R+\phi R+{\rm{\Phi }}(\phi )\right],\end{eqnarray}$
where we have defined Φ(φ) ≡ −φF(φ) + f(F(φ)). In this article we shall apply a BRST quantization method to this Lagrangian (To close this section, we would like to comment on two remarks. First, as a classically equivalent Lagrangian to the f(R) gravity, the Lagrangian (4 ) is sometimes utilized. Actually, taking the variation of χ yields4 ) we recover (1 ). Note that the condition f″(χ) ≠ 0 corresponds to the condition f″(R) ≠ 0, which is needed for the inverse function theorem. Incidentally, there are some references which insist that the equivalence between (1 ) and (4 ) is valid even at the one-loop level on shell in arbitrary space–time dimensions [24, 25].
$\begin{eqnarray}{f}^{{\prime\prime} }(\chi )(R-\chi )=0.\end{eqnarray}$
As long as f″(χ) ≠ 0, this equation gives us χ = R, and then substituting this relation into (As the second remark, when f(R) is quadratic in R, that is f(R) = αR2 with a constant α, it is possible to rewrite (1 ) by introducing a scalar field φ with mass squared dimension into a quantum mechanically equivalent form1 ) and (7 ) (or (9 )) is guaranteed. The Lagrangian (9 ) will be considered in section 5 when analyzing the physical content of f(R) gravity.
$\begin{eqnarray}{{ \mathcal L }}_{{R}^{2}}=\sqrt{-g}\left(\frac{1}{2}R+\phi R-\frac{1}{4\alpha }{\phi }^{2}\right).\end{eqnarray}$
Thus, for quadratic gravity without the conformal tensor squared the quantum equivalence between the Lagrangian (3. BRST transformation and field equations
Let us fix the general coordinate symmetry by the de Donder gauge condition
$\begin{eqnarray}{\partial }_{\mu }{\tilde{g}}^{\mu \nu }=0,\end{eqnarray}$
where we have defined ${\tilde{g}}^{\mu \nu }=\sqrt{-g}{g}^{\mu \nu }$. Then, the BRST transformation has the form $\begin{eqnarray}\begin{array}{rcl}{\delta }_{B}{\bar{c}}_{\rho } & = & \,\rm{i}\,{B}_{\rho },\quad {\delta }_{B}{c}^{\rho }=-{c}^{\lambda }{\partial }_{\lambda }{c}^{\rho },\quad {\delta }_{B}\phi =-{c}^{\lambda }{\partial }_{\lambda }\phi ,\\ {\delta }_{B}{g}_{\mu \nu } & = & -({{\rm{\nabla }}}_{\mu }{c}_{\nu }+{{\rm{\nabla }}}_{\nu }{c}_{\mu })\\ & = & -({c}^{\alpha }{\partial }_{\alpha }{g}_{\mu \nu }+{\partial }_{\mu }{c}^{\alpha }{g}_{\alpha \nu }+{\partial }_{\nu }{c}^{\alpha }{g}_{\mu \alpha }),\\ {\delta }_{B}{\tilde{g}}^{\mu \nu } & = & \sqrt{-g}({{\rm{\nabla }}}^{\mu }{c}^{\nu }+{{\rm{\nabla }}}^{\nu }{c}^{\mu }-{g}^{\mu \nu }{{\rm{\nabla }}}_{\rho }{c}^{\rho }).\end{array}\end{eqnarray}$
Using this BRST transformation, the Lagrangian for the gauge-fixing condition and Faddeev–Popov (FP) ghosts can be constructed in a standard manner12 ) can be cast to the form
$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal L }}_{{\rm{G}}{\rm{F}}+{\rm{F}}{\rm{P}}} & = & {\delta }_{B}(\,\rm{i}\,{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{\bar{c}}_{\nu })\\ & = & -{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{B}_{\nu }-\,\rm{i}\,{\partial }_{\mu }{\bar{c}}_{\nu }\\ & & \times \left[[,{\tilde{g}}^{\mu \rho }{\partial }_{\rho }{c}^{\nu }+{\tilde{g}}^{\nu \rho }{\partial }_{\rho }{c}^{\mu }-{\partial }_{\rho }({\tilde{g}}^{\mu \nu }{c}^{\rho })\right].\end{array}\end{eqnarray}$
To simplify this expression, let us introduce a new auxiliary field bρ defined as $\begin{eqnarray}{b}_{\rho }={B}_{\rho }-\,\rm{i}\,{c}^{\lambda }{\partial }_{\lambda }{\bar{c}}_{\rho },\end{eqnarray}$
and its BRST transformation reads $\begin{eqnarray}{\delta }_{B}{b}_{\rho }=-{c}^{\lambda }{\partial }_{\lambda }{b}_{\rho }.\end{eqnarray}$
Then, the Lagrangian ( $\begin{eqnarray}\begin{array}{rcl}{{ \mathcal L }}_{{\rm{G}}{\rm{F}}+{\rm{F}}{\rm{P}}} & = & -{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{b}_{\nu }-\,\rm{i}\,{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{\bar{c}}_{\rho }{\partial }_{\nu }{c}^{\rho }\\ & & +\rm{i}\,{\partial }_{\rho }({\tilde{g}}^{\mu \nu }{\partial }_{\mu }{\bar{c}}_{\nu }\cdot {c}^{\rho }).\end{array}\end{eqnarray}$
As a result, up to a total derivative, the gauge-fixed and BRST-invariant quantum Lagrangian is given by4 A , it turns out that the field equation for bρ field takes the form
$\begin{eqnarray}\begin{array}{rcl}{ \mathcal L } & = & \sqrt{-g}\left[\frac{1}{2}R+\phi R+{\rm{\Phi }}(\phi )\right]\\ & & -{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{b}_{\nu }-\,\rm{i}\,{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{\bar{c}}_{\rho }{\partial }_{\nu }{c}^{\rho }.\end{array}\end{eqnarray}$
From this Lagrangian, we can obtain field equations by taking the variation with respect to gμν, φ, bν, ${\bar{c}}_{\rho }$ and cρ, in order $\begin{eqnarray}\begin{array}{l}\left(\frac{1}{2}+\phi \right){G}_{\mu \nu }-({{\rm{\nabla }}}_{\mu }{{\rm{\nabla }}}_{\nu }-{g}_{\mu \nu }\square )\phi \\ \,-\,\frac{1}{2}{g}_{\mu \nu }{\rm{\Phi }}(\phi )-\frac{1}{2}\left({E}_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }E\right)=0,\\ R+{{\rm{\Phi }}}^{{\prime} }(\phi )=0,\quad {\partial }_{\mu }{\tilde{g}}^{\mu \nu }=0,\\ {g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{c}^{\rho }=0,\quad {g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{\bar{c}}_{\rho }=0,\end{array}\end{eqnarray}$
where ${G}_{\mu \nu }\equiv {R}_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R$ is the Einstein tensor and we have defined $\begin{eqnarray}\begin{array}{rcl}{E}_{\mu \nu } & = & {\partial }_{\mu }{b}_{\nu }+\,\rm{i}\,{\partial }_{\mu }{\bar{c}}_{\rho }{\partial }_{\nu }{c}^{\rho }+(\mu \leftrightarrow \nu ),\\ E & = & {g}^{\mu \nu }{E}_{\mu \nu }.\end{array}\end{eqnarray}$
From the derivations in Appendix $\begin{eqnarray}{g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{b}_{\rho }=0.\end{eqnarray}$
Hence, together with space–time coordinates xμ5 , the bρ field, the ghost field cρ and the anti-ghost field ${\bar{c}}_{\rho }$ all satisfy the d’Alembert equation10 ), this equation provides us with two kinds of conserved currents
$\begin{eqnarray}{g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{X}^{M}=0,\end{eqnarray}$
where we have denoted ${X}^{M}=\{{x}^{\mu },{b}_{\mu },{c}^{\mu },{\bar{c}}_{\mu }\}$. With the help of the de Donder gauge condition ( $\begin{eqnarray}\begin{array}{rcl}{{ \mathcal P }}^{\mu M} & = & {g}^{\mu \nu }{\partial }_{\nu }{X}^{M}={g}^{\mu \nu }(1{\overleftrightarrow{\partial }}_{\nu }{X}^{M}),\\ {{ \mathcal M }}^{\mu MN} & = & {g}^{\mu \nu }({X}^{M}{\overleftrightarrow{\partial }}_{\nu }{Y}^{N}),\end{array}\end{eqnarray}$
and the corresponding charges $\begin{eqnarray}\begin{array}{rcl}{{ \mathcal P }}^{M} & = & \displaystyle \int {{\rm{d}}}^{3}x\,{{ \mathcal P }}^{0M},\\ {{ \mathcal M }}^{MN} & = & \displaystyle \int {{\rm{d}}}^{3}x\,{{ \mathcal M }}^{0MN},\end{array}\end{eqnarray}$
where ${X}^{M}{\overleftrightarrow{\partial }}_{\mu }{Y}^{N}\equiv {X}^{M}{\partial }_{\mu }{Y}^{N}-({\partial }_{\mu }{X}^{M}){Y}^{N}$. The charges ${{ \mathcal P }}^{M}$ and ${{ \mathcal M }}^{MN}$ have 16 and 128 generators, respectively, and they constitute a 16-dimensional global Poincaré-like symmetry IOSp(8, 8) as in general relativity [11, 12]. In other words, the quantum f(R) gravity under consideration possesses a very huge IOSp(8, 8) symmetry. Of course, the existence of such a global symmetry is not inconsistent with the black hole no-hair theorem [22] since the present global symmetry is of quantum origin. Actually, the whole generator of the IOSp(8, 8) symmetry is in essence constructed out of quantum fields such as the bμ field and FP ghosts.4. Equal-time commutation relations
In this section, after introducing canonical commutation relations (CCRs) we wish to evaluate various ETCRs among the fundamental variables in f(R) gravity. In particular, we are interested in the ETCR between the metric field and its time derivative. In previous works on quadratic gravity [8–10], it was mentioned that in contrast to the case of Einstein’s general relativity [11, 12] its ETCR vanishes identically. f(R) gravity involves higher-derivative terms as in quadratic gravity so it is of interest whether this feature is taken over by f(R) gravity or not. We will show clearly that the ETCR between the metric field and its time derivative is not only nonvanishing but also has a more complicated structure than that of general relativity.
To simplify various expressions, we follow the abbreviations adopted in the textbook by Nakanishi and Ojima6
$\begin{eqnarray}\begin{array}{rcl}[A,{B}^{{\prime} }] & = & [A(x),B({x}^{{\prime} })]{| }_{{x}^{0}={x}^{{\prime} 0}},\,{\delta }^{3}=\delta (\overrightarrow{x}-{\overrightarrow{x}}^{{\prime} }),\\ \tilde{f} & = & \frac{1}{\tilde{{g}^{00}}}=\frac{1}{\sqrt{-g}{g}^{00}},\end{array}\end{eqnarray}$
where ${\tilde{g}}^{00}$ is assumed to be invertible.To remove second-order derivatives of the metric involved in R, we perform the integration by parts once and rewrite the Lagrangian (16 ) as
$\begin{eqnarray}\begin{array}{rcl}{ \mathcal L } & = & -\,\left(\frac{1}{2}+\phi \right){\tilde{g}}^{\mu \nu }({{\rm{\Gamma }}}_{\sigma \alpha }^{\alpha }{{\rm{\Gamma }}}_{\mu \nu }^{\sigma }-{{\rm{\Gamma }}}_{\sigma \nu }^{\alpha }{{\rm{\Gamma }}}_{\mu \alpha }^{\sigma })\\ & & -\,{\partial }_{\mu }\phi ({\tilde{g}}^{\alpha \beta }{{\rm{\Gamma }}}_{\alpha \beta }^{\mu }-{\tilde{g}}^{\mu \nu }{{\rm{\Gamma }}}_{\nu \alpha }^{\alpha })\\ & & +\,\sqrt{-g}{\rm{\Phi }}(\phi )+{\partial }_{\mu }{\tilde{g}}^{\mu \nu }{b}_{\nu }-\,\rm{i}\,{\tilde{g}}^{\mu \nu }{\partial }_{\mu }{\bar{c}}_{\rho }{\partial }_{\nu }{c}^{\rho }+{\partial }_{\mu }{{ \mathcal V }}^{\mu },\end{array}\end{eqnarray}$
where a surface term ${{ \mathcal V }}^{\mu }$ is defined as $\begin{eqnarray}{{ \mathcal V }}^{\mu }=\left(\frac{1}{2}+\phi \right)({\tilde{g}}^{\alpha \beta }{{\rm{\Gamma }}}_{\alpha \beta }^{\mu }-{\tilde{g}}^{\mu \nu }{{\rm{\Gamma }}}_{\nu \alpha }^{\alpha })-{\tilde{g}}^{\mu \nu }{b}_{\nu }.\end{eqnarray}$
For later convenience, let us take account of the de Donder gauge condition (10 ), from which we have the identities
$\begin{eqnarray}{g}^{\mu \nu }{{\rm{\Gamma }}}_{\mu \nu }^{\lambda }=0,\quad {g}^{\lambda \mu }{\partial }_{\lambda }{g}_{\mu \nu }={{\rm{\Gamma }}}_{\lambda \nu }^{\lambda }.\end{eqnarray}$
These identities and the de Donder gauge condition will be often utilized to arrive at final expressions of various results. Moreover, since the equation ${g}^{\mu \nu }{{\rm{\Gamma }}}_{\mu \nu }^{\lambda }=0$ reads $\begin{eqnarray}(2{g}^{\lambda \mu }{g}^{\nu \rho }-{g}^{\mu \nu }{g}^{\lambda \rho }){\partial }_{\rho }{g}_{\mu \nu }=0,\end{eqnarray}$
it is possible to express the time derivative of the metric field in terms of its spatial one as $\begin{eqnarray}{{ \mathcal D }}^{\lambda \mu \nu }{\dot{g}}_{\mu \nu }=(2{g}^{\lambda \mu }{g}^{\nu k}-{g}^{\mu \nu }{g}^{\lambda k}){\partial }_{k}{g}_{\mu \nu },\end{eqnarray}$
where the operator ${{ \mathcal D }}^{\lambda \mu \nu }$ is defined by $\begin{eqnarray}{{ \mathcal D }}^{\lambda \mu \nu }={g}^{0\lambda }{g}^{\mu \nu }-2{g}^{\lambda \mu }{g}^{0\nu }.\end{eqnarray}$
Now let us set up the canonical (anti)commutation relations
$\begin{aligned} {\left[\pi_{g}^{\rho \lambda}, g_{\mu \nu}^{\prime}\right] } & =-\mathrm{i} \frac{1}{2}\left(\delta_{\mu}^{\rho} \delta_{\nu}^{\lambda}+\delta_{\mu}^{\lambda} \delta_{\nu}^{\rho}\right) \delta^{3} \\ {\left[\pi_{\phi}, \phi^{\prime}\right] } & =-\mathrm{i} \delta^{3} \\ \left\{\pi_{c \lambda}, c^{\sigma^{\prime}}\right\} & =\left\{\pi_{\bar{c}}^{\sigma}, \bar{c}_{\lambda}^{\prime}\right\}=\mathrm{i} \delta_{\lambda}^{\sigma} \delta^{3} . \end{aligned}$
Here the canonical variables are ${g}_{\mu \nu },\phi ,{c}^{\rho },{\bar{c}}_{\rho }$ and the corresponding canonical conjugate momenta are ${\pi }_{g}^{\mu \nu },{\pi }_{\phi },{\pi }_{c\rho },{\pi }_{\bar{c}}^{\rho }$, respectively; the bμ field is regarded as not a canonical variable but a conjugate momentum of ${\tilde{g}}^{0\mu }$.Based on the Lagrangian (24 ), the concrete expressions for canonical conjugate momenta read
$\begin{eqnarray}\begin{array}{rcl}{\pi }_{g}^{\mu \nu } & = & \frac{\partial { \mathcal L }}{\partial {\dot{g}}_{\mu \nu }}\\ & = & -\,\frac{1}{2}\sqrt{-g}\,\left(\frac{1}{2}+\phi \right)\left[\Space{0ex}{2.75ex}{0ex}-{g}^{0\lambda }{g}^{\mu \nu }{g}^{\sigma \tau }\right.\\ & & -\,{g}^{0\tau }{g}^{\mu \lambda }{g}^{\nu \sigma }-{g}^{0\sigma }{g}^{\mu \tau }{g}^{\nu \lambda }+{g}^{0\lambda }{g}^{\mu \tau }{g}^{\nu \sigma }\\ & & \left.+\,{g}^{0\tau }{g}^{\mu \nu }{g}^{\lambda \sigma }+\frac{1}{2}\left({g}^{0\mu }{g}^{\nu \lambda }+{g}^{0\nu }{g}^{\mu \lambda }\right){g}^{\sigma \tau }\right]{\partial }_{\lambda }{g}_{\sigma \tau }\\ & & -\,\sqrt{-g}\left[\frac{1}{2}({g}^{0\mu }{g}^{\rho \nu }+{g}^{0\nu }{g}^{\rho \mu })-{g}^{\mu \nu }{g}^{\rho 0}\right]{\partial }_{\rho }\phi \\ & & -\,\frac{1}{2}\sqrt{-g}\left({g}^{0\mu }{g}^{\nu \rho }+{g}^{0\nu }{g}^{\mu \rho }-{g}^{0\rho }{g}^{\mu \nu }\right){b}_{\rho },\\ {\pi }_{\phi } & = & \frac{\partial { \mathcal L }}{\partial \dot{\phi }}=-\left({\tilde{g}}^{\alpha \beta }{{\rm{\Gamma }}}_{\alpha \beta }^{0}-{\tilde{g}}^{0\nu }{{\rm{\Gamma }}}_{\nu \alpha }^{\alpha }\right),\\ {\pi }_{c\sigma } & = & \frac{\partial { \mathcal L }}{\partial {\dot{c}}^{\sigma }}=-\,\rm{i}\,{\tilde{g}}^{0\mu }{\partial }_{\mu }{\bar{c}}_{\sigma },\\ {\pi }_{\bar{c}}^{\sigma } & = & \frac{\partial { \mathcal L }}{\partial {\dot{\bar{c}}}_{\sigma }}=\,\rm{i}\,{\tilde{g}}^{0\mu }{\partial }_{\mu }{c}^{\sigma },\end{array}\end{eqnarray}$
where differentiation of ghosts is taken from the right and the other (anti)commutation relations vanish. Here the dot denotes time derivative, for example ${\dot{g}}_{\mu \nu }={\partial }_{0}{g}_{\mu \nu }=\frac{\partial {g}_{\mu \nu }}{\partial {x}^{0}}=\frac{\partial {g}_{\mu \nu }}{\partial t}$.From now on we would like to evaluate various nontrivial ETCRs. In particular, we will exhibit that the ETCR between the metric field and its time derivative takes a nonvanishing and complicated expression.
For this purpose, let us start with the CCR32 ) produces
$\begin{eqnarray}[{\pi }_{g}^{\alpha 0},{g}_{\mu \nu }^{{\prime} }]=-\,\rm{i}\,\frac{1}{2}({\delta }_{\mu }^{\alpha }{\delta }_{\nu }^{0}+{\delta }_{\mu }^{0}{\delta }_{\nu }^{\alpha }){\delta }^{3}.\end{eqnarray}$
The canonical conjugate momentum ${\pi }_{g}^{\alpha 0}$ has the structure $\begin{eqnarray}{\pi }_{g}^{\alpha 0}={A}^{\alpha }+{B}^{\alpha \beta }{\partial }_{\beta }\phi +{C}^{\alpha \beta }{b}_{\beta },\end{eqnarray}$
where Aα, Bαβ and ${C}^{\alpha \beta }\equiv -\frac{1}{2}{\tilde{g}}^{00}{g}^{\alpha \beta }$ have no ${\dot{g}}_{\mu \nu }$ and Bαβ∂βφ does not include $\dot{\phi }$. Then, we find that the CCR ( $\begin{eqnarray}[{g}_{\mu \nu },{b}_{\rho }^{{\prime} }]=-\,\rm{i}\,\tilde{f}({\delta }_{\mu }^{0}{g}_{\rho \nu }+{\delta }_{\nu }^{0}{g}_{\rho \mu }){\delta }^{3}.\end{eqnarray}$
From this ETCR, it is easy to derive the following two ETCRs: $\begin{aligned} {\left[g^{\mu \nu}, b_{\rho}^{\prime}\right] } & =\mathrm{i} \tilde{f}\left(g^{0 \mu} \delta_{\rho}^{\nu}+g^{0 \nu} \delta_{\rho}^{\mu}\right) \delta^{3} \\ {\left[\tilde{g}^{\mu \nu}, b_{\rho}^{\prime}\right] } & =\mathrm{i} \tilde{f}\left(\tilde{g}^{0 \mu} \delta_{\rho}^{\nu}+\tilde{g}^{0 \nu} \delta_{\rho}^{\mu}-\tilde{g}^{\mu \nu} \delta_{\rho}^{0}\right) \delta^{3} \end{aligned}$
Next, we see that the CCR $[{\pi }_{\phi },{\phi }^{{\prime} }]=-\,\rm{i}\,{\delta }^{3}$ directly leads to the relation28 ) and (36 ). Indeed, it turns out that equation (28 ) fixes ${a}_{2}=\frac{2}{{g}^{00}}$ and equation (36 ) fixes ${a}_{1}=-\frac{\,\rm{i}\,}{3}\tilde{f}$, so we have the ETCR
$\begin{eqnarray}({\tilde{g}}^{0\mu }{g}^{0\nu }-{\tilde{g}}^{00}{g}^{\mu \nu })[{\dot{g}}_{\mu \nu },{\phi }^{{\prime} }]=\,\rm{i}\,{\delta }^{3}.\end{eqnarray}$
From the symmetry argument, the ETCR, $[{\dot{g}}_{\mu \nu },{\phi }^{{\prime} }]$ should have the following expression: $\begin{eqnarray}[{\dot{g}}_{\mu \nu },{\phi }^{{\prime} }]={a}_{1}({g}_{\mu \nu }+{a}_{2}{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}){\delta }^{3},\end{eqnarray}$
where the coefficients a1 and a2 can be determined from equations ( $\begin{eqnarray}[{\dot{g}}_{\mu \nu },{\phi }^{{\prime} }]=-\frac{\,\rm{i}\,}{3}\tilde{f}\left({g}_{\mu \nu }+\frac{2}{{g}^{00}}{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}\right){\delta }^{3}.\end{eqnarray}$
We are now ready to evaluate the ETCR between the metric field and its time derivative, i.e. $[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]$. To determine this ETCR, we look for conditions imposed on it. First of all, let us consider the CCR $[{\pi }_{\phi },{g}_{\mu \nu }^{{\prime} }]=0$. From equation (31 ), πφ is written as 28 ) stemming from the de Donder gauge condition gives rise to the equation 40 ) and (42 ) together give us the equations
$\begin{eqnarray}\begin{array}{rcl}{\pi }_{\phi } & = & ({\tilde{g}}^{00}{g}^{\rho \sigma }-{\tilde{g}}^{0\rho }{g}^{0\sigma }){\dot{g}}_{\rho \sigma }\\ & & +\,({\tilde{g}}^{0k}{g}^{\rho \sigma }-{\tilde{g}}^{0\rho }{g}^{k\sigma }){\partial }_{k}{g}_{\rho \sigma }.\end{array}\end{eqnarray}$
Using this expression, $[{\pi }_{\phi },{g}_{\mu \nu }^{{\prime} }]=0$ provides us with the equation $\begin{eqnarray}({g}^{00}{g}^{\rho \sigma }-{g}^{0\rho }{g}^{0\sigma })[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0.\end{eqnarray}$
Moreover, the ${ \mathcal D }$ equation ( $\begin{eqnarray}[{{ \mathcal D }}^{\lambda \rho \sigma }{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0,\end{eqnarray}$
from which we have the equation for λ = 0: $\begin{eqnarray}({g}^{00}{g}^{\rho \sigma }-2{g}^{0\rho }{g}^{0\sigma })[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0.\end{eqnarray}$
Equations ( $\begin{eqnarray}{g}^{\rho \sigma }[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0,\quad {g}^{0\rho }{g}^{0\sigma }[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0.\end{eqnarray}$
Next, putting λ = i in equation (41 ), we have43 ) and (44 ) are then summarized to two equations
$\begin{eqnarray}({g}^{0i}{g}^{\rho \sigma }-2{g}^{i\rho }{g}^{0\sigma })[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0.\end{eqnarray}$
Equations ( $\begin{eqnarray}{g}^{\rho \sigma }[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0,\quad {g}^{0\sigma }[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0.\end{eqnarray}$
Here let us note that $[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]$ has a symmetry under the simultaneous exchange of (μν) ↔ (ρσ) and primed ↔ unprimed in addition to the usual symmetry μ ↔ ν and ρ ↔ σ, because the time derivative ∂0 of $[{g}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=0$ leads to $[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]=[{\dot{g}}_{\mu \nu }^{{\prime} },{g}_{\rho \sigma }]$. We assume that this ETCR is proportional to δ3. Then, we can write down its generic expression 45 ) on (46 ), we find that ci(i = 2, ⋯ , 5) can be expressed only in terms of c1 as
$\begin{eqnarray}\begin{array}{rcl}[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }] & = & \left\{\{,{c}_{1}{g}_{\rho \sigma }{g}_{\mu \nu }+{c}_{2}({g}_{\rho \mu }{g}_{\sigma \nu }+{g}_{\rho \nu }{g}_{\sigma \mu })\right.\\ & & +\,\sqrt{-g}\tilde{f}[{c}_{3}({\delta }_{\rho }^{0}{\delta }_{\sigma }^{0}{g}_{\mu \nu }+{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}{g}_{\rho \sigma })+{c}_{4}({\delta }_{\rho }^{0}{\delta }_{\mu }^{0}{g}_{\sigma \nu }\\ & & +\,{\delta }_{\rho }^{0}{\delta }_{\nu }^{0}{g}_{\sigma \mu }+{\delta }_{\sigma }^{0}{\delta }_{\mu }^{0}{g}_{\rho \nu }+{\delta }_{\sigma }^{0}{\delta }_{\nu }^{0}{g}_{\rho \mu })]\\ & & +\,\left.{(\sqrt{-g}\tilde{f})}^{2}{c}_{5}{\delta }_{\rho }^{0}{\delta }_{\sigma }^{0}{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}\right\}{\delta }^{3},\end{array}\end{eqnarray}$
where ci(i = 1, ⋯ , 5) are suitable coefficients to be fixed shortly. By imposing equation ( $\begin{eqnarray}\begin{array}{rcl}{c}_{2} & = & -\frac{3}{2}{c}_{1},\quad {c}_{3}=-{c}_{1},\\ {c}_{4} & = & \frac{3}{2}{c}_{1},\quad {c}_{5}=-2{c}_{1}.\end{array}\end{eqnarray}$
In order to determine the coefficient c1, we need to calculate the ETCR $[{\dot{g}}_{kl},{g}_{mn}^{{\prime} }]$ explicitly. To do so, let us first consider the CCR31 ), ${\pi }_{g}^{kl}$ can be expanded as
$\begin{eqnarray}[{\pi }_{g}^{kl},{g}_{mn}^{{\prime} }]=-\,\rm{i}\,\frac{1}{2}({\delta }_{m}^{k}{\delta }_{n}^{l}+{\delta }_{m}^{l}{\delta }_{n}^{k}){\delta }^{3}\equiv -\,\rm{i}\,{\delta }_{m}^{(k}{\delta }_{n}^{l)}{\delta }^{3}.\end{eqnarray}$
Next, from equation ( $\begin{eqnarray}{\pi }_{g}^{kl}={\hat{A}}^{kl}+{\hat{B}}^{kl\rho }{b}_{\rho }+{\hat{C}}^{klmn}{\dot{g}}_{mn}+{\hat{D}}^{kl}\dot{\phi }.\end{eqnarray}$
Here ${\hat{A}}^{kl},{\hat{B}}^{kl\rho },{\hat{C}}^{klmn}$ and ${\hat{D}}^{kl}$ commute with gmn, and ${\hat{C}}^{klmn}$ and ${\hat{D}}^{kl}$ are defined as7 $\begin{eqnarray}\begin{array}{rcl}{\hat{C}}^{klmn} & = & \frac{1}{2}\sqrt{-g}\left(\frac{1}{2}+\phi \right){K}^{klmn},\\ {\hat{D}}^{kl} & = & {\tilde{g}}^{00}{g}^{kl}-{\tilde{g}}^{0k}{g}^{0l},\end{array}\end{eqnarray}$
where the definition of Kklmn and its properties is given by $\begin{eqnarray}\begin{array}{l}{K}^{klmn}=\left|\begin{array}{rrr}{g}^{00} & {g}^{0l} & {g}^{0n}\\ {g}^{k0} & {g}^{kl} & {g}^{kn}\\ {g}^{m0} & {g}^{ml} & {g}^{mn}\\ \end{array}\right|,\\ {K}^{klmn}\frac{1}{2}{({g}^{00})}^{-1}({g}_{ij}{g}_{mn}-{g}_{im}{g}_{jn}-{g}_{in}{g}_{jm})\\ \quad =\,\frac{1}{2}({\delta }_{i}^{k}{\delta }_{j}^{l}+{\delta }_{i}^{l}{\delta }_{j}^{k})\equiv {\delta }_{i}^{(k}{\delta }_{j}^{l)}.\end{array}\end{eqnarray}$
From equation (49 ), we can calculate 34 ), (38 ) and (48 ). Since we can calculate 46 ) we have the ETCR54 ) with (55 ), we can deduce that 47 ), ${c}_{2}=-\frac{3}{2}{c}_{1}$, which gives us a nontrivial verification of our calculation. In this way, we have succeeded in getting the following ETCR:
$\begin{eqnarray}\begin{array}{rcl}[{\dot{g}}_{kl},{g}_{mn}^{{\prime} }] & = & {\hat{C}}_{klpq}^{-1}\left((,[{\pi }_{g}^{pq},{g}_{mn}^{{\prime} }]-{\hat{B}}^{pq\rho }[{b}_{\rho },{g}_{mn}^{{\prime} }]-{\hat{D}}^{pq}[\dot{\phi },{g}_{mn}^{{\prime} }]\right)\\ & = & {\hat{C}}_{klpq}^{-1}\left[-\,\rm{i}\,{\delta }_{m}^{(p}{\delta }_{n}^{q)}+\,\rm{i}\,\frac{1}{3}{g}_{mn}({\tilde{g}}^{00}{g}^{pq}-{\tilde{g}}^{0p}{g}^{0q})\right]{\delta }^{3},\end{array}\end{eqnarray}$
where we have used equations ( $\begin{eqnarray}{\hat{C}}_{klpq}^{-1}=\tilde{f}{\left(\frac{1}{2}+\phi \right)}^{-1}({g}_{kl}{g}_{pq}-{g}_{kp}{g}_{lq}-{g}_{kq}{g}_{lp}),\end{eqnarray}$
we can eventually arrive at the result $\begin{eqnarray}[{\dot{g}}_{kl},{g}_{mn}^{{\prime} }]=-\,\rm{i}\,\tilde{f}{\left(\frac{1}{2}+\phi \right)}^{-1}\left(\frac{2}{3}{g}_{kl}{g}_{mn}-{g}_{km}{g}_{ln}-{g}_{kn}{g}_{lm}\right){\delta }^{3}.\end{eqnarray}$
Meanwhile, from equation ( $\begin{eqnarray}[{\dot{g}}_{kl},{g}_{mn}^{{\prime} }]=\left[{c}_{1}{g}_{kl}{g}_{mn}+{c}_{2}({g}_{km}{g}_{ln}+{g}_{kn}{g}_{lm})\right]{\delta }^{3}.\end{eqnarray}$
Hence, comparing ( $\begin{eqnarray}{c}_{1}=-\frac{2}{3}\,\rm{i}\,\tilde{f}{\left(\frac{1}{2}+\phi \right)}^{-1},\quad {c}_{2}=\,\rm{i}\,\tilde{f}{\left(\frac{1}{2}+\phi \right)}^{-1}.\end{eqnarray}$
Note that these values satisfy the relation in equation ( $\begin{eqnarray}\begin{array}{rcl}[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }] & = & \,\rm{i}\,\tilde{f}{\left(\frac{1}{2}+\phi \right)}^{-1}\left\{-\frac{2}{3}{g}_{\rho \sigma }{g}_{\mu \nu }+{g}_{\rho \mu }{g}_{\sigma \nu }\right.\\ & & +\,{g}_{\rho \nu }{g}_{\sigma \mu }+\sqrt{-g}\tilde{f}\left[\frac{2}{3}({\delta }_{\rho }^{0}{\delta }_{\sigma }^{0}{g}_{\mu \nu }+{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}{g}_{\rho \sigma })\right.\\ & & \left.-\,{\delta }_{\rho }^{0}{\delta }_{\mu }^{0}{g}_{\sigma \nu }-{\delta }_{\rho }^{0}{\delta }_{\nu }^{0}{g}_{\sigma \mu }-{\delta }_{\sigma }^{0}{\delta }_{\mu }^{0}{g}_{\rho \nu }-{\delta }_{\sigma }^{0}{\delta }_{\nu }^{0}{g}_{\rho \mu }\Space{0ex}{3.25ex}{0ex}\right]\\ & & +\,\left.{(\sqrt{-g}\tilde{f})}^{2}\frac{4}{3}{\delta }_{\rho }^{0}{\delta }_{\sigma }^{0}{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}\right\}{\delta }^{3}.\end{array}\end{eqnarray}$
It is valuable to compare this ETCR with those of the other gravitational theories. In general relativity with only Einstein–Hilbert action [11, 12], the ETCR, $[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]$ is nonvanishing but it takes a simpler expression than equation (57 ). As for conformal gravity [13], this ETCR is also nonvanishing but has a rather simple structure compared with that of general relativity and equation (57 ). On the other hand, in quadratic gravity, which symbolically takes the Lagrangian form $\sqrt{-g}(R+{R}^{2}+{C}_{\mu \nu \rho \sigma }^{2})$ with the conventional conformal tensor Cμνρσ, it was mentioned that this ETCR identically vanishes [8–10]. However, as shown in the present study, in f(R) gravity equation (57 ) has a nonvanishing and complicated expression. Since f(R) gravity includes the R + R2 gravitational theory in equation (9 ) as a special case, the present result suggests that the ${C}_{\mu \nu \rho \sigma }^{2}$ term might play an important role in making this ETCR vanishing in quadratic gravity. As a final comment, the ETCR (57 ) does not reduce to the corresponding ETCR in general relativity in the limit φ → 0. This fact suggests that the canonical structure for the gravitational sector is quite distinct between f(R) gravity and general relativity.
Now we would like to present the remaining ETCRs. The CCR $[{\pi }_{g}^{\alpha 0},{\phi }^{{\prime} }]=0$ leads to
$\begin{eqnarray}[{b}_{\rho },{\phi }^{{\prime} }]=0.\end{eqnarray}$
From the CCR $[{\pi }_{g}^{ij},{\phi }^{{\prime} }]=0$ we can show that $\begin{eqnarray}[\dot{\phi },{\phi }^{{\prime} }]=\,\rm{i}\,\frac{1}{3}\tilde{f}\left(\frac{1}{2}+\phi \right){\delta }^{3}.\end{eqnarray}$
As for the ETCRs relevant to FP ghosts, we find that the anti-CCRs $\{{\pi }_{c\lambda },{c}^{\sigma ^{\prime} }\}=\{{\pi }_{\bar{c}}^{\sigma },{\bar{c}}_{\lambda }^{{\prime} }\}=\,\rm{i}\,{\delta }_{\lambda }^{\sigma }{\delta }^{3}$ yield the ETCRs
$\begin{eqnarray}\{{\dot{\bar{c}}}_{\lambda },{c}^{{\prime} \sigma }\}=-\{{\dot{c}}^{\sigma },{\bar{c}}_{\lambda }^{{\prime} }\}=-\tilde{f}{\delta }_{\lambda }^{\sigma }{\delta }^{3}.\end{eqnarray}$
Moreover, it is easy to see that the CCRs $\{{\pi }_{g}^{\alpha 0},{c}^{{\prime} \sigma }\}\,=\{{\pi }_{g}^{\alpha 0},{\bar{c}}_{\lambda }^{{\prime} }\}=0$ produce $\begin{eqnarray}[{b}_{\rho },{c}^{{\prime} \sigma }]=[{b}_{\rho },{\bar{c}}_{\lambda }^{{\prime} }]=0.\end{eqnarray}$
The CCRs $[{g}_{\mu \nu },{\pi }_{c\lambda }^{{\prime} }]=[{g}_{\mu \nu },{\pi }_{\bar{c}}^{{\prime} \sigma }]=0$ directly yield $\begin{eqnarray}[{g}_{\mu \nu },{\dot{c}}^{{\prime} \sigma }]=[{g}_{\mu \nu },{\dot{\bar{c}}}_{\lambda }^{{\prime} }]=0.\end{eqnarray}$
Similarly, the CCRs, $[\phi ,{\pi }_{c\lambda }^{{\prime} }]=[\phi ,{\pi }_{\bar{c}}^{{\prime} \sigma }]=0$ lead to $\begin{eqnarray}[\phi ,{\dot{c}}^{{\prime} \sigma }]=[\phi ,{\dot{\bar{c}}}_{\lambda }^{{\prime} }]=0.\end{eqnarray}$
Next, let us evaluate the type of ETCRs $[\dot{{\rm{\Psi }}},{b}_{\rho }^{{\prime} }],$ where $\Psi$ is a set of variables ${\rm{\Psi }}=\{{g}_{\mu \nu },\phi ,{c}^{\sigma },{\bar{c}}_{\lambda },{b}_{\rho }\}$. A somewhat complicated ETCR is given by 62 ). Namely, taking the BRST transformation of $[{g}_{\mu \nu },{\dot{\bar{c}}}_{\rho }^{{\prime} }]=0$, we have62 ) and ${\ddot{\bar{c}}}_{\rho }=-\tilde{f}(2{\tilde{g}}^{0k}{\partial }_{k}{\dot{\bar{c}}}_{\rho }+{\tilde{g}}^{kl}{\partial }_{k}{\partial }_{l}{\bar{c}}_{\rho })$ which is obtained from the field equation ${g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{\bar{c}}_{\rho }=0$. To calculate the second term on the RHS of equation (65 ), we need to make use of $[{\dot{g}}_{\mu \nu },{\dot{\bar{c}}}_{\rho }^{{\prime} }]=0$, equation (60 ) and the equation 60 ) and the field equation for ${\bar{c}}_{\rho }$. As a result, we can show that 34 ), equation (68 ) produces (64 ). In a similar manner, the BRST transformation makes it possible to derive the following ETCRs:
$\begin{eqnarray}\begin{array}{rcl}[{\dot{g}}_{\mu \nu },{b}_{\rho }^{{\prime} }] & = & -\,\rm{i}\,\left\{\tilde{f}({\partial }_{\rho }{g}_{\mu \nu }+{\delta }_{\mu }^{0}{\dot{g}}_{\rho \nu }+{\delta }_{\nu }^{0}{\dot{g}}_{\rho \mu }){\delta }^{3}\right.\\ & & +\left.[({\delta }_{\mu }^{k}-2{\delta }_{\mu }^{0}\tilde{f}{\tilde{g}}^{0k}){g}_{\rho \nu }+(\mu \leftrightarrow \nu )]{\partial }_{k}(\tilde{f}{\delta }^{3})\right\}.\end{array}\end{eqnarray}$
This ETCR has been previously derived by using the fact that the translation generator is given by ${P}_{\rho }=\int {{\rm{d}}}^{3}x\,{\tilde{g}}^{0\lambda }{\partial }_{\lambda }{b}_{\rho }$. Here we present an alternative derivation based on the BRST transformation of the latter equation in equation ( $\begin{eqnarray}\begin{array}{rcl}[{g}_{\mu \nu },{\dot{b}}_{\rho }^{{\prime} }] & = & -\,\rm{i}\,[{g}_{\mu \nu },{\partial }_{0}({c}^{{\prime} \lambda }{\partial }_{\lambda }{\bar{c}}_{\rho }^{{\prime} })]\\ & & -\,\rm{i}\,\{{c}^{\alpha }{\partial }_{\alpha }{g}_{\mu \nu }+{\partial }_{\mu }{c}^{\alpha }{g}_{\alpha \nu }+{\partial }_{\nu }{c}^{\alpha }{g}_{\mu \alpha },{\dot{\bar{c}}}_{\rho }^{{\prime} }\}.\end{array}\end{eqnarray}$
The first term on the RHS turns out to be vanishing since $\begin{eqnarray}-\,\rm{i}\,[{g}_{\mu \nu },{\partial }_{0}({c}^{{\prime} \lambda }{\partial }_{\lambda }{\bar{c}}_{\rho }^{{\prime} })]=-\,\rm{i}\,{c}^{{\prime} 0}[{g}_{\mu \nu },{\ddot{\bar{c}}}_{\rho }^{{\prime} }]=0,\end{eqnarray}$
where we have used equation ( $\begin{eqnarray}\{{\dot{c}}^{\sigma },{\dot{\bar{c}}}_{\rho }^{{\prime} }\}={\delta }_{\rho }^{\sigma }[{\partial }_{0}\tilde{f}+2\tilde{f}{\tilde{g}}^{0k}{\partial }_{k}(\tilde{f}{\delta }^{3})],\end{eqnarray}$
which is easily proved by using equation ( $\begin{eqnarray}\begin{array}{rcl}[{g}_{\mu \nu },{\dot{b}}_{\rho }^{{\prime} }] & = & \,\rm{i}\,\left\{[\tilde{f}{\partial }_{\rho }{g}_{\mu \nu }-{\partial }_{0}\tilde{f}({\delta }_{\mu }^{0}{g}_{\rho \nu }+{\delta }_{\nu }^{0}{g}_{\rho \mu })]{\delta }^{3}\right.\\ & & +\,\left.[({\delta }_{\mu }^{k}-2{\delta }_{\mu }^{0}\tilde{f}{\tilde{g}}^{0k}){g}_{\rho \nu }+(\mu \leftrightarrow \nu )]{\partial }_{k}(\tilde{f}{\delta }^{3})\right\}.\end{array}\end{eqnarray}$
Then, with the help of equation ( $\begin{aligned} {\left[\dot{\phi}, b_{\rho}^{\prime}\right] } & =-\mathrm{i} \tilde{f} \partial_{\rho} \phi \delta^{3}, \\ {\left[\dot{\bar{c}}_{\sigma}, b_{\rho}^{\prime}\right] } & =-\mathrm{i} \tilde{f} \partial_{\rho} \bar{c}_{\sigma} \delta^{3}, \\ {\left[\dot{c}^{\sigma}, b_{\rho}^{\prime}\right] } & =-\mathrm{i} \tilde{f} \partial_{\rho} c^{\sigma} \delta^{3} . \end{aligned}$
Note that the second and third equations can be also obtained from the CCRs $[{\pi }_{c\lambda },{\pi }_{g}^{\alpha 0^{\prime} }]=[{\pi }_{\bar{c}}^{\sigma },{\pi }_{g}^{\alpha 0^{\prime} }]=0$.Finally, let us comment on the ETCR $[{b}_{\mu },{\dot{b}}_{\nu }^{{\prime} }]$. First, let us note that the BRST transformation of $[{b}_{\mu },{\bar{c}}_{\nu }^{{\prime} }]=0$ provides us with the ETCRB we can deduce that
$\begin{eqnarray}[{b}_{\mu },{b}_{\nu }^{{\prime} }]=0.\end{eqnarray}$
Moreover, from two different derivations in Appendix $\begin{eqnarray}[{b}_{\rho },{\dot{b}}_{\lambda }^{{\prime} }]=\,\rm{i}\,\tilde{f}({\partial }_{\rho }{b}_{\lambda }+{\partial }_{\lambda }{b}_{\rho }){\delta }^{3}.\end{eqnarray}$
5. Unitarity of the physical S-matrix
In this section we analyze asymptotic fields under the assumption that all fields have their own asymptotic fields and there is no bound state. We also assume that all asymptotic fields are governed by the quadratic part of the quantum Lagrangian apart from possible renormalization. We will prove the unitarity of the physical S-matrix of f(R) gravity on the basis of the BRST quartet mechanism [6].
Let us expand the gravitational field gμν around a flat Minkowski metric ημν as 16 ) reads9 ), makes a contribution to this quadratic order equation. To put it differently, only an R2 term in f(R) contributes to kinetic terms: higher-order terms contribute more than the quadratic terms (f(R) = c1R3 + c2R4 + …, with c1, c2 being constants) to interaction terms since R starts with linear terms in φμν. In this section, the space–time indices μ, ν, … are raised or lowered by the Minkowski metric ημν = ημν = diag( − 1, 1, 1, 1), and we define ≡ ημν∂μ∂ν and φ ≡ ημνφμν.
$\begin{eqnarray}{g}_{\mu \nu }={\eta }_{\mu \nu }+{\varphi }_{\mu \nu },\end{eqnarray}$
where φμν denotes fluctuations. For the sake of simplicity we use the same notation for the other asymptotic fields as for the interacting fields. Then, up to surface terms the quadratic part of the quantum Lagrangian ( $\begin{eqnarray}\begin{array}{rcl}{{ \mathcal L }}_{q} & = & \frac{1}{2}\left(\frac{1}{4}{\varphi }_{\mu \nu }\square {\varphi }^{\mu \nu }-\frac{1}{4}\varphi \square \varphi \right.\\ & & -\,\left.\frac{1}{2}{\varphi }^{\mu \nu }{\partial }_{\mu }{\partial }_{\rho }{\varphi }_{\nu }{\,}^{\rho }+\frac{1}{2}{\varphi }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }\varphi \right)\\ & & -\,\phi (\square \varphi -{\partial }_{\mu }{\partial }_{\nu }{\varphi }^{\mu \nu })-\frac{1}{4\alpha }{\phi }^{2}\\ & & +\,\left({\varphi }^{\mu \nu }-\frac{1}{2}{\eta }^{\mu \nu }\varphi \right){\partial }_{\mu }{b}_{\nu }-\,\rm{i}\,{\partial }_{\mu }{\bar{c}}_{\rho }{\partial }^{\mu }{c}^{\rho }.\end{array}\end{eqnarray}$
Here note that only the quadratic term in Φ(φ), which is represented by $-\frac{1}{4\alpha }{\phi }^{2}$ as in equation (From this Lagrangian, it is straightforward to derive the following linearized field equations:
$\begin{eqnarray}\begin{array}{l}\frac{1}{2}\left(\frac{1}{2}\square {\varphi }_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\square \varphi -{\partial }_{\rho }{\partial }_{(\mu }{\varphi }_{\nu )}{\,}^{\rho }\right.\\ \quad +\,\left.\frac{1}{2}{\partial }_{\mu }{\partial }_{\nu }\varphi +\frac{1}{2}{\eta }_{\mu \nu }{\partial }_{\rho }{\partial }_{\sigma }{\varphi }^{\rho \sigma }\right)\\ \quad +\,(-{\eta }_{\mu \nu }\square +{\partial }_{\mu }{\partial }_{\nu })\phi +{\partial }_{(\mu }{b}_{\nu )}-\frac{1}{2}{\eta }_{\mu \nu }{\partial }_{\rho }{b}^{\rho }=0.\end{array}\end{eqnarray}$
$\begin{eqnarray}\square \varphi -{\partial }_{\mu }{\partial }_{\nu }{\varphi }^{\mu \nu }+\frac{1}{2\alpha }\phi =0.\end{eqnarray}$
$\begin{eqnarray}{\partial }^{\nu }{\varphi }_{\mu \nu }-\frac{1}{2}{\partial }_{\mu }\varphi =0.\end{eqnarray}$
$\begin{eqnarray}\square {c}^{\rho }=\square {\bar{c}}_{\rho }=0.\end{eqnarray}$
Here, for instance, ${\partial }_{(\mu }{b}_{\nu )}\equiv \frac{1}{2}({\partial }_{\mu }{b}_{\nu }+{\partial }_{\nu }{b}_{\mu })$.Now we are ready to simplify the field equations obtained above. First, operating ∂μ on the linearized Einstein equation (74 ), we have 19 ). This equation can be also obtained by taking the linearized BRST transformation ${\delta }_{B}^{(L)}{\bar{c}}_{\mu }=i{b}_{\mu }$ of $\square {\bar{c}}_{\mu }=0$ in equation (77 ).
$\begin{eqnarray}\square {b}_{\mu }=0,\end{eqnarray}$
which is a linearized analog of equation (Next, taking the trace of the linearized Einstein equation (74 ) and using equation (75 ) produces 78 ), we arrive at 75 ) and (76 ), it is easy to see that 80 ), this equation implies that
$\begin{eqnarray}(\square -{m}^{2})\phi +\frac{1}{3}{\partial }_{\rho }{b}^{\rho }=0,\end{eqnarray}$
where we have defined ${m}^{2}\equiv \frac{1}{12\alpha }\gt 0$. Moreover, acting on this equation and using equation ( $\begin{eqnarray}\square (\square -{m}^{2})\phi =0.\end{eqnarray}$
From equations ( $\begin{eqnarray}\phi =-\alpha \square \varphi .\end{eqnarray}$
Then, together with equation ( $\begin{eqnarray}{\square }^{2}(\square -{m}^{2})\varphi =0.\end{eqnarray}$
Finally, let us focus on the linearized Einstein equation (74 ). After some calculations using several equations, it turns out that equation (74 ) can be rewritten into a more compact form: 78 ), (80 ) and (83 ), we can obtain the equation for the gravitational field φμν: 82 ). Equation (84 ) implies that there are both massless and massive modes in φμν. In order to disentangle these two modes, let us introduce the following two fields: 81 ) was utilized in the former equation. With these fields, we find that $\tilde{\phi }$ satisfies the massive Klein–Gordon equation
$\begin{eqnarray}\frac{1}{4}\square {\varphi }_{\mu \nu }+\left({\partial }_{\mu }{\partial }_{\nu }+\frac{1}{2}{\eta }_{\mu \nu }\square \right)\phi +{\partial }_{(\mu }{b}_{\nu )}=0.\end{eqnarray}$
Using equations ( $\begin{eqnarray}{\square }^{2}\left((,\square -{m}^{2}\right){\varphi }_{\mu \nu }=0.\end{eqnarray}$
Note that the trace part of this equation reduces to equation ( $\begin{eqnarray}\begin{array}{rcl}\tilde{\phi } & = & \phi -\frac{1}{3{m}^{2}}{\partial }_{\rho }{b}^{\rho }=-\alpha \square \varphi -\frac{1}{3{m}^{2}}{\partial }_{\rho }{b}^{\rho },\\ {h}_{\mu \nu } & = & {\varphi }_{\mu \nu }+2{\eta }_{\mu \nu }\phi +48\alpha {\partial }_{\mu }{\partial }_{\nu }\phi -8\alpha {\eta }_{\mu \nu }{\partial }_{\rho }{b}^{\rho },\end{array}\end{eqnarray}$
where equation ( $\begin{eqnarray}(\square -{m}^{2})\tilde{\phi }=0,\end{eqnarray}$
while hμν obeys both the massless dipole ghost equation and the de Donder condition $\begin{eqnarray}{\square }^{2}{h}_{\mu \nu }=0,\quad {\partial }^{\mu }{h}_{\mu \nu }-\frac{1}{2}{\partial }_{\nu }h=0.\end{eqnarray}$
In fact, it can be verified shortly by calculating the four-dimensional commutation relation between $\tilde{\phi }$ and hμν that these two fields are independent modes involved in φμν. Later we will see that the massive scalar field $\tilde{\phi }$ and two transverse components of hμν, which is nothing but a massless spin-2 graviton, are physical modes in the theory at hand.Next, following the standard technique, let us calculate the four-dimensional (anti)commutation relations (4D CRs) between asymptotic fields. The point is that the simple pole fields, for instance the Nakanishi–Lautrup field bμ(x) obeying bμ = 0, can be expressed in terms of the invariant delta function D(x) as 88 ) is independent of z0, and this fact will be used in evaluating 4D CRs via the ETCRs in what follows.
$\begin{eqnarray}{b}_{\mu }(x)=-\int {{\rm{d}}}^{3}zD(x-z){\overleftrightarrow{\partial }}_{0}^{z}{b}_{\mu }(z).\end{eqnarray}$
Here the invariant delta function D(x) for massless simple pole fields and its properties are described as $\begin{eqnarray}\begin{array}{rcl}D(x) & = & -\frac{\,\rm{i}\,}{{(2\pi )}^{3}}\displaystyle \int {{\rm{d}}}^{4}k\,\epsilon ({k}^{0})\delta ({k}^{2}){\rm{e}\,}^{\,\rm{i}kx},\\ & & \square D(x)=0,\\ D(-x) & = & -D(x),\quad D(0,\overrightarrow{x})=0,\\ & & {\partial }_{0}D(0,\overrightarrow{x})=-{\delta }^{3}(x),\end{array}\end{eqnarray}$
where $\epsilon ({k}^{0})\equiv \frac{{k}^{0}}{| {k}^{0}| }$. With these properties, it is easy to see that the right-hand side (RHS) of equation (To illustrate the detail of the calculation, let us evaluate a 4D CR, [hμν(x), bρ(y)] explicitly. Using equation (88 ), it can be described as 88 ) is independent of z0, we can put z0 = x0 and use relevant ETCRs91 ) into equation (90 ), we can easily obtain
$\begin{eqnarray}\begin{array}{l}[{h}_{\mu \nu }(x),{b}_{\rho }(y)]=-\displaystyle \int {{\rm{d}}}^{3}zD(y-z)\\ \times \,{\overleftrightarrow{\partial }}_{0}^{z}[{h}_{\mu \nu }(x),{b}_{\rho }(z)]=-\displaystyle \int {{\rm{d}}}^{3}z\left(D(y-z)\right.\\ \times \,\left.[{h}_{\mu \nu }(x),{\dot{b}}_{\rho }(z)]-{\partial }_{0}^{z}D(y-z)[{h}_{\mu \nu }(x),{b}_{\rho }(z)]\right).\end{array}\end{eqnarray}$
As mentioned above, since the RHS of equation ( $\begin{aligned} {\left[h_{\mu \nu}(x), b_{\rho}(z)\right] } & =\mathrm{i}\left(\delta_{\mu}^{0} \eta_{\rho \nu}+\delta_{\nu}^{0} \eta_{\rho \mu}\right) \delta^{3}(x-z), \\ {\left[h_{\mu \nu}(x), \dot{b}_{\rho}(z)\right] } & =-\mathrm{i}\left(\delta_{\mu}^{k} \eta_{\rho \nu}+\delta_{\nu}^{k} \eta_{\rho \mu}\right) \partial_{k} \delta^{3}(x-z). \end{aligned}$
Substituting equation ( $\begin{eqnarray}[{h}_{\mu \nu }(x),{b}_{\rho }(y)]=-\,\rm{i}\,({\eta }_{\mu \rho }{\partial }_{\nu }+{\eta }_{\nu \rho }{\partial }_{\mu })D(x-y).\end{eqnarray}$
In a similar manner, we can calculate the four-dimensional (anti)commutation relations among $\tilde{\phi },{h}_{\mu \nu },{b}_{\mu },{c}^{\mu }$ and ${\bar{c}}_{\mu }$. To do this, let us note that since $\tilde{\phi }$ obeys the massive Klein–Gordon equation (86 ), it can be expressed in terms of the invariant delta function Δ(x; m2) for massive simple pole fields as
$\begin{eqnarray}\tilde{\phi }(x)=-\int {{\rm{d}}}^{3}z{\rm{\Delta }}(x-z;{m}^{2}){\overleftrightarrow{\partial }}_{0}^{z}\tilde{\phi }(z),\end{eqnarray}$
where Δ(x; m2) is defined as $\begin{eqnarray}\begin{array}{rcl}{\rm{\Delta }}(x;{m}^{2}) & = & -\frac{\,\rm{i}\,}{{(2\pi )}^{3}}\displaystyle \int {{\rm{d}}}^{4}k\,\epsilon ({k}^{0})\delta ({k}^{2}+{m}^{2}){\rm{e}\,}^{\,\rm{i}kx},\\ & & (\square -{m}^{2}){\rm{\Delta }}(x;{m}^{2})=0,\\ {\rm{\Delta }}(-x;{m}^{2}) & = & -{\rm{\Delta }}(x;{m}^{2}),\quad {\rm{\Delta }}(0,\overrightarrow{x};{m}^{2})=0,\\ {\partial }_{0}{\rm{\Delta }}(0,\overrightarrow{x};{m}^{2}) & = & -{\delta }^{3}(x),\quad {\rm{\Delta }}(x;0)=D(x).\end{array}\end{eqnarray}$
As for hμν, since hμν is a massless dipole ghost field as can be seen in equation (87 ), it can be described as 79 ), (83 ) and (85 ), and we have introduced the invariant delta function E(x) for massless dipole ghost fields and its properties are given by 88 ), we can also show that the RHSs of both (93 ) and (95 ) are independent of z0.
$\begin{eqnarray}\begin{array}{rcl}{h}_{\mu \nu }(x) & = & -\displaystyle \int {{\rm{d}}}^{3}z\Space{0ex}{0.25ex}{0ex}D(x-z){\overleftrightarrow{\partial }}_{0}^{z}{h}_{\mu \nu }(z)\\ & & +\,E(x-z){\overleftrightarrow{\partial }}_{0}^{z}\square {h}_{\mu \nu }(z)\Space{0ex}{0.25ex}{0ex}\\ & = & -\displaystyle \int {{\rm{d}}}^{3}z\Space{0ex}{0.25ex}{0ex}D(x-z){\overleftrightarrow{\partial }}_{0}^{z}{h}_{\mu \nu }(z)-4E(x-z){\overleftrightarrow{\partial }}_{0}^{z}\\ & & \times \,\left.\left({\partial }_{(\mu }{b}_{\nu )}+\frac{1}{3{m}^{2}}{\partial }_{\mu }{\partial }_{\nu }{\partial }_{\rho }{b}^{\rho }\right)(z)\right],\end{array}\end{eqnarray}$
where we have used the equation $\begin{eqnarray}\square {h}_{\mu \nu }=-4\left({\partial }_{(\mu }{b}_{\nu )}+\frac{1}{3{m}^{2}}{\partial }_{\mu }{\partial }_{\nu }{\partial }_{\rho }{b}^{\rho }\right),\end{eqnarray}$
which can be derived from equations ( $\begin{eqnarray}\begin{array}{rcl}E(x) & = & -\frac{\,\rm{i}\,}{{(2\pi )}^{3}}\displaystyle \int {{\rm{d}}}^{4}k\,\epsilon ({k}^{0}){\delta }^{{\prime} }({k}^{2}){\rm{e}\,}^{\,\rm{i}kx},\\ & & \square E(x)=D(x),\\ E(-x) & = & -E(x),\quad E(0,\overrightarrow{x})={\partial }_{0}E(0,\overrightarrow{x})\\ & = & {\partial }_{0}^{2}E(0,\overrightarrow{x})=0,\\ {\partial }_{0}^{3}E(0,\overrightarrow{x}) & = & {\delta }^{3}(x),E(x)=\frac{\partial }{\partial {m}^{2}}{\rm{\Delta }}(x;{m}^{2}){| }_{{m}^{2}=0}.\end{array}\end{eqnarray}$
As in equation (After straightforward calculations, we find the following 4D CRs: 100 ) implies that two fields hμν and $\tilde{\phi }$ are independent fields.
$\begin{eqnarray}[\tilde{\phi }(x),\tilde{\phi }(y)]=\,\rm{i}\,\frac{1}{6}{\rm{\Delta }}(x-y;{m}^{2}).\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}[{h}_{\mu \nu }(x),{h}_{\sigma \tau }(y)] & = & -2\,\rm{i}\,\left[\Space{0ex}{2.65ex}{0ex}{\eta }_{\mu \nu }{\eta }_{\sigma \tau }-{\eta }_{\mu \sigma }{\eta }_{\nu \tau }\right.\\ & & -\,{\eta }_{\mu \tau }{\eta }_{\nu \sigma }+\frac{2}{3{m}^{2}}({\eta }_{\mu \nu }{\partial }_{\sigma }{\partial }_{\tau }+{\eta }_{\sigma \tau }{\partial }_{\mu }{\partial }_{\nu })\\ & & \left.-\,\frac{4}{3{m}^{4}}{\partial }_{\mu }{\partial }_{\nu }{\partial }_{\sigma }{\partial }_{\tau }\right]D(x-y)\\ & & -\,2\,\rm{i}\,\left(\Space{0ex}{2.25ex}{0ex}{\eta }_{\mu \sigma }{\partial }_{\nu }{\partial }_{\tau }+{\eta }_{\mu \tau }{\partial }_{\nu }{\partial }_{\sigma }+{\eta }_{\nu \sigma }{\partial }_{\mu }{\partial }_{\tau }\right.\\ & & +\,\left.{\eta }_{\nu \tau }{\partial }_{\mu }{\partial }_{\sigma }+\frac{4}{3{m}^{2}}{\partial }_{\mu }{\partial }_{\nu }{\partial }_{\sigma }{\partial }_{\tau }\right)E(x-y).\end{array}\end{eqnarray}$
$\begin{eqnarray}[{h}_{\mu \nu }(x),\tilde{\phi }(y)]=0.\end{eqnarray}$
$\begin{eqnarray}[\tilde{\phi }(x),{b}_{\rho }(y)]=0.\end{eqnarray}$
$\begin{eqnarray}[{h}_{\mu \nu }(x),{b}_{\rho }(y)]=-\,\rm{i}\,({\eta }_{\mu \rho }{\partial }_{\nu }+{\eta }_{\nu \rho }{\partial }_{\mu })D(x-y).\end{eqnarray}$
$\begin{eqnarray}\{{c}^{\mu }(x),{\bar{c}}_{\nu }(y)\}={\delta }_{\nu }^{\mu }D(x-y).\end{eqnarray}$
Here let us note that equation (As usual, the physical Hilbert space ∣phys⟩ is defined by the Kugo–Ojima subsidiary conditions [6]
$\begin{eqnarray}{{\rm{Q}}}_{{\rm{B}}}| {\rm{phys}}\rangle =0,\end{eqnarray}$
where QB is the BRST charge associated with the general coordinate transformation (GCT).Now we would like to discuss the issue of the unitarity of the theory at hand. To do that, it is convenient to perform the Fourier transformation of equations (98 )–(103 ). However, for the dipole field we cannot use the three-dimensional Fourier expansion to define the creation and annihilation operators. We therefore make use of the four-dimensional Fourier expansion [12]8 98 )–(103 ):
$\begin{eqnarray}\begin{array}{rcl}{\varphi }_{\mu \nu }(x) & = & \frac{1}{{(2\pi )}^{\frac{3}{2}}}\displaystyle \int {{\rm{d}}}^{4}p\,\theta ({p}^{0})\\ & & \times \,[{\varphi }_{\mu \nu }(p){\rm{e}\,}^{\,\rm{i}px}+{\varphi }_{\mu \nu }^{\dagger }(p){\rm{e}\,}^{-\,\rm{i}px}],\end{array}\end{eqnarray}$
where θ(p0) is the unit step function. For any simple pole field we adopt the same Fourier expansion, for instance $\begin{eqnarray}\begin{array}{rcl}{\beta }_{\mu }(x) & = & \frac{1}{{(2\pi )}^{\frac{3}{2}}}\displaystyle \int {{\rm{d}}}^{4}p\,\theta ({p}^{0})\\ & & \times \,[{\beta }_{\mu }(p){\rm{e}\,}^{\,\rm{i}px}+{\beta }_{\mu }^{\dagger }(p){\rm{e}\,}^{-\,\rm{i}px}].\end{array}\end{eqnarray}$
Incidentally, for a generic, massless simple pole field Φ, the three-dimensional Fourier expansion is defined as $\begin{eqnarray}\begin{array}{rcl}{\rm{\Phi }}(x) & = & \frac{1}{{(2\pi )}^{\frac{3}{2}}}\displaystyle \int {{\rm{d}}}^{3}p\,\frac{1}{\sqrt{2| \overrightarrow{p}| }}\\ & & \times [{\rm{\Phi }}(\overrightarrow{p}){\rm{e}\,}^{-\,\rm{i}| \overrightarrow{p}| {x}^{0}+\,\rm{i}\,\overrightarrow{p}\cdot \overrightarrow{x}}+{{\rm{\Phi }}}^{\dagger }(\overrightarrow{p}){\rm{e}\,}^{\,\rm{i}| \overrightarrow{p}| {x}^{0}-\,\rm{i}\,\overrightarrow{p}\cdot \overrightarrow{x}}],\end{array}\end{eqnarray}$
whereas the four-dimensional Fourier expansion reads $\begin{eqnarray}\begin{array}{rcl}{\rm{\Phi }}(x) & = & \frac{1}{{(2\pi )}^{\frac{3}{2}}}\displaystyle \int {{\rm{d}}}^{4}p\,\theta ({p}^{0})\\ & & \times \,[{\rm{\Phi }}(p){\rm{e}\,}^{\,\rm{i}px}+{{\rm{\Phi }}}^{\dagger }(p)(p){\rm{e}\,}^{-\,\rm{i}px}].\end{array}\end{eqnarray}$
Thus, the annihilation operator Φ(p) in the four-dimensional Fourier expansion has a connection with the annihilation operator ${\rm{\Phi }}(\overrightarrow{p})$ in the three-dimensional Fourier expansion like9 $\begin{eqnarray}{\rm{\Phi }}(p)=\theta ({p}^{0})\delta ({p}^{2})\sqrt{2| \overrightarrow{p}| }{\rm{\Phi }}(\overrightarrow{p}).\end{eqnarray}$
Note that the Fourier transform Φ(p) takes the form for a massless simple pole field $\begin{eqnarray}{\rm{\Phi }}(p)=\frac{\,\rm{i}\,}{{(2\pi )}^{\frac{3}{2}}}\theta ({p}^{0})\delta ({p}^{2})\int {{\rm{d}}}^{3}z\,{\rm{e}\,}^{-\,\rm{i}pz}{\overleftrightarrow{\partial }}_{0}^{z}{\rm{\Phi }}(z).\end{eqnarray}$
Based on these Fourier expansions, we can calculate the Fourier transform of equations ( $\begin{eqnarray}[\tilde{\phi }(p),{\tilde{\phi }}^{\dagger }(q)]=\frac{1}{6}\theta ({p}^{0})\delta ({p}^{2}+{m}^{2}){\delta }^{4}(p-q).\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}[{h}_{\mu \nu }(p),{h}_{\sigma \tau }^{\dagger }(q)] & = & \frac{1}{2}\theta ({p}^{0}){\delta }^{4}(p-q)\\ & & \times \,\left\{\Space{0ex}{2.25ex}{0ex}\delta ({p}^{2})\left[\Space{0ex}{2.25ex}{0ex}{\eta }_{\mu \sigma }{\eta }_{\nu \tau }+{\eta }_{\mu \tau }{\eta }_{\nu \sigma }-{\eta }_{\mu \nu }{\eta }_{\sigma \tau }\right.\right.\\ & & +\,\frac{2}{3{m}^{2}}({\eta }_{\mu \nu }{p}_{\sigma }{p}_{\tau }+{\eta }_{\sigma \tau }{p}_{\mu }{p}_{\nu })\\ & & +\,\left.\frac{4}{3{m}^{4}}{p}_{\mu }{p}_{\nu }{p}_{\sigma }{p}_{\tau }\right]+3{\delta }^{{\prime} }\\ & & \times \,({p}^{2})\left(\Space{0ex}{2.25ex}{0ex}{\eta }_{\mu \sigma }{p}_{\nu }{p}_{\tau }+{\eta }_{\nu \sigma }{p}_{\mu }{p}_{\tau }+{\eta }_{\mu \tau }{p}_{\nu }{p}_{\sigma }\right.\\ & & +\,\left.\left.{\eta }_{\nu \tau }{p}_{\mu }{p}_{\sigma }-\frac{4}{3{m}^{2}}{p}_{\mu }{p}_{\nu }{p}_{\sigma }{p}_{\tau }\right)\right\}.\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}[{h}_{\mu \nu }(p),{b}_{\rho }^{\dagger }(q)] & = & -\,\rm{i}\,\frac{1}{2}({\eta }_{\mu \rho }{p}_{\nu }+{\eta }_{\nu \rho }{p}_{\mu })\\ & & \times \,\theta ({p}^{0})\delta ({p}^{2}){\delta }^{4}(p-q).\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\{{c}^{\mu }(p),{\bar{c}}_{\nu }^{\dagger }(q)\} & = & -\,\rm{i}\,{\delta }_{\nu }^{\mu }\theta ({p}^{0})\\ & & \times \,\delta ({p}^{2}){\delta }^{4}(p-q).\end{array}\end{eqnarray}$
Next, let us turn our attention to the linearized field equations. In the Fourier transformation, the de Donder condition in equation (87 ) takes the form
$\begin{eqnarray}{p}^{\nu }{h}_{\mu \nu }-\frac{1}{2}{p}_{\mu }h=0.\end{eqnarray}$
Since this equation gives us four independent relations in ten components of hμν(p), the number of independent components of hμν(p) is six. To deal with these six independent components it is convenient to take a specific Lorentz frame which is defined by pμ = (p, 0, 0, p) with p > 0, and choose the six components as follows: $\begin{eqnarray}\begin{array}{l}{h}_{1}(p)={h}_{11}(p),\quad {h}_{2}(p)={h}_{12}(p),\\ {\omega }_{0}(p)=\frac{1}{2p}[{h}_{00}(p)-{h}_{11}(p)],\\ {\omega }_{I}(p)=\frac{1}{p}{h}_{0I}(p),\\ {\omega }_{3}(p)=-\frac{1}{2p}[{h}_{11}(p)+{h}_{33}(p)],\end{array}\end{eqnarray}$
where the index I takes the transverse components I = 1, 2. The other four components are expressible by the above six ones.In this respect, it is worthwhile considering the BRST transformation for asymptotic fields. Since the BRST transformation for the Fourier transform of the asymptotic fields is given by 116 ) takes the form
$\begin{eqnarray}\begin{array}{l}{\delta }_{B}\tilde{\phi }(p)=0,\quad {\delta }_{B}{h}_{\mu \nu }(p)=-\,\rm{i}\,[{p}_{\mu }{c}_{\nu }(p)+{p}_{\nu }{c}_{\mu }(p)],\\ {\delta }_{B}{b}_{\mu }(p)=0,\\ {\delta }_{B}{\bar{c}}_{\mu }(p)=\,\rm{i}\,{b}_{\mu }(p),\\ {\delta }_{B}{c}^{\mu }(p)=0,\end{array}\end{eqnarray}$
the BRST transformation in terms of the components in ( $\begin{eqnarray}\begin{array}{l}{\delta }_{B}\tilde{\phi }(p)=0,\quad {\delta }_{B}{h}_{I}(p)=0,\\ {\delta }_{B}{\omega }_{\mu }(p)=\,\rm{i}\,{c}_{\mu }(p),\\ {\delta }_{B}{\bar{c}}_{\mu }(p)=\,\rm{i}\,{b}_{\mu }(p),\\ {\delta }_{B}{c}^{\mu }(p)={\delta }_{B}{b}_{\mu }(p)=0.\end{array}\end{eqnarray}$
This BRST transformation implies that $\tilde{\phi }(p)$ and hI(p) could be physical observables while a set of fields $\{{\omega }_{\mu }(p),{b}_{\mu }(p),{c}_{\mu }(p),{\bar{c}}_{\mu }(p)\}$ might belong to a BRST quartet, which are dropped from the physical state by the Kugo–Ojima subsidiary condition, QB∣phys⟩ = 0 [6]. However, note that bμ(p), cμ(p) and ${\bar{c}}_{\mu }(p)$ are simple pole fields obeying ${p}^{2}{b}_{\mu }(p)={p}^{2}{c}_{\mu }(p)={p}^{2}{\bar{c}}_{\mu }(p)=0$, whereas hμν(p) is a dipole field satisfying ${({p}^{2})}^{2}{h}_{\mu \nu }(p)=0$, so that a naive Kugo–Ojima quartet mechanism does not work in a straightforward manner.To clarify the BRST quartet mechanism, let us present 4D CRs in terms of the components in (116 ). From equations (111 )–(114 ) and the definition (116 ) it is straightforward to derive the following 4D CRs: 96 ). It is then easy to verify the equation116 ), and we redefine ωμ by ${\hat{\omega }}_{\mu }$ as
$\begin{eqnarray}[\tilde{\phi }(p),{\tilde{\phi }}^{\dagger }(q)]=\frac{1}{6}\theta ({p}^{0})\delta ({p}^{2}+{m}^{2}){\delta }^{4}(p-q).\end{eqnarray}$
$\begin{eqnarray}[{h}_{I}(p),{h}_{J}^{\dagger }(q)]=\frac{1}{2}{\delta }_{IJ}\theta ({p}^{0})\delta ({p}^{2}){\delta }^{4}(p-q).\end{eqnarray}$
$\begin{eqnarray}[{h}_{I}(p),{\omega }_{\mu }^{\dagger }(q)]=[{h}_{I}(p),{b}_{\mu }^{\dagger }(q)]=[{b}_{\mu }(p),{b}_{\nu }^{\dagger }(q)]=0.\end{eqnarray}$
$\begin{eqnarray}[{\omega }_{\mu }(p),{b}_{\nu }^{\dagger }(q)]=\,\rm{i}\,\frac{1}{2}{\eta }_{\mu \nu }\theta ({p}^{0})\delta ({p}^{2}){\delta }^{4}(p-q).\end{eqnarray}$
$\begin{eqnarray}\{{c}^{\mu }(p),{\bar{c}}_{\nu }^{\dagger }(q)\}=-\,\rm{i}\,{\delta }_{\nu }^{\mu }\theta ({p}^{0})\delta ({p}^{2}){\delta }^{4}(p-q).\end{eqnarray}$
In addition to them, we have a rather complicated expression for $[{\omega }_{\mu }(p),{\omega }_{\nu }^{\dagger }(q)]$ because hμν(p) is a dipole field, but luckily this expression is unnecessary for our purpose [6]. It is known how to take out a simple pole field from a dipole field, which amounts to using an operator defined by [6] $\begin{eqnarray}{{ \mathcal D }}_{p}=\frac{1}{2| \overrightarrow{p}{| }^{2}}{p}_{0}\frac{\partial }{\partial {p}_{0}}+c,\end{eqnarray}$
where c is a constant. Using this operator, we can define a simple pole field ${\hat{h}}_{\mu \nu }(p)$ from the dipole field hμν(p), which obeys ${({p}^{2})}^{2}{h}_{\mu \nu }(p)=0$, as $\begin{eqnarray}\begin{array}{rcl}{\hat{h}}_{\mu \nu }(p) & \equiv & {h}_{\mu \nu }(p)-{{ \mathcal D }}_{p}({p}^{2}{h}_{\mu \nu }(p))\\ & = & {h}_{\mu \nu }(p)-4\,\rm{i}\,{{ \mathcal D }}_{p}\\ & & \times \,\left({p}_{(\mu }{b}_{\nu )}(p)-\frac{1}{3{m}^{2}}{p}_{\mu }{p}_{\nu }{p}_{\rho }{b}^{\rho }(p)\right),\end{array}\end{eqnarray}$
where in the last equality we have used the Fourier transform of the linearized field equation ( $\begin{eqnarray}{p}^{2}{\hat{h}}_{\mu \nu }(p)=0.\end{eqnarray}$
Then, we replace hμν of ωμ with ${\hat{h}}_{\mu \nu }$ in equation ( $\begin{eqnarray}\begin{array}{rcl}{\hat{\omega }}_{0}(p) & = & \frac{1}{2p}[{\hat{h}}_{00}(p)-{\hat{h}}_{11}(p)],\\ {\hat{\omega }}_{I}(p) & = & \frac{1}{p}{\hat{h}}_{0I}(p),\\ {\hat{\omega }}_{3}(p) & = & -\frac{1}{2p}[{\hat{h}}_{11}(p)+{\hat{h}}_{33}(p)].\end{array}\end{eqnarray}$
The key point is that with this redefinition from ωμ to ${\hat{\omega }}_{\mu }$, the BRST transformation and the 4D CRs remain unchanged owing to δBbμ = 0 and $[{b}_{\mu }(p),{b}_{\nu }^{\dagger }(q)]=[{h}_{I}(p),{b}_{\mu }^{\dagger }(q)]=0$, those are $\begin{aligned} \delta_{B} \hat{\omega}_{\mu}(p) & =\mathrm{i} c_{\mu}(p), \\ {\left[\hat{\omega}_{\mu}(p), b_{\nu}^{\dagger}(q)\right] } & =\left[\omega_{\mu}(p), b_{\nu}^{\dagger}(q)\right], \\ {\left[h_{I}(p), \hat{\omega}_{\mu}^{\dagger}(q)\right] } & =\left[h_{I}(p), \omega_{\mu}^{\dagger}(q)\right] . \end{aligned}$
Since it turns out that all the fields, $\{\tilde{\phi },{h}_{I},{\hat{\omega }}_{\mu },{b}_{\mu },{c}_{\mu },{\bar{c}}_{\mu }\}$ are simple pole fields,10 we can obtain the standard creation and annihilation operators in the three-dimensional Fourier expansion from those in the four-dimensional one through the relation (109 ). As a result, the three-dimensional (anti)commutation relations, which are denoted as $[{\rm{\Phi }}(\overrightarrow{p}),{{\rm{\Phi }}}^{\dagger }(\overrightarrow{q})\}$ with ${\rm{\Phi }}(\overrightarrow{p})\equiv \{\tilde{\phi }(\overrightarrow{p}),{h}_{I}(\overrightarrow{p}),{\hat{\omega }}_{\mu }(\overrightarrow{p}),{b}_{\mu }(\overrightarrow{p}),{c}_{\mu }(\overrightarrow{p}),{\bar{c}}_{\mu }(\overrightarrow{p})\}$, are given by11 The (anti)commuatation relations (129 ) have in essence the same structure as those of the Yang–Mills theory [6]. In particular, the positive coefficients $\frac{1}{6}$ and $\frac{1}{2}{\delta }_{IJ}$ mean that $\tilde{\phi }$ and hI have a positive norm. Hence, we find that $\tilde{\phi }$ and hI, which correspond to transverse gravitons, are physical observables of a positive norm while a set of fields $\{{\hat{\omega }}_{\mu },{b}_{\mu },{c}_{\mu },{\bar{c}}_{\mu }\}$ belong to a BRST quartet. This quartet appears in the physical subspace only as zero norm states by the Kugo–Ojima subsidiary conditions (104 ). It is worthwhile stressing that both the massive scalar and the massless graviton of the positive norm appear in the physical Hilbert space so the physical S-matrix is unitary in the present theory.
6. Conclusions
In this paper, we have performed the manifestly covariant quantization of f(R) gravity in the de Donder gauge condition, clarified its global symmetry and calculated the whole equal-time (anti)commutation relation (ETCR) among fundamental variables on the basis of the BRST formalism. Via this quantization procedure, we have clearly demonstrated that the physical modes are a massive scalar and the massless graviton, both of which have a positive norm, and the other dynamical degrees of freedom such as FP ghosts emerge in the physical subspace defined by the physical condition QB∣phys⟩ = 0 only as zero norm states. Thus, we can definitely conclude that the physical S-matrix is manifestly unitary in f(R) gravity, although it is not renormalizable perturbatively.
One of motivations behind the present study is to understand the canonical structure of the gravitational sector in gravitational theories possessing higher-derivative terms. As shown in the present paper, the ETCR between the metric tensor and its time derivative $[{\dot{g}}_{\rho \sigma },{g}_{\mu \nu }^{{\prime} }]$ takes a nontrivial expression while that in quadratic gravity was mentioned to be vanishing [8–10]. Although the physical meaning of this vanishing ETCR is not understood yet, it is certain that this triviality comes from the squared term of the conformal tensor $\sqrt{-g}\,{C}_{\mu \nu \rho \sigma }^{2}$ in the action of quadratic gravity as shown in the present article (in this context, we consider a specific case, f(R) = αR2). Indeed, in our previous work [13] the BRST formalism of conformal gravity was investigated and it was found that the ETCR between the metric and its time derivative has a very simple expression, although it is not vanishing.
Related to this fact, it is useful to recall that quadratic gravity is renormalizable by power counting, but it is not unitary because of the existence of a massive ghost [7]. This breakdown of the unitarity arises from the squared term of the conformal tensor since f(R) gravity is manifestly unitary, as seen in this article. We are in a somewhat tantalizing situation. We need the squared term of the conformal tensor for ensuring the renormalizability of a theory, but this term pays a massive price for the issue of unitarity. In any case, in order to obtain a consistent quantum gravity which is renormalizable and unitary at the same time, it seems to be necessary to have a deep understanding of the squared term of conformal tensor.
Appendix A. Two derivations of equation (19)
In this appendix, we present two different derivations of the field equation for the bμ field since bμ plays an important role as the generator of global symmetry in the formalism at hand. In the first derivation we make use of field equations while in the second we rely on the BRST transformation.
Let us first present a derivation based on field equations. We take a covariant derivative of the Einstein equation, i.e. the first equation in equation (17 ). The result reads 17 ), i.e. $R+{{\rm{\Phi }}}^{{\prime} }(\phi )=0$, the three terms except for the last term on the LHS of the above equation identically vanish. Thus, we can obtain
$\begin{eqnarray}\begin{array}{l}{{\rm{\nabla }}}^{\mu }\phi \,{G}_{\mu \nu }-{R}_{\mu \nu }{{\rm{\nabla }}}^{\mu }\phi -\frac{1}{2}{{\rm{\nabla }}}_{\nu }{\rm{\Phi }}\\ \,-\,\frac{1}{2}{{\rm{\nabla }}}^{\mu }\left({E}_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }E\right)=0,\end{array}\end{eqnarray}$
where we have used the Bianchi identity ∇μGμν = 0 and the relation ∇μ(∇μ∇ν − gμν)φ = Rμν∇μφ. Then, using the field equation for φ in equation ( $\begin{eqnarray}{{\rm{\nabla }}}_{\nu }\left({E}_{\mu }^{\nu }-\frac{1}{2}{\delta }_{\mu }^{\nu }E\right)=0.\end{eqnarray}$
Generally for a symmetric tensor Sμν we have a formulaA2 ), with the help of the other field equations in equation (17 ), is seen to lead to the field equation (19 ) for the bμ field.
$\begin{eqnarray}{{\rm{\nabla }}}_{\nu }{S}_{\mu }^{\nu }=\frac{1}{\sqrt{-g}}{\partial }_{\nu }(\sqrt{-g}\,{S}^{\nu }{\,}_{\mu })+\frac{1}{2}{S}_{\nu \rho }{\partial }_{\mu }{g}^{\nu \rho }.\end{eqnarray}$
Using this, we can show an equality $\begin{eqnarray}\begin{array}{rcl}{{\rm{\nabla }}}_{\nu }{E}_{\mu }^{\nu } & = & \frac{1}{\sqrt{-g}}{\partial }_{\nu }{\tilde{g}}^{\nu \rho }\cdot {E}_{\rho \mu }+{\partial }^{2}{b}_{\mu }\\ & & +\,\rm{i}\,\left({\partial }^{2}{\bar{c}}_{\lambda }\cdot {\partial }_{\mu }{c}^{\lambda }+{\partial }_{\mu }{\bar{c}}_{\lambda }\cdot {\partial }^{2}{c}^{\lambda }\right)+\frac{1}{2}{\partial }_{\mu }E,\end{array}\end{eqnarray}$
where ∂2 ≡ gμν∂μ∂ν. Then equation (Next, let us prove the same equation (19 ) by using the BRST transformation. This derivation is a bit simpler than the above one. Let us start with the field equation ${g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{\bar{c}}_{\rho }=0$. Taking the BRST transformation of this field equation produces13 ) in this equation, and using the field equations for FP (anti)ghosts, we can easily arrive at the desired equation (19 ).
$\begin{eqnarray}\begin{array}{l}(-{c}^{\lambda }{\partial }_{\nu }{g}^{\mu \nu }+{g}^{\mu \lambda }{\partial }_{\lambda }{c}^{\nu }+{g}^{\nu \lambda }{\partial }_{\lambda }{c}^{\mu }){\partial }_{\mu }{\partial }_{\nu }{\bar{c}}_{\rho }\\ \quad +\,\rm{i}\,{g}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }{B}_{\rho }=0.\end{array}\end{eqnarray}$
Then, rewriting Bρ in terms of bρ via equation (Appendix B. Two derivations of $[{b}_{\mu },{\dot{b}}_{\nu }^{{\prime} }]$
In this appendix we present two different derivations of $[{b}_{\mu },{\dot{b}}_{\nu }^{{\prime} }]$ in equation (71 ), one of which is based on field equations and the other on BRST transformation.
In the first derivation let us first solve ${\dot{b}}_{\rho }$ from equation (18 )17 ) in the formB1 ) as
$\begin{eqnarray}\begin{array}{rcl}{\dot{b}}_{0} & = & \frac{1}{2}{E}_{00}-\,\rm{i}\,{\dot{\bar{c}}}_{\rho }{\dot{c}}^{\rho },\\ {\dot{b}}_{k} & = & {E}_{0k}-{\partial }_{k}{b}_{0}-\,\rm{i}\,({\dot{\bar{c}}}_{\rho }{\partial }_{k}{c}^{\rho }+{\partial }_{k}{\bar{c}}_{\rho }{\dot{c}}^{\rho }).\end{array}\end{eqnarray}$
Next, let us rewrite the Einstein equation in equation ( $\begin{eqnarray}\begin{array}{rcl}{E}_{\nu }^{\mu }-\frac{1}{2}{\delta }_{\nu }^{\mu }E & = & 2\left(\frac{1}{2}+\phi \right){G}_{\nu }^{\mu }\\ & & -2({{\rm{\nabla }}}^{\mu }{{\rm{\nabla }}}_{\nu }-{\delta }_{\nu }^{\mu }\square )\phi -{\delta }_{\nu }^{\mu }{\rm{\Phi }}(\phi ).\end{array}\end{eqnarray}$
Since we find that ${R}_{00}\sim -\frac{1}{2}{g}^{ij}{\ddot{g}}_{ij},{R}_{0i}\sim \frac{1}{2}{g}^{0j}{\ddot{g}}_{ij}$ and ${R}_{ij}\sim -\frac{1}{2}{g}^{00}{\ddot{g}}_{ij}$, it turns out that ${G}_{\nu }^{0}={R}_{\nu }^{0}-\frac{1}{2}{\delta }_{\nu }^{0}R$ contains no ${\ddot{g}}_{ij}$. Then, we can rewrite ${\dot{b}}_{\rho }$ of equation ( $\begin{eqnarray}\begin{array}{rcl}{\dot{b}}_{0} & = & \frac{2}{{g}^{00}}\left[\left(\frac{1}{2}+\phi \right){G}_{0}^{0}-({{\rm{\nabla }}}^{0}{{\rm{\nabla }}}_{0}-\square )\phi \right.\\ & & -\,\left.\frac{1}{2}{\rm{\Phi }}(\phi )+\frac{1}{4}{g}^{ij}{E}_{ij}\right]-\,\rm{i}\,{\dot{\bar{c}}}_{\rho }{\dot{c}}^{\rho },\\ {\dot{b}}_{k} & = & \frac{2}{{g}^{00}}\left[\left(\frac{1}{2}+\phi \right){R}_{k}^{0}-{{\rm{\nabla }}}^{0}{{\rm{\nabla }}}_{k}\phi -\frac{1}{2}{g}^{0j}{E}_{kj}\right]\\ & & -\,{\partial }_{k}{b}_{0}-\,\rm{i}\,({\dot{\bar{c}}}_{\rho }{\partial }_{k}{c}^{\rho }+{\partial }_{k}{\bar{c}}_{\rho }{\dot{c}}^{\rho }).\end{array}\end{eqnarray}$
After some calculations, we can show that 58 ) and (69 ) we can also show that 64 ) was used. Then, it is straightforward to prove the ETCR in equation (71 ).
$\begin{eqnarray}[{G}_{\nu }^{0},{b}_{\rho }^{{\prime} }]=\,\rm{i}\,\tilde{f}({\delta }_{\rho }^{0}{R}_{\nu }^{0}-{\delta }_{\nu }^{0}{R}_{p}^{0}){\delta }^{3}.\end{eqnarray}$
Furthermore, from equations ( $\begin{eqnarray}[{{\rm{\nabla }}}_{k}{{\rm{\nabla }}}_{\mu }\phi ,{b}_{\rho }^{{\prime} }]=-\,\rm{i}\,\tilde{f}{\delta }_{\mu }^{0}{{\rm{\nabla }}}_{k}{{\rm{\nabla }}}_{\rho }\phi {\delta }^{3}.\end{eqnarray}$
To derive this equation, we need to use the equation $\begin{eqnarray}\begin{array}{rcl}[{{\rm{\Gamma }}}_{\mu \nu }^{\rho },{b}_{\lambda }^{{\prime} }] & = & \,\rm{i}\,\tilde{f}({\delta }_{\lambda }^{\rho }{{\rm{\Gamma }}}_{\mu \nu }^{0}-{\delta }_{\mu }^{0}{{\rm{\Gamma }}}_{\lambda \nu }^{\rho }-{\delta }_{\nu }^{0}{{\rm{\Gamma }}}_{\mu \lambda }^{\rho }){\delta }^{3}\\ & & +\,\rm{i}\,{\delta }_{\lambda }^{\rho }(2{\delta }_{\mu }^{0}{\delta }_{\nu }^{0}\tilde{f}{\tilde{g}}^{0k}-{\delta }_{\mu }^{0}{\delta }_{\nu }^{k}-{\delta }_{\nu }^{0}{\delta }_{\mu }^{k}){\partial }_{k}(\tilde{f}{\delta }^{3}),\end{array}\end{eqnarray}$
where equation (In the second derivation of equation (71 ) we make use of the BRST transformation. To do this, let us begin with the ETCR $[{\pi }_{c\mu },{b}_{\nu }^{{\prime} }]=0$, which can be easily proved. Taking the BRST transformation of this equation, it is straightforward to obtain the following equation: 71 ).
$\begin{eqnarray}[{b}_{\mu },{\dot{b}}_{\nu }^{{\prime} }]=\,\rm{i}\,\tilde{f}({\partial }_{\mu }{b}_{\nu }+{\partial }_{\nu }{b}_{\mu }){\delta }^{3}+{A}_{\mu \nu }+{B}_{\mu \nu }+{C}_{\mu \nu }.\end{eqnarray}$
Here Aμν, Bμν and Cμν are defined as $\begin{eqnarray}\begin{array}{rcl}{A}_{\mu \nu } & = & -\,\rm{i}\,\tilde{f}[({\tilde{g}}^{\rho \sigma }{{\rm{\nabla }}}_{\sigma }{c}^{0}+{\tilde{g}}^{0\sigma }{{\rm{\nabla }}}_{\sigma }{c}^{\rho }\\ & & -\,{\tilde{g}}^{0\rho }{{\rm{\nabla }}}_{\lambda }{c}^{\lambda }){\partial }_{\rho }{\bar{c}}_{\mu },{b}_{\nu }^{{\prime} }],\\ {B}_{\mu \nu } & = & i\tilde{f}[{\tilde{g}}^{0\rho }{\partial }_{\rho }({c}^{\lambda }{\partial }_{\lambda }{\bar{c}}_{\mu }),{b}_{\nu }^{{\prime} }],\\ {C}_{\mu \nu } & = & -\tilde{f}{c}^{{\prime} \lambda }[{\pi }_{c\mu },{\partial }_{\lambda }{b}_{\nu }^{{\prime} }].\end{array}\end{eqnarray}$
For their evaluation we utilize the following ETCRs: $\begin{aligned} {\left[\nabla_{\sigma} c^{\rho}, b_{\nu}^{\prime}\right]=} & \mathrm{i} \tilde{f}\left(-\delta_{\sigma}^{0} \nabla_{\nu} c^{\rho}+\delta_{\nu}^{\rho} \Gamma_{\sigma \lambda}^{0} c^{\lambda}-\Gamma_{\sigma \nu}^{\rho} c^{0}\right) \delta^{3} \\ & +\mathrm{i} \delta_{\nu}^{\rho}\left(2 \delta_{\sigma}^{0} c^{0} \tilde{f} \tilde{g}^{0 k}-\delta_{\sigma}^{0} c^{k}-\delta_{\sigma}^{k} c^{0}\right) \partial_{k}\left(\tilde{f} \delta^{3}\right) \\ {\left[\ddot{\bar{c}}_{\mu}, b_{\nu}^{\prime}\right]=} & -2 \mathrm{i} \tilde{f}\left[\partial_{\nu} \dot{\overline{\bar{c}}}_{\mu} \delta^{3}-\tilde{g}^{0 k} \partial_{\nu} \bar{c}_{\mu} \partial_{k}\left(\tilde{f} \delta^{3}\right)\right]. \end{aligned}$
After some calculations, we find that they are concretely given by $\begin{eqnarray}\begin{array}{rcl}{A}_{\mu \nu } & = & {\tilde{f}}^{2}[-{\tilde{g}}^{\rho \sigma }({\delta }_{\nu }^{0}{\partial }_{\sigma }{c}^{0}+{{\rm{\Gamma }}}_{\sigma \nu }^{0}{c}^{0})-{\tilde{g}}^{0\sigma }({\delta }_{\nu }^{\rho }{\partial }_{\sigma }{c}^{0}+{{\rm{\Gamma }}}_{\sigma \nu }^{\rho }{c}^{0})\\ & & +\,{\tilde{g}}^{\rho 0}({\partial }_{\nu }{c}^{0}+{{\rm{\Gamma }}}_{\nu \lambda }^{\lambda }{c}^{0})]{\partial }_{\rho }{\bar{c}}_{\mu }{\delta }^{3}\\ & & +\,\tilde{f}[-{\tilde{g}}^{\rho \sigma }{\delta }_{\nu }^{0}{\delta }_{\sigma }^{k}{c}^{0}+{\tilde{g}}^{0\sigma }{\delta }_{\nu }^{\rho }(2{\delta }_{\sigma }^{0}{c}^{0}\tilde{f}{\tilde{g}}^{0k}\\ & & -\,{\delta }_{\sigma }^{0}{c}^{k}-{\delta }_{\sigma }^{k}{c}^{0})\\ & & +\,{\tilde{g}}^{0\rho }{\delta }_{\nu }^{k}{c}^{0}]{\partial }_{\rho }{\bar{c}}_{\mu }{\partial }_{k}(\tilde{f}{\delta }^{3}),\\ {B}_{\mu \nu } & = & \tilde{f}({c}^{0}{\partial }_{\nu }{\dot{\bar{c}}}_{\mu }+\tilde{f}{\tilde{g}}^{0\rho }{\partial }_{\rho }{c}^{0}{\partial }_{\nu }{\bar{c}}_{\mu }+\tilde{f}{\tilde{g}}^{0k}{c}^{0}{\partial }_{k}{\partial }_{\nu }{\bar{c}}_{\mu }){\delta }^{3}\\ & & +\,\tilde{f}(-{\tilde{g}}^{0k}{c}^{0}+{\tilde{g}}^{00}{c}^{k}){\partial }_{\nu }{\bar{c}}_{\mu }{\partial }_{k}(\tilde{f}{\delta }^{3}),\\ {C}_{\mu \nu } & = & -{\tilde{f}}^{2}{c}^{0}({\partial }_{\nu }{\tilde{g}}^{0\rho }{\partial }_{\rho }{\bar{c}}_{\mu }+{\tilde{g}}^{00}{\partial }_{\nu }{\dot{\bar{c}}}_{\mu }+{\tilde{g}}^{0k}{\partial }_{k}{\partial }_{\nu }{\bar{c}}_{\mu }){\delta }^{3}\\ & & +\,\tilde{f}({\tilde{g}}^{\rho k}{\delta }_{\nu }^{0}-{\tilde{g}}^{0\rho }{\delta }_{\nu }^{k}){\partial }_{k}(\tilde{f}{c}^{0}{\delta }^{3}).\end{array}\end{eqnarray}$
Then, we see that Aμν + Bμν + Cμν = 0, thereby proving equation (