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# 物理代考|量子场论代考QUANTUM FIELD THEORY代考|PHYS4125

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## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Covariant derivatives

In order not to affect our counting of degrees of freedom, the interactions in the Lagrangian must respect gauge invariance. For example, you might try to add an interaction
$$\mathcal{L}=\cdots+A_\mu \phi \partial_\mu \phi$$
but this is not invariant. Under the gauge transformation

$$A_\mu \phi \partial_\mu \phi \rightarrow A_\mu \phi \partial_\mu \phi+\left(\partial_\mu \alpha\right) \phi \partial_\mu \phi$$
In fact, it is impossible to couple $A_\mu$ to any field with only one degree of freedom, such as the scalar field $\phi$. We must be able to make $\phi$ transform to compensate for the gauge transformation of $A_\mu$, in order to cancel the $\partial_\mu \alpha$ term. But if there is only one field $\phi$, it has nothing to mix with so it cannot transform.

Thus, we need at least two fields $\phi_1$ and $\phi_2$. It is easiest to deal with such a doublet by putting them together into a complex field $\phi=\phi_1+i \phi_2$, and then to work with $\phi$ and $\phi^{\star}$. Under a gauge transformation, $\phi$ can transform as
$$\phi \rightarrow e^{-i \alpha(x)} \phi,$$
which makes $m^2 \phi^{\star} \phi$ gauge invariant. But what about the derivatives? $\left|\partial_\mu \phi\right|^2$ is not invariant.

We can in fact make the kinetic term gauge invariant using something we call a covariant derivative. Adding a conventional constant $e$ to the transformation of $A_\mu$, so $A_\mu \rightarrow A_\mu+$ $\frac{1}{e} \partial_\mu \alpha$, we find
$$\left(\partial_\mu+i e A_\mu\right) \phi \rightarrow\left(\partial_\mu+i e A_\mu+i \partial_\mu \alpha\right) e^{-i \alpha(x)} \phi=e^{-i \alpha(x)}\left(\partial_\mu+i e A_\mu\right) \phi .$$
This leads us to define the covariant derivative as
$$D_\mu \phi \equiv\left(\partial_\mu+i e A_\mu\right) \phi \rightarrow e^{-i \alpha(x)} D_\mu \phi,$$
which transforms just like the field does. Thus
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2+\left(D_\mu \phi\right)^{\star}\left(D_\mu \phi\right)-m^2 \phi^{\star} \phi$$
is gauge invariant. This is the Lagrangian for scalar QED.

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Gauge symmetries and conserved currents

Symmetries parametrized by a function such as $\alpha(x)$ are called gauge or local symmetries, while if they are only symmetries for constant $\alpha$ they are called global symmetries. For gauge symmetries, we can pick a separate transformation at each point in space-time. A gauge symmetry automatically implies a global symmetry. Global symmetries imply conserved currents by Noether’s theorem. For example, the Lagrangian $\mathcal{L}=-\phi^{\star} \square \phi$ of a free complex scalar field is not gauge invariant, but it does have a symmetry under which $\phi \rightarrow e^{-i \alpha} \phi$ for a constant $\alpha$ and it does have an associated Noether current.

Let us see how the Noether current changes when the gauge field is included. Expanding out the scalar QED Lagrangian, Eq. (8.52), gives
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^2+\partial_\mu \phi^{\star} \partial_\mu \phi+i e A_\mu\left(\phi \partial_\mu \phi^{\star}-\phi^{\star} \partial_\mu \phi\right)+e^2 A_\mu^2 \phi^{\star} \phi-m^2 \phi^{\star} \phi .$$
The equations of motion are
\begin{aligned} \left(\square+m^2\right) \phi & =-2 i e A_\mu \partial_\mu \phi+e^2 A_\mu^2 \phi, \ \left(\square+m^2\right) \phi^{\star} & =2 i e A_\mu \partial_\mu \phi^{\star}+e^2 A_\mu^2 \phi^{\star} . \end{aligned}
The Noether current associated with the global symmetry for which $\frac{\delta \phi}{\delta \alpha}=-i \phi$ and $\frac{\delta \phi^{\star}}{\delta \alpha}=$ $i \phi^{\star}$ is (using Eq. (3.23))
$$J_\mu=\sum_n \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_n\right)} \frac{\delta \phi_n}{\delta \alpha}=-i\left(\phi \partial_\mu \phi^{\star}-\phi^{\star} \partial_\mu \phi\right)-2 e A_\mu \phi^{\star} \phi .$$
The first term on the right-hand side is the Noether current in the free theory $(e=0)$. You should check this full current is also conserved on the equations of motion.

By the way, you might have noticed that the term in the scalar QED Lagrangian linear in $A_\mu$ is just $-e A_\mu J_\mu$. There is a quick way to see why this will happen in general. Define $\mathcal{L}0$ as the limit of a gauge-invariant Lagrangian when $A\mu=0$ (or equivalently $e=0$ ). $\mathcal{L}0$ will still be invariant under the global symmetry for which $A\mu$ is the gauge field, since $A_\mu$ does not transform when $\alpha$ is constant. If we then let $\alpha$ be a function of $x$, the transformed $\mathcal{L}0$ can only depend on $\partial\mu \alpha$. Thus, for infinitesimal $\alpha(x)$,
$$\delta \mathcal{L}0=\left(\partial\mu \alpha\right) J_\mu+\mathcal{O}\left(\alpha^2\right)$$
for some $J_\mu$. For example, in scalar QED with $A_\mu=0, \mathcal{L}0=\left(\partial\mu \phi\right)^{\star}\left(\partial_\mu \phi\right)-m^2 \phi^{\star} \phi$ and
$$\delta \mathcal{L}0=\left(\partial\mu \alpha\right) J_\mu+\left(\partial_\mu \alpha\right)^2 \phi^{\star} \phi$$
with $J_\mu$ given by Eq. (8.57). Returning to the general theory, after integration by parts the term linear in $\alpha$ is $\delta \mathcal{L}0=\alpha \partial\mu J_\mu$. Since the variation of the Lagrangian vanishes on the equations of motion for any transformation, including this one parametrized by $\alpha$, we must have $\partial_\mu J_\mu=0$ implying that $J_\mu$ is conserved. In fact, $J_\mu$ is the Noether current, since we have just rederived Noether’s theorem a different way. To make the Lagrangian invariant without using the equations of motion, we can add a field $A_\mu$ with $\delta A_\mu=\partial_\mu \alpha$ and define $\mathcal{L}=\mathcal{L}0-A\mu J_\mu$ so that
$$\delta \mathcal{L}=\mathcal{L}0-\delta A\mu J_\mu=\left(\partial_\mu \alpha\right) J_\mu-\left(\partial_\mu \alpha\right) J_\mu=0 .$$

## 物理代考|量子场论代考QUANTUM FIELD THEORY代考|Covariant derivatives

$$\mathcal{L}=\cdots+A_\mu \phi \partial_\mu \phi$$

$$A_\mu \phi \partial_\mu \phi \rightarrow A_\mu \phi \partial_\mu \phi+\left(\partial_\mu \alpha\right) \phi \partial_\mu \phi$$

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

$$J_\mu=\sum_n \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \phi_n\right)} \frac{\delta \phi_n}{\delta \alpha}=-i\left(\phi \partial_\mu \phi^{\star}-\phi^{\star} \partial_\mu \phi\right)-2 e A_\mu \phi^{\star} \phi .$$

$$\delta \mathcal{L}0=\left(\partial\mu \alpha\right) J_\mu+\mathcal{O}\left(\alpha^2\right)$$

$$\delta \mathcal{L}0=\left(\partial\mu \alpha\right) J_\mu+\left(\partial_\mu \alpha\right)^2 \phi^{\star} \phi$$

$$\delta \mathcal{L}=\mathcal{L}0-\delta A\mu J_\mu=\left(\partial_\mu \alpha\right) J_\mu-\left(\partial_\mu \alpha\right) J_\mu=0 .$$

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