CAF123 said:
Di Francesco's ... He writes that the most general form of the tensor is $$S_{\mu \nu \rho \sigma} = (x^2)^{-4} \left\{ A_1 g_{\mu \nu} g_{\rho \sigma} (x^2)^2 + A_2 (g_{\mu \rho}g_{\nu \sigma} + g_{\mu \sigma}g_{\nu \rho})(x^2)^2 + A_3(g_{\mu \nu}x_{\rho}x_{\sigma} + g_{\rho \sigma}x_{\mu}x_{\nu})x^2 + A_4 x_{\mu}x_{\nu}x_{\rho}x_{\sigma}\right\}$$
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.
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: $$S_{\mu \nu \rho\sigma} = A_5 (x^2)^{-3} (g_{\mu \sigma} x_{\rho}x_{\nu} + g_{\mu \rho}x_{\sigma}x_{\nu} + g_{\nu \sigma}x_{\rho}x_{\mu} + g_{\nu \rho}x_{\sigma}x_{\mu})$$
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
