Let us consider following integral $$\int \int dx dy f(x)g(y)\delta(x-y).\tag{1}$$ We can bring integral with respect to $x$ to the front or $y$ to the front and integrate out to get the same answer $$\int dx f(x) \int dy g(y) \delta(x-y) = \int dx f(x) g(x),$$ $$\int dy g(y) \int dx f(x) \delta(x-y) = \int dyg(y)f(y).$$

However, it seems different for operators. Consider $$\int \int dx dy \hat{A_x}\hat{B_y}\delta(x-y)\psi(x,y).\tag{2}$$ and integrate in two ways $$\int dx \hat{A_x} \int dy \hat{B_y} \delta(x-y)\psi(x,y) = \int dx \hat{A_x} \hat{B_x}\psi(x,x),$$ $$\int dy \hat{B_y} \int dx \hat{A_x} \delta(x-y)\psi(x,y) = \int dy\hat{B_y}\hat{A_y}\psi(y,y) ,$$ where $\psi(x,y)$ is a function, and this differs by $[\hat{A},\hat{B}]$.

I suspect that $$[\hat{A_x}, \delta(x-y)] \neq 0,$$ but I can't utilize this insight to get a consistent conclusion. Can someone shed some light on this?

  • 3
    $\begingroup$ In the "second way" to integrate the operators you are changing their order, which is wrong. $\endgroup$
    – fqq
    Commented Nov 16, 2019 at 19:49
  • $\begingroup$ I thought $[\hat{A_x}, \hat{B_y}]=0.$ Is it false? $\endgroup$
    – Nugi
    Commented Nov 16, 2019 at 19:50
  • $\begingroup$ How can generic (one-parameter families of) operators commute? $\endgroup$
    – fqq
    Commented Nov 16, 2019 at 19:53
  • $\begingroup$ $\left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right] =0 $ is one example. $\endgroup$
    – Nugi
    Commented Nov 16, 2019 at 19:54
  • $\begingroup$ Can you specify one-parameter families of operators? $\endgroup$
    – Nugi
    Commented Nov 16, 2019 at 19:55

1 Answer 1

  1. Here we assume $A_x$ and $B_y$ are differential operators in the $x$ and $y$ variables, respectively. They commute since they depend on different variables.

  2. OP's previous expression (v3) $$\iint_{\mathbb{R}^2}\! dx~dy~ \hat{A}_x\hat{B}_y\delta(x-y)\tag{i}$$ is not mathematically well-defined: A distribution (such as the Dirac delta distribution) should have a way to act on test functions.

  3. In later versions OP includes a test function: $$\begin{align}\iint_{\mathbb{R}^2}\! dx~dy~\psi(x,y)~ \hat{A}_x\hat{B}_y\delta(x-y) ~=~&\iint_{\mathbb{R}^2}\! dx~dy~\delta(x-y)~\hat{A}^T_x\hat{B}^T_y\psi(x,y) \cr ~=~&\int_{\mathbb{R}}\! dz~\hat{A}^T_1\hat{B}^T_2\psi(z,z) ,\end{align}\tag{ii}$$ where the superscript "$T$" denotes the transposed differential operator, and the subscripts "$1$" and "$2$" indicate which entries of the test function $\psi$ that the differential operator acts on. In the first equality of eq. (ii) we integrated by parts. We repeat (to rule out any possible misunderstanding) that at any step the two differential operators commute.

  • $\begingroup$ My interpretation of the expression would be that $A_x$ and $B_y$ are one-parameter families of operators (operator-valued functions/distributions), with the same domain for the parameter, which is also the domain of integration. This would make sense, right? If instead $A_x$ is a differential operator in $x$ etc, I agree that it's not clear what it means. $\endgroup$
    – fqq
    Commented Nov 16, 2019 at 20:12
  • $\begingroup$ I have edited by putting in a general function $\psi(x,y)$. Can you and @fqq check if it is now mathematically valid? $\endgroup$
    – Nugi
    Commented Nov 16, 2019 at 20:23
  • $\begingroup$ In section 3, second term, why can't we commute $\hat{A_x}^T$ and $\hat{B_y}^T$ and yield the same result (third term) with $A$ and $B$ swapped? $\endgroup$
    – Nugi
    Commented Nov 16, 2019 at 22:08
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Nov 16, 2019 at 22:22

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