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The $\delta$ function has the following property: $$\int \text{d}x\; f(x)\delta(a-x)=f(a)$$ This actually answers both of your questions. First, the non-square-integrability: $$\int \text{d}x\;\delta(x-y)\delta(x-y)=\delta(y-y)=\delta(0)=\infty$$ according to the rule above if you choose one of the $\delta$'s to be the $f$. Your second question is the ...

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The Dirac delta is defined by the equation $$\int_a^b\mathrm{d}x \,\delta(x)f(x) = f(0)$$ for $a < 0 < b$. By direct application of this definition we get $$\int \mathrm{d}x\, \delta(x - y) \delta(x - y^\prime) = \delta(y - y^\prime)$$ If we let $a < c < 0$ we can write ...

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What happens when you integrate a function multiplied by a delta function? You get: $$\int f(x) \delta (x-y) dx=f(y)$$ (Because the delta function is zero everywhere, except at zero where its integral gives 1.) So, when integrating over $\delta^2 (x-y)$ we get: $$\int \delta^2 (x-y) dx=\int \delta (x-y) \delta (x-y)dx = \delta (y-y) = \infty$$

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The integral without delta square would converge to 1, but squaring the function somehow breaks it? Yes. Recall the sifting property: $$\int_{-\infty}^{\infty}f(x)\delta(x - a)dx = \int_{-\infty}^{\infty}f(a)\delta(x - a)dx = f(a)$$ Then, it follows formally that $$\int_{-\infty}^{\infty}\delta^2(x - a)dx = \int_{-\infty}^{\infty}\delta(x - ... 4 Yes, operators in quantum mechanics can be understood basically as infinite matrices, |r\rangle as basis vectors and \psi(r)\equiv \langle\psi|r\rangle \sim \psi_r \sim "\psi_i" as components of the state vector numbered by a continuous index r. \langle r|F|r'\rangle are indeed just matrix components of the operator F. Generally \langle ... 3 In a certain sense, what you said that every operator might be represented in position representation as a integral operator may be true for many operators only if you allow distributions to be used, and even though that's not always the case. You are kind of confusing things when you say about the dual of distributions. What is a distribution is the ... 3 Unfortunately, I am not so sure what ⟨r|F|r′⟩ is ... in order to evaluate this expectation value it's not an expectation, it is a matrix element; think of it as the components of the operator F on the position basis. If the operator is 'diagonal' on the position basis then \langle r|F|r' \rangle is zero except when r = r'. Thus, for example, ... 4 It seems OP's question (v4) is related to the proper handling of derivatives of Dirac delta distributions. Reductions are performed with the help of (the appropriate 3D generalizations of) the following formulas:$$\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 ...

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Yes, it's enough! Require $S \le 2$ is the similar as require percent of heads to be always $\le 1/2$ when flipping a coin. It should be always $1/2$ in theory, but obviously not in practice. Even 100 000 repetitions of coin flipping won't guarantee you $\le$ 50000 heads. Most or results will be 50000 $\pm$ 800 ($5\sigma$). Returning to the S, 100000 ...

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