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).

  • 1
    $\begingroup$ I think you are on the right track. You now have to compute the energy of the vacuum with this propagator. $\endgroup$ – Adam May 11 '14 at 1:43
  • $\begingroup$ @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 ? $\endgroup$ – user35952 May 11 '14 at 14:59
  • $\begingroup$ 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 $m^2\to\infty$). $\endgroup$ – Adam May 11 '14 at 16:58
  • $\begingroup$ Then what about the case of massless scalar field ? $\endgroup$ – user35952 May 11 '14 at 17:30
  • $\begingroup$ Then the calculation should be pretty much the same than that of photons. $\endgroup$ – Adam May 11 '14 at 17:46

From your edit 3, I think that what you need then is to do the subtraction of the free energy without the plates, namely

$$ \text{Tr} \log \Delta - \text{Tr} \log \Delta_{free} $$

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.


You can simply calculate the modification on the propagator due to casimir forces by evaluation one loop Feynman diagram or to any order just by putting appropriate current sources. It is believed that Casimir forces results due to virtual creation and annihilation of virtual particles in the photon propagator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.