# Quantizing the complex Klein-Gordon field

I'm self-studying QFT and attempting exercise 2.2 on Peskin & Schroeder. First off, I'm a bit confused on the logic the authors use in the quantization process. They first expand the fields in terms of these ##a_{\vec{p}},a_{\vec{p}}^\dagger## operators which, if I understand correctly, is simply because the Klein-Gordon equation looks like the harmonic oscillator equation for each fixed ##\vec{p}##. Though, I'm not sure where the exponentials come from exactly. They then impose the commutation relations ##[a_{\vec{p}},a_{\vec{p}'}^\dagger] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{p}')## by analogy with the harmonic oscillator and use that to find a nice expression for the Hamiltonian. Now my first concrete question is: what do they mean exactly by "find the commutation relations" in part (a)? Like I said, it seems like they simply impose those relations on the creation and annihilation operators and those determine the commutation relations between ##\phi## and ##\pi##. My approach to this was to treat the complex field as ##\frac{1}{\sqrt{2}}(\phi_1 + i\phi_2)##, where the ##\phi_i## real scalar fields. I assumed that these real fields could be expressed in terms of their individual creation and annihilation operators as in the book and used those expressions to confirm the commutation relations were as expected. I was then able to do part (b) and get the correct expression for the Hamiltonian in terms of creation and annihilation operators, but when I check other sources, they seem to take a very different approach, which makes me feel that I may have oversimplified the problem. Any hints to point me in the right direction would be greatly appreciated!

## Answers and Replies

I'm self-studying QFT and attempting exercise 2.2 on Peskin & Schroeder. First off, I'm a bit confused on the logic the authors use in the quantization process. They first expand the fields in terms of these ##a_{\vec{p}},a_{\vec{p}}^\dagger## operators which, if I understand correctly, is simply because the Klein-Gordon equation looks like the harmonic oscillator equation for each fixed ##\vec{p}##. Though, I'm not sure where the exponentials come from exactly. They then impose the commutation relations ##[a_{\vec{p}},a_{\vec{p}'}^\dagger] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{p}')## by analogy with the harmonic oscillator and use that to find a nice expression for the Hamiltonian. Now my first concrete question is: what do they mean exactly by "find the commutation relations" in part (a)? Like I said, it seems like they simply impose those relations on the creation and annihilation operators and those determine the commutation relations between ##\phi## and ##\pi##. My approach to this was to treat the complex field as ##\frac{1}{\sqrt{2}}(\phi_1 + i\phi_2)##, where the ##\phi_i## real scalar fields. I assumed that these real fields could be expressed in terms of their individual creation and annihilation operators as in the book and used those expressions to confirm the commutation relations were as expected. I was then able to do part (b) and get the correct expression for the Hamiltonian in terms of creation and annihilation operators, but when I check other sources, they seem to take a very different approach, which makes me feel that I may have oversimplified the problem. Any hints to point me in the right direction would be greatly appreciated!
I think what you have got is the charged boson model which I have only seen in
Elements of Advanced Quantum Theory , A.L. Ziman, Cambridge (2002) page 28.

I think what you have got is the charged boson model which I have only seen in
Elements of Advanced Quantum Theory , A.L. Ziman, Cambridge (2002) page 28.

Well I noticed that with what I have, I can rewrite the fields in the way most online sources I’ve seen have them (with e.g the field phi being a sum of a annihilation operator of a certain type and a creation operator of a different type). These new operators end up being linear combinations of the scalar field’s operators.I just don’t see why I was able to get the correct Hamiltonian the other way.