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I've been puzzling with the problem below for more than a hour since it is misleadingly discussed in some textbooks, so I believe it deserves a solution here. Any comments are welcome.

I'm trying to calculate the gauge field strength tensor from the circulation of the gauge potential around an infinitesimal loop. The basic idea is to expand the Wilson line operator (path ordering transport operator) $$ U\left(b,a\right)=P\exp\left[-\mathbf{i}g\int_{a}^{b}A\left(x\right)\cdot dx\right] $$ which becomes $$ U\left(a+da,a\right)\approx1-\mathbf{i}gA\left(a\right)da $$ at first level in the infinitesimal $da$. Then we suppose a two-dimensional manifold (differentiable, ... ), and we want to calculate the circulation around a close square parametrised by the four point $$ A:\left(a,b\right)\;;\; B:\left(a+\delta a,b\right)\;;\; C:\left(a+\delta a,b+\delta b\right)\;;\; D:\left(a,b+\delta b\right) $$ in the direction $A\rightarrow B\rightarrow C\rightarrow D\rightarrow A$. Owing to the formula $\oint A\cdot dx=\iint F\cdot dS$, one expects to find the differential form of the field $$ F_{12}=\partial_{1}A_{2}-\partial_{2}A_{1}+\mathbf{i}g\left[A_{1},A_{2}\right] $$ but I didn't... Here what I've done: I calculate $$ U_{1}^{\dagger}\left[\left(a,b\right);\left(a,b+\delta b\right)\right]U_{2}^{\dagger}\left[\left(a,b+\delta b\right),\left(a+\delta a,b+\delta b\right)\right] $$ $$ \times U_{3}\left[\left(a+\delta a,b+\delta b\right);\left(a+\delta a,b\right)\right]U_{4}\left[\left(a+\delta a,b\right);\left(a,b\right)\right] $$ wich has to be read $U\left[\left(x,y\right);\left(a,b\right)\right]$ : the transport from the point $\left(a,b\right)$ to the point $\left(x,y\right)$. The indices $1,\;2,\;3,\;4$ have no mathematical meaning, they allow me to keep track of the operator without repeating the arguments. Note the dagger on $U_{1,2}$, since the transportation is opposite to the convention. Using the infinitesimal path ordering transport, one has $$ U_{1}^{\dagger}=1+\mathbf{i}gA_{2}\left(a,b+\delta b\right)\delta b=1+\mathbf{i}gA_{2}\left(a,b\right)\delta b+\mathbf{i}g\partial_{2}A_{2}\delta b\delta b $$ $$ U_{2}^{\dagger}=1+\mathbf{i}gA_{1}\left(a+\delta a,b+\delta b\right)\delta a=1+\mathbf{i}gA_{1}\delta a+\mathbf{i}g\partial_{1}A_{1}\delta a\delta a+\mathbf{i}g\partial_{2}A_{1}\delta b\delta a $$ $$ U_{3}=1-\mathbf{i}gA_{2}\left(a+\delta a,b\right)\delta b=1-\mathbf{i}gA_{2}\left(a,b\right)\delta b-\mathbf{i}g\partial_{1}A_{2}\delta a\delta b $$ $$ U_{4}=1-\mathbf{i}gA_{1}\left(a,b\right)\delta a $$ up to the second order in $\delta\left(a,b\right)$. One has to keep the second order, because the first order terms (proportional to $\delta a$ or $\delta b$) cancel multiplying all the $U_i$'s. Nevertheless, in addition to the gauge field strength proportional to $-\mathbf{i}g\delta a\delta b$, one has some $A_{1}A_{1}$, $A_{2}A_{2}$, $\partial_{1}A_{1}$ and $\partial_{2}A_{2}$ terms.

So, why the infinitesimal circulation of the gauge potential does not give the infintesimal gauge field strength, as excepted from basic differential geometry ?

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1 Answer 1

Well, the infinitesimal circulation of the gauge potential indeed gives the infinitesimal gauge field strength. The point is that a second order term in perturbation must be calculated from second order expansion ! Otherwise the calculation (as the one in the question) is meaningless.

In general, the expansion of the path-ordered transport operator up to the second order term reads $$ U\left(b,a\right)\approx1-\mathbf{i}g\int_{a}^{b}dx\cdot A\left(x\right)-g^{2}\int_{a}^{b}dx_{1}\int_{a}^{x_{1}}dx_{2}\cdot\left[A\left(x_{1}\right)A\left(x_{2}\right)\right] $$ (note the interval for $x_{2}$, due to the path ordering). In general the calculation of the second term is troublesome, but here we need only to know it along a straight line (the one or the second coordinate $q^{1}$ or $q^{2}$ in our 2-dimensonal manifold). Then, we have $$ U\left[\left(q^{1}+dq^{1},q^{2}\right);\left(q^{1},q^{2}\right)\right]\approx1-\mathbf{i}gA_{1}\left(q^{1}\right)dq^{1}-\mathbf{i}\dfrac{g}{2}\partial_{1}A_{1}dq^{1}dq^{1}-\dfrac{g^{2}}{2}A_{1}A_{1}dq^{1}dq^{1} $$ and the same for a transport along the $q^{2}$ coordinate. It looks much more complicated, but a straightforward calculation gives (I keep the same notations for the $U_{i}$'s, the same path, the same orientation, ... from the question) $$ U_{1}^{\dagger}=1+\mathbf{i}gA_{2}\delta b+\mathbf{i}g\partial_{2}A_{2}\delta b\delta b-\mathbf{i}\dfrac{g}{2}\partial_{2}A_{2}\delta b\delta b-\dfrac{g^{2}}{2}A_{2}A_{2}\delta b\delta b $$ Note that the term $\partial_{2}A_{2}\delta b\delta b$ do not change sign, since it contains two $\delta b$ and that the path in the inverse direction means sending $\delta b$ to $-\delta b$. This is obviously not the same for $A_{2}\left(a,b+\delta b\right)\approx A_{2}+\partial_{2}A_{2}\delta b$ from usual expansion. $$ U_{2}^{\dagger}=1+\mathbf{i}gA_{1}\delta a+\mathbf{i}g\partial_{1}A_{1}\delta a\delta a+\mathbf{i}g\partial_{2}A_{1}\delta a\delta b-\mathbf{i}\dfrac{g}{2}\partial_{1}A_{1}\delta a\delta a-\dfrac{g^{2}}{2}A_{1}A_{1}\delta a\delta a $$ $$ U_{3}=1-\mathbf{i}gA_{2}\delta b-\mathbf{i}g\partial_{1}A_{2}\delta a\delta b-\mathbf{i}\dfrac{g}{2}\partial_{2}A_{2}\delta b\delta b-\dfrac{g^{2}}{2}A_{2}A_{2}\delta b\delta b $$ $$ U_{4}=1-\mathbf{i}gA_{1}\delta a-\mathbf{i}\dfrac{g}{2}\partial_{1}A_{1}\delta a\delta a-\dfrac{g^{2}}{2}A_{1}A_{1}\delta a\delta a $$ Now multiply all these terms up to the second order, we get the correct gauge field strength tensor, since all the nasty terms proportionnal to $\delta a$, $\delta b$, $\delta a\delta a$ and $\delta b\delta b$ vanish. Keeping track of the gauge potential ordering we get the non-Abelian gauge field strength tensor, too.

Have fun looking in textbooks which calculate second order terms (like the curvature tensor / gauge field strength tensor) using only first order expansion of some quantities :-).

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