Klein Gordon eqn, decoupling degrees of freedom

AI Thread Summary
The discussion centers on the Klein-Gordon equation for a real scalar field and the process of decoupling degrees of freedom using the Fourier transform. The transformed equation reveals that \(\phi(\overline{p}, t)\) satisfies a second-order differential equation, which is crucial for understanding the behavior of the field. A key point is that each term in the Fourier-transformed equation must be zero due to the linear independence of the exponential functions, leading to the conclusion that the integral of the transformed equation must also equal zero. The participants clarify the reasoning behind this decoupling and the handling of integrals in the context of quantum field theory. This exchange highlights the importance of Fourier transforms in simplifying complex equations in quantum field theory.
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Having some trouble following my notes in QFT. Any help greatly appreciated.

We have the Klein Gordon equation for a real scalar field \phi\left(\overline{x},t\right); \partial_{\mu}\partial^{\mu}\phi + m^{2}\phi = 0.

To exhibit the coordinates in which the degrees of freedom decouple from each other, we take the Fourier transform, \phi\left(\overline{x},t\right)= \int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{i \overline{p} .\overline{x}}\phi\left(\overline{p},t\right).

Then \phi\left(\overline{p},t\right) satisfies \left(\frac{\partial^{2}}{\partial t^{2}}+\left(\overline{p}^{2} + m^{2}\right)\right)\phi\left(\overline{p},t\right) = 0.

If you do it by brute force you get \int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} .\overline{x}}\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) - \int\frac{d^{3}p}{\left(2\pi\right)^{3}}\partial^{2}_{i}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) + m^{2}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) = 0

then \int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} .\overline{x}}\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \int\frac{d^{3}p}{\left(2\pi\right)^{3}} \overline{p}^{2} e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) + m^{2}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) = 0

Now I don't see how to get rid of the intergrals. I can see its similar to a delta function, but you can't just take the \phi\left(\overline{p},t\right) out of the integrals because the measure is p.

Thanks for helping me with this, please let me know if I haven't been clear.
 
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Your last line can be rewritten as
\int\frac{d^{3}p}{\left(2\pi\right)^{3}}[\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \overline{p}^{2} \phi\left(\overline{p},t\right) + m^{2}\phi\left(\overline{p},t\right)]e^{ i \overline{p} .\overline{x}} = 0
Each term of e^{ip.x} is linearly independent (as they form a basis) and thus each term must be identically 0, giving the desired relation. Identically you can say the Fourier transform of \frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \overline{p}^{2} \phi\left(\overline{p},t\right) + m^{2}\phi\left(\overline{p},t\right) is 0. So it must be 0 as well.
 
Ah, that's why. Thanks very much, much appreciated.
 
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