Quantizing the Klein-Gordon equation

  • Thread starter Chopin
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  • #1
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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?
 
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Answers and Replies

  • #2
strangerep
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[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. [...]
Try thinking about:

[tex]
\varphi(x) ~=~ \int\! d^4k ~\Big( a(k) e^{ik\cdot x} + a^*(k) e^{-ik \cdot x} \Big)
\delta(k^2 - m^2)
[/tex]

where the nonbold k,x, etc are now 4-vectors. This expression is now more obviously
Lorentz-invariant, and the delta distribution constrains it to be a KG solution.
Now perform the t-integration. This is nontrivial contour integration so if you haven't
done it before, you'll need a textbook. One of Greiner's texts does it slowly, iirc, but
there's probably others. You'll end up with the 3D integral that has an energy term
in the denominator.

Another way to see Lorentz-invariance of the 3D term directly is to examine the
transformation properties of [itex]d^3{\textbf k}[/itex] . Peskin & Schroeder do this, iirc.
 
  • #3
Fredrik
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I posted the "physicist's derivation" of a delta function identity that can be used to evaluate the integral in strangerep's post here.
 
  • #4
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Ok, that clears things up a bit. The text I'm using (Srednicki) basically does the first approach, although I don't think it explains it all that well. It just says that this is because

[tex]\int{dx\:\delta(g(x))} = \sum_{i}\frac{1}{g'(x_i)}[/tex]

Where [tex]x_i[/tex] are the zeros of [tex]g(x)[/tex]. But it doesn't give any proof of this. Fredrik's derivation basically makes sense--it's just an integration by substitution, along with a little prayer that all of that stuff still works with distributions.

If I wanted to go the second route, and examine the transformation properties of [tex]d^3\textbf{k}[/tex], is it correct to just say that under a boost, [tex]d^3\textbf{k}[/tex] transforms by a factor of [tex]\gamma[/tex], so I just need to divide by something else which transforms by the same amount, like [tex]\omega[/tex]?
 
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  • #5
strangerep
Science Advisor
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996
[...]
If I wanted to go the second route, and examine the transformation properties of [tex]d^3\textbf{k}[/tex], is it correct to just say that under a boost, [tex]d^3\textbf{k}[/tex] transforms by a factor of [tex]\gamma[/tex], so I just need to divide by something else which transforms by the same amount, like [tex]\omega[/tex]?
Yes, that's the basic idea. A volume element gets divided by [itex]\gamma[/itex] in the
boosted frame due to Lorentz contraction, so this needs to be compensated somehow.
 

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