Expectation of the Wilson Loop

[□h=-16πT]

Hey, I've got this problem from Peskin & Schroeder (chapter 15). I'm not particularly confident with functional integration, as I'm pretty new to it, and working through such a book by myself is pretty tricky in places. Well here goes

 

The Wilson Loop for QED is defined as

 

[math]U_p(z, z)=\exp \left[-ie\oint_pdx^{\mu}A_{\mu}\right][/math]

 

With the Wilson line defined similarly (just change it so that there's not a closed contour integral and with the end points (z,z) changed to (z, y), or whatever you like).

 

Where A is the photon field, the gauge connection asociated with transformations in U(1).

 

Now it says: using functional integration, show that the expectation of the Wilson loop for the electromagnetic field free of fermions is

 

[math]\langle U_p(z, z)\rangle =\exp \left[-e^2\oint_pdx^{\mu}\oint_pdy^{\nu}\frac{g_{\mu\nu}}{8\pi^2(x-y)^2}\right][/math]

 

Where x and y are integrated over the closed loop P.

 

I think the Feynman propogator might be useful here, so to save anyone having to look it up,

 

[math]D_F^{\mu\nu}(x-y)=\int\frac{d^4q}{(2\pi )^4}\frac{-ig^{\mu\nu}e^{-ip\cdot (x-y)}}{q^2+i\epsilon }[/math]

 

(The imaginary term in the denominator of the integrand is the application of the Feynman boundary conditions, ensuring the convergence of the Gaussian integral involved in the derivation of the propogator.)

 

I have a vague idea of how to go about it, but I'm not particularly confident about it, it's finding the relevant starting point that's causing me problems, i.e. putting together and computing the functional integral for the expectation. I'm just going through this to gain some confidence in functional integration etc. so if anyone can give a few pointers as to going about this it'd be much appreciated. This isn't a homework question, if that puts anyone off helping me, I doubt my A level teacher would set something like this .

 

Cheers

[□h=-16πT]

*shameless bump*

 

Anyone?

[Severian]

I can see that it is true, but my proof isn't very rigorous.

 

[math] \langle U_P(z,z) \rangle = \langle 0 | \exp \left[ -ie\oint_P dx^\mu A_\mu \right] |0 \rangle[/math]

 

Now expand the exponential:

 

[math] \langle U_P(z,z) \rangle = \langle 0 | 1 -ie \oint_P dx^\mu A_\mu -e^2 \oint_P dx^\mu A_\mu (x) \oint_P dy^\nu A_\nu (y) + \cdots |0 \rangle[/math]

 

But we know that [math]\langle A_\mu \rangle = 0 [/math] since each term is either a creation or annihilation operator which will cancel with either the vacuum to the left or right. Thus every odd term in A will be zero.

 

Also, I can write [math]\langle A_\mu (x) A_\nu (y) \rangle = \langle \left[ A_\mu (x),A_\nu (y) \right] \rangle [/math] since the extra term is zero.

 

Resumming the exponential:

 

[math] \langle U_P(z,z) \rangle = \langle 0 | \exp \left[ -e^2 \oint_P dx^\mu \oint_P dy^\nu \left[ A_\mu(x), A_\nu (y) \right] \right] |0 \rangle[/math]

 

Putting in the photon propagator (in position space) for [math]\left[ A_\mu (x),A_\nu (y) \right] [/math] gives the desired result.

 

There must be a nicer way of doing this without expanding the exponential I think.... possibly by constructing the Wilson Loop out of Wilson lines...

[□h=-16πT]

I was trying to approach it, as the question suggested, by constructing it out of Wilson lines and using a functional integral.

 

There should be factorials in the exponent that you've missed out, which will muck up the end result.

[□h=-16πT]

Or have you done something that gets rid of that factorial?

 

Cheers, man.

[Severian]

no - that is a mistake. I will have to think about it some more.

[□h=-16πT]

no - that is a mistake. I will have to think about it some more.

 

Danke, I wish it did work though . I hate functional integrals, as elegant as the whole procedure is for deriving Feynman rules.