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weirdoguy
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I'm reading Klaubers QFT book and I stuck with his derivation of Hamiltonian of scalar field on page 53. To derive it one needs to deal with integrals like this: $$\int\dot{\phi}\dot{\phi}^\dagger d^3x$$ He is using discrete plane-wave solutions and after plugging them in, we end up with integrals such as this one:
$$\int e^{2ikx}d^3x$$
It is said in his book, and also here:
https://physicspages.com/pdf/Klauber%20QFT/Klauber%20Problems%2003.08.pdf
that since we are integrating over finite domain and because of the boundary conditions, this integral is zero. And this is the thing that I don't understand. Also in the link above there is another comment about this boundary conditions:
In this case, we’re dealing with discrete solutions over a finite volume V , such that the values of k are determined by the condition that an integral number of wavelengths fits into V.
I've done this calculation with continuous solutions and I had no problems with that, but I really don't understand these boundary conditions and how they relate to this integral being zero. Thanks for any help
EDIT:
Ok, I went to the shop and I got enlightened xD So it's zero for the same reason $$\int_0^{2k\pi}\sin x dx$$ is zero for integer k.
$$\int e^{2ikx}d^3x$$
It is said in his book, and also here:
https://physicspages.com/pdf/Klauber%20QFT/Klauber%20Problems%2003.08.pdf
that since we are integrating over finite domain and because of the boundary conditions, this integral is zero. And this is the thing that I don't understand. Also in the link above there is another comment about this boundary conditions:
In this case, we’re dealing with discrete solutions over a finite volume V , such that the values of k are determined by the condition that an integral number of wavelengths fits into V.
I've done this calculation with continuous solutions and I had no problems with that, but I really don't understand these boundary conditions and how they relate to this integral being zero. Thanks for any help
EDIT:
Ok, I went to the shop and I got enlightened xD So it's zero for the same reason $$\int_0^{2k\pi}\sin x dx$$ is zero for integer k.
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