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Casimir forces due to scalar field using Path integrals

+ 3 like - 0 dislike
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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).

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user user35952
asked May 3, 2014 in Theoretical Physics by user35952 (155 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Adam
@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).

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user user35952
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.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Adam
@Adam : But as I said, I have included $Z_0$ to make sure $Z[J=0]=1$. A choice of normalisation

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user user35952
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.)

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Adam
@Adam : My bad, indeed it is "just" the generating functional. With this, how do I now go about calculating the forces

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user user35952
I don't really know (or I'd have written an answer). My guess would be to compute the determinant (which corresponds to the vacuum energy) by taking into account the boundary conditions (the fact that there is an "inside" and an "outside" defined by the plates). Another possibility would be to compute the log of $Z[J]$ (without forgetting $Z_0$), for some sources playing the role of the role of the plates.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Adam

1 Answer

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user TwoBs
answered May 3, 2014 by TwoBs (265 points) [ no revision ]
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

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user user35952
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.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user TwoBs

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