Saturday, February 6, 2016

quantum field theory - Casimir forces and its associated Feynman propagator



This is a continuation to my previous question, in which I began an attempt solve the Casimir Force problem using path integrals. As one of the answers there suggest I solve the Feynman propagator subject to the boundary conditions $x=0$ and $x=L$ at the plate boundaries. The equation for Feynman propagator is $$ (\Box^2+m^2)\Delta_F(x-x') = -\delta(x-x') $$


The solution to the free field is


$$ \Delta_F(x-x') = \lim_{\epsilon\rightarrow0}\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip_{\mu}(x^{\mu}-x'^{\mu})}}{p^2-m^2+i\epsilon} $$


What would be the boundary conditions that I have to exactly impose ?


Imposing a boundary condition would mean, I think we might have to introduce the a new function (I don't if am right, but this is in general true for Green's function I guess) $$ \Delta_F(x-x') \rightarrow \Delta_F(x-x') + F(x-x') $$


where $F(x-x')$ is such that it satisfies the Boundary condition.


Now my question is in case I have boundary condition (like below) how do I solve the differential equation for the boundary conditions like, (take plates to be at $z=0$ and $z=L$) $$ \Delta_F(x-x')\bigg|_{z=0} = \Delta_F(x-x')\bigg|_{z=L} = 0 $$


EDIT 1: It just occurred to me that there might be short route to this problem with some conceptual reasoning, I gave this a try..


Considering the region between the plates, I know the momentum is quantised in the z-direction, so I have (which is some sense imposed by the boundary condtions) $$ p_z = \frac{n\pi}{L} $$ Now using the Feynman propagator in momentum representation, which is $$\widetilde\Delta_F(p) = \frac{1}{(p^0)^2-(\textbf p^2+m^2)+i\epsilon}$$


In this I can substitute for, $p_z$, which will give me $$ \widetilde\Delta_F(p) = \frac{1}{(p^0)^2-(p_x^2+p_y^2+\Big(\frac{n\pi}{L}\Big)^2)+m^2)+i\epsilon} $$



Now can I get back to position representation, but with integral on $p_z$ replaced by a sum over $n$. Am I right in doing this procedure ?


EDIT 2 : Following the procedure that I have mentioned, for a simple (1+1) case of the Feynman propagator in position representation, I have


$$ \Delta_F(x-x') = \sum_{n=1}^\infty\int\frac{dp_0}{(2\pi)^2}\frac{e^{ip_0(x^0-x'^0)}e^{i\frac{n\pi}{L}(z-z')}}{(p^0)^2-\big(\big(\frac{n\pi}{L}\big)^2+m^2\big)} $$


EDIT 3 : $$ \text{Tr}\log{\Delta} = - \sum_n \int dp_0 \log{\bigg(p_0^2 - \bigg(\frac{n\pi}{L}\bigg)^2 + m^2\bigg)} $$


But this term seems to diverge, how does one obtain a cutoff in the context of this problem. (A cutoff for $p_0$ integral is also needed I guess).




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