Deriving Massive spin 2 propagator from polarization vectors

[BradC]

In A. Zee "QFT in a nutshell" in chapter I.5 Exercise 1.5.1 on page 39 for spin 2 massive propagator. I know I’m missing something very simple (self-taught beginner). I'm trying to derive equation (13) on page 35, which is
$$G_{\mu\nu,\lambda\sigma} = G_{\mu\lambda}G_{\nu\sigma} + G_{\mu\sigma}G_{\nu\lambda} - 2/3 * G_{\mu\nu}G_{\lambda\sigma}$$
I created something similar to equation (22) by multiplying the three unique forms(eliminating symmetric permutations) of
$$(g_{\mu\nu}-k_{\mu}k_{\nu})$$
Then I use
$$k^{\mu} G_{\mu\nu,\lambda\sigma}= 0$$
for
$$k^{\mu}, k^{\nu}, k^{\lambda}, k^{\sigma}.$$
Then for equation (22), I get B = -A/2, D = -C/2, E = 0, which brings me to:
$$G_{\mu\nu,\lambda\sigma} = G_{\mu\lambda}G_{\nu\sigma} + G_{\mu\sigma}G_{\nu\lambda} - 2 G_{\mu\nu}G_{\lambda\sigma} - 2G_{\mu\nu}k_{\lambda}k_{\sigma} -2 G_{\lambda\sigma}k_{\mu}k_{\nu} + G_{\nu\sigma}k_{\mu}k_{\lambda} + G_{\nu\lambda}k_{\mu}k_{\sigma}+ G_{\mu\lambda}k_{\nu}k_{\sigma}+ G_{\mu\sigma}k_{\nu}k_{\lambda}$$
I'm lost as to the next step to get to:
$$G_{\mu\nu,\lambda\sigma} = G_{\mu\lambda}G_{\nu\sigma} + G_{\mu\sigma}G_{\nu\lambda} - 2/3 * G_{\mu\nu}G_{\lambda\sigma}$$
MANY thanks in advance for filling in some missing gap in knowledge or very simple error in arithmetic.

 

[BradC]

To show more of my work, so someone can help me figure out the gap in my knowledge.

I'm trying to solve "spin 2 massive propagator." Exercise 1.5.1 chapter I.5 on page39 in A. Zee "QFT in a nutshell" 2nd edition in

I know I’m missing something very simple (self-taught beginner). I created something similar to equation (22) by multiplying the 3 variations (eliminating symmetric permutations)
$$(G_{\mu\nu}-k_{\mu}k_{\nu}) * (G_{\mu\lambda}-k_{\mu}k_{\lambda}) * (G_{\mu\sigma}-k_{\mu}k_{\sigma})$$
In order to find the factors I then get:
$$G_{\mu\nu,\lambda\sigma} = a G_{\mu\nu}G_{\lambda\sigma} + b G_{\mu(\lambda}G_{\sigma)\nu} + c (G_{\mu\nu}k_{\lambda}k_{\sigma} + G_{\lambda\sigma}k_{\mu}k_{\nu}) + d ( G_{\mu(\lambda}k_{\sigma)}k_{\nu} + G_{\nu(\lambda}k_{\sigma)}k_{\mu}) + e k_\mu k_\nu k_\lambda k_\sigma$$
Then using the variations of
$$k^{\mu} \epsilon_{\mu\nu} = 0 ,$$
to determine the coefficients,
$$k^{\mu} * k^{\nu} * k^{\lambda} * k^{\sigma} * G_{\mu\nu,\lambda\sigma} = 0$$
then I end up with:
$$0 = a*k^4 + 2b* k^4 + 2c * k^6 + 4d* k^6 + e * k^8$$
which simply leads me to: b = -a/2, d = -c/2, e = 0, Which is
$$G_{\mu\nu,\lambda\sigma} = G_{\mu\lambda}G_{\nu\sigma} + G_{\mu\sigma}G_{\nu\lambda} - 2 G_{\mu\nu}G_{\lambda\sigma} - 2G_{\mu\nu}k_{\lambda}k_{\sigma} -2 G_{\lambda\sigma}k_{\mu}k_{\nu} + G_{\nu\sigma}k_{\mu}k_{\lambda} + G_{\nu\lambda}k_{\mu}k_{\sigma}+ G_{\mu\lambda}k_{\nu}k_{\sigma}+ G_{\mu\sigma}k_{\nu}k_{\lambda}$$
How do I get from here to derive equation below {(13) in 2nd ed book} on page 35?
$$G_{\mu\nu,\lambda\sigma} = G_{\mu\lambda}G_{\nu\sigma} + G_{\mu\sigma}G_{\nu\lambda} - 2/3 * G_{\mu\nu}G_{\lambda\sigma}$$
Many thanks in advance for filling in missing gap in knowledge or error in arithmetic.If you don’t have 2nd edition, Here is the URL first edition in pdf(exercise on pages 36,37 other info on p33).
http://press.princeton.edu/chapters/s7573.pdf

 

[samalkhaiat]

Zee’s exercise is a stupid one, and you made a total mess in trying to solve it. Before I show you how to solve the exercise, let me explain why it is “stupid”. Suppose you have at your disposal two INDEPENDENT objects, say the metric tensor $\eta_{\mu \nu}$ and the vector $k_{\mu}$. Now I ask you to construct, out of $\eta_{\mu \nu}$ and $k_{\rho}$, the most general rank-4 tensor $T_{\mu \nu \rho \sigma}$ which is symmetric under exchanging $( \mu \leftrightarrow \nu )$, $( \rho \leftrightarrow \sigma )$ and $( \mu \nu \leftrightarrow \rho \sigma )$. The answer, as you can easily check, is given by equation (22) with $G_{\mu \nu}$ replaced by the metric tensor $\eta_{\mu \nu}$. But, in Zee’s exercise $G_{\mu \nu}$ and $k_{\mu}$ ARE NOT independent of each other. Indeed, since $$G_{\mu \nu} = \eta_{\mu \nu} - \frac{k_{\mu} k_{\nu}}{k^{2}} ,$$ we have the following relations $$k^{\mu} G_{\mu \nu} = k^{\nu} G_{\mu \nu} = 0 . \ \ \ \ (1)$$ So, it is rather idiotic to include the last three terms in equation (22), because (i) repeated use of (1) leads to $C = D = E = 0$ and (ii) terms of the form $(k \ k \ G)$ and $(k \ k \ k \ k)$ are already included in the terms of the form $(G \ G)$ which are there in the first two terms. So, to construct the tensor $T_{\mu \nu \rho \sigma}$, only $G_{\mu \nu}$ is needed: $$T_{\mu \nu \rho \sigma} = A G_{\mu \nu} G_{\rho \sigma} + B ( G_{\mu \rho} G_{\nu \sigma} + G_{\mu \sigma} G_{\nu \rho} ) .$$ Before we go any further, please notice that in D-dimension we have $$\eta^{\mu \nu} G_{\mu \nu} = D - 1 . \ \ \ \ \ \ \ (2)$$ You can also check that $$G^{\mu}{}_{\rho} G_{\mu \sigma} = G_{\rho \sigma} . \ \ \ \ \ (3)$$ Now, in order to determine the coefficients $A$ and $B$, we impose two further conditions relevant to the spin-2 nature of the massive field:
(1) Tracelessness, $$\eta^{\mu \nu} T_{\mu \nu \rho \sigma} = \eta^{\rho \sigma} T_{\mu \nu \rho \sigma} = 0 . \ \ \ \ \ \ (4)$$ Using the relations (2) and (3) in (4), we find $$A = - \frac{2}{D - 1} B . \ \ \ \ \ \ (5)$$
Thus, our tensor becomes $$T_{\mu \nu \rho \sigma} = B \left( G_{\mu \rho} G_{\nu \sigma} + G_{\mu \sigma} G_{\nu \rho} - \frac{2}{D - 1} G_{\mu \nu} G_{\rho \sigma} \right) . \ \ \ (6)$$
(2) In order to represent a spin-2 sum, i.e. in order to write $$T_{\mu \nu \rho \sigma} = \sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon^{(m)}_{\rho \sigma} ,$$ where $\epsilon^{(m)}_{\mu \nu}$ is the orthonormal polarisation tensor $$\epsilon^{(m)}_{\mu \nu} \epsilon^{\mu \nu}_{(n)} = \delta^{m}_{n} , \ \ \Rightarrow \sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon_{(m)}^{\mu \nu} = 2 ( 2 ) + 1 = 5 ,$$ we must have $$T_{\mu \nu}{}^{\mu \nu} \equiv \eta^{\mu \rho} \eta^{\nu \sigma} T_{\mu \nu \rho \sigma} = 5 . \ \ \ \ (7)$$ This is the number of degrees of freedom in massive spin-2 field (i.e. the dimension of the irreducible representation space). Now, contracting (6) with $\eta^{\mu \rho} \eta^{\nu \sigma}$ and using (2), (3) and (7) we find $$B = \frac{5}{D ( D - 1) - 2} .$$ So, in $D = 4$ spacetime, we have $$T_{\mu \nu \rho \sigma} = \sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon^{(m)}_{\rho \sigma} = \frac{1}{2} \left( G_{\mu \rho} G_{\nu \sigma} + G_{\mu \sigma} G_{\nu \rho} - \frac{2}{3} G_{\mu \nu} G_{\rho \sigma} \right) .$$
Sam

 

[BradC]

Hi Sam,

THANKS! Sorry I made a mess of it.

I do want to thank you for taking the time to clear this all up for me, I learned a few things from your writeup.
I think I've got it now, but I may have a question or two in a few days.

One quick question is there a particular reason why you do the sum -2 to 2 instead of just 1 to 5?
$$T_{\mu \nu \rho \sigma} = \sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon^{(m)}_{\rho \sigma} ,$$
I'm guessing it highlights the difference of the one polarization tensor (the one with the different normalization factor).
But didn't know if there was some insight I was missing by use of this convention.

 

[samalkhaiat]

Hi Sam,

THANKS! Sorry I made a mess of it.

I do want to thank you for taking the time to clear this all up for me, I learned a few things from your writeup.
I think I've got it now, but I may have a question or two in a few days.

One quick question is there a particular reason why you do the sum -2 to 2 instead of just 1 to 5?
$$T_{\mu \nu \rho \sigma} = \sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon^{(m)}_{\rho \sigma} ,$$
I'm guessing it highlights the difference of the one polarization tensor (the one with the different normalization factor).
But didn't know if there was some insight I was missing by use of this convention.

The labelling, in the method I used to solve Zee’s exercise, was irrelevant. I could have done the sum from 7 to 12. The important thing was the number (5), which represents the dimension of the representation space, i.e., the trace of $5 \times 5$ unit matrix. However, I wanted to connect you with group theory. I assume you are familiar with the expansion $$| j m \rangle = \sum_{m_{1} = - j_{1}}^{j_{1}} \sum_{m_{2} = - j_{2}}^{j_{2}} \langle j_{1} m_{1} ; j_{2} m_{2} | j m \rangle \ | j_{1} m_{1} \rangle \otimes | j_{2} m_{2} \rangle .$$ In our case, $j_{1} = j_{2} = 1$ and $j = 2$. So, $m \in \{ - 2 , \cdots , +2 \}$ and the above expansion becomes $$\epsilon^{(m)}_{\mu \nu} = \sum_{(m_{1} , m_{2}) = - 1}^{+1} C^{(m)}_{(m_{1}) (m_{2})} \ \epsilon_{\mu}^{(m_{1})} \ \epsilon_{\nu}^{(m_{2})} ,$$ where $\epsilon^{(m_{i})}_{\rho} \equiv | 1 , m_{i} \rangle_{\rho}$ are the two spin-1 polarisation vectors which form the basis for the spin-2 polarisation tensor $\epsilon_{\mu \nu}^{(m)} \equiv | 2 , m \rangle_{\mu \nu}$, and $$C^{(m)}_{(m_{1}) (m_{2})} = \langle 1 , m_{1} ; 1 , m_{2} | 2 , m \rangle ,$$ are the usual Clebsch-Gordan coefficients. Now, if you are brave enough, you can solve the same exercise. You only need to choose the polarisation vector properly and take the sum $$\sum_{m = - 2}^{2} \epsilon^{(m)}_{\mu \nu} \epsilon^{(m)}_{\rho \sigma} .$$

 

[BradC]

Hi Sam,

Thanks, I thought the reason for the sum a connection to group theory.

I went back and reviewed my group theory and tried the same exercise as you suggested. This helps. thanks again!