In QED and the basic Higgs mechanism, there is a local gauge transformation where a scalar field $\phi$ is transformed as:

$e^{i\theta\eta(x)} \phi$

The partial derivative of this however makes the above not invariant, and so a covariant derivative is introduced in this way:

$D_\mu e^{i\theta\eta(x)} \phi$=$(\partial_\mu- i \theta A_\mu)$$e^{i\theta\eta(x)} \phi$

So, the derivative remains invariant. However, what if the scalar field is transformed by TWO U(1) symmetries like this:

$e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} \phi$

This may be a strange symmetry transformation, but I wonder how would one make the derivative of this invariant like this:

$e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} D_\mu \phi$

For the derivative is now of three different functions which differentiate by the product rule like this:


Thus, the derivative of the function would be:

$i \lambda_1 \eta_1' (x)e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} \phi$ +$i \lambda_2 \eta_2' (x)e^{i\lambda_2\eta_2(x)} e^{i\lambda_1\eta_1(x)} \phi$+$(\partial_\mu \phi) e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)}$

So, how would the local gauge invariance derivative be applied in this situation? Would another gauge field be introduced such as $B_\mu$ along with $A_\mu$ ?


1 Answer 1


Yes, you would have to introduce another gauge field. For example in the Standard Model there is gauge invariance under $SU(3)\times SU(2) \times U(1)$, and so there are three gauge fields: the gluons, the $W^\pm, Z$ weak gauge bosons and the photon.

In general terms, it is simpler to argue like this: if you have gauge invariance under a Lie group $G$, the covariant derivative will include a 1-form taking values in the Lie algebra $\mathfrak g$ of $G$. Since the Lie algebra of $G \times H$ is $\mathfrak g \oplus \mathfrak h$, 1-form taking values in this Lie algebra can be decomposed into one 1-form taking values in $\mathfrak g$ and one 1-form taking values in $\mathfrak h$. In your case this would be your $A_\mu$ and $B_\mu$.


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