# Correlation functions and Ward identities

1. Jul 21, 2014

### CAF123

The definition I have for a 2-point correlator is $$\langle \phi_1(x_1)\phi_2(x_2)\rangle = \frac{1}{Z} \int \mathcal D \phi \,\,\phi_1(x_1)\phi_2(x_2) \exp-S[\phi],$$ where $Z = \int \mathcal D \phi \,\,\exp-S[\phi]$. I am trying to understand the overall physical meaning of such a quantity and the motivation for why we define an object like this.

The LHS looks to be the expectation value of the product of two fields at two different positions. So it is a measure of the average value the product of these two states would have if we sampled the product over all possible $x_i$ in space? I don't think this is quite correct because it does not really make the RHS sensible. There, we seem to be integrating over a configuration of fields. Perhaps $Z$ is the integral over all permissible classical field configurations satisfying some classical theory and each configuration brings with it its effect on the action that is manifest in the $\exp-S[\phi]$ term. (i.e there exists a configuration where the action is minimized).

The definition looks analogous to how the expectation of the product of two random or stochastic variables would be defined, with the $\exp-S[\phi]$ being the weighting factor.
Thanks for any clarity.

2. Jul 21, 2014

### WannabeNewton

3. Jul 22, 2014

### CAF123

Hi WannabeNewton,
I haven't really studied much statistics, other than what comes in a statistical mechanics setting or probability.

So is it correct to regard $\langle \phi_1(x_1)\phi_2(x_2)\rangle$ as a propagator in the path integral sense? So is it a measure of the transition amplitude for the field $\phi_1(x_1)$ at $t_1$ to resume the value of $\phi_2(x_2)$ at a later time $t_2$?

4. Jul 22, 2014

### vanhees71

It's the Wick rotated (Euclidean) version of QFT. It is of utmost importance to keep in mind that a path integral always gives the time ordered products (here ordering meant for Euclidean time), i.e., on the left-hand side you have $\langle T \phi(x_1) \phi(x_2) \rangle$. For more details (also on Ward-Takahashi identities), but using the real-time formalism, see my QFT manuscript:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

5. Jul 22, 2014

### WannabeNewton

Yes, see: http://www.physics.umd.edu/courses/Phys851/Luty/notes/diagrams.pdf and in particular take note of equation (1.10) relating the 2-point correlation function to the VEV (2-point Green's function).

6. Jul 22, 2014

### CAF123

ok, many thanks. I have a quick question about this used in practice. Let $X$ denote a set of $n$ fields. The Ward identity associated with Lorentz invariance is then

$$\partial_{\mu} \langle (T^{\mu}x^{\rho} - T^{\mu \rho}x^{\nu})X\rangle = \sum_i \delta(x-x_i)\left[ x^{\nu}_i \partial^{\rho}_i - x^{\rho}_i\partial^{\nu}_i\langle X \rangle - iS^{\nu \rho}_i \langle X \rangle\right]. (1)$$
This is then equal to
$$\langle (T^{\rho \nu} - T^{\nu \rho})X \rangle = -i\sum_i \delta (x-x_i)S^{\nu \rho}_i\langle X \rangle,$$ which states that the stress tensor is symmetric within correlation functions, except at the position of the other fields of the correlator.

My question is: how is this last equation and statement derived?

I think the Ward identity associated with translation invariance is used after perhaps splitting (1) up like so:
$$\sum_i^n x^{\nu}_i \sum_i^n \delta(x-x_i)\partial^{\rho}_i \langle X \rangle - \sum_i^n x^{\rho}_i \sum_i^n \delta(x-x_i)\partial^{\nu}_i \langle X \rangle - i\sum_i^n\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$ and then replacing $$\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X \rangle = -\sum_i \delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$
for example. The result I am getting is that $$\partial_{\mu}\langle ((T^{\mu \nu})x^{\rho} - (T^{\mu \rho})x^{\nu}) X\rangle + \langle (T^{\rho \nu} - T^{\nu \rho})X \rangle = \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle + \sum_i x^{\rho}_i \partial_{\mu} \langle T^{\mu \nu} X \rangle - i\sum_i\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$ To obtain the required result, this means that e.g$$\sum_i x^{\nu}_i \partial_{\mu} \langle T^{\mu \rho}X \rangle = \partial_{\mu}\langle(T^{\mu \rho})x^{\nu} X \rangle,$$ but why is this the case? (the brackets around $T^{\mu \rho}$ on the RHS just indicate the derivative is only acting on this term)

As for the statement, I see that when $x \neq x_i$ we recover the classical result, namely that $T^{\mu \rho}$ is symmetric for a Lorentz invariant theory. But why do we recover this result for $x$ not being equal to $x_i$? And what is the physical interpretation of the divergency when $x$ happens to coincide with $x_i$?
Thanks very much.

7. Jul 23, 2014

### samalkhaiat

This equation is not correct. If T is the Belinfate tensor, then the RHS of (1) would contain more junks which vanish only in the integrated Ward identity. This is because the Belinfante Lorentz current fails to generate the correct Lorentz transformations on the fields.

Multiply both sides of
$$\partial_{ \mu } \langle T^{ \mu \nu } X \rangle = - \sum_{ i } \delta ( x - x_{ i } ) \partial_{ i }^{ \nu } \langle X \rangle$$
with $x^{ \rho }$ and use
$$x^{ \rho } \delta ( x - x_{ i } ) = x_{ i }^{ \rho } \delta ( x - x_{ i } )$$

Sam

8. Jul 23, 2014

### vanhees71

9. Jul 23, 2014

### CAF123

Hi samalkhaiat,
Hmm, this equation I wrote from the book by Di Francesco. Maybe I should write out the whole paragraph for context:
'We consider now the Ward identity associated with Lorentz invariance. Once the energy-momentum tensor has been made symmetric, the associated $j^{\mu \nu \rho}$ has the form given in Eq. (2.172):* $j^{\mu \nu \rho} = T^{\mu \nu}x^{\rho} - T^{\mu \rho}x^{\nu}$. The generator of the Lorentz transformations is given by Eq. (2.134)**. Consequently, the Ward identity is $$\partial_{\mu}\langle (T^{\mu \nu}x^{\rho} - T^{\mu \rho}x^{\nu})X \rangle = \sum_i \delta (x-x_i) \left[(x^{\nu}_i \partial^{\rho}_i - x^{\rho}_I\partial^{\nu}_i)\langle X\rangle - iS^{\nu \rho}_i \langle X \rangle \right]$$
* is $j^{\mu \nu \rho} = T^{\mu \nu}x^{\rho} - T^{\mu \rho}x^{\nu} + i \frac{\partial \mathcal L}{\partial (\partial_{\mu}\Phi)}S^{\mu \nu}\Phi$ which doesn't seem to match with the expression they gave. However, (2.173) is $j^{\mu \nu \rho} = T^{\mu \nu}_Bx^{\rho} - T^{\mu \rho}_Bx^{\nu}$, where B stands for Belinfante.

** is $L^{\rho \nu} = i(x^{\rho}\partial^{\nu} - x^{\nu}\partial^{\rho}) + S^{\rho \nu}$

Yes, I think that is perhaps what I am not getting. The equation I needed to be true so that I could get the result was: $$\sum_i x^{\nu}_i \partial_{\mu} \langle T^{\mu \rho}X \rangle = \partial_{\mu}\langle(T^{\mu \rho})x^{\nu} X \rangle,$$ which seems similar to your last equation there. (the brackets around $T^{\mu \rho}$ on the RHS just indicate the derivative is only acting on this term). Why is this last equation true? The reason why I ask is because the LHS is a summation over all the points at which the fields in X are evaluated that. But should this be equal to $x^{\nu}$?

Did you have any comments regarding what I wrote about the interpretation of the Ward identity:
Thanks Sam.

Last edited: Jul 23, 2014
10. Jul 23, 2014

### samalkhaiat

You aren’t doing yourself a favour by learning this stuff from that book. Their treatment of this subject uses misleading, inaccurate and sometimes wrong mathematical reasoning. I remember calling their “derivation” of the Hilbert energy-momentum tensor and their “proof” of Noether theorem as mathematical comedy.
As I said before, the Belinfante modified Lorentz current DOES NOT generate the correct Lorentz transformation on the fields,
$$[ J_{ B }^{ 0 \nu \rho } ( x ) , \phi ( y ) ] \neq \delta^{ 3 } ( x - y ) L^{ \rho \nu } ( x ) \phi ( x ) = \delta^{ 3 } ( x - y ) \delta \phi ( x ) ,$$
but the integrated charge does
$$[ \int d^{ 3 } x J_{ B }^{ 0 \nu \rho } ( x ) , \phi ( y ) ] = L^{ \rho \nu } ( y ) \phi ( y ) .$$
Work on the LHS of eq(1)
$$\partial_{ \mu } \langle x^{ \rho } T^{ \mu \nu } X \rangle - ( \rho \leftrightarrow \nu ) = \delta^{ \rho }_{ \mu } \langle T^{ \mu \nu } X \rangle + x^{ \rho } \partial_{ \mu } \langle T^{ \mu \nu } X \rangle - ( \rho \leftrightarrow \nu ) .$$
The second term on the right can be written as
$$x^{ \rho } \partial_{ \mu } \langle T^{ \mu \nu } X \rangle = - \sum_{ i } x^{ \rho } \delta^{ 4 } ( x - x_{ i } ) \partial^{ \nu }_{ i } \langle X \rangle = - \sum_{ i } \delta^{ 4 } ( x - x_{ i } ) x^{ \rho }_{ i } \partial^{ \nu }_{ i } \langle X \rangle .$$
No. They use the ward identity for all wrong purposes. Okay look, as I said in my last post, that “Ward-Takahashi” identity [eq(1)] makes sense only when integrated, i.e., when you get rid of all contact terms (delta function and its derivatives).

Sam

Last edited: Jul 23, 2014
11. Jul 23, 2014

### strangerep

I think you are too kind, calling it merely a "comedy". I'd call it "disgracefully pathetic".

IIUC, CAF123 is in a difficult position, having been referred to that book by his professor (sigh).

12. Jul 24, 2014

### CAF123

Thanks for the heads up. strangerep has also been discouraging me from using this book, but as he writes above, I was assigned it to assist me in parts of my project on CFT. When that is done, yes, the book will be gone. However, I have also tried to seek other treatments of topics in the book. I have found many additional sources with regard to the Noether current treatment, but none whatsoever on the Ward identities above. The ones I did find were a bit too advanced for me at the moment or were applying the formulism, rather than introducing it. Please recommend a source that is more reliable as an introduction to this area!

Yes, that appears to be close enough to eqn (2.163) of Francesco's book, namely $[Q_a, \Phi] = -iG_a \Phi$

Ok, my only remaining question would be why is the last equality valid, that is why is $\sum_i x^{\rho}_i = x^{\rho}$. The i-index runs to the number of points at which the fields in the correlator are evaluated at.

many thanks.

Last edited: Jul 24, 2014
13. Jul 24, 2014

### samalkhaiat

How about "comedy" in higher dimensions and "pathetic" in two dimensions.

14. Jul 24, 2014

### strangerep

It's not. But consider that, effectively, $a\, \delta(a-b) = b\, \delta(a-b)$, since the delta distribution constrains us to $a=b$.

15. Jul 24, 2014

### samalkhaiat

You should try different soueces yourself, after all, I don't know how much you know about the subject, and in what context? In the PDF below I gave simple but thorough study of Noether theorem. The Ward-Takahashi identity is also derived in the operator formalism, i.e., in the same way Takahashi derived it in his classic paper. If you need to derive it in the Path Integral formalism, let me know and I can do it for you in one page. Okay?

Come on! this is elementary algebra

$$x^{ \rho } \sum \delta ( x - x_{ i } ) \partial_{ i } (..) = \delta ( x - x_{1} ) x^{ \rho } \partial_{ 1 } (..) + \delta ( x - x_{2} ) x^{ \rho } \partial_{ 2 } (..) + ...$$
Now use the delta function identity and resum
$$\delta ( x - x_{ 1 } ) \ x_{ 1 }^{ \rho } \ \partial_{ 1 } (..) + \delta ( x - x_{ 2 } ) \ x_{ 2 }^{ \rho } \ \partial_{ 2 } (..) + \ ... = \sum \delta ( x - x_{ i } ) \ x^{ \rho }_{ i } \ \partial_{ i } (..)$$

What is this? Who said that is true?

Sam

#### Attached Files:

• ###### Ch3 Noether.pdf
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16. Jul 24, 2014

### strangerep

Is this part of a larger book? I think I'd like to get a copy of the whole thing...

17. Jul 24, 2014

### samalkhaiat

Yes it is. I'm still working on it, Noether charge, symmetry breakning, etc. still to be finished . I Will make it available for everybody in here when I complete the work.

18. Jul 25, 2014

### CAF123

If that is okay, thanks.

...
Thanks, it makes sense now.

19. Jul 26, 2014

### samalkhaiat

20. Jul 26, 2014

### MathematicalPhysicist

It's a pity I purchased this book a few years ago (I haven't yet started reading from it, cause I still have to grasp QFT and GR).:uhh:

What other mistakes or bad things do you have to say on this book? (are there any good things as well?).