Charge conjugation of Complex Klein Gordon Lagrangian

In summary, the conversation discusses the complex Klein-Gordon Lagrangian density and how it is invariant under charge conjugation. The transformation of the fields \phi(x) and \phi^\dagger (x) can be written as \phi(x)\rightarrow C\phi(x)C^{-1}=\eta_c \phi^\dagger (x). The solution is valid and it can also be shown that C a(\mathbf{k})C^{-1}=\eta_c b(\mathbf{k}) and C b(\mathbf{k})C^{-1}=\eta_c^* a(\mathbf{k}). The second question is resolved by considering the fact that applying charge conjugation twice takes one back to the original state.
  • #1
orentago
27
0

Homework Statement



Show that the complex Klein-Gordon Lagrangian density:

[tex]L=N\left(\partial_\alpha\phi^{\dagger}(x)\partial^\alpha\phi(x)-\mu^2\phi^{\dagger}(x)\phi(x)\right)[/tex]

is invariant under charge conjugation:

[tex]\phi(x)\rightarrow C\phi(x)C^{-1}=\eta_c \phi^\dagger (x)[/tex]

Where [tex]C[/tex] is a unitary operator and [tex]\eta_c[/tex] is a phase factor.

Homework Equations





The Attempt at a Solution



The transformation can also be written as follows: [tex]\phi^\dagger (x) \rightarrow \eta_c^{-1} \phi(x)[/tex]

Hence performing the transformations on [tex]\phi(x)[/tex] and [tex]\phi^\dagger (x)[/tex] gives:

[tex]N\left(\partial_\alpha(\eta_c^{-1}\phi(x))\partial^\alpha(\eta_c\phi^\dagger (x))-\mu^2(\eta_c^{-1}\phi(x))(\eta_c \phi^\dagger(x))\right)=N\left(\partial_\alpha\phi(x)\partial^\alpha \phi^\dagger (x)-\mu^2\phi(x) \phi^\dagger(x)\right)=N\left(\partial^\alpha\phi(x)\partial_\alpha \phi^\dagger (x)-\mu^2\phi(x) \phi^\dagger(x)\right)[/tex]

Where the final step can be made fairly easily by raising and lowering indices. I'm a little unsure over my first assumption about how [tex]\phi^\dagger (x)[/tex] transforms, but otherwise I'm fairly confident in the rest of my steps. Is this solution valid?
 
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  • #2
I have another question. It can be shown that [tex]C a(\mathbf{k})C^{-1}=\eta_c b(\mathbf{k})[/tex] and [tex]C b(\mathbf{k})C^{-1}=\eta_c^* a(\mathbf{k})[/tex].

I can show the first one fairly easily by just substituting the full expressions of the fields into the transformation condition above. I'm a little uncertain about the second one. I can't see how to derive it using the same method as the first one. My other idea was that it could be derived from the first expression, by multiplying both sides from the left by [tex]C^{-1}[/tex] and from the right by [tex]C[/tex], to get rid of the operators around the [tex]a(\mathbf{k})[/tex], but then this wouldn't give the right expression on the RHS.

Any thoughts?
 
Last edited:
  • #3
Does anyone have any ideas for the second question, or should I supply more information?
 
  • #4
Actually no worries I've sorted it: applying charge conjugation twice takes one back to the original state. It's so simple I don't know why I didn't think of it!
 
  • #5




Your solution is valid and follows the correct steps for showing that the complex Klein-Gordon Lagrangian density is invariant under charge conjugation. Your assumption about how \phi^\dagger (x) transforms is correct, as it is consistent with the definition of charge conjugation. Overall, your response is clear and well-explained. Great job on your solution!
 

Related to Charge conjugation of Complex Klein Gordon Lagrangian

What is charge conjugation?

Charge conjugation is a type of symmetry operation in physics that involves swapping particles with their corresponding antiparticles, while also reversing their electric charge. This operation is commonly denoted by the letter C.

What is a Complex Klein Gordon Lagrangian?

A Complex Klein Gordon Lagrangian is a mathematical expression that describes the dynamics of a complex scalar field in the context of quantum field theory. It is based on the Klein-Gordon equation, which is a relativistic wave equation that describes spinless particles.

How does charge conjugation of Complex Klein Gordon Lagrangian work?

Charge conjugation of Complex Klein Gordon Lagrangian involves applying the charge conjugation operation to the fields and parameters in the Lagrangian. This results in a new Lagrangian that is equivalent to the original one, but with the particles and antiparticles swapped and their charges reversed.

Why is charge conjugation important in physics?

Charge conjugation is important in physics because it is a fundamental symmetry operation that helps us understand the behavior of particles and antiparticles. It is one of the three CPT transformations (charge conjugation, parity, and time reversal) that are crucial for understanding the laws of physics.

Are there any applications of charge conjugation of Complex Klein Gordon Lagrangian?

Yes, charge conjugation of Complex Klein Gordon Lagrangian has various applications in theoretical and experimental physics. For example, it is used in the study of particle-antiparticle interactions, as well as in the development of theories such as quantum chromodynamics and the standard model of particle physics.

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