Minimal substitution on the Lagrangian of the complex KG field

[JD_PM]

Homework Statement

By making the minimal substitution

$$\partial_{\alpha} \phi(x) \rightarrow D_{\alpha} \phi(x) = \Big[\partial_{\alpha} + \frac{ie}{\hbar c} A_{\alpha} (x) \Big] \phi(x)$$

$$\partial_{\alpha} \phi^{\dagger}(x) \rightarrow [D_{\alpha} \phi(x)]^{\dagger} = \Big[\partial_{\alpha} - \frac{ie}{\hbar c} A_{\alpha} (x) \Big] \phi^{\dagger}(x)$$

In the Lagrangian density $\mathscr{L} = \partial_{\alpha} \phi \partial^{\alpha} \phi^{\dagger} - \mu^2 \phi \phi^{\dagger}$ (where we'll have to normal-order* the operators $\phi(x), \phi^{\dagger}(x)$) of the complex KG scalar field $\phi(x)$.

a) Derive the Lagrangian density $\mathscr{L}_I (x)$ for the interaction of the charged bosons, described by the field $\phi(x)$, with the electromagnetic field $A^{\alpha} (x)$

b) Assuming that $\mathscr{L}_I (x)$ is invariant under the charge conjugation transformation $\mathscr{C}$, show that

$$\mathscr{C} A^{\alpha} (x) \mathscr{C}^{-1}=-A^{\alpha}(x)$$

* Let me clarify what I mean by 'normal-order': we place all absorption operators to the right of all creation operators.

Relevant Equations

$$\partial_{\alpha} \phi(x) \rightarrow D_{\alpha} \phi(x) = \Big[\partial_{\alpha} + \frac{ie}{\hbar c} A_{\alpha} (x) \Big] \phi(x)$$

$$\partial_{\alpha} \phi^{\dagger}(x) \rightarrow [D_{\alpha} \phi(x)]^{\dagger} = \Big[\partial_{\alpha} - \frac{ie}{\hbar c} A_{\alpha} (x) \Big] \phi^{\dagger}(x)$$

More below

a) I think I got this one right. Please let me know otherwise

We have (let's leave the $x$ dependence of the fields implicit )

$$\mathscr{L} = N \Big(\partial_{\alpha} \phi \partial^{\alpha} \phi^{\dagger} - \mu^2 \phi \phi^{\dagger} \Big) = \partial_{\alpha} \phi^{\dagger} \partial^{\alpha} \phi - \mu^2 \phi^{\dagger} \phi$$ $$=\Big[\partial_{\alpha} - \frac{ie}{\hbar c} A_{\alpha} \Big]\phi^{\dagger}\Big[\partial^{\alpha} + \frac{ie}{\hbar c} A^{\alpha} \Big]\phi - \mu^2 \phi^{\dagger} \phi$$

We now expand it out to get

$$\mathscr{L} = \partial_{\alpha} \phi^{\dagger}\partial^{\alpha} \phi - \Big(\frac{ie}{\hbar c}\Big)^2 A_{\alpha}A^{\alpha} \phi^{\dagger} \phi + \Big(\frac{ie}{\hbar c}\Big) A^{\alpha} (\partial_{\alpha} \phi^{\dagger}) \phi - \Big(\frac{ie}{\hbar c}\Big) \phi^{\dagger} A^{\alpha} (\partial_{\alpha} \phi)- \mu^2 \phi^{\dagger} \phi$$ $$=(\partial_{\alpha}\partial^{\alpha} - \mu^2)\phi^{\dagger} \phi + \frac{ie}{\hbar c} A_{\alpha}\Big( (\partial^{\alpha} \phi^{\dagger})\phi - \phi^{\dagger} (\partial^{\alpha} \phi) \Big)+\Big(\frac{e}{\hbar c}\Big)^2 A_{\alpha}A^{\alpha} \phi^{\dagger} \phi = \mathscr{L}_0 + \mathscr{L}_I$$

Where

$$\mathscr{L}_0 = (\partial_{\alpha}\partial^{\alpha} - \mu^2)\phi^{\dagger} \phi$$

$$\mathscr{L}_I = \frac{ie}{\hbar c} A_{\alpha}\Big( (\partial^{\alpha} \phi^{\dagger})\phi - \phi^{\dagger} (\partial^{\alpha} \phi) \Big)+\Big(\frac{e}{\hbar c}\Big)^2 A_{\alpha}A^{\alpha} \phi^{\dagger} \phi \tag{*}$$

So $(*)$ is the answer.

b) Here I assumed that $(*)$ is correct. I know from exercise $(3.5)$, which I have previously solved, that

$$\phi \rightarrow \mathscr{C} \phi \mathscr{C}^{-1}=\eta_c \phi^{\dagger} = \phi^{\dagger}$$

$$\phi^{\dagger} \rightarrow \mathscr{C} \phi^{\dagger} \mathscr{C}^{-1}=\eta_c^* \phi = \phi$$

Where

- $\mathscr{C}$ is a unitary operator which leaves the vacuum invariant $\mathscr{C} |0 \rangle = |0 \rangle$

- The phase factor $\eta_c$ is arbitrary and it is customary to set it equal to $1$.

My idea was to sandwich $\mathscr{L}_I$ with the unitary operator but I am having troubles to get the desired answer. Before posting the whole detailed attempt I'd like to kindly ask

1) Is the following correct

$$\mathscr{C} \phi^{\dagger} \phi \mathscr{C} = \mathscr{C} \phi^{\dagger} \mathscr{C} + \mathscr{C} \phi \mathscr{C} = \phi + \phi^{\dagger}\rm \,\,\,?$$

2) Is $(*)$ correct?

Thank you

 

[Nate810]

Yes, your answer to (a) is correct. For (b), you have the right idea of sandwiching $\mathscr{L}_I$ with the unitary operator, and your equation $$\mathscr{C} \phi^{\dagger} \phi \mathscr{C} = \mathscr{C} \phi^{\dagger} \mathscr{C} + \mathscr{C} \phi \mathscr{C} = \phi + \phi^{\dagger}$$ is correct.Now, applying the unitary operator on $\mathscr{L}_I$ gives\begin{align}\mathscr{C}\mathscr{L}_I\mathscr{C}^{-1} &= \frac{ie}{\hbar c} A_{\alpha}\Big(\mathscr{C}(\partial^{\alpha} \phi^{\dagger})\phi \mathscr{C}^{-1} - \mathscr{C}\phi^{\dagger}(\partial^{\alpha} \phi)\mathscr{C}^{-1} \Big)+\Big(\frac{e}{\hbar c}\Big)^2 \mathscr{C}A_{\alpha}A^{\alpha} \phi^{\dagger} \phi\mathscr{C}^{-1} \\&= \frac{ie}{\hbar c} A_{\alpha}\Big(\mathscr{C}(\partial^{\alpha} \phi^{\dagger})\mathscr{C}\mathscr{C}\phi \mathscr{C}^{-1} - \mathscr{C}\phi^{\dagger}\mathscr{C}\mathscr{C}(\partial^{\alpha} \phi)\mathscr{C}^{-1} \Big)+\Big(\frac{e}{\hbar c}\Big)^2 \mathscr{C}A_{\alpha}A^{\alpha} \phi^{\dagger} \phi\mathscr{C}^{-1}\\&= \frac{ie}{\hbar c} A