Due to a previous question, I am confused with the expectation value of the stress-energy tensor in a 2-D conformal field theory.

Let's take the example of string theory, to sketch the problem. Defining the action by $S = \frac{1}{2 \pi \alpha'}\int d^2z \partial X^\mu \bar \partial X_\mu$, the (euclidean) expectation value of an operator is :

$$\langle\mathcal F[X]\rangle=\int [dX]~ exp(-S)~F[X]\tag{0}$$

My questions are :

1) In string theory, what is : $$\langle \partial X^\mu \partial X_\mu\rangle \tag{1}$$ 2) In string theory, what is : $$\langle :\partial X^\mu \partial X_\mu:\rangle \tag{2}$$ 3) In a general 2-D conformal field theory, what is : $$\langle T^{\alpha \beta}\rangle \tag{3}$$


1 Answer 1


First, the two $\partial$ derivatives in the action should be $\partial^\alpha$ and $\partial_\alpha$ or $\partial $ and $\bar\partial$.

In a general 2D CFT, the expectation value of $T_{\alpha\beta}$ – note that you must have confused world sheet and spacetime indices; the stress energy tensor only has world sheet indices – is of course zero. The stress-energy tensor is a tensor whose dimension is nonzero so any nonzero expectation value in the vacuum would be dimensionful and would define a preferred length scale, and therefore violated the scaling (and conformal) symmetry.

The trace of the stress energy tensor, $T_{\alpha}^\alpha$, is zero even as an operator due to the scaling symmetry (it's the generator of the scaling symmetry). So all correlators including this trace of the tensor as a factor will vanish. This is not true for the remaining two components of the tensor, $T_{zz}$ and $T_{\bar z\bar z}$. Their expectation values vanish but the more complicated correlators of these components with other tensors don't vanish.

The arguments above were general, for any CFT. You may also see how it works "constructively" for the simple $X$ free boson CFT. The stress energy tensor is a normally ordered expression containing $\partial X\partial X$ terms contracted in two ways, see 2.3.15b or 2.4.4 in Polchinski I.

The normally ordered part of $\partial X\partial X$ is defined as $\partial X\partial X$ minus whatever you need to subtract to eliminate the $c$-number part, see equation 2.2.4 for the formula for normal ordering in the $X$ free boson 2D CFT. The first term on the right hand side – which becomes $1/(z\bar z)$ times the same numerical factor if you add the two world sheet derivatives – is exactly the (nonzero) expectation value of $X X$ or $\partial X\partial X$ (depending on whether you differentiate twice). The expectation values of the normal-ordered $:\partial X\partial X:$ expressions vanish pretty much by construction, and those are the operators that appear in the stress-energy tensor whose expectation value vanishes, too.

The normal ordering of operators that depend on $z,\bar z$ may sound unfamiliar but it's really isomorphic to the "undergraduate courses of QFT" normal ordering which puts creation operators on the left from annihilation operators. When you do so, it's guaranteed that the annihilation operators on the right annihilate the ket vacuum or, e.g. if there are none, the creation operators on the left side annihilate the bra vacuum. So the expectation value of any normal-ordered polynomial of operators in free field theories, except for $:\langle 1 \rangle:=1$ itself, vanishes. (Well, sometimes for zero modes, like those of the $bc$ ghosts, are defined in a more symmetric way so the normal ordering of $b_0 c_0$ and its normal ordering may deserve a special treatment. But no subtleties arise for $X$ CFT unless you would consider things like $:X_0 P_0:$ for the zero modes.)

  • $\begingroup$ +1 : Thanks for the corrections and you remarks. So, I don't understand this recent answer, where one has non-null expectation values for the stress-energy tensor. $\endgroup$
    – Trimok
    Commented Sep 6, 2013 at 9:29
  • 1
    $\begingroup$ @Trimok: Lubos' post is valid for flat-space CFT. On the cylinder (on in higher dimensions, for a finite-temperature field theory) the stress tensor can get a vev. $\endgroup$
    – Vibert
    Commented Sep 6, 2013 at 9:39
  • $\begingroup$ @Vibert: by vev you mean $<T^{\mu}_{\;\mu}>\neq 0$ as in the conformal anomaly when defined on a cylinder? $\endgroup$ Commented Sep 6, 2013 at 11:22
  • $\begingroup$ Exactly, vev = vacuum expectation value. $\endgroup$
    – Vibert
    Commented Sep 6, 2013 at 12:04
  • $\begingroup$ Ok, my interrogation was vaccum expectation of which field. And you answered: trace of the stress-energy tensor. $\endgroup$ Commented Sep 6, 2013 at 12:56

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