Onamor
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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.
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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