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

Let’s start with the general form:

\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}

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