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Concrete example

gauging the complex scalar field

$\mathcal{L}=(\partial_\mu \phi)(\partial^\mu \phi^*)+m^2 \phi^*\phi$

$\phi(x) \rightarrow e^{-i\Lambda(x)}\phi(x)$

$A_\mu \rightarrow A_\mu + \frac{1}{q}\partial_\mu \Lambda$

The covariant derivative is then

$D_\mu = \partial_\mu + iq A_\mu$

I am having trouble reconciling this with a more general formula for the covariant derivative in a gauge theory from Chapter 11 of Freedman and Van Proeyen’s supergravity textbook which reads

The covariant derivative of any covariant quantity is given by the operator

$$\mathcal{D}_\mu = \partial_\mu - \delta(B_\mu) \tag{11.39}$$

acting on that quantity. The instruction $\delta(B_\mu)$ means compute all gauge transformations of the quantity, with the potential $B_\mu^A$ as the gauge parameter.

We were given previously in the text, the formula for a symmetry transformation on the gauge field,

the parameters $\epsilon^A(x)$ are arbitrary functions on spacetime. To realize local symmetry in a Lagrangian field theory, one needs a gauge field, which we will generically denote by $B_\mu^A(x)$, for every gauged symmetry. The gauge fields transform as

$$\delta(\epsilon){B_\mu}^A \equiv \partial_\mu \epsilon^A + \epsilon^C {B_\mu}^B {f_{BC}}^A . \tag{11.24}$$

but I am struggling to rectify the covariant derivative expression with this prescription of the symmetry transformation on the gauge field.

For details on the nomenclature of this textbook, please see my previous post, Gauge theory formalism.

The only relation I can make with my concrete $U(1)$ example from above, is that the formula for the symmetry transformation on the gauge field from this textbook matches up if I take the coupling $q=1$, since $\Lambda$ takes the place of the symmetry transformation's parameter $e^A$ in the textbook.

However, the formula for the covariant derivative in the $U(1)$ case IS NOT

$\begin{eqnarray*} D_\mu &=& \partial_\mu - \delta(A_\mu) \\ &=& \partial_\mu - \partial_\mu \Lambda \end{eqnarray*}$

as the formula from the textbook prescribes.

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    $\begingroup$ Please type out the question yourself instead of using images. Avoid using images as they make the question less accessible and images might not look great in mobile devices. $\endgroup$
    – user243267
    Dec 19, 2019 at 14:48

2 Answers 2

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You are misreading all formulas in a maximally disruptive way.

For starters, it is not gauge fields (photons) that carry the charge q of the arbitrary charge covariant quantity, as it has to gauge all quantities with all charges! The charge is a property of the representation of the covariant quantity itself, instead, $$\mathcal{L}=(\partial_\mu \phi)(\partial^\mu \phi^*)+m^2 \phi^*\phi \\ \phi(x) \rightarrow e^{-iq\Lambda(x)}\phi(x)\\ A_\mu \rightarrow A_\mu + \partial_\mu \Lambda \\ D_\mu = \partial_\mu + iq A_\mu ,\\ \implies D_\mu \phi \to e^{-iq\Lambda(x)} D_\mu \phi . $$ So the covariant derivative of the covariant quantity transforms like the quantity itself: this is its very defining function.

Now, the only piece of the nonabelian 11.24 that survives upon abelian reduction (suppression of the structure constant f) is the first, gradient term, $$ \delta A_\mu = \partial_\mu \Lambda , $$ so (11.39) collapses, for your given infinitesimal abelian gauge transformation on $\phi\to \phi - iq\delta\Lambda ~\phi$ to but $$ D_\mu = \partial_\mu - (-iqA_\mu) = \partial_\mu +iqA_\mu . $$

The nontrivial, and delightful part is when you carry this out in the nonabelian case, and, e.g., apply it to gauge fields instead of matter representations!

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My confusion resides in adapting the author's notation to my own. Where the authors wrote $\delta(\epsilon)\phi$, I would write $\delta_\epsilon (\phi)$. This lead me to see the quantity in question $\delta(B_\mu)$ as the variation of the gauge field's transformation, when in fact it is merely denoting that I ought to use the gauge field itself as the parameter of the symmetry transformation. I.e. where the original symmetry transformation read $\delta(\epsilon)\phi= \epsilon^A T_A \phi,$ now we have $\delta(B_\mu)\phi = B_\mu{}^A T_A \phi.$

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  • $\begingroup$ I gather my answer made that clear. It is not acceptable? $\endgroup$ Dec 19, 2019 at 23:09

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