Noether's Theorem and Energy-Momentum Tensor
Energy-Momentum tensor is Neother current with translation.
We can understand this as generalized energy (or momentum).
See Wikipedia first.
2.4 Symmetries and Conservation Laws
2.4.1 Continuous Symmetry Transformations
Fomulation
Consider a collection of fields, which we collectively denote by $\Phi$. The action functional will depend in general on $\Phi$ and its first derivatives: $$\begin{align}S=\int d^dx\, \mathcal{L}(\Phi,\partial_\mu\Phi)\end{align}$$ Then action change or invariant by position and fields transformation: $$\begin{align}\begin{matrix}x &\rightarrow& x'\\ \Phi(x)&\rightarrow& \Phi'(x')\end{matrix}\end{align}$$ In these transformations the new filed $\Phi'$ at $x'$ is expressed as a function of the old field $\Phi$ at $x$: $$\begin{align}\Phi'(x')=\mathcal{F}(\Phi(x))\end{align}$$ The field $\Phi$, considered as a mapping from space-time to some target space $\mathcal{M}$ ($\Phi:\mathbb{R}^d\rightarrow \mathcal{M}$), is affected by the tranfromation (2) in two ways:
1. By the functional change $\Phi'(x')=\mathcal{F}(\Phi(x))$ "active"
2. By the attaching ' like $x\rightarrow x'$ (& $\Phi(x)\rightarrow \Phi'(x')$): "passive"
The change of the action functional under the field transformation in same point $\Phi(x)\rightarrow \Phi'(x)$ is obtained by substituting the new function $\Phi'(x)$ for the function $\Phi(x)$. In other words, the new action is $$\begin{align}S'=\int d^dx\, \mathcal{L}(\Phi'(x),\partial_\mu \Phi'(x))=\int d^dx'\, \mathcal{L}(\Phi'(x'),\partial'_\mu\Phi'(x'))\nonumber\\ =\int d^dx'\, \mathcal{L}(\mathcal{F}(\Phi(x)),\partial'_\mu\mathcal{F}(\Phi(x)))\nonumber\\ =\int d^dx\, \left| \frac{\partial x'}{\partial x}\right| \mathcal{L}(\mathcal{F}(\Phi(x)),(\partial x^\nu/\partial {x'}^\mu)\partial_\nu \mathcal{F}(\Phi(x)))\end{align}$$ In the first line, we have performed a change of integration variables $x\rightarrow x'$ according to the transformation (2), which allows us to express $\Phi'(x')$ in terms of $\Phi(x)$ in the second line. In the last line, we express $x'$ in terms of $x$.
Translation
Translation defined as $$\begin{align}\begin{matrix}x' &=& x+a\\ \Phi'(x')&=& \Phi(x)\end{matrix}\end{align}$$ Here $\partial x^\nu/\partial {x'}^\mu=\delta^\nu_\mu$ and $\mathcal{F}$ is trivial. It follows that $S'=S$. The action is invariant under translations, unless it depends explicitly on position.
Lorentz transformation
It takes the following form: $$\begin{align}\begin{matrix}{x'}^\mu&=&\Lambda^\mu_\nu x^\nu\\ \Phi'(x') &=& L_\Lambda \Phi(x)\end{matrix}\end{align}$$ where $\Lambda$ is a matrix satisfying $$\begin{align}\eta_{\mu\nu}\Lambda^\mu_\rho \Lambda^\nu_\sigma=\eta_{\rho\sigma}\end{align}$$ and where $L_\Lambda$ is another matrix, depending on $\Lambda$ and acting on $\Phi$ if the latter has more than one component. Matrices $\Lambda$ form the Lorentz group, $L_\Lambda$ forma representation of the Lorentz group.
Because of (7), the Jacobian $|\partial x'/\partial x|$ is unity and the transformed action is $$\begin{align}S'=\int d^dx\, \mathcal{L}(L_\Lambda\Phi,\Lambda^{-1}\cdot \partial(L_\Lambda \Phi))\end{align}$$ For a scalar field $\varphi$ the representation is trivial ($L_\Lambda = 1$) and the action is invariant under Lorentz transformations ($S' = S$) if the derivatives $\partial_\mu$ appear in a Lorentzinvariant way. The most general Lorentz-invariant Lagrangian containing at most two derivatives is then $$\begin{align}\mathcal{L}(\varphi,\partial_\mu\varphi)=f(\varphi)+g(\varphi)\partial_\mu \varphi \partial^\mu\varphi\end{align}$$ where $f$ and $g$ are arbitrary functions.
Scale transformations
They are defined as $$\begin{align}\begin{matrix}x' &=& \lambda x\\ \Phi'(\lambda x) &=& \lambda^{-\Delta}\Phi(x)\end{matrix}\end{align}$$ where $\lambda$ is the dilation factor and where $\Delta$ is the scaling dimension of the field $\Phi$. Since the Jacobian of this transformation is $|\partial x'/\partial x|=\lambda^d$, the transformed action is $$\begin{align}S'=\lambda^d\int d^dx\, \mathcal{L}(\lambda^{-\Delta}\Phi,\lambda^{-1-\Delta}\partial_\mu\Phi)\end{align}$$ We condier in particular the action of a massless scalar field $\varphi$ in space-time dimension $d$: $$\begin{align}S[\varphi]=\int d^dx\, \partial_\mu\varphi\partial^\mu\varphi\end{align}$$ We check that this action is scale invariant provided we make the choice $$\begin{align}\Delta=\frac{1}{2}d-1\end{align}$$
Internal Transformation
Finally, various transformations may be defined that affect only the field $\Phi$ and not the coordinates. The simplest example is that of a complex field with an action invariant under global phase transformations $\Phi'(x)=e^{i\theta}\Phi(x)$. A more complicated example is that of a multi-component field $\Phi$ transforming as $\Phi'(x)=R_\omega \Phi(x)$ where $R_\omega$ belongs to some representation of a Lie group parametrized by the group coordinate $\omega$.
2.4.2 Infinitesimal Transformations and Noether's Theorem
Transformation at Same Point: Generator
Infinitesimal transformation is generally written as (definition of derivative $\delta/\delta\omega_a$) $$\begin{align}\begin{matrix}{x'}^\mu&=&x^\mu+\omega_a\frac{\delta x^\mu}{\delta \omega_a}\\ \Phi'(x')&=&\Phi(x)+\omega_a\frac{\delta\mathcal{F}}{\delta \omega_a}(x)\end{matrix}\end{align}$$ We defined the generator $G_a$ of a symmetry transformation by the following expression for the infinitesimal transformation at a same point: $$\begin{align}\delta_\omega\Phi(x)\equiv \Phi'(x)-\Phi(x)\equiv -i\omega_a G_a\Phi(x)\end{align}$$ Put above infinitesimal transformation, $$\begin{align}\Phi'(x')=\Phi(x)+\omega_a\frac{\delta\mathcal{F}}{\delta \omega_a}(x)=\Phi(x')-\omega_a\frac{\delta x^\mu}{\delta \omega_a}\partial_\mu \Phi(x')+\omega_a\frac{\delta \mathcal{F}}{\delta \omega_a}(x')\end{align}$$ Then the explicit expression for the generator is therefore $$\begin{align}iG_a\Phi=\frac{\delta x^\mu}{\delta \omega_a}\partial_\mu\Phi-\frac{\delta\mathcal{F}}{\delta \omega_a}\end{align}$$
If we suppose for the moment that the fields are unaffected by the transformation (i.e., $\mathcal{F}(\Phi)=\Phi$).
1. Translation by a vector $\omega^\mu$, one has $$\delta x^\mu/\delta \omega^\nu=\delta^\mu_\nu$$ and $\delta\mathcal{F}/\delta \omega^\nu=0$. Then generator is $$\begin{align}P_\nu=-i\partial_\nu.\end{align}$$
2. Lorentz transformation $$\begin{align}{x'}^\mu=x^\mu+\omega^\mu_\nu x^\nu=x^\mu+\omega_{\rho\nu}\eta^{\rho\mu}x^\nu\end{align}$$ Substitution into the condition (7) yields the antisymmetry property $\omega_{\rho\nu}=-\omega_{\nu\rho}$. Using this antisymmetry, one may write the variation of the coordinate under an infinitesimal Lorentz transformation as $$\begin{align}\frac{\delta x^\mu}{\delta \omega_{\rho\nu}}=\frac{1}{2}(\eta^{\rho\mu}x^\nu-\eta^{\nu\mu}x^\rho)\end{align}$$ Its effect on the generic field $\Phi$ is $$\begin{align}\mathcal{F}(\Phi)=L_\Lambda \Phi\quad L_\Lambda\approx 1-\frac{1}{2}i\omega_{\rho\nu}S^{\rho\nu}\end{align}$$ where $S^{\rho\nu}$ is some Hermitian matrix obeying the Lorentz algebra (spin representation). From (18), one therefore writes $$\begin{align}\frac{1}{2}i\omega_{\rho\nu}L^{\rho\nu}\Phi=\frac{1}{2}\omega_{\rho\nu}(x^\nu\partial^\rho-x^\rho \partial^\nu)\Phi+\frac{1}{2}i\omega_{\rho\nu}S^{\rho\nu}\Phi\end{align}$$ where $L^{\rho\nu}$ is the generator. The factor of $\frac{1}{2}$ preceding $\omega_{\rho\nu}$ in the definitions of $L^{\rho\nu}$ and $S^{\rho\nu}$ compensates for the double counting of transformation parameters caused by the full contraction of indices. The generators of Lorentz transformations are thus $$\begin{align}L^{\rho\nu}=i(x^\rho\partial^\nu-x^\nu\partial^\rho)+S^{\rho\nu}\end{align}$$
(Non rigid) Transformation every point: Noether's Theorem
We now demonstrate Noether's theorem, which states that to every continuous symmetry of the action one may associate a current that is classically conserved. Given such a symmetry, the action is invariant under the transformation (14) only if the transformation is rigid, that is, if the parameters $\omega_a$ are independent of position. However, an especially elegant way to derive Noether's theorem is to suppose, as we will, that the infinitesimal transformation (14) is not rigid, with $\omega_a$ depending on the position. Or, just we can think transformation of every point to see whole space to get action.
From the last of Eqs. (4), we may write the effect on the action of the infinitesimal transformation (4). To first order, the Jacobian matrix is $$\begin{align}\frac{\partial {x'}^\nu}{\partial x^\mu}=\delta^\nu_\mu+\partial_\mu\left( \omega_a\frac{\delta x^\nu}{\delta \omega_a}\right)\end{align}$$ The determinant of this matrix may be calculated to first order form the formula $$\begin{align}\det (1+E)\approx 1+\mbox{Tr}E\quad (E \mbox{ small})\end{align}$$ We obtain $$\begin{align}\left|\frac{\partial x'}{\partial x}\right| \approx 1+\partial_\mu \left(\omega_a\frac{\delta x^\mu}{\delta \omega_a}\right)\end{align}$$ The inverse Jacobian matrix may be obtained to first order simply by reversing the sign of the transformation parameter: $$\begin{align}\frac{\partial x^\nu}{\partial {x'}^\mu}=\delta^\nu_\mu -\partial_\mu\left(\omega_a\frac{\delta x^\nu}{\delta \omega_a}\right)\end{align}$$ With the help of these preliminary steps, the transformed action $S'$ may be written as $$\begin{align}S'=\int d^dx\, \left(1+\partial_\mu\left( \omega_a\frac{\delta x^\mu}{\delta \omega_a}\right)\right) \times \mathcal{L}\left(\Phi+\omega_a\frac{\delta\mathcal{F}}{\delta\omega_a},\left[\delta^\nu_\mu-\partial_\mu(\omega_a(\delta x^\nu/\delta\omega_a))\right]\left( \partial_\nu\Phi+\partial_\nu\left[\omega_a(\delta\mathcal{F}/\delta\omega_a)\right]\right)\right)\end{align}$$ The variation $\delta S=S'-S$ of the action contains terms with no derivatives of $\omega_a$. These sum up to zero if the action is symmetric under rigid transformations. Then $\delta S$ involves only the first derivatives of $\omega_a$, obtained by expanding the Lagrangian. We write $$\begin{align}\delta S=-\int dx\, j^\mu_a\partial_\mu\omega_a\end{align}$$ where $$\begin{align}j^\mu_a=\left\{ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\partial_\nu\Phi-\delta^\mu_\nu\mathcal{L}\right\}\frac{\delta x^\nu}{\delta \omega_a}-\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\frac{\delta\mathcal{F}}{\delta\omega_a}\end{align}$$ The quantity $j^\mu_a$ is called the current associated with the infinitesimal transformation (14). Integration by parts yields $$\begin{align}\delta S=\int d^dx\, \partial_\mu j^\mu_a\, \omega_a\end{align}$$ Now comes Noether's theorem: if the field configuration obeys the classical equations of motion, the action is stationary against any variation of the fields. In other words, $\delta S$ should vanish for any position-dependent parameters $\omega_a(x)$. This implies the conservation law $$\begin{align}\partial_\mu j^\mu_a=0\end{align}$$ In words, every continuous symmetry implies the existence of a current given by (30), which is classically conserved.
The conserved charge associated with $j^\mu_a$ is $$\begin{align}Q_a=\int d^{d-1}x\, j^0_a\end{align}$$ where $j^0_a$ is the time component of $j^\mu_a$, and $d^{d-1}x$ stands for the purely spatial integration measure. Its time derivative indeed vanishes: $$\begin{align}\dot{Q}_a=\int d^{d-1}x\, \partial_0 j^0_a =-\int d^{d-1}x\, \partial_ij^i=-\int_\infty j^id\sigma^i\end{align}$$ where $d\sigma^i$ is a surface element at spatial infinity. Therefore $\dot{Q}_a=0$, provided the current $j^i$ vanishes sufficiently rapidly as $x\rightarrow \infty$.
The expression (20) for the conserved current is termed "canonical", implying that there are other admissible expressions. In fact we may freely add to it the divergence of an antisymmetric tensor without affecting its conservation: $$\begin{align}j^\mu_a\rightarrow j^\mu_a+\partial_\nu B^{\nu\mu}_a,\quad B^{\nu\mu}_a=-B^{\mu\nu}_a\end{align}$$ Indeed, $\partial_\mu\partial_\nu B^{\nu\mu}_a=0$ by antisymmetry. The definition of $j^\mu_a$ is therefore ambiguous to some extent.
2.4.3 Transformation of the Correlation Functions
Consider a theory involving a collection of fields $\Phi$ with an action $S[\Phi]$ invariant under a transformation of the type (2). Consider then the general correlation function $$\begin{align}\langle \Phi(x_1)\cdots\Phi(x_n)\rangle=\frac{1}{Z}\int [d\Phi]\Phi(x_1)\cdots\Phi(x_n)\exp -S[\Phi]\end{align}$$ where $Z$ is the vacuum functional. The consequence of the symmetry of the action and of the invariance of the functional integration measure under the transformation (2) is the following identity: $$\begin{align}\langle \Phi(x'_1)\cdots\Phi(x'_n)\rangle=\langle \mathcal{F}(\Phi(x_1))\cdots\mathcal{F}(\Phi(x_n))\rangle\end{align}$$ where the mapping $\mathcal{F}$ describes the functional change of the field under the transformation, as in Eq. (3). The demonstration of the identity is straightforward: $$\begin{align}\langle \Phi(x'_1)\cdots\Phi(x'_n)\rangle=\frac{1}{Z}\int [d\Phi]\Phi(x'_1)\cdots\Phi(x'_n)\exp -S[\Phi]\nonumber\\ \frac{1}{Z}\int [d\Phi']\Phi'(x'_1)\cdots\Phi'(x'_n)\exp -S[\Phi']\nonumber\\ =\frac{1}{Z}\int [d\Phi]\mathcal{F}(\Phi(x_1))\cdots\mathcal{F}(\Phi(x_n))\exp -S[\Phi]\nonumber\\ =\langle \mathcal{F}(\Phi(x_1))\cdots\mathcal{F}(\Phi(x_n))\rangle\end{align}$$ An explanation is in order. In going from the first to the second line of Eq. (38) we have just renamed the dummy integration variable $\Phi\rightarrow \Phi'$, without performing a real change of integration variables. In going from the second to the third line we have performed a change of functional integration variables, in which $\Phi'(x')$ is expressed in terms of $\Phi(x)$.
Translation $x'=x+a$, $$\begin{align}\langle \Phi(x_1+a)\cdots\Phi(x_n+a)\rangle=\langle \Phi(x_1)\cdots\Phi(x_n)\rangle\end{align}$$
Lorentz invariace $$\begin{align}\langle\Phi(\Lambda^\mu_\nu x^\nu_1)\cdots\Phi(\Lambda^\mu_\nu x^\nu_n)\rangle=\langle\Phi(x_1^\mu)\cdots\Phi(x^\mu_n)\rangle\end{align}$$
Scaling dimensions $$\begin{align}\langle\phi_1(\lambda x_1)\cdots\phi_n(\lambda x_n)\rangle=\lambda^{-\Delta_1}\cdots\lambda^{-\Lambda_n}\langle\phi_1(x_1)\cdots\phi_n(x_n)\rangle\end{align}$$
2.4.4 Ward Identities
The consequence of a symmetry of the action and the measure on correlation functions may also be expressed via the so-called Ward identities. An infinitesimal transformation may be written in terms of the generators as $$\begin{align}\Phi'(x)=\Phi(x)-i\omega_a G_a \Phi(x)\end{align}$$ where $\omega_a$ is a collection of infinitesimal, constant parameters. We put this change in (36). Action changed by (21). Denoting by $X$ the collection $\Phi(x_1),\cdots\Phi(x_n)$ of fields. (Think as $\Phi^n$.) $$\begin{align}\langle X\rangle=\frac{1}{Z}\int [d\Phi'](X+\delta X)\exp -\left\{ S[\Phi]+\int dx\, (\partial_\mu j^\mu_a)\omega_a(x)\right\}\end{align}$$ We again assume that the functional integration measure is invariant under the local transformation (i.e., $[d\Phi']=[d\Phi]$). When expand $\exp$ to first order in $\omega_a(x)$, the above yields ($\partial_\mu$ is derivative for $x$, so it can get out.) $$\begin{align}\langle\delta X\rangle=\int dx \, \partial_\mu\langle j^\mu_a(x)X\rangle\omega_a(x)\end{align}$$ The variation $\delta X$ is explicitly given by $$\begin{align}\delta X=-i\sum^n_{i=1}\left(\Phi(x_1)\cdots G_a\Phi(x_i)\cdots \Phi(x_n)\right)\omega_a(x_i)\nonumber\\ =-i\int dx\, \omega_a(x)\sum^n_{i=1}\left\{ \Phi(x_1)\cdots G_a\Phi(x_i)\cdots\Phi(x_n)\right\} \delta (x-x_i)\end{align}$$ Since (44) holds for any infinitesimal function $\omega_a(x)$, we may write the following local relation: $$\begin{align}\frac{\partial}{\partial x^\mu}\langle j^\mu_a(x)\Phi(x_1)\cdots\Phi(x_n)\rangle=-i\sum^n_{i=1}\delta(x-x_i)\langle \Phi(x_1)\cdots G_a\Phi(x_i)\cdots\Phi(x_n)\rangle\end{align}$$ This is the Ward identity for the current $j^\mu_a$. ($x$ is connect to each $x_i$, it seems like 'ward'.) Note that the form of the current may be modified from the canonical definition (30) without affecting the Ward identity, if one adds to $j^\mu_a$ a quantity that is divergenceless identically (i.e., without using the equations of motion), such as in Eq. (35).
Integrate $x$ whole space: Correlator invariance
We integrate the Ward identity (46) over a region of space-time that includes all the points $x_i$. On the left-hand side (l.h.s.), we obtain a surface integral $$\begin{align}\int_\Sigma ds_\mu\langle j^\mu_a(x)\Phi(x_1)\cdots\Phi(x_n)\rangle\end{align}$$ which since the hypersurface $\Sigma$ may be sent to infinity without affecting the integral: indeed, the divergence a $\partial_\mu\langle j^\mu_aX\rangle$ vanishes away from the points $x_i$ and the correlator $\langle j^\mu_a(x)X\rangle$ goes to zero sufficiently fast as $x\rightarrow \infty$, by hypothesis. For the right-hand side (r.h.s.) of Eq. (46), this implies $$\begin{align}\delta_\omega\langle\Phi(x_1)\cdots\Phi(x_n)\rangle\equiv -i\omega_a\sum^n_{i=1}\langle \Phi(x_1)G_a\Phi(x_i)\cdots\Phi(x_n)\rangle=0\end{align}$$ In other words, the variation of the correlator under an infinitesimal transformation vanishes. This is simply the infinitesimal version of Eq. (38) (see also the definition (15)).
Integrate $x$ time pill box & spatial infinity: Charge is generator
The Ward identity allows us to identify the conserved charge $$\begin{align}Q_a=\int d^{d-1}x\, j^0_a(x)\end{align}$$ as the generator of the symmetry transformation in the Hilbert space of quantum states. Let $Y=\Phi(x_2)\cdots \Phi(x_n)$ and suppose that the time $t=x^0_1$ is differnet form all the times in $Y$. We integrate the Ward identity (46) in a very thin "pill box" bounded by $t_-<t$, by $t_+>t$, and by spatial infinity, which excludes all the other points $x_2,\cdots,x_n$. The integral of the l.h.s. of (26) is converted into a surface integral and yields $$\begin{align}\langle Q_a(t_+)\Phi(x_1)Y\rangle-\langle Q_a(t_-)\Phi(x_1)Y\rangle=-i\langle G_a\Phi(x_1)Y\rangle\end{align}$$ Remembering that a correlation function is the vacuum expectation value of a time-ordered product in the operator formalism, and assuming, for the sake of argument, that all other times $x^0_i$ are greater than $t$, we write, in the limit $t_-\rightarrow t_+$, $$\begin{align}\langle0|[Q_a,\Phi(x_1)]Y|0\rangle=-i\langle0|G_a\Phi(x_1)Y|0\rangle\end{align}$$ This being true for an arbitrary set of fields $Y$, we conclude that $$\begin{align}[Q_a,\Phi]=-iG_a\Phi\end{align}$$ In other words, the conserved charge $Q_a$ is the generator of the infinitesimal symmetry transformations in the operator formalism. Of course, these identities are obtained in the Euclidian formalism. An easy way to go back to Minkowski spacetime is to replace the charge $Q$ by $-iQ$, since it is the outcome of an integration of the time-like component of a vector.
2.5 The Energy-Momentum Tensor
The conserved current associated with translation invariance is the energy-momentum tensor, whose components are the density and flux density of energy and momentum. The infinitesimal translation ${x'}^\mu\rightarrow x^\mu+\epsilon^\mu$ induces the following variations in the coordinates and the fields (see Eq. (14)): $$\begin{align}\frac{\delta x^\mu}{\delta \epsilon^\nu}=\delta^\mu_\nu,\quad \frac{\delta\Phi}{\delta \epsilon^\nu}=0\end{align}$$ Consequently the corresponding canonical conserved current is $$\begin{align}T^{\mu\nu}_c=-\eta^{\mu\nu}\mathcal{L}+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\partial^\nu\Phi\end{align}$$ and the conservation law is $\partial_\mu T^{\mu\nu}_c=0$. The conserved charge is the four-momentum $$\begin{align}P^\nu=\int d^{d-1}x\, T^{0\nu}_c\end{align}$$ In particular, the energy is $$\begin{align}P^0=\int d^{d-1}x\left\{\frac{\partial\mathcal{L}}{\partial\dot{\Phi}}\dot{\Phi}-\mathcal{L}\right\}\end{align}$$ which is the usual definition of the Hamiltonian. As an operator, the conserved charge $P_\mu$. has therefore the following effect in Euclidian time, according to Eq. (52): $$\begin{align}[P_\mu,\Phi]=-\partial_\mu\Phi\end{align}$$ In real time, this relation becomes $[P_\mu,\Phi]=-i\partial_\mu\Phi$, which is the well-known commutator of an $x$-dependent operator with momentum in ordinary quantum mechanics.
2.5.1 The Belinfante Tensor
In general, the canonical energy-momentum tensor $T^{\mu\nu}_c$ is not symmetric. However, we have the freedom to modify this tensor by adding the divergence of a tensor $B^{\rho\mu\nu}$ antisymmtric in the first two indices: $$\begin{align}T^{\mu\nu}_B=T^{\mu\nu}_c+\partial_\rho B^{\rho\mu\nu},\quad B^{\rho\mu\nu}=-B^{\mu\rho\nu}\end{align}$$ This addition does not affect the classical conservation law nor the Ward identity. Indeed, the variation of the action under a nonuniform translation with position-dependent parameter $\epsilon^\mu(x)$ is still given by $$\begin{align}\delta S=-\int d^dx\, \partial_\mu T^{\mu\nu}_B\epsilon_\nu\end{align}$$ since $\partial_\mu T^{\mu\nu}_B=\partial_\mu T^{\mu\nu}_c$ identically. If we succeed in finding $B^{\rho\mu\nu}$ such that the new tensor $T^{\mu\nu}_B$ is symmetric, then the latter is called the Belinfante energy-momentum tensor. In order to accomplish this, we consider the conserved currents associated with Lorentz transformations. (Because Lorentz transformation Noether current shows antisymmetric part of energy momentum tensor.)
Lorentz transformation: Antisymmetric part of EM tensor
Form (20) and (21), the variantions of the coordinates and fields under an infinitesimal Lorentz gransformation are $$\begin{align}\frac{\delta x^\rho}{\delta \omega_{\mu\nu}}=\frac{1}{2}(\eta^{\rho\mu}x^\nu-\eta^{\rho\nu}x^\mu),\quad \frac{\delta \mathcal{F}}{\delta \omega_{\mu\nu}}=-i\frac{1}{2}S^{\mu\nu}\Phi\end{align}$$ and the associated canonical conserved current is $$\begin{align}j^{\mu\nu\rho}=T^{\mu\nu}_cx^\rho-T^{\mu\rho}_cx^\nu+\frac{1}{2}i\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}S^{\nu\rho}\Phi\end{align}$$ We look for $B^{\rho\mu\nu}$ such that this current may be expressed as $$\begin{align}j^{\mu\nu\rho}=T^{\mu\nu}_Bx^\rho-T^{\mu\rho}_Bx^\nu\end{align}$$ This relation ensures that $T^{\mu\nu}_B=T^{\nu\mu}_B$, as is easily seen by applying the conservation laws $\partial_\mu j^{\mu\nu\rho}=0$ and $\partial_\mu T^{\mu\nu}_B=0$. However, this impleis only that $T^{\mu\nu}_B$ is symmetric classically (i.e., for field configurations obeying the equations of motions).
Belinfante tensor
An explicit expression for $B^{\rho\mu\nu}$ can be found by inspection: $$\begin{align}B^{\mu\rho\nu}=\frac{1}{4}i\left\{ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}S^{\nu\rho}\Phi+\frac{\partial\mathcal{L}}{\partial(\partial_\rho\Phi)}S^{\mu\nu}\Phi+\frac{\partial\mathcal{L}}{\partial(\partial_\nu\Phi)}S^{\mu\rho}\Phi\right\}\end{align}$$ We chekc that this expression is indeed antisymmteric in the first two indices, since $S^{\mu\nu}=-S^{\nu\mu}$. In order to show that the above has the right form, we calculate its antisymmetric part in $(\rho\nu)$: $$\begin{align}B^{\mu\rho\nu}-B^{\mu\nu\rho}=\frac{1}{2}i\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}S^{\nu\rho}\Phi\end{align}$$ On the other hand the antisymmtric part of $T^{\rho\nu}_c$ in a classical configuration is obtained by aplying the conservation laws to Eq. (61): $$\begin{align}T^{\rho\nu}_c-T^{\nu\rho}_c=-\frac{1}{2}i\partial_\mu\left\{ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}S^{\nu\rho}\Phi\right\}\end{align}$$ We see that the antisymmtric part of $T^{\rho\nu}_c+\partial_\mu B^{\mu\rho\nu}$ vanishes, that is, $T^{\mu\nu}_B$ is indeed symmtreic in a classical confihuration. Note that the form given in Eq. (63) for $B^{\mu\rho\nu}$ is not unique; further modifications of the energy-momentum tensor are possible.
Example: Electromagnetism
We can illustrate this with an example. Consider the following Lagrangian for a massive vector field $A_\mu$ (in Euclidian space-time): $$\begin{align}\mathcal{L}=\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}+\frac{1}{2}m^2A^\alpha A_\alpha\end{align}$$ wherein $F_{\alpha\beta}=\partial_\alpha A_\beta-\partial_\beta A_\alpha$. The canonical energy-momentum tensor is $$\begin{align}T^{\mu\nu}_c=F^{\mu\alpha}\partial^\nu A_\alpha-\eta^{\mu\nu}\mathcal{L}\end{align}$$ and it is not symmetric. We now calculate $T^{\mu\nu}_B$ as defined in Eq. (58), with $$\begin{align}B^{\alpha\mu\nu}=F^{\alpha\mu}A^\nu\end{align}$$ We end up with $$\begin{align}T^{\mu\nu}_B=T^{\mu\nu}_c+F^{\alpha\mu}\partial_\alpha A^\nu+\partial_\alpha F^{\alpha\mu}A^\nu\end{align}$$ This tensor is classically symmetric, as may be seen from the following: we define the identically symmetric tensor $$\begin{align}\tilde{T}^{\mu\nu}_B=F^{\mu\alpha}F^\nu_\alpha-\frac{1}{4}\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}+m^2\left[ A^\mu A^\nu-\frac{1}{2}\eta^{\mu\nu}A^\alpha A_\alpha\right]\nonumber\\ =T^{\mu\nu}_B-(\partial_\alpha F^{\alpha\mu}-m^2A^\mu)A^\nu\end{align}$$ The two tensors $\tilde{T}^{\mu\nu}_B$ and $T^{\mu\nu}_B$ coincide for classical configurations, since the equations of motion are $$\begin{align}\partial_\alpha F^{\alpha\mu}-m^2A^\mu=0\end{align}$$ It is $\tilde{T}^{\mu\nu}_B$ which is written down in standard texts, whereas it is $T^{\mu\nu}_B$ which a priori appears in the Ward identity: $$\begin{align}\partial_\mu\langle T^{\mu\nu}_BX\rangle=-\sum_j\delta(x-x_j)\langle\Phi(x_1)\cdots\partial^\nu\Phi(x_j)\cdots\Phi(x_n)\rangle\end{align}$$ If we wish to use a symmetric tensor in the Ward identity, we must replace $T^{\mu\nu}_B$ by $\tilde{T}^{\mu\nu}_B$ therein, but this modifies the Ward identity. However, as we shall see presently, the modification to the Ward identity coming from this substitution has no effect and may be ignored in general.
Adding EOM no effect to correlators
Indeed, the Ward identity in terms of $\tilde{T}^{\mu\nu}_B$ is $$\begin{align}\partial_\mu\langle\tilde{T}^{\mu\nu}_BX\rangle=-\sum_j\delta(x-x_j)\langle\Phi(x_1)\cdots\partial^\nu\Phi(x_j)\cdots\Phi(x_n)\rangle-\partial_\mu\langle[\partial_\alpha F^{\alpha\mu}(x)-m^2A^\mu(x)]A^\nu(x)X\rangle\end{align}$$ We wish to show that the last term is of no consequence. For this we need to use the following relation. written here in Euclidian time. which is a consequence of the equations of motion on correlation functions (see Ex. 2.2): $$\begin{align}\left\langle\frac{\delta Y}{\delta \Phi(x)}\right\rangle=\left\langle Y\frac{\delta S}{\delta \Phi(x)}\right\rangle\end{align}$$ Here, $Y$ is a product of local fields. We apply this relation to our system, with $Y=A_\nu(y)X$ ($X$ is again a product of local fields) and $\Phi(x)\rightarrow A_\mu(x)$. We find $$\begin{align}\frac{\delta S}{\delta A_\mu(x)}=-\partial_\alpha F^{\alpha\mu}(x)+m^2 A^\mu(x)\end{align}$$ Therefore, $$\begin{align}\left\langle\frac{\delta X}{\delta A_\mu(x)}A_\nu(y)\right\rangle+\delta(x-y)\delta_{\mu\nu}\langle X\rangle=\langle(-\partial_\alpha F^{\alpha\mu}(x)+m^2A^\mu(x))A^\nu(y)X\rangle\end{align}$$ We take the limit $x\rightarrow y$ and ignore the delta function $\delta(x-y)$, which is automatically subtracted if normal order is used for the product $[\partial_\alpha F^{\alpha\mu}-m^2A^\mu]A^\nu$. We find $$\begin{align}\langle[(-\partial_\alpha F^{\alpha\mu}+m^2A^\mu)A^\nu]|_xX\rangle=\left\langle\frac{\delta X}{\delta A_\mu(x)}A_\nu(y)\right\rangle_{x\rightarrow y}\end{align}$$ This last expression will vanish for all $x$ except at the isolated points $x_i$ the positions of the fields appearing in the product $X$. For instance, if $X=A_\rho(x_1)A_\sigma(x_2)$, then $$\begin{align}\left\langle\frac{\delta X}{\delta A_\mu(x)}A_\nu(y)\right\rangle_{x\rightarrow y}=\delta^\mu_\rho\delta (x-x_1)\langle A_\sigma(x_2)A_\nu(x)\rangle+\delta^\mu_\sigma\delta(x-x_2)\langle A_\rho(x_1)A_\nu(x)\rangle\end{align}$$ In general, the additional contribution to the Ward identity will have the following form: $$\begin{align}\partial_\mu\langle [\partial_\alpha F^{\alpha\mu}(x)-m^2A^\mu(x)]A^\nu(x)X\rangle=\partial_\mu\sum_j \delta(x-x_j)f^\mu_j(x_1,\cdots,x_n)\end{align}$$ The reason such an addition is of no consequence is that the Ward identity, like any other expression involving delta functions, has a precise meaning only after integration through some arbitrary volume. The added term is a total divergence containing delta functions and can thus be converted into a surface integral, which receives no contribution from the delta functions.
In summary, provided the theory has rotation symmetry, we may define a new energy-momentum tensor $T^{\mu\nu}_B$, which is conserved, classically symmetric, and plays the same role in Ward identities as $T^{\mu\nu}_c$ In fact, one may use the equations of motion to bring $T^{\mu\nu}_B$ into another form (noted $\tilde{T}^{\mu\nu}_B$ above) which is now identically symmetric, still conserved, and still plays the same role as $T^{\mu\nu}_c$ in the Ward identity, except for terms that may be ignored. Consequently, we shall no longer distinguish between $T^{\mu\nu}_B$ and $\tilde{T}^{\mu\nu}_B$ (as far as Ward identities are concerned) in the remainder of this work.
2.5.2 Alternate Definition of the Energy-Momentum Tensor
We now consider a general infinitesimal transformation of the coordinates $x^\mu\rightarrow {x'}^\mu=x^\mu+\epsilon^\mu(x)$. This can be considered as a translation with an $x$-dependent parameter $\epsilon^\mu(x)$. According to (31) the induced change in the action is $$\begin{align}\delta S=\int d^dx\, T^{\mu\nu}\partial_\mu\epsilon_\nu=\frac{1}{2}\int d^dx\, T^{\mu\nu}(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)\end{align}$$ where we have assumed that $T^{\mu\nu}$ is identically symmetric. If the diffeomorphism $x'=x+\epsilon$ is considered as an infinitesimal change of coordinates, the corresponding change in the metric tensor $g_{\mu\nu}$ is (to first order in $\epsilon$ $$\begin{align}g'_{\mu\nu}=\frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}=(\delta^\alpha_\mu-\partial_\mu\epsilon^\alpha)(\delta^\beta_\nu-\partial_\nu\epsilon^\beta)g_{\alpha\beta}=g_{\mu\nu}-(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)\end{align}$$ This prompts for an alternate definition of the energy-momentum tensor, as the functional derivative of the action with respect to the metric, evaluated in flat space: $$\begin{align}\delta S=-\frac{1}{2}\int d^dx\, T^{\mu\nu}\delta g_{\mu\nu}\end{align}$$
For instance, on a general manifold, the action for a free scalar field $\varphi$ is $$\begin{align}S=\int d^dx\, \sqrt{g}\mathcal{L}=\frac{1}{2}\int d^dx\, \sqrt{g}\left\{ g^{\mu\nu}\partial_\mu\varphi\partial_\nu\varphi+m^2\varphi^2\right\}\end{align}$$ where $g\equiv \det g_{\mu\nu}$ and the factor $\sqrt{g}$ is required for the invariance of the space-time integration measure. Using the identities $$\begin{align}\det A=e^{\mbox{Tr}\ln A}\quad \mbox{and}\quad \delta g^{\mu\nu}=-g^{\alpha\mu}g^{\beta\nu}\delta g_{\alpha\beta}\end{align}$$ we find $$\begin{align}\delta\sqrt{g}=\frac{1}{2}\sqrt{g}g^{\mu\nu}\delta g_{\mu\nu}\end{align}$$ and the definition (82) yields $$\begin{align}T^{\mu\nu}=-g^{\mu\nu}\mathcal{L}+\partial^\mu\varphi\partial^\nu\varphi\end{align}$$which coincides with the canonical definition (54). The advantage of the new definition (82) is that the energy-momentum tensor is identically symmetric. However, obtaining an explicit expression for $T^{\mu\nu}$ from (82) requires more involved calculations than going through the canonical definition, or its Belinfante generalization.
If a tetrad (Apendix 2.C) $e^a_\mu$ is used instead of a metric then the energy-momentum tensor is endowed with a Lorentz index and an Einstein index: Since $g_{\mu\nu}=e^a_\mu e^a_\nu$, we easily find that $$\begin{align}\delta S=-\int d^dx\, e\, T^\mu_a\delta e^a_\mu\end{align}$$ where $e=\det e^a_\mu$.
In the quantum theory, the alternate definition (82) of the energy-momentum
tensor takes the following meaning. Let $\Phi$ represent the set of dynamical fields of
the theory, and $g$ the metric. On a general manifold the action is a functional $S[\Phi,g]$ of both quantities. The vacuum functional $Z[g]$ and the functional integration
measure $[d\Phi
Appendix 2.C. Tetrads
At each point of a manifold, coordinate differentials $dx^\mu$ span a local vector space (the cotangent space). Under a change of coordinate system $x\rightarrow x'$, the differentials transform as follows: $$\begin{align}d{x'}^\mu=\frac{\partial {x'}^\mu}{\partial x^\nu}dx^\nu\end{align}$$ The only requirement imposed on the Jacobian matrix $\partial {x'}^\mu/\partial x^\nu$ is invertibility: It should be an element of the group $GL(d)$ of invertible $d$-dimensional matrices. For general coordinate system, we introduce at each point a local orthogonal frame of basis vectors for the cotangent space: $$\begin{align}e^a=e^a_\mu dx^\mu\quad a=1,\cdots,d\end{align}$$ where the frame vectors $e^a$ form a tetrad, or vierbein. A natural choice for the tetrad is detennined by the conditions $$\begin{align}e^a_\mu e^b_\nu g^{\mu\nu}=\eta^{ab}\quad g_{\mu\nu}=\eta_{ab}e^a_\mu e^b_\nu\end{align}$$ which express the orthogonality of the tetrad. The lower (Greek) index of $e^a_\mu$ is called an Einstein index, while the upper (Latin) index is called a Lorentz index.
In order to compare vectors belonging to different (but nearby) cotangent spaces, we need to introduce a prescription for parallel transport, specified by the so-called spin connection $\omega^{ab}_\mu$: $$\begin{align}V^a\rightarrow V^a-\omega^{ab}_\mu dx^\mu V^b\end{align}$$ where $dx^\mu$ is the amount of transport. The covariant derivative is defined as $$\begin{align}(D_\mu V)^a=\partial_\mu V^a+\omega^{ab}_\mu V^b\end{align}$$ and results from the comparison of a vector at $x$ with a vector parallel-transported from $x +dx$. Since parallel transport changes only the direction of a vector and not its length, the spin-connection is antisymmetric in its Lorentz indices: $\omega^{ab}_\mu=-\omega^{ba}_\mu$.
The tetrad $e^a_\mu$ may be used to convert between Lorentz and Einstein indices: $V^a=e^a_\mu V^\mu$. The Christoffel symbols $\Gamma^\mu_{\nu\lambda}$ are used to specify the parallel transport in a tetrad-free language: $$\begin{align}\Gamma^\mu_{\nu\lambda}e^a_\mu=\partial_\lambda e^a_\nu+\omega^{ab}_\lambda e^b_\nu\end{align}$$
The curvature of a manifold manifests itself when a vector is parallel-transported around a closed path. Around an infinitesimal "square" loop of sides $dx$ and $dy$, the difference between the initial vector va and the transported vector ${V'}^a$ is $$\begin{align}{V'}^a-V^a=-[D_\mu,D_\nu]^{ab}V^b dx^\mu dy^\nu= R^{ab}_{\mu\nu}V^b dx^\mu dy^\nu\end{align}$$ More explicitly, the curvature tensor $R^{ab}_{\mu\nu}$ is $$\begin{align}R^{ab}_{\mu\nu}=\partial_\mu \omega^{ab}_\nu -\partial_\nu \omega^{ab}_\mu+\omega^{ac}_\mu\omega^{cb}_\nu-\omega^{ac}_\nu\omega^{cb}_\mu\end{align}$$ This tensor is related to the usual Riemann tensor $R^{\rho\sigma}_{\mu\nu}$ by contraction with $e^a_\rho e^b_\sigma$.
The connection is determined by the metric $g_{\mu\nu}$ together with the torsion-free condition $e^a_\rho e^b_\sigma$. The latter condition is natural if we define the manifold as embedded in a higher-dimensional Euclidian space, as a hypersurface $X(x)$. Then, the metric is given by $$\begin{align}g_{\mu\nu}=\partial_\mu X\cdot \partial_\nu X\end{align}$$ and the Christoffel symbols are easily derived to be $$\begin{align}\Gamma^\mu_{\nu\lambda}=\partial^\mu X\cdot \partial_\nu\partial_\lambda X=\frac{1}{2}g^{\mu\rho}(\partial_\nu g_{\rho\lambda}+\partial_\lambda g_{\rho\nu}-\partial_\rho g_{\nu\lambda})\end{align}$$
On a two-dimensional manifold, the spin-connection can be expressed in terms of a single-covariant vector $\omega_\mu$: $$\begin{align}\omega^{ab}_\mu=\epsilon^{ab}\omega_\mu\end{align}$$ while the curvature tensor is $$\begin{align}R^{ab}_{\mu\nu}=\epsilon^{ab}(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu)=\sqrt{g}\epsilon^{ab}\epsilon_{\mu\nu}R\end{align}$$ where $R$ is the scalar curvature.
Reference
Francesco - Conformal Field Theory 2.5