Self-Interactions of the unphysical gauge bosons $W_1, W_2, W_3$ are written within the gauge term

$L_\mathrm{Gauge}=-\frac{1}{4} W_{\mu \nu} W^{\mu \nu}$ with $W_{\mu \nu}= \partial_\mu W_\nu - \partial_\nu W_\mu + i g [W_\mu,W_\nu]$.

Using now the transformations for Mass matrix diagonalization

$W_{3 \mu} = \mathrm{cw} Z_\mu + \mathrm{sw} A_\mu$,

$W_{2 \mu} = \frac{i}{\sqrt{2}}(W^+_\mu - W^-_\mu)$,

$W_{1 \mu} = \frac{1}{\sqrt{2}}(W^+_\mu + W^-_\mu)$,

one is able to derive e.g. the vertex of WWA. It contains terms of the form $(\partial A) W^+ W^-, A (\partial W^+) W^-, A W^+ (\partial W^-)$. To be precise it should read \begin{align} -i e [ \partial^\mu W^{\nu +} (A_\nu W_\mu^- - A_\mu W_\nu^-) + \partial^\mu W^{\nu -} (A_\mu W_\nu^+ - A_\nu W_\mu^+ ) + \partial^\mu A^\nu (W_\nu^+ W_\mu^- - W_\mu^+ W_\nu^-)] \end{align}

The Photon gradient term can be integrated by parts, then I get \begin{align} -i e \partial^\mu A^\nu (W_\nu^+ W_\mu^- - W_\mu^+ W_\nu^-) = i e A^\nu (\partial^\mu W_\nu^+ W_\mu^- - W_\mu^+ \partial^\mu W_\nu^-) \end{align} and in total this would mean for the WWA vertex \begin{align} -i e &[ \partial^\mu W^{\nu +} (A_\nu W_\mu^- - A_\mu W_\nu^-) + \partial^\mu W^{\nu -} (A_\mu W_\nu^+ - A_\nu W_\mu^+ ) - A^\nu (\partial^\mu W_\nu^+ W_\mu^- - W_\mu^+ \partial^\mu W_\nu^-)] \\ = -i e &[ -\partial^\mu W^{\nu +} A_\mu W_\nu^- + \partial^\mu W^{\nu -} A_\mu W_\nu^+ ] \end{align}

Now I tried to rewrite the gauge boson self interactions in terms of the covariant derivative using directly the physical fields by writing down \begin{align} L_\mathrm{cov} = -\frac{1}{2}(D_\mu W_\nu^+ - D_\nu W_\mu^+) (D^\mu W^{\nu -} - D^\nu W^{\mu -}) - \frac{1}{4}(\partial_\mu A_\nu - \partial_\nu A_\mu)^2 - \frac{1}{4}(\partial_\mu Z_\nu - \partial_\nu Z_\mu)^2 \end{align} with

$D_\mu W_\nu^+ = (\partial_\mu + i e A_\mu) W_\nu^+$,

$D_\mu W_\nu^- = (\partial_\mu - i e A_\mu) W_\nu^+$.

Collecting all contributions to WWA results in \begin{align} -i e [ \partial^\mu W^{\nu +} (A_\nu W_\mu^- - A_\mu W_\nu^-) + \partial^\mu W^{\nu -} (A_\mu W_\nu^+ - A_\nu W_\mu^+ )] \end{align}

Obviously, this is not yet the same as with $L_\mathrm{Gauge}$, so what am I missing? In the end, both approaches should lead to equivalent vertices...

It already would be helpful if somebody knows literature where these terms are calculated explicitly using the covariant derivative.

  • $\begingroup$ Have you integrated by parts the gauge action piece? $\endgroup$ Nov 24, 2023 at 12:55
  • $\begingroup$ Have added it and get a really short form for the WWA vertex, but not the same as with the L cov. $\endgroup$
    – Jan
    Nov 24, 2023 at 13:20
  • $\begingroup$ Your photon gradient term is missing from your guess for the covariant action, which is, ipso facto, not a mere rewriting of the gauge action! The correct covariant action should include it. The German WP and Schwartz’s book, where it came from, are right on this…. $\endgroup$ Nov 25, 2023 at 12:32

1 Answer 1


Your covariant action guess is incomplete. It is missing the gauge invariant $$\propto \bbox[yellow]{(\partial^\mu A^\nu - \partial^\nu A^\mu) (W_\mu^+ W_\nu^- - c.c.)}\\ \propto [D_\mu,D_\nu](W_\mu^+ W_\nu^- - c.c.),$$ giving the Ws their characteristic magnetic moment.

In both cases, EM gauge invariance is paramount, and should obtain. In both cases, the $W\partial W$ current coupling to the photon should be hermitean (with the i; or antihermitean without it!).

The gauge piece, available in good texts, e.g. M Schwartz, (29.9), is proportional to the top expression in $$ -ieA_\nu [ \partial_\mu (-W^{\mu~+} W^{\nu~-}+ W^{\nu~+}W^{\mu~-}) \\-W^{\mu~+} \partial_\nu W^{\mu~-}+ W^{\mu~-}\partial_\nu W^{\mu~+ } + W^{\mu~+} \partial_\mu W^{\nu~-}- W^{\mu~-}\partial_\mu W^{\nu~+}]\\ =... $$ You may halve these terms by subtracting the c.c.s! Proceed to rewrite it to your form.

Where did the extra term come from? you might ask. Schematicaly, and with conventional (unlike your) normalizations, recall the gauge action is the square of $$ \vec W_{\mu\nu}= \partial_\mu \vec W_\nu - \partial_\nu \vec W_\mu + g \vec W _\mu \times \vec W _\nu \leadsto \\ \vec W_{\mu\nu}^+= \partial_\mu \vec W_\nu^+ - \partial_\nu \vec W_\mu^+ -i g \vec ( W _\mu^3 \vec W _\nu^+ - \hbox{c.c.}),\\ \vec W_{\mu\nu}^3= \partial_\mu \vec W_\nu^3 - \partial_\nu \vec W_\mu^3 +i g \vec ( W _\mu^+ \vec W _\nu^- - \hbox{c.c.}), $$ where you might adjust potential wrong signs to your liking.

The key point is that for $$ gW_\mu^3= eA_\mu + g\cos\theta_W ~Z_\mu, $$ ignoring Z, this yields the now complete EM gauge-invariant action piece, $$ -ieA_\nu [ (W_\mu^+(\partial^\mu W^{\nu ~ -}- \partial^\nu W^{\mu ~ -} )- \hbox{c.c.})+ W_\nu^+ W_\mu^-(\partial^\mu A^\nu -\partial^\nu A^\mu )] $$ with the new, formerly missing, term originating in $(W^3_{\mu\nu})^2$, now put in the end.

  • $\begingroup$ Thanks! I thought one could rewrite everything using only covariant derivatives, but this is not the case. $\endgroup$
    – Jan
    Nov 26, 2023 at 16:45
  • $\begingroup$ It is the case. Updated my answer. The photon part of the extra term is the commutator of two covariant derivatives on the rest, which is gauge-invariant (0 charge!) by itself. $\endgroup$ Nov 26, 2023 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.