I have been reading Introduction to Conformal Field Theory by Blumenhagen and Plauschinn. Equation (2.19) on page 19 states that if our theory is invariant under a general conformal transformation $x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu(x)$, then the conserved current is given by

$$ j^\mu = \epsilon^\nu T^\mu_{\ \nu}\tag{2.19}.$$

I have been trying to show this myself but I can't. The general procedure of Noether's theorem is to evaluate the off-shell variation and see whether it is given by the integral of a total derivative. We then equate this with the on-shell variation to yield the conserved current.

My attempt

Let the action be of the form

$$ S[\phi] = \int \mathrm{d}^3 x \mathcal{L}(\phi,\partial \phi).$$

We know that if the theory is translationally invariant under $x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu$, where $\epsilon^\mu$ here is a constant, the corresponding conserved currents are given by the energy-momentum tensor $T^\mu_{\ \nu}$ which takes the form

$$ T^\mu_{\ \nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_{\nu} \mathcal{L}, $$

where $\partial_\mu T^\mu_{\ \nu} = 0$. Now I promote $\epsilon \rightarrow \epsilon(x)$. First, I compute the on-shell variation:

$$ \delta S_\text{on-shell} = \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu(\delta \phi) \\ =\int \mathrm{d}^3 x \bigg( \frac{\partial \mathcal{L}}{\partial \phi}- \partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \bigg)\delta \phi + \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi \bigg) \\ = \int \mathrm{d}^3 x \partial_\mu \bigg(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi \bigg).$$

The translation when viewed as an active transformation on the fields means $\delta \phi = - \epsilon^\mu(x) \partial_\mu \phi(x)$. Plugging this into the on-shell variation yields

$$ \delta S_\text{on-shell}= - \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} (\partial_\mu \epsilon^\nu)( \partial_\nu \phi) + \epsilon^\nu \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi \bigg) . $$

The first term in the integrand would vanish if $\epsilon^\mu$ was a constant. Now we evaluate the off-shell variation:

$$ \delta S_\text{off-shell} = \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu(\delta \phi) \\ = - \int \mathrm{d}^3 x\frac{\partial \mathcal{L}}{\partial \phi} \epsilon^\nu \partial_\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} (\partial_\mu \epsilon^\nu \partial_\nu \phi + \epsilon^\nu \partial_\nu \partial_\mu \phi) \\ = - \int \mathrm{d}^3 x \epsilon^\nu \bigg( \frac{\partial \mathcal{L}}{\partial \phi} \partial_\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\nu \partial_\mu \phi \bigg) + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu \epsilon^\nu \partial_\nu \phi \\ = -\int \mathrm{d}^3 x \epsilon^\nu \partial_\nu \mathcal{L} + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu \epsilon^\nu \partial_\nu \phi. $$

Equating the on and off-shell variations, we find the terms with $\partial_\mu \epsilon^\nu$ cancel, yielding

$$\int \mathcal{d}^3 x \epsilon^\nu \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_{\nu} \mathcal{L} \bigg) = 0, $$


$$ \int \mathcal{d}^3 x \epsilon^\nu \partial_\mu T^\mu_{\ \nu} = 0. $$

The integrand is not of the form $\partial_\mu j^\mu$ where $j^\mu = \epsilon^\nu T^\mu_{\ \nu}$. In fact, we already know that $\partial_\mu T^\mu_{\ \nu} = 0$ from our assumption that the theory was translationally invariant already so this statement is trivially true. I am unable to pass the $\epsilon^\nu$ into the derivative too because it depends on the coordinates.

What am I missing? Have I made a mistake in my analysis?


2 Answers 2


In these types of problems, it is very useful to work on a generic spacetime with metric $g$. For instance, we can promote the usual free-field Lagrangian


to one on a general curved spacetime as


The general recipe here is to replace the volume form with $\mathrm{d}^dx\,\sqrt{g}$ and replace all derivatives with covariant derivatives.

Once this is done, the energy-momentum tensor is simply defined as

$$T_{\mu\nu}=\frac{1}{\sqrt{g}}\frac{\delta\mathcal{L}}{\delta g^{\mu\nu}}$$

(up to a possible uninteresting factor of 2).

For any coordinate (infinitesimal) transformation $x\to x+\epsilon(x)$, the variation of the metric tensor takes the form

$$\delta g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}.$$

A vector is called a Killing vector if this variation vanishes. Plugging this in to the definition of $T_{\mu\nu}$, this tells us that the variation of $\mathcal{L}$ under this coordinate transformation is

$$\delta\mathcal{L}=\sqrt{g}\,T_{\mu\nu}\delta g^{\mu\nu}=2\sqrt{g}\,T_{\mu\nu}\nabla^{\mu}\epsilon^{\nu}=2\sqrt{g}\nabla_{\mu}\left(\epsilon^{\nu}T^{\mu}_{\,\,\nu}\right)=2\partial_{\mu}\left(\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}\right),$$

where we have used the fact that the energy-momentum tensor is conserved. Note that this expression vanishes if the field $\epsilon(x)$ is such that $\delta g_{\mu\nu}=0$ -- that is, the current


is conserved if $\epsilon$ is a Killing vector of $g$.

Now, to answer your question, a conformal transformation is one such that $g'(x)=\Omega(x)g(x)$. If we take an infinitesimal version $x\to x+\epsilon(x)$, this tells us that

$$\delta g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}=\omega(x)g_{\mu\nu}.$$

A vector field $\epsilon(x)$ which satisfies the requirement that $\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}=\omega(x)g_{\mu\nu}$ for some smooth $\omega$ is a conformal Killing vector. If $\epsilon$ is a conformal Killing vector, we then have

$$\delta\mathcal{L}=\sqrt{g}\,T_{\mu\nu}\,\delta g^{\mu\nu}= \begin{cases} 2\partial_{\mu}\left(\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}\right)\\ \sqrt{g}\,\omega(x)T_{\mu\nu}g^{\mu\nu}. \end{cases}$$

Since conformal symmetry implies that the trace of $T$ vanishes, we see that


is conserved. This becomes the statement you want to prove when we take the flat limit.

  • $\begingroup$ With regards to your last equation, I cannot conclude this because it would require $\delta \mathcal{L} = \epsilon^\mu \partial_\mu \mathcal{L} = \partial_\mu \epsilon^\mu \mathcal{L}$ which isn’t true if $\epsilon $ is not a constant. Am I missing something? $\endgroup$ Jul 5, 2019 at 15:13
  • $\begingroup$ You are 100% right -- the steps in the second half of the answer are wrong, unless the coordinate transformation is generated by a killing vector of flat space. I'm editing them out $\endgroup$ Jul 5, 2019 at 15:24
  • $\begingroup$ I have always been rather uncomfortable with the Hilbert energy-momentum tensor as it doesn’t seem to arise from Noether’s theorem. I assume this is the energy-momentum tensor that is used in CFT rather than the canonical one. Does it have the same interpretation however? And why would I use that one over the canonical one? $\endgroup$ Jul 6, 2019 at 7:36
  • $\begingroup$ One way to get around using the Hilbert tensor is to define the energy-momentum tensor implicitly through the current $J$ defined in your question. Indeed, we do this for rigid translations in Minkowski space already. This can be shown to be equivalent to the Hilbert tensor (up to the usual ambiguity in the definition of the energy-momentum tensor). $\endgroup$ Jul 6, 2019 at 23:17

TL;DR: The conclusions surrounding eq. (2.19) are not true in general.

  1. First of all, Ref. 1 makes it clear on top of p. 19 that $T^{\mu\nu}$ denotes the metric/Hilbert stress-energy-momentum (SEM) tensor, not some other SEM tensor.

  2. Now it is true that if $\epsilon=\epsilon^{\mu}\partial_{\mu}$ is a Killing vector field (KVF), then the current (2.19) satisfies $$ \nabla_{\mu} j^{\mu}~\stackrel{m}{\approx}~0,\tag{A}$$ cf. Ref. 2. [Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eom. The connection $\nabla$ is the Levi-Civita connection.]

  3. Next let us for simplicity in the rest of this answer assume that the metric $g_{\mu\nu}$ is constant in the coordinates $x^{\mu}$ that we use. Ref. 1 is interested in the case where $\epsilon$ generates a conformal transformation, i.e. it is a conformal Killing vector field (CKVF). [This transformation is assumed to be a quasi-symmetry of the action, so that Noether's 1st theorem applies.] Concerning conformal transformations, see e.g. this Phys.SE post. In particular, the CKVF $\epsilon$ is not arbitrary, in contrast to what Ref. 1 incorrectly claims on the bottom of p. 19.

  4. According to Refs. 3 & 4, the corresponding conserved Noether current is not eq. (2.19) but instead on the form $$ j^{\mu}~=~ \epsilon_{\nu}T^{\nu\mu} + (\partial\cdot \epsilon) K^{\prime \mu} + \partial_{\nu}(\partial\cdot \epsilon)~L^{\nu\mu}, \tag{4/3.4}$$ where $K^{\prime \mu}$ and $L^{\nu\mu}$ are some $x$-local expressions. [It is reassuring that when the CKVF $\epsilon$ is a KVF, then $\partial\cdot \epsilon=0$ so that eq. (4/3.4) reduces to eq. (2.19).]

  5. More importantly, Refs. 3 & 4 state that scale-invariance does not imply that $T^{\mu\nu}$ is traceless (as Ref. 1's incorrectly claims on the bottom of p.19), but rather that $$T^{\mu}{}_{\mu} ~\stackrel{m}{\approx}~-\partial_{\mu}K^{\mu}.\tag{Ca/3.6}$$ [Tracelessness of $T^{\mu\nu}$ follows instead from Weyl-invariance. See also e.g. this Phys.SE post.]


  1. R. Blumenhagen and E. Plauschinn, Intro to CFT, Lecture Notes in Physics 779, 2009; eq. (2.19).

  2. S.W. Hawking and G.F.R. Ellis, The Large Scale Structure of Space-Time, Section 3.2.

  3. J. Polchinski, Scale and conformal invariance in QFT, Nucl. Phys. B303 (1988) 226; eqs. (Ca) & (4).

  4. K. Farnsworth, M.A. Luty & V. Prilepina, Weyl versus Conformal Invariance in QFT, JHEP (2017) 170, arXiv:1702.07079; eqs. (3.4) & (3.6).


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