Hi all. I'm taking my first foray into QFT, and have a question which is hopefully pretty basic. I think I understand the concept itself, I just don't quite get how the math works out.

I'm right at the beginning, in the discussion of how to set up the creation/annihilation operators for a spin-0 field. The text starts with the Klein-Gordon equation:

[tex](\partial^2 + m^2)\varphi(\textbf{x}, t) = 0[/tex]

A real-valued solution to this equation is a combination of plane waves:

[tex]\varphi(\textbf{x}, t) = a\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*} \textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}[/tex]

Where [tex]\omega = +\sqrt{\textbf{k}^2 + m^2}[/tex]. Next, we try to generalize this by summing over all possible momentum vectors. A naive solution would be this:

[tex]\varphi(\textbf{x}, t) = \int{d^3\textbf{k}\:a(\textbf{k})\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*}(\textbf{k})\textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}}[/tex]

But this isn't Lorentz invariant. I think I get that--the Lorentz transform will change the size of the volume element, sort of like what happens when you take a volume integral in spherical coordinates. Is that correct?

To get around this, we do the following:

[tex]\varphi(\textbf{x}, t) = \int{\frac{d^3\textbf{k}}{(2\pi)^{3}2\omega}\: a(\textbf{k})\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*}(\textbf{k})\textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}}[/tex]

That sort of makes sense--we're basically scaling down each plane wave by its corresponding energy, which rises in proportion to the volume element as we move up the momentum scale. I don't quite understand how to show mathematically that this ends up being Lorentz invariant, though. The text basically just handwaves it, so it's no help--can anyone explain how one could arrive at that normalizing factor if one were starting from scratch?

I'm right at the beginning, in the discussion of how to set up the creation/annihilation operators for a spin-0 field. The text starts with the Klein-Gordon equation:

[tex](\partial^2 + m^2)\varphi(\textbf{x}, t) = 0[/tex]

A real-valued solution to this equation is a combination of plane waves:

[tex]\varphi(\textbf{x}, t) = a\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*} \textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}[/tex]

Where [tex]\omega = +\sqrt{\textbf{k}^2 + m^2}[/tex]. Next, we try to generalize this by summing over all possible momentum vectors. A naive solution would be this:

[tex]\varphi(\textbf{x}, t) = \int{d^3\textbf{k}\:a(\textbf{k})\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*}(\textbf{k})\textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}}[/tex]

But this isn't Lorentz invariant. I think I get that--the Lorentz transform will change the size of the volume element, sort of like what happens when you take a volume integral in spherical coordinates. Is that correct?

To get around this, we do the following:

[tex]\varphi(\textbf{x}, t) = \int{\frac{d^3\textbf{k}}{(2\pi)^{3}2\omega}\: a(\textbf{k})\textit{e}^{\textit{i}\textbf{k}\cdot\textbf{x}-{\omega}t} + a^{*}(\textbf{k})\textit{e}^{-\textit{i}\textbf{k}\cdot\textbf{x}+{\omega}t}}[/tex]

That sort of makes sense--we're basically scaling down each plane wave by its corresponding energy, which rises in proportion to the volume element as we move up the momentum scale. I don't quite understand how to show mathematically that this ends up being Lorentz invariant, though. The text basically just handwaves it, so it's no help--can anyone explain how one could arrive at that normalizing factor if one were starting from scratch?

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