I am stuck with this problem for quite sometime. I have a propagator in the momentum representation (from this Phys.SE question), which looks like $$ \widetilde\Delta_F(p) = \frac{1}{(p^0)^2-\left(\left(n\pi/L\right)^2+m^2\right)+i\epsilon} $$

I wish to know how do go about calculating $\mathrm{Tr}[\log \Delta_F]$ for in general these kind of propagators.The propagator in the position representation would look like,

$$ \Delta_F(x-x') = \sum_{n=1}^\infty\int\frac{dp_0}{(2\pi)^2}e^{ip_0(x^0-x'^0)}e^{i\frac{n\pi}{L}(z-z')}\frac{1}{(p^0)^2-\left(\left((n\pi/L\right)^2+m^2\right)} $$ where I have replaced the integral over $p_z$ with a sum over $n$.

EDIT 1 : With the given propagator I can write the Trace to be, $$ \text{Tr}\log{\Delta} = - \sum_n \int dp_0 \log{\bigg(p_0^2 - \bigg(\frac{n\pi}{L}\bigg)^2 + m^2\bigg)} $$ but this is divergent in both the limits of $p_0$ I suppose. I have not introduced any cut-off too. How do I renormalise this given the context of this problem.

PS: Sorry, I am a beginner with QFT and path integral calculations. It would be helpful if I could get quite an explicit answer. More precisely, I wish to know what is the meaning of $\mathrm{Tr}[\log \Delta_F]$.

  • $\begingroup$ I just remembered that this calculation is made in this appendix A of arxiv:1303.6559. The matsubara summation is equivalent to the discretzation in the L direction (this is done in Euclidian space). I think that should do the job. $\endgroup$ – Adam May 14 '14 at 15:38
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    $\begingroup$ The trlog corresponds to the contribution to the free energy (i.e. to the log of the parition function) of a gaussian (free) mode. You can for example compute the partition function of a single harmonic oscillator that way. $\endgroup$ – Adam May 14 '14 at 16:43
  • $\begingroup$ You are calculating such a quantity because you wanted to extract the Casimir force. Just take (minus) the derivative wrt to $L$ of your expression in momentum space, the result is finite. Btw, I don't think your propagator is correct since it depends only on the difference $x-x^\prime$ while you have two boundaries that break translations so that it should be a separate function of $x$ and $x^\prime$ $\endgroup$ – TwoBs May 16 '14 at 4:51
  • $\begingroup$ LaTeX tips (see edit): Try to avoid using \frac inside other denominators; if your fraction is small enough, a backslash is fine and looks better. If you need larger brackets, use \left( ... \right) rather than the \big command. Avoid nested brackets of the same type; if appropriate, have one be square, and the other normal. The symbol for a trace should not be italicized, just as $Re(z)$ should be $\mathrm{Re}(z)$, or even $\Re (z)$. $\endgroup$ – JamalS Oct 7 '14 at 8:56

$\DeclareMathOperator\Tr{Tr}$ From what I've read from your original post, it seems that you're trying to calculate the Casimir force from the vacuum energy. Recall the relationship between the path integral and vacuum energy $$\int e^{-S[\phi]}{\cal D}\phi = e^{-\beta E_0}$$ in the limit $\beta\rightarrow\infty$, where I've gone to Euclidean space. For a free scalar theory in (1 + 1) dimensions, for the Casimir effect, we've got to impose Dirichlet boundary conditions on the spatial dimension. Explicitly, we've got to do the integral $$\int\exp{\bigg(-\frac{1}{2}\int\phi(x)(-\partial^2 + m^2)\phi(x)d^2x\bigg)}{\cal D}\phi$$ This is an easy Gaussian integral, given by $$\frac{1}{\sqrt{\det(-\partial^2 +m^2)}} = e^{-\frac{1}{2}\Tr\log(-\partial^2 +m^2)}$$ To calculate $\Tr\log(-\partial^2 + m^2)$, it's easiest to sum over momentum states (Remember we get a factor of $\beta$ to ensure we're counting states with the correct weight) $$\beta\sum_{n=1}^{\infty}\int\log\bigg((p^0)^2+\Big(\frac{\pi n}{L}\Big)^2+m^2\bigg)\frac{dp^0}{2\pi}$$ Breaking up the log and sending the sum in as a product, we get $$\beta\int\log\bigg(\prod_{n=1}^{\infty}\Big(\frac{\pi n}{L}\Big)^2\bigg)\frac{dp^0}{2\pi}+\beta\int\log\bigg(\prod_{n=1}^{\infty}\bigg(1+\bigg(\frac{L\sqrt{(p^0)^2+m^2}}{\pi n}\bigg)^2\bigg)\bigg)\frac{dp^0}{2\pi}$$ The last term is a convergent product (hyperbolic sine), we only need to regulate the second term. The best method for this is Zeta function regularization, which I'll briefly outline. If we have a product of the form $\prod_{n}a_{n}$, we can form the associated Zeta function $\zeta_{A}(s)=\sum_{n}a_{n}^{-s}$. Just take the derivative and do a line of algebra to show that $\prod_{n}a_{n} = e^{-\zeta_{A}'(0)}$. The basic idea of the regulator is that while the product may diverge, the Zeta function may have an analytic continuation to $s=0$ which gives a finite value. For $a_{n}=(\pi n/L)^2$, we get $$\zeta_{A}(s)= \bigg(\frac{L}{\pi}\bigg)^{2s}\zeta(2s)$$ where a Zeta with no subscript is just the Riemann Zeta function. Taking the derivative and exponentiating, the regulator gives us $\prod_{n=1}^{\infty}a_{n}=2L$. Putting it all together, we get $$\Tr\log(-\partial^2+m^2)=\beta\int\log\bigg(2\sinh\bigg(L\sqrt{(p^0)^2+m^2}\bigg)\frac{dp^0}{2\pi}$$ plus some $L$ independent constant that we don't care about. You should be able to calculate the ground state energy, and then the Casimir force by differentiating, from comparison with the path integral. Hope this helped!

Edit: My original argument was a bit sloppy, so I cleaned it up. Also, you will have to do a trick where you introduce a 3rd plate to prevent the infinite ground state contributions from the field oscillators from rendering the force infinite. For details, see Zee's QFT in a Nutshell, Section 1.9.

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Since you want to extract the Casimir force, just take (minus) the derivative wrt to L of your expression in momentum space, the result is finite. This is actually a general mechanism to regularize the theory by taking derivatives of some parameters that lower the divergence degree. You can integrate back after you have gotten to the finite expression.

Btw, I don't think your propagator is correct since it depends only on the difference x−x′ while you have two boundaries that break translations so that it should be a separate function of x and x′ – TwoBs 16 hours ago

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  • $\begingroup$ So you say, my first expression for the propagator in the momentum space is correct ?? $\endgroup$ – user35952 May 17 '14 at 6:01
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    $\begingroup$ I am just saying that there is no reason why $\langle T \phi(x)\phi(x^\prime)\rangle$ should be a function of $x-x^\prime$ only, since the translation operator $P_\mu$ does not annihilate the vacuum anymore (and in fact, it is the very reason why you have Casimir effect in the first place, moving the boundary changes the energy of the vacuum). I do not know how you have gotten to that propagator and I do not have time to check it for you, but I think I told you already the way to go: solve the (non-homog.) KG eq. for $\Delta(x,x^\prime)$ for each variables imposing the proper boundary cond. $\endgroup$ – TwoBs May 17 '14 at 7:46
  • $\begingroup$ @TwoBs: I think the situation only breaks $SO(4)$. All 4 translations are still symmetries (eg: theory on a cylinder, a la finite temperature). It's not quite a Casimir force calculation; it does compute the free energy. $\endgroup$ – Siva Sep 8 '15 at 13:59
  • $\begingroup$ @Siva In fact, a cylinder does break translations down to a discrete subgroup of it. The momenta along the compact dimensions get quantized. $\endgroup$ – TwoBs Sep 9 '15 at 22:57
  • $\begingroup$ No, a UV cutoff would break translational invariance down to a discrete subgroup. An IR cutoff leads to quantized momenta, and does not lead to a broken translational symmetry. $\endgroup$ – Siva Sep 10 '15 at 2:42

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