# 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).

• I think you are on the right track. You now have to compute the energy of the vacuum with this propagator. – Adam May 11 '14 at 1:43
• @Adam : In that case, the energy of the vacuum can be calculated from the first order term in $Z[J]$, which is given by $$\bigg(i\int \mathrm d^4x \;\mathrm d^4x'J(x')\Delta_F(x-x') J(x) \bigg) \qquad \qquad (1)$$ with sources being replaced as delta functions ? – user35952 May 11 '14 at 14:59
• No. The energy of the vacuum is given by $\frac{1}{2}Tr\log\Delta$. This can be computed by differentiating with respect to $m^2$, doing the integral over momenta, integrating with respect to $m^2$ (with boundary condition that the integral vanishes for $m^2\to\infty$). – Adam May 11 '14 at 16:58
• Then what about the case of massless scalar field ? – user35952 May 11 '14 at 17:30
• Then the calculation should be pretty much the same than that of photons. – Adam May 11 '14 at 17:46

$$\text{Tr} \log \Delta - \text{Tr} \log \Delta_{free}$$
This difference should yield a finite answer. The physical meaning is clear: the energy stored in vacuum is obviously divergent due to the $$\frac{1}{2}\hbar \omega$$ term in the zero-point energy from the canonical approach. What is observable is the shift in vacuum energy due to interaction, which in your formalism is signified by the boundary condition.