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I would like to know how to take the functional derivative of the holonomy, or Wilson line. I have tried it and I will show what I have done below, but before I wanted to say that I also have seen and done this with the characteristic deifferential equation for the holonomy $$ \frac{\partial U}{\partial s}+\dot{\gamma}^a A_{a} U=0 $$ with $\dot{\gamma}$ a tangent vector to the curve and $A$ the connection. By varying this equation I can find what $\frac{\delta U}{\delta A}$ is, but I would like to know how to do it from the expression for $U$ $$ U=\mathcal{P}\exp \left[ -\int_{\gamma} \dot{\gamma}^a(s) A_a(\gamma(s)) ds \right] $$ with $\dot{\gamma}^a=\frac{dx^a}{ds}$ as before. Now I have tried to directly vary this with respect to $A_b$ $$ \frac{\delta U}{\delta A_b(x)}=\mathcal{P} \exp \left[ -\int_{\gamma} \dot{\gamma}^a A_a ds \right] \cdot \frac{\delta}{\delta A_b}\left[ -\int_{\gamma} \dot{\gamma}^a A_a ds \right] $$ Now if $A_a=A_{a}^{i}\tau^i$ then $$ \frac{\delta}{\delta A_{b}^i }\left[ -\int_{\gamma} \dot{\gamma}^a A_{a}^j \tau^j ds \right]=-\int_{\gamma} \dot{\gamma}^a \delta _{ab}\delta_{ij} \delta^3(\gamma(s)-x) \tau^j ds=-\int_{\gamma}\dot{\gamma}^b \delta^3(\gamma(s)-x) \tau^j ds $$ So I end with $$ \frac{\delta U}{\delta A_{b}^j}=U(\gamma)\left[ -\int_{\gamma}\dot{\gamma}^b \delta^3(\gamma(s)-x) \tau^j ds \right] $$ Which isn't right.. Can someone point me in a better direction, Thanks.

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Qmechanic's detailed answer amounts to: put the integral inside the path ordering symbol. – Ron Maimon Aug 29 '12 at 4:00

1) Let us write the Wilson-line of a simple open curve $\gamma: [s_i,s_f]\to \mathbb{R}^4$ as

$$\tag{1} U(s_f,s_i) ~=~ \mathcal{P}\exp \left[ i\int_{\gamma} A_{\mu}~ dx^{\mu} \right]. $$

2) The path-ordering $\mathcal{P}$ becomes important if the gauge potential

$$\tag{2}A_{\mu}~=~A^a_{\mu} T_a$$

is non-abelian. Here $T_a$ are the generators of the corresponding Lie algebra.

3) The Wilson-line has groupoid properties, e.g.,

$\tag{3}U(s_3,s_2)U(s_2,s_1)~=~ U(s_3,s_1), \qquad U(s,s) ~=~ {\bf 1}.$

4) If one differentiates wrt. the final point $s_f$, one gets

$$\tag{4}\frac {dU(s_f,s_i)}{ds_f} ~=~ i\dot{\gamma}^{\mu}(s_f)~A_{\mu}(\gamma(s_f)) ~U(s_f,s_i). $$

5) If one differentiates wrt. the initial point $s_i$, one gets

$$\tag{5} \frac {dU(s_f,s_i)}{ds_i} ~=~ -U(s_f,s_i)~i\dot{\gamma}^{\mu}(s_i)~A_{\mu}(\gamma(s_i)) . $$

6) OP wants to differentiate the Wilson-line $U(s_f,s_i)$ functionally wrt. the gauge potential components $A^a_{\mu}(x)$. One gets

$$\tag{6} \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(x)} ~=~\int_{s_i}^{s_f}\! ds~ U(s_f,s)~ i\dot{\gamma}^{\mu}(s)\delta^4(x-\gamma(s))T_a~U(s,s_i). $$

7) Heuristic proof of (6). Since we have already used the letter $x\in\mathbb{R}^4$ in (6) as a fixed space-time point, let us call an arbitrary spacetime point for $y\in\mathbb{R}^4$.

  1. Imagine that $\tilde{A}(y)=A(y)+\delta A(y)$ is an infinitesimal variation of the gauge potential $A(y)$.

  2. Imagine that $\delta A(y)$ only differs from zero in an infinitesimally small neighborhood $\Omega$ of the fixed space-time point $x$.

  3. Assume that the curve $\gamma$ intersects the neighborhood $\Omega$ at the parametervalue interval $[s_x-\varepsilon,s_x+\varepsilon]\subseteq [s_i,s_f]$. (If the curve $\gamma$ does not intersects the neighborhood $\Omega$, then the equation (6) becomes trivially correct: $0=0$.)

On one hand, such infinitesimal variation of the gauge potential yields

$$\tag{7}\delta U(s_f,s_i)~=~U(s_f,s_x+\varepsilon)~\delta U(s_x+\varepsilon,s_x-\varepsilon)~U(s_x-\varepsilon,s_i), $$


$$\delta U(s_x+\varepsilon,s_x-\varepsilon)~\approx~2\varepsilon i~ \dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(\gamma(s_x)) $$ $$~=~\int_{\Omega} \!d^4y~\delta^4(y-\gamma(s_x))~2\varepsilon i\dot{\gamma}^{\mu}(s_x)~\delta A_{\mu}(y)$$ $$\tag{8}~\approx~ \int_{\Omega} \!d^4y~\int_{s_x-\varepsilon}^{s_x+\varepsilon}\! ds~\delta^4(y-\gamma(s))~i\dot{\gamma}^{\mu}(s)~\delta A_{\mu}(y).$$

On the other hand, the defining property of a functional derivative yields

$$\tag{9}\delta U(s_f,s_i) ~=~\int_{\mathbb{R}^4} \!d^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y)~=~\int_{\Omega} \!d^4y~ \frac {\delta U(s_f,s_i)}{\delta A^a_{\mu}(y)} ~\delta A^a_{\mu}(y).$$

An comparison of eqs. (7), (8) and (9) yields eq. (6).

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Yes, that is what I have as the answer also, but can you give me another hint as to how you get $U$ to branch into $U(s_f ,s)$ and $U(s,s_i)$? – kηives Jul 29 '12 at 19:22
I updated the answer. – Qmechanic Jul 29 '12 at 20:52

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