Let's have constraints for Yang-Mills theory: $$ \varphi_{a} = \partial_{i}\pi^{i}_{a} - f_{abc}\pi^{b}_{i}A^{c}_{i}. $$ I have read the statement that $$ \tag 1 [\varphi_{a}(\mathbf x), \varphi_{b}(\mathbf y )] = f_{abc}\varphi^{c}(\mathbf x) \delta (\mathbf x - \mathbf y). $$ $(1)$ can be computed by using canonical Poisson brackets $$ [A_{a}^{i}(\mathbf x ), \pi_{b}^{j}(\mathbf y )] = \delta_{ab}\delta^{ij}\delta (\mathbf x - \mathbf y ), \quad [\pi_{a}^{i}(\mathbf x ), \pi_{b}^{j}(\mathbf y )] = [A_{a}^{i}(\mathbf x ), A_{b}^{j}(\mathbf y )] = 0. $$ But I can't get $(1)$. The problem is in two summands which have the form $$ \tag 2 -\varepsilon_{klm}[\partial_{i}\pi^{i}_{a}(\mathbf x ), A_{m}^{j}(\mathbf y )]\pi_{l}^{j}(\mathbf y) = -\varepsilon_{kla}\pi_{l}^{i}(\mathbf y)\partial_{i}^{\mathbf x}\delta (x - y) = -\varepsilon_{kla}\pi_{l}^{i}(\mathbf y)\partial_{i}^{\mathbf y}\delta(\mathbf x - \mathbf y) $$ For getting $(1)$ I need to move derivative away from delta-function, but it can be done (in my opinion) only if there also is integration of $(2)$ over $x, y$. Where did I make the mistake? How to get $(1)$?
Answer
It seems OP's question (v4) is related to the proper handling of derivatives of Dirac delta distributions. Reductions are performed with the help of (the appropriate 3D generalizations of) the following formulas:
$$\tag{A} \{\partial_x+\partial_y\}\delta (x-y)~=~ 0,$$
$$\tag{B} \{f(x)-f(y)\}~\delta (x-y)~=~ 0,$$
$$\tag{C} \{f(y)-f(x)\}~\partial_x \delta (x-y)~=~ \delta (x-y) ~f^{\prime}(x),$$
which may be derived e.g. using test functions. Pay attention to the non-zero right-hand side of eq. (C), which we suspect is the culprit of OP's question.
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