This question is about how the normal ordering in the energy momentum tensor for a free field is consistent with a non-vanishing vacuum expectation value implied by the transformation rules for a CFT.

The energy momentum tensor for a free field on a plane is $$T(z)= -2\pi:\partial_z \phi(z)\partial_z \phi(z):$$ where the normal ordering is implemented by point splitting and subtracting out the propagator. $$:\partial_z \phi(z)\partial_z \phi(z):\quad\equiv \lim_{\delta\rightarrow 0} \partial_z \phi(z+\delta)\partial_z \phi(z)+\frac {1}{4\pi \delta^2}$$ Here I am using the conventions of Di Francesco et al's textbook on CFT, e.g. around equation (5.136) in section 5.4.

The point of the normal ordering is that when we take the expectation value we get the propagator minus the propagator, so it is simply zero $$\langle T(z)\rangle=0.$$

Now my question arises when we try to transform the energy momentum tensor. A standard mapping in CFT relates a theory on the plane with coordinate $z$ to that on a cylinder with circumference $L$ and coordinate $w$.

Due to the central charge $c$, the energy momentum tensor $T(z)$ on the plane is related to that on the cylinder $T'(w)$ like $$T'(w)=\left(\frac{2\pi}{L}\right)^2\left(T(z)z^2-\frac{c}{24}\right)$$ This appears e.g. in the same textbook at equation (5.138).

Given $\langle T(z)\rangle=0$ this implies $$\langle T'(w)\rangle=-\frac{c\pi^2}{6L^2}$$ from which physical results about the free field with periodic boundary conditions are derived.

Now the problem is that we might expect that the energy momentum tensor for a free field theory with periodic boundary conditions takes the same form $$T'(w)= -2\pi\lim_{\delta\rightarrow 0} \left(\partial_w \phi'(w+\delta)\partial_w \phi'(w)+\frac {1}{4\pi \delta^2}\right)$$ But if that is the case, it seems its expectation value must be zero not something involving $c$ and $L$!

Possibly the resolution is that $T'$ does not have this naive form. But in Di Francesco's textbook they use this form to show the transformation law above involving $c$ at equation (5.136). This is a simple enough derivation that if you don't have the textbook you can derive it yourself. The central charge term comes directly from the subtracted propagator terms in the normal ordering. If $T'$ doesn't have the normal ordered form why would this argument work?

The other thing I considered is that the propagator is slightly different with periodic boundary conditions. But this difference is only seen at scales on the order of $L$, and the normal ordering involves the short distance limit.

So the question is simply how can $T'$ have a non-vanishing expectation value?

  • $\begingroup$ i don't understand your last sentence, can you elaborate? $\endgroup$ Aug 16, 2019 at 12:47
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    $\begingroup$ ok yes, the answer is here: physics.stackexchange.com/a/483863/83405 You need to be careful with the change of coordinates in the normal ordering of $T'(w)$ .. Your last expression for $T'(w)$ is wrong I believe. $\endgroup$ Aug 16, 2019 at 13:37
  • $\begingroup$ @Wakabaloola, Your answer to that question looks promising (actually maybe it is more suited to my question than the one you were answering). But it is quite long and I am having trouble seeing where exactly your answer differs from the yellow book quick derivation. It seems in (5.136), the central charge term also comes from a coordinate change in the normal ordering. Would you mind pointing out where that derivation goes wrong, and how your way of thinking about it leads to a nonvanishing vacuum expectation value for $T'$? $\endgroup$
    – octonion
    Aug 16, 2019 at 14:18
  • $\begingroup$ in the equations you have written you are using the same $\delta$ for both z and w T’s, whereas they are related nontrivially - you need to transform the subtractions too in order to obtain the w energy momentum T’ $\endgroup$ Aug 16, 2019 at 14:25
  • $\begingroup$ @Wakabaloola, Thanks again for the discussion. I deleted some of my comments here because it's always annoying when a mod moves everything to chat. I think I've resolved my own question. I don't think the issue is due to naive normal ordering, I think it is due to the non-invariance of the vacuum state under local conformal transformations $\endgroup$
    – octonion
    Aug 17, 2019 at 15:35

1 Answer 1


The real quick answer to my question is that the normal ordered form of $T'(w)$ in the statement of my question is exactly right, but the primes above the fields $\phi'(w)$ need to be taken seriously. The meaning is that $\phi'(w)=\phi(z)$, and after making this substitution (noting that we do not change anything about the commutator subtracted in the normal ordering) we indeed see that the expectation value of $T'(w)$ is nonzero and what it should be.

This answer is maybe a little too quick, because when I asked my question I was still a little unsure that $T'(w)$ had the correct form. So let me go through a slightly longer reasoning so that there are no doubts.

Starting on the plane, with coordinate $z$, we have an uncontroversial definition of normal ordering $$:\partial_z\phi(z)\partial_{z'}\phi(z'): \quad=\quad \partial_z\phi(z)\partial_{z'}\phi(z')-\langle\partial_z\phi(z)\partial_{z'}\phi(z')\rangle $$ The first term is a product of two primary fields at different points so we know exactly how it transforms. The second term is not an operator at all (or rather proportional to the identity operator) so we know exactly how it transforms.

There are different ways to think about the transformation, in particular we could think about field configurations in the path integral, but I find it helpful to think in terms of operators. From an operator perspective there will be some unitary operator $U$ associated with the transformation which can ultimately be written in terms of $T(z)$ or the generators of the Virasoro algebra. Its effect on the scalar field is simple. $$U\phi(z)U^{-1}=\phi(w)$$ Note this is the exact same field appearing on both sides, only the argument has changed. So writing our transformation law: $$U:\partial_z\phi(z)\partial_{z'}\phi(z'):U^{-1} =\partial_z w(z)\partial_{z'}w(z')\partial_{w}\phi(w)\partial_{w'}\phi(w')-\langle\partial_z\phi(z)\partial_{z'}\phi(z')\rangle$$ Now since $\phi(w)$ is indeed the same operator, we can rewrite this expression in terms of normal ordered $:\partial_{w}\phi(w)\partial_{w'}\phi(w'):$ the ordinary way. When we take the limit where $z, z'$ approach each other we get the extra central charge term on the left hand side, following the exact same reasoning as Di Francesco et al do in their textbook at (5.136).

Moving the $U$ to the other side we have $$T(z)=\frac{1}{z^2}\left(\left(\frac{L}{2\pi}\right)^2 U^{-1}T(w)U +\frac{c}{24}\right)$$ Now the key thing from the operator point of view is that when we take the vacuum expectation value $\langle U^{-1} T(w)U\rangle$ is not the same as $\langle T(w)\rangle$ which vanishes. They would be the same thing if $U$ were a global conformal transformation, since that would leave the vacuum invariant, but this particular transformation will involve Virasoro generators which do not annihilate the vacuum.

So the non-invariance of the vacuum is really what opens up the possibility of a non-vanishing $\langle U^{-1}T(w)U\rangle\equiv \langle T'(w)\rangle$. And not at all coincidentally the conformal transformations which do change the vacuum are those that have non-vanishing Schwarzian derivative, and thus an extra inhomogeneous central charge term.

  • $\begingroup$ I don't really follow you reasoning, but the important thing is that you are happy with it since you asked the question. As a comment, note that $\phi(z)$ transforms as a scalar under $z\rightarrow w(z)$, which by definition means: $\phi(z)\rightarrow \phi'(w)=\phi(z)$, so I don't understand what happened to the primes. (E.g., it is not true that $\phi(w)=\phi(z)$.) $\endgroup$ Aug 17, 2019 at 19:38
  • $\begingroup$ $U^{-1}\phi(w)U\equiv \phi'(w)=\phi(z)$ $\endgroup$
    – octonion
    Aug 17, 2019 at 19:40

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