# Zee QFT – Part I.5

#### Main text

In this section I will follow some derivations made in the text.

##### Integration by parts of the lagrangian

$\displaystyle S(A) = \int d^4x \mathcal{L} = \int d^4x \left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu + A_\mu J^\mu\right)$

Separately analyzing the integral of the $F_{\mu\nu}F^{\mu\nu}$ term:

$\displaystyle \int d^4x\,F_{\mu\nu}F^{\mu\nu} = \int d^4x\,(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu)$

$\displaystyle = \int d^4x\,\left[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu) + (\partial_\nu A_\mu)(\partial^\nu A^\mu)\right]$

Changing dummy indices and reordering:

$\displaystyle = \int d^4x\,\left[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu) + (\partial_\mu A_\nu)(\partial^\mu A^\nu)\right]$

$\displaystyle = 2\int d^4x\,\left[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu)\right]$

$\displaystyle = 2\int d^4x\,\left[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial^\mu A^\nu)(\partial_\nu A_\mu)\right]$

$\displaystyle = 2\int d^4x\,\left[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\nu A_\mu)(\partial^\mu A^\nu)\right]$

Integrating by parts:

$\displaystyle = 2\int d^4x\,\left[-A_\nu \partial_\mu\partial^\mu A^\nu + A_\mu \partial_\nu\partial^\mu A^\nu\right]$

$\displaystyle = 2\int d^4x\,\left[-A_\nu \partial^2 g^{\mu\nu} A_\mu + A_\mu \partial_\nu\partial^\mu A^\nu\right]$

Changing dummy indices:

$\displaystyle = 2\int d^4x\,\left[-A_\nu \partial^2 g^{\mu\nu} A_\mu + A_\nu \partial_\mu\partial^\nu A^\mu\right]$

It’s a contraction between $\partial_\mu$ and $A^\mu$:

$\displaystyle = 2\int d^4x\,\left[-A_\nu \partial^2 g^{\mu\nu} A_\mu + A_\nu \partial^\mu \partial^\nu A_\mu\right]$

Factoring:

$\displaystyle = 2\int d^4x\,A_\nu \left[-\partial^2 g^{\mu\nu} + \partial^\mu \partial^\nu\right]A_\mu$

Expressing the bilinear terms of the lagrangian in this form we get:

$\displaystyle -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} = \frac{1}{2}A_\nu \left[\partial^2 g^{\mu\nu} - \partial^\mu \partial^\nu\right]A_\mu$

$\displaystyle \frac{1}{2}m^2A_\mu A^\mu = \frac{1}{2}A_\nu A^\nu = \frac{1}{2}A_\nu\left[m^2 g^{\mu\nu}\right]A_mu$

Putting the whole integral together:

$\displaystyle S(A) = \int d^4x \mathcal{L} = \int d^4x\left\{ \frac{1}{2}A_\nu \left[(\partial^2 + m^2) g^{\mu\nu} - \partial^\mu \partial^\nu\right]A_\mu + A_\mu J^\mu\right\}$,

matching (1).

##### Propagator

Let $M^{\mu\nu}(x)$ be the differential operator:

$\displaystyle M^{\mu\nu} = (\partial^2 + m^2) g^{\mu\nu} - \partial^\mu \partial^\nu$.

We can define it’s inverse $D_{\nu\rho}(x)$ by the condition:

$\displaystyle M^{\mu\nu} D_{\nu\rho}(x) = \delta^\mu_\rho \delta^{(4)}(x)$

Doing a Fourier transform:

$\displaystyle \widetilde{M}^{\mu\nu}(k) \widetilde{D}_{\nu\rho}(k) = \delta^\mu_\rho$

$\displaystyle \left[(-k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu\right] \widetilde{D}_{\nu\rho}(k) = \delta^\mu_\rho$

As $\widetilde{D}_{\nu\rho}(k)$ must be Lorentz invariant, we can write it as:

$\displaystyle \widetilde{D}_{\nu\rho}(k) = A\,g_{\nu\rho} + B\,k_\nu k_\rho$.

Putting this definition in the previous expression:

$\displaystyle \left[-(k^2 - m^2) g^{\mu\nu} + k^\mu k^\nu\right] (A\,g_{\nu\rho} + B\,k_\nu k_\rho) = \delta^\mu_\rho$

$\displaystyle -A\,g_{\nu\rho} (k^2 - m^2) g^{\mu\nu} + A\,g_{\nu\rho} k^\mu k^\nu - B\,k_\nu k_\rho (k^2 - m^2) g^{\mu\nu} + B\,k_\nu k_\rho k^\mu k^\nu = \delta^\mu_\rho$

$\displaystyle -A (k^2 - m^2) \delta^\mu_\rho + A\,k^\mu k_\rho - B\,k^\mu k_\rho (k^2 - m^2) + B\,k^2 k_\rho k^\mu = \delta^\mu_\rho$

$\displaystyle -A (k^2 - m^2) \delta^\mu_\rho + A\,k^\mu k_\rho + B\,k^\mu k_\rho m^2 = \delta^\mu_\rho$

$\displaystyle -A (k^2 - m^2) \delta^\mu_\rho + (A + B\,m^2) \,k^\mu k_\rho = \delta^\mu_\rho$

This gives us two equations:

$\displaystyle -A (k^2 - m^2) = 1 \to A = -\frac{1}{k^2-m^2}$

$\displaystyle A + B\,m^2 = 0 \to B = -\frac{A}{m^2} = \frac{1}{m^2(k^2 - m^2)}$

Putting all together:

$\displaystyle \widetilde{D}_{\nu\rho}(k) = -\frac{1}{k^2-m^2}g_{\nu\rho} + \frac{1}{m^2(k^2 - m^2)} k_\nu k_\rho$

$\displaystyle = \frac{-g_{\nu\rho} + \frac{k_\nu k_\rho}{m^2}}{k^2-m^2}$

Why this doesn’t work?

$\displaystyle \left[(-k^2 + m^2) g^{\mu\nu}k_\mu k_\nu + k^\mu k^\nu k_\mu k_\nu\right] \widetilde{D}_{\nu\rho}(k) = \delta^\mu_\rho k_\mu k_\nu$

$\displaystyle \left[(-k^2 + m^2) k^2 + k^4\right] \widetilde{D}_{\nu\rho}(k) = k_\rho k_\nu$

$\displaystyle \left[m^2 k^2 \right] \widetilde{D}_{\nu\rho}(k) = k_\rho k_\nu$

$\displaystyle \widetilde{D}_{\nu\rho}(k) = \frac{k_\rho k_\nu}{m^2 k^2}$

It’s easy, the derivation is incoherent since the first step:

$\displaystyle g^{\mu\nu}k_\mu k_\nu \widetilde{D}_{\nu\rho}$

This term doesn’t make sense.

##### Bypassing Maxwell – Propagator

$\displaystyle \sum_i \epsilon_\nu^{(i)}(k) \epsilon_\lambda^{(i)}(k) = A g_{\nu\lambda} + B k_\nu k_\lambda$

If we multiply both sides by $k^\nu k^\lambda$ and apply (9), $k^\mu \epsilon_\mu^{(a)} = 0$, we get

$\displaystyle \sum_i k^\nu k^\lambda \epsilon_\nu^{(i)}(k) \epsilon_\lambda^{(i)}(k) = A k^\nu k^\lambda g_{\nu\lambda} + B k^\nu k^\lambda k_\nu k_\lambda$

$\displaystyle 0 = A m^2 + B m^4$

$\displaystyle -\frac{A}{m^2} = B$.

So the equation has the form

$\displaystyle \sum_i \epsilon_\nu^{(i)}(k) \epsilon_\lambda^{(i)}(k) = A \left(g_{\nu\lambda} - \frac{k_\nu k_\lambda}{m^2}\right)$.

If we now work in the rest frame of the particle for $\nu = \lambda = 1$:

$\displaystyle \sum_i \epsilon_1^{(i)}(k) \epsilon_1^{(i)}(k) = A \left(g_{11} - \frac{k_1 k_1}{m^2}\right)$

$\displaystyle 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = A \left(-1 - \frac{0 \cdot 0}{m^2}\right)$

$\displaystyle 1 = -A$

$\displaystyle A = -1$.

So the numerator in the propagator is

$\displaystyle -\left(g_{\nu\lambda} - \frac{k_\nu k_\lambda}{m^2}\right) = -g_{\nu\lambda} + \frac{k_\nu k_\lambda}{m^2}$,

as we found starting from the Maxwell Equations.

##### I.5.1

Definition of $G_{\mu\nu}$:

$G_{\mu\nu}(k) = g_{\mu\nu} - \frac{k_\mu k_\nu}{m^2}$

Let’s first start writing all terms with four covector indices $\mu$, $\nu$, $\lambda$ and $\sigma$ that are symmetric under $\mu\nu \leftrightarrow \lambda\sigma$, $\mu \leftrightarrow \nu$ and $\lambda \leftrightarrow \sigma$:

1. $G_{\mu\nu} G_{\lambda\sigma}$
2. $G_{\mu\sigma} G_{\lambda\nu} + G_{\mu\lambda} G_{\nu\sigma}$
3. $G_{\mu\nu}k_\lambda k_\sigma + G_{\lambda\sigma}k_\mu k_\nu$
4. $G_{\mu\sigma} k_\lambda k_\nu + G_{\nu\sigma} k_\lambda k_\mu + G_{\mu\lambda} k_\sigma k_\nu + G_{\lambda\nu} k_\mu k_\sigma$
5. $k_\mu k_\nu k_\lambda k_\sigma$

This means that our general expression for the numerator will be:

$\displaystyle \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = A G_{\mu\nu} G_{\lambda\sigma} + B(G_{\mu\sigma} G_{\lambda\nu}\,+ G_{\mu\lambda} G_{\nu\sigma})$

$\displaystyle +\,C(G_{\mu\nu}k_\lambda k_\sigma + G_{\lambda\sigma}k_\mu k_\nu)$

$\displaystyle +\,D(G_{\mu\sigma} k_\lambda k_\nu + G_{\nu\sigma} k_\lambda k_\mu + G_{\mu\lambda} k_\sigma k_\nu + G_{\lambda\nu} k_\mu k_\sigma) + Ek_\mu k_\nu k_\lambda k_\sigma$

Eq. 13 for reference:

$\displaystyle \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = (G_{\mu\lambda} G_{\nu\sigma} + G_{\mu\sigma} G_{\nu\lambda}) - \frac{2}{3} G_{\mu\nu} G_{\lambda\sigma}$

Now we need to apply two conditions to get (13):

• $k^\mu \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = 0$
• $g^{\mu\nu} \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = 0$

Let’s calculate the result of multiplying $G_{\mu\nu}$ by $k^\mu$ and $g^{\mu\nu}$:

$\displaystyle k^\mu G_{\mu\nu} = k^\mu \left(g_{\mu\nu} - \frac{k_\mu k_\nu}{m^2}\right)$

$= k^\mu g_{\mu\nu} - \frac{k^\mu k_\mu k_\nu}{m^2}$

$= k_\nu - \frac{m^2 k_\nu}{m^2}$

$= k_\nu - k_\nu$

$= 0$

$\displaystyle g^{\mu\nu} G_{\mu\nu} = g^{\mu\nu} \left(g_{\mu\nu} - \frac{k_\mu k_\nu}{m^2}\right)$

$= g^{\mu\nu} \left(g_{\mu\nu} - \frac{k_\mu k_\nu}{m^2}\right)$

$= g^{\mu\nu} g_{\mu\nu} - \frac{g^{\mu\nu} k_\mu k_\nu}{m^2}$

$= 4 - \frac{m^2}{m^2}$

$= 4 - 1$

$= 3$

Now let’s analyze the results of products that aren’t completely contracted (OK, I could have started with those 😀 ):

$\displaystyle g^{\lambda\mu} G_{\mu\nu} = g^{\lambda\mu} \left(g_{\mu\nu} - \frac{k_\mu k_\nu}{m^2}\right)$

$\displaystyle = g^{\lambda\mu} g_{\mu\nu} - \frac{g^{\lambda\mu} k_\mu k_\nu}{m^2}$

$\displaystyle = \delta^\lambda_\nu - \frac{k^\lambda k_\nu}{m^2}$

Analyzing the product between $k^\mu$ and $\sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k)$:

$\displaystyle k^\mu \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = 0$

$\displaystyle A k^\mu G_{\mu\nu} G_{\lambda\sigma} + B(k^\mu G_{\mu\sigma} G_{\lambda\nu}\,+ k^\mu G_{\mu\lambda} G_{\nu\sigma})$

$\displaystyle +\,C(k^\mu G_{\mu\nu}k_\lambda k_\sigma + k^\mu G_{\lambda\sigma}k_\mu k_\nu)$

$\displaystyle +\,D(k^\mu G_{\mu\sigma} k_\lambda k_\nu + k^\mu G_{\nu\sigma} k_\lambda k_\mu + k^\mu G_{\mu\lambda} k_\sigma k_\nu + k^\mu G_{\lambda\nu} k_\mu k_\sigma)$

$\displaystyle +\,E k^\mu k_\mu k_\nu k_\lambda k_\sigma = 0$

$\displaystyle C m^2 G_{\lambda\sigma} k_\nu + D m^2 (G_{\nu\sigma} k_\lambda + G_{\lambda\nu} k_\sigma) +\,E m^2 k_\nu k_\lambda k_\sigma = 0$

Replacing $G_{\mu\nu}$ by its definition:

$\displaystyle C (m^2 g_{\lambda\sigma} + k_\lambda k_\sigma) k_\nu + D \left[ (m^2 g_{\nu\sigma} + k_\nu k_\sigma) k_\lambda + (m^2 g_{\lambda\nu} + k_\lambda k_\nu ) k_\sigma \right]$

$\displaystyle +\,E m^2 k_\nu k_\lambda k_\sigma = 0$

$\displaystyle C m^2 g_{\lambda\sigma} k_\nu + D m^2 (g_{\nu\sigma} k_\lambda + g_{\lambda\nu} k_\sigma) + (C + 2D + m^2 E) k_\lambda k_\sigma k_\nu = 0$

Let’s try the values $\lambda = \sigma = 1$ and $\nu = 0$, using $k$ in a reference frame where it is at rest:

$\displaystyle C m^2 g_{11} k_0 + D m^2 (g_{01} k_1 + g_{10} k_1) + (C + 2D + m^2 E) k_1 k_1 k_0 = 0$

$\displaystyle C m^2 \cdot 1 \cdot (-m) + D m^2 (0 \cdot 0 + 0 \cdot 0) + (C + 2D + m^2 E) 0 \cdot 0 \cdot (-m) = 0$

$\displaystyle -C m^3 = 0$

$\displaystyle C = 0$

Now substituting $\lambda = 0$ and $\nu = \sigma = 1$:

$\displaystyle D m^2 (g_{11} k_0 + g_{01} k_1) + (2D + m^2 E) k_0 k_1 k_1 = 0$

$\displaystyle D m^2 (1 \cdot (-m) + 0 \cdot 0) + (2D + m^2 E) (-m) \cdot 0 \cdot 0 = 0$

$\displaystyle -D m^3 = 0$

$\displaystyle D = 0$

Finally, using $\lambda = \nu = \sigma = 0$ we get:

$\displaystyle m^2 E k_0 k_0 k_0 = 0$

$\displaystyle m^2 E (-m)(-m)(-m) = 0$

$\displaystyle -m^5 E = 0$

$\displaystyle E = 0$

Now we know better the form of the numerator:

$\displaystyle \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = A G_{\mu\nu} G_{\lambda\sigma} + B(G_{\mu\sigma} G_{\lambda\nu}\,+ G_{\mu\lambda} G_{\nu\sigma})$

We cannot learn anything more multiplying $k^\mu$, as this terms go to zero for any value of $k^\mu$. But we can use $g^{\mu\nu}$:

$\displaystyle g^{\mu\nu} \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = 0$

$\displaystyle A g^{\mu\nu} G_{\mu\nu} G_{\lambda\sigma} + B(g^{\mu\nu}G_{\mu\sigma} G_{\lambda\nu}\,+ g^{\mu\nu} G_{\mu\lambda} G_{\nu\sigma}) = 0$

$\displaystyle 3 A G_{\lambda\sigma} + B(g^{\mu\nu}G_{\mu\sigma} G_{\lambda\nu}\,+ g^{\mu\nu} G_{\mu\lambda} G_{\nu\sigma}) = 0$

$\displaystyle 3 A G_{\lambda\sigma} + B\left[\left(\delta^\nu_\sigma - \frac{k^\nu k_\sigma}{m^2}\right) G_{\lambda\nu}\,+ \left(\delta^\nu_\lambda - \frac{k^\nu k_\lambda}{m^2}\right) G_{\nu\sigma}\right] = 0$

$\displaystyle 3 A G_{\lambda\sigma} + B\left[\delta^\nu_\sigma G_{\lambda\nu} - \frac{k^\nu k_\sigma}{m^2} G_{\lambda\nu} \,+ \delta^\nu_\lambda G_{\nu\sigma} - \frac{k^\nu k_\lambda}{m^2} G_{\nu\sigma} \right] = 0$

$\displaystyle 3 A G_{\lambda\sigma} + B\left[G_{\lambda\sigma} + G_{\lambda\sigma}\right] = 0$

$\displaystyle (3 A + 2 B) G_{\lambda\sigma} = 0$

$\displaystyle 3 A + 2 B = 0$

$\displaystyle B = -\frac{3}{2}A$

Then, up to an overall proportionality constant, we have:

$\displaystyle \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = A \left[G_{\mu\nu} G_{\lambda\sigma} -\frac{3}{2}\left(G_{\mu\sigma} G_{\lambda\nu}\,+ G_{\mu\lambda} G_{\nu\sigma}\right)\right]$

Evaluating the components of this equation at rest for $\mu = \lambda = 1$ and $\nu = \sigma = 2$:

$\displaystyle \sum_a \epsilon_{12}^{(a)} \epsilon_{12}^{(a)} = A \left[G_{12} G_{12} -\frac{3}{2}\left(G_{12} G_{12}\,+ G_{11} G_{22}\right)\right]$

$\displaystyle 1 = A \left[0 -\frac{3}{2}\left(0 + 1\right)\right]$

$\displaystyle 1 = A \left[-\frac{3}{2}\right]$

$\displaystyle A = -\frac{2}{3}$.

Finally:

$\displaystyle \sum_a \epsilon_{\mu\nu}^{(a)}(k) \epsilon_{\lambda\sigma}^{(a)}(k) = G_{\mu\sigma} G_{\lambda\nu} + G_{\mu\lambda} G_{\nu\sigma} -\frac{2}{3}G_{\mu\nu}G_{\lambda\sigma}$