Minimal substitution on the Lagrangian of the complex KG field

In summary: 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)+\cdots$$$$\mathscr{C}\mathscr{L}_I\mathscr{C}^{-1} &= \frac{ie}{\hbar c} A_{\alpha}\Big(\
  • #1
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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 :wink:)

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

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 :biggrin:
 
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  • #2
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
 

Related to Minimal substitution on the Lagrangian of the complex KG field

What is minimal substitution on the Lagrangian of the complex KG field?

Minimal substitution is a mathematical technique used in quantum field theory to incorporate the dynamics of a charged particle into the Lagrangian of a complex Klein-Gordon (KG) field. It replaces the mass term in the Lagrangian with the mass of the particle and introduces a coupling term between the field and the particle's charge.

Why is minimal substitution important in quantum field theory?

Minimal substitution is important because it allows us to describe the interactions between charged particles and fields in a consistent and mathematically tractable way. It is a crucial step in the development of quantum field theories that incorporate electromagnetism.

How does minimal substitution affect the equations of motion for the KG field?

Minimal substitution introduces an additional term in the equations of motion for the KG field, representing the interaction between the field and the charged particle. This modifies the dynamics of the field, leading to new phenomena such as particle creation and annihilation.

Can minimal substitution be applied to other fields besides the KG field?

Yes, minimal substitution can be applied to any field theory that includes charged particles. It is a general technique that has been used in various areas of physics, including quantum electrodynamics and the Standard Model of particle physics.

Are there any limitations or criticisms of minimal substitution?

One criticism of minimal substitution is that it is not a fully self-consistent approach, as it does not take into account the back-reaction of the charged particle on the field. This can lead to inconsistencies in certain situations, such as when the particle's charge becomes large. Additionally, some researchers have proposed alternative approaches to describing the interaction between fields and particles, such as the covariant derivative method.

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