I have just started learning QFT. I have just completed scalar fields, which I learnt in using Canonical Quantisation and Path integrals. I did calculation of Casimir force between two metal plates using just free scalar field theory (using the vacuum energy). However, I am not able to find a way to do this thing using Path integrals and propagators. The partition function for the case of free scalar field (i.e KG field) turns out to be,

$$ Z[J] = \text{exp}\bigg(i\int \mathrm d^4x \;\mathrm d^4x'J(x')\Delta_F(x-x') J(x) \bigg) \qquad \qquad (1) $$

which after setting the $Z[J=0] =1$. I wish to know, how to approach my problem from here.

PS : I have not learnt vector or spinor fields yet. Most of the references or notes that I checked either assumed a prior knowledge of that or did not say how to quantise scalar fields.

EDIT : This is the integral to begin with right $$ Z[J] = \frac{1}{Z_0} \int [d \phi] \text{exp}\bigg(-i\int d^4x \bigg[ \frac{1}{2}\phi (\Box + m^2 - i\epsilon)\phi - \phi J\bigg]\bigg) $$

All I did was to introduce $\phi \rightarrow \phi + \phi_0 $ and demand that

$$ (\Box + m^2 - i\epsilon)\phi_0 = J(x) $$ and $\Delta_F(x-x') $ is the Green's function involved in solving this equation.

Then I obtain the equation (1).

  • $\begingroup$ This is not the partition function of the free field. When the scalar field is integrated out, the is a contribution $\left(\det\Delta_F\right)^{-1/2}$, which will correspond to the vacuum energy. If one is just interested in the correlation function, one can forget about it (it does not contribute), but here it is important to keep it. $\endgroup$
    – Adam
    Commented May 3, 2014 at 15:00
  • $\begingroup$ @Adam : It would be really helpful if you could tell me how one integrates out $\phi$, I have added some edit in that regard (I hope). $\endgroup$
    – user35952
    Commented May 3, 2014 at 15:51
  • $\begingroup$ First, there is no $Z_0$ in you new equation. Second, to do in the functional integral, think of it as a gaussian integral, where the inverse propagator is a "functional matrix". By fourier transform, you obtain a diagonal matrix $p^2+m^2$, and you can perform the path integral, that gives $\left(\Pi_p (p^2+m^2)\right)^{-1/2}$. Have a look at any good textbook, for example Zinn-Justin's book. $\endgroup$
    – Adam
    Commented May 3, 2014 at 17:35
  • $\begingroup$ @Adam : But as I said, I have included $Z_0$ to make sure $Z[J=0]=1$. A choice of normalisation $\endgroup$
    – user35952
    Commented May 3, 2014 at 19:17
  • $\begingroup$ Yes, but by doing that, you are losing some information, which is included into $Z_0$. (What I mean is that your $Z[J]$ is not the partition function anymore, it is "just" a generating functional.) $\endgroup$
    – Adam
    Commented May 3, 2014 at 21:32

1 Answer 1


What I would do is to calculate the effective action from integrating at one loop the propagator in a space with boundaries. The result is quite simple, schematically of the form $\mathrm{Tr}\log \Delta$ where $\Delta(x_1,x_2)$ is the propagator in position space. Indeed, the free action is quadratic in the field, $ S\sim -\frac{1}{2}\phi(\partial^2-m^2)\phi$, which makes the path integral Gaussian and hence explicitly calculable. Since $\Delta$ will depend on the geometry of your space, say the distance $L$ between two parallel planes, you will get that the vacuum energy depends on such a separation too. Taking minus the derivative w.r.t L gives the force. Of course, you need to have calculated what the propagator in such a non trivial space is, by solving e.g. the Klein- Gordon equations with boundary conditions at $y=L$ and $y=0$, $y$ being the coordinates orthogonal to the plates . This is not the usual Feynman propagator because of the non trivial boundary conditions (far away from the boundary you should recover Feynman). Note also that th effective action will be UV divergent (therefore you will need to regularize the trace above) but its derivative w.r.t to $L$, the force, is finite and calculable.

  • 2
    $\begingroup$ I really appreciate your answer, but I just wish it were more self-explanative. I don't understand what this "boundaries at one loop" means. I haven't done any prior calculation on Casimir forces, therefore some explicit details would be helpful. Thank you so much $\endgroup$
    – user35952
    Commented May 4, 2014 at 1:36
  • $\begingroup$ I have added a few extra comments. What I meant is that the calculation is at one loop in a space time that has boundaries. What you need to do is to calculate the propagator in such a space, then integrate the action in $d\phi$ to get to the result, that's all. $\endgroup$
    – TwoBs
    Commented May 4, 2014 at 5:57
  • $\begingroup$ You mean to say that I have to solve the equation $$ (\Box^2 + m^2) \Delta(x-y) = -\delta(x-y)$$ subject to the boundary condition $y = 0, y =L$ or do you imply I take the regular Feynman propagator and impose the boundary conditions to the inhomogeneous solution ? $\endgroup$
    – user35952
    Commented May 9, 2014 at 14:24
  • 1
    $\begingroup$ Yes, you should solve that equation you have written explicitly except that $\Delta$ is actually a function of $x$ and $y$ separately since translations are generically broken by the presence of the boundary. In particular you will need to impose the boundary conditions on both variables. $\endgroup$
    – TwoBs
    Commented May 10, 2014 at 19:29
  • $\begingroup$ Please check this question, my try at such a solution. Although I have used a slightly different procedure there. $\endgroup$
    – user35952
    Commented May 10, 2014 at 19:32

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