# Derivation of Schwinger function in Di Francesco

1. Sep 14, 2014

### CAF123

This is a question for anyone who is familiar with Di Francesco's book on Conformal Field theory. In particular, on P.108 when he is deriving the general form of the 2-point Schwinger function in two dimensions. He writes that the most general form of the tensor is $$S_{\mu \nu \rho \sigma} = (x^2)^{-4} \left\{ A_1 g_{\mu \nu} g_{\rho \sigma} (x^2)^2 + A_2 (g_{\mu \rho}g_{\nu \sigma} + g_{\mu \sigma}g_{\nu \rho})(x^2)^2 + A_3(g_{\mu \nu}x_{\rho}x_{\sigma} + g_{\rho \sigma}x_{\mu}x_{\nu})x^2 + A_4 x_{\mu}x_{\nu}x_{\rho}x_{\sigma}\right\}$$ This I understand and have obtained this result myself. What I don't understand however, is why he has neglected the following term since it seems to satisfy all the constraints presented on P.108: $$S_{\mu \nu \rho\sigma} = A_5 (x^2)^{-3} (g_{\mu \sigma} x_{\rho}x_{\nu} + g_{\mu \rho}x_{\sigma}x_{\nu} + g_{\nu \sigma}x_{\rho}x_{\mu} + g_{\nu \rho}x_{\sigma}x_{\mu})$$ I wondered whether this could be reduced to terms already present in the form Di Francesco gave, but I quickly realized this to not be the case. So, if anyone is familiar with his book and would be willing to clarify this it would be great. I asked a professor at my university and he was not sure either why it has been neglected. He suggested that I impose the conservation law and then see that $A_5$ vanishes, but I tried this and it wasn't the case. This also would not explain why the tensor is not present in eqn (4.74) or in the first eqn I posted above.

Many thanks.

2. Sep 15, 2014

### samalkhaiat

No, it is not the most general form for the correlation function. If you impose conservation on this tensor you will be able to determine 3 out of the 4 constants and this in turn leads to a very wrong conclusion about the trace of the correlation function. That is, you get an-everywhere-vanishing trace which is very wrong. As a general rule, if the trace does not produce singular contact term, then something is wrong.

Yes. You have to include this tensor because it satisfies all symmetry constraints. As to why the book does not include it, you need to ask the authors. I can only say that I have seen few incorrect mathematical reasoning in that book.
Any way, when you include that tensor and impose conservation, you will be left with 2 unconstrained constants. It is a lot easier to do it in momentum space though. This is because the conservation can be imposed without carrying out any differentiations.
So, let us do it. We start by writing the Fourier transform of the position space correlation function as
$$S_{ \mu \nu \rho \sigma } ( q ) \equiv \langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle . \ \ \ \ (1)$$
1) Translation invariance implies energy-momentum conservation. In momentum space, this simply says
$$q^{ \mu } \langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle =0 \ \ \ \ \ (2) .$$
2) Poincare’ invariance allows us to construct symmetric energy momentum tensor. The correlation function will therefore inherit all the symmetry properties of $T_{ \mu \nu }$,
$$S_{ \mu \nu \rho \sigma } = S_{ \mu \nu \sigma \rho } = S_{ \nu \mu \rho \sigma } = S_{ \nu \mu \sigma \rho } .$$
3) Translation and parity invariance imply
$$S_{ \mu \nu \rho \sigma } ( q ) = S_{ \rho \sigma \mu \nu } ( q ) .$$
4) In two dimensions, scale invariance implies the scaling law
$$S_{ \mu \nu \rho \sigma } ( \alpha q ) = \alpha^{ 2 } S_{ \mu \nu \rho \sigma } ( q ) .$$
Now we can list all possible tensors that satisfy the conditions 2-4. These are
$$T^{ 1 }_{ \mu \nu \rho \sigma } = \frac{ 1 }{ q^{ 2 } } q_{ \mu } q_{ \nu } q_{ \rho } q_{ \sigma } , \ \ T^{ 2 }_{ \mu \nu \rho \sigma } = ( q_{ \mu } q_{ \rho } g_{ \nu \sigma } + q_{ \nu } q_{ \sigma } g_{ \mu \rho } ) + ( \rho \leftrightarrow \sigma ) ,$$
$$T^{ 3 }_{ \mu \nu \rho \sigma } = q^{ 2 } ( g_{ \mu \rho } g_{ \nu \sigma } + g_{ \mu \sigma } g_{ \nu \rho } ) , \ \ \ T^{ 5 }_{ \mu \nu \rho \sigma } = q^{ 2 } g_{ \mu \nu } g_{ \rho \sigma } ,$$
and finally
$$T^{ 4 }_{ \mu \nu \rho \sigma } = q_{ \mu } q_{ \nu } g_{ \rho \sigma } + q_{ \rho } q_{ \sigma } g_{ \mu \nu } .$$
Notice that $T^{ 2 }$ has the same tonsorial structure as the one you mentioned.

Now, the most general form for the correlation function (1) can be written as
$$\langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle = \sum_{ i = 1 }^{ 5 } a_{ i } T^{ i }_{ \mu \nu \rho \sigma } . \ \ \ \ (3)$$
Contracting Eq(3) with $q^{ \mu }$ and using Eq(2), we find
$$( a_{ 1 } + 2 a_{ 2 } + a_{ 4 } ) q_{ \nu } q_{ \rho } q_{ \sigma } + ( a_{ 2 } + a_{ 3 } ) q^{ 2 } ( q_{ \rho } g_{ \nu \sigma } + q_{ \sigma } g_{ \nu \rho } ) + ( a_{ 4 } + a_{ 5 } ) q^{ 2 } q_{ \nu } g_{ \rho \sigma } = 0 .$$
Since the three tensors making up this equation are linearly independent (prove it!), their coefficients must all vanish separately. Thus, we may set
$$a_{ 2 } = - a_{ 3 } \equiv - A / 2 , \ \ \ a_{ 4 } = - a_{ 5 } \equiv B , \ \ (4a)$$
and deduce
$$a_{ 1 } = - 2 a_{ 2 } - a_{ 4 } = A + B . \ \ \ \ (4b)$$
Thus, energy momentum conservation leaves us with two unconstraint constants, $A$ and $B$. Substituting all the tensors $T^{ i }_{ \mu \nu \rho \sigma }$ and the values of all constants $a_{ i }$ in equation (3), we get one long and ugly looking equation. However, with some algebra, we can transform it into the following nice compact form,
$$\langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle = \frac{ A }{ 2 } q^{ 2 } ( P_{ \mu \rho } P_{ \nu \sigma } + P_{ \mu \sigma } P_{ \nu \rho } ) + B q^{ 2 } P_{ \mu \nu } P_{ \rho \sigma } , \ \ \ (5)$$
where $P_{ \mu \nu }$ is the projection tensor
$$P_{ \rho \sigma } = g_{ \rho \sigma } - \frac{ q_{ \rho } q_{ \sigma } }{ q^{ 2 } } . \ \ \ \ (6)$$
You can easily verify the following relations
$$g^{ \mu \nu } P_{ \mu \nu } = P^{ \rho }_{ \rho } = 1 , \ \ \ P_{ \mu \nu } P^{ \mu \nu } = 1 , \ \ \ (7a)$$
$$P_{ \mu }^{ \rho } P_{ \nu \rho } = P_{ \mu \nu } , \ \ \ q^{ \mu } P_{ \mu \nu } = q^{ \nu } P_{ \mu \nu } = 0 . \ \ \ (7b)$$
Now, we can use these relations to show that
$$g^{ \rho \sigma } \langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle = \langle T_{ \mu \nu } ( q ) \ T_{ \rho }^{ \rho } ( - q ) \rangle = ( A + B ) q^{ 2 } P_{ \mu \nu } , \ \ (8)$$
and
$$\langle T_{ \mu }^{ \mu } ( q ) \ T_{ \rho }^{ \rho } ( - q ) \rangle = ( A + B ) q^{ 2 } . \ \ \ \ (9)$$
Since the right-hand sides of (8) and (9) are polynomials in momentum we expect singular contact terms in position space. Indeed, transforming back to position space we find
$$\langle T_{ \mu \nu } ( x ) \ T_{ \rho }^{ \rho } ( 0 ) \rangle \sim ( A + B ) ( g_{ \mu \nu } \partial^{ 2 } - \partial_{ \mu } \partial_{ \nu } ) \delta^{ 2 } ( x ) , \ \ \ (8a)$$
and
$$\langle T_{ \mu }^{ \mu } ( x ) \ T_{ \rho }^{ \rho } ( 0 ) \rangle \sim ( A + B ) \partial^{ 2 } \delta^{ 2 } ( x ) . \ \ \ \ (9b)$$
Thus, at coincident points, the correlation function may fail to vanish. In this case, the presence of the contact terms, on the right hand sides, represents an anomaly known by the name “The Trace Anomaly”. On the other hand, Eq(9b) means that the correlation function vanishes at separated points
$$\langle T_{ \mu }^{ \mu } ( x ) \ T_{ \rho }^{ \rho } ( 0 ) \rangle = 0 , \ \ \mbox{at} \ \ x \neq 0 .$$
In unitary theories, the separating property of the vacuum then implies that $T^{ \mu }_{ \mu } ( x ) = 0$ holds as local operator equation. But this is exactly the condition of invariance under the full conformal group in 2D. Indeed, we have a theorem which says: In 2D, scale invariant unitary QFT’s are necessarily invariant under the full conformal group.
In $D > 2$, it is not known if the theorem is valid. The embarrassing situation is that we know no counterexample yet we don't have proof for the theorem in higher dimensions.

One last thing, In higher dimensions, Eq(5) can be generalized to

$$\langle T_{ \mu \nu } ( q ) \ T_{ \rho \sigma } ( - q ) \rangle = \frac{ A ( q ) }{ 2 } ( q^{ 2 } )^{ \frac{ D }{ 2 } } ( P_{ \mu \rho } P_{ \nu \sigma } + P_{ \mu \sigma } P_{ \nu \rho } ) + \frac{ B ( q ) }{ D - 1 } ( q^{ 2 } )^{ \frac{ D }{ 2 } } P_{ \mu \nu } P_{ \rho \sigma } , \ \ \ (5)$$

where $A(q)$ and $B(q)$ can now be very complicated functions of the momentum.

Sam

Last edited: Sep 15, 2014
3. Sep 16, 2014

### CAF123

Hi Sam,
Before I completely digest what you wrote, I went ahead and included the missing tensor in my calculations. This, as you said, gives the $A_i$ in terms of two unknown constants. When I put these constants back into the tensor and take the trace of that tensor, the result is such that everything cancels. (I.e I get that the trace is identically zero). This seems to be in conflict with what you wrote about Di Francesco's claim that an everywhere-vanishing trace is wrong. Is it then likely that I made a mistake? (I could post my exact calculation if that is preferable, it would just be heavy going with the latex)

4. Sep 16, 2014

### samalkhaiat

Did you do $\partial \langle ... \rangle = \langle \partial ... \rangle$? If you do that, you never get singular term. You need to be careful when you take the derivative inside the time-ordered product. This is why one choses to do the calculation in momentum space.

5. Sep 16, 2014

### Avodyne

Why is that? Isn't there still a scaling symmetry that forces A and B to be constant?

6. Sep 17, 2014

### CAF123

I didn't use any time ordered products in my calculation, I just wrote it out explicity like $$S_{\mu \nu \rho \sigma} = (x^2)^{-4} \left\{A_1 g_{\mu \nu}g_{\rho \sigma}(x^2)^2 + ....\right\}$$ this time including the tensor with $A_5$ premultiplying it. Then I took the divergence of the tensor, imposing the conservation law: $$\partial^{\mu}S_{\mu \nu \rho \sigma} = 0 = \dots$$ Collect all linearly independent terms and for the resulting expression to be identically zero, each coefficient must vanish. This gave me the following three independent equations (and since I have five constants, I will have them all expressed in terms of each other, as you surmised). I get the equations $$A_1 = 3A - A_5,\,\,\,\, A_2 = -A,\,\,\,\, A_3 = -4A + 2A_5, \,\,\,\,\,A_4 = 8A - 8A_5$$ Now if I set $A_5 = 0$ here I recover Di Francesco's equations on P.108 which is a good check.

Then I subbed these back into the tensor, took the trace and everything falls into place so that the trace vanishes identically, that is $S^{\mu\,\,\,\rho}_{\,\,\,\,\mu\,\,\,\,\rho} = 0$.

I posted the question as well on the stack exchange before I obtained a reply here. It somehow made it to overflow and here is the response I got. http://www.physicsoverflow.org/2353...tensor-in-di-francescos-cft?show=23743#a23743 As you can see, I get the same equations as the responder there but I am missing a factor of 2 in one of my equations which I cannot recover. Edit: Since I posted this, an edit has been made to that thread, the responder now agrees with my equations which seems to imply an everywhere vanishing trace.

Last edited: Sep 17, 2014
7. Sep 17, 2014

### samalkhaiat

No, in QFT scale transformation is not “directly” a symmetry transformation, because the same theory in a larger scale corresponds to a different value of the renormalized coupling, i.e., the symmetry is broken by renormalization effects. So, you expect terms with RG $\beta ( g )$-functions to be present in $T^{ \mu }_{ \mu }$. Indeed
$$T^{ \mu }_{ \mu } = \beta_{ i } ( g ) \frac{ \delta \mathcal{ L } }{ \delta g_{ i } } .$$
So a classical scale-invariant theory preserves this symmetry at the quantum level if $\beta ( g ) = 0$.

8. Sep 17, 2014

### samalkhaiat

You are doing what the book does. That is, you are treating the QUANTUM object $\langle T_{ \mu \nu } ( x ) T_{ \rho \sigma } ( y ) \rangle$ as if it is CLASSICAL field theory object. In classical field theory, scale invariance requires $T^{ \mu }_{ \mu } ( x ) = 0$ everywhere. So classically, if you take the object $S_{ \mu \nu \rho \sigma } = T_{ \mu \nu } ( x ) T_{ \rho \sigma } ( 0 )$ and demands scale invariance, you also get zero traces.
In QFT, scale invariance means that
$$\langle T^{ \mu }_{ \mu } ( x ) \mathcal{ O } ( y ) \rangle \sim \delta ( x - y ) \langle \mathcal{ O } \rangle .$$
Does this vanish at $x = y$? No, it does not. It vanishes when $x \neq y$ not EVERYWHERE. In fact if you take $\mathcal{ O } = T_{ \rho \sigma }$ and carefully derive the Ward identities, you find
$$\langle T_{ \mu \nu } ( x ) T^{ \rho }_{ \rho } ( y ) \rangle \sim ( \eta_{ \mu \nu } \partial^{ 2 } - \partial_{ \mu } \partial_{ \nu } ) \delta ( x - y ) ,$$
$$\langle T^{ \mu }_{ \mu } ( x ) T^{ \rho }_{ \rho } ( y ) \rangle \sim \partial^{ 2 } \delta ( x - y ) .$$
Did you read the paragraph after equation (9b)?

9. Sep 18, 2014

### Avodyne

I thought the discussion was about conformal field theory, for which indeed $\beta(g)=0$, e.g. N=4 SYM in d=4.

10. Sep 18, 2014

### CAF123

Does that mean that Di Francesco's result is then just plain and simply wrong? Since his result gives a trace free function everywhere and yours is manifestly not. Is there any merit in his derivation at all? Gosh, I didn't think such a book would have been warranted publishing if there are all these mistakes ;)