Skip to main content
Mathematics

Return to Answer

Post Timeline

deleted 285 characters in body
Source Link
Cryo
Cryo
  • 918
  • 4
  • 14

You could try to tackle it by regularizing two delta-s. Let them be some gaussians of vanishing width $\alpha$, $\beta$. Then

$$ \begin{align} \int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta(x-y)\delta(\epsilon-|x-y|)\to I(\alpha,\beta)\\ I(\alpha,\beta,\epsilon)=\int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta_\alpha(x-y)\partial_y\delta_\beta(\epsilon-|x-y|) \end{align} $$

Replace coordinates to $q=x+y$ and $p=x-y$ and $\partial_y\delta_\beta(\epsilon-|x-y|)=\partial_\epsilon\delta_\beta(\epsilon-|x-y|)\,\text{sign}_\beta(x-y)$:

$$ \begin{align} 2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \\ &=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \end{align} $$

In the limit $\alpha \ll \beta,\epsilon$:

$$ \begin{align} \lim_{\alpha \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dq\, f(\frac{q}{2},-\frac{q}{2})\delta_\beta(\epsilon)\text{sign}_\beta(0) \\ &=0 \end{align} $$

In the limit $\beta \ll \alpha,\epsilon$:

$$ \begin{align} \lim_{\beta \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\left[\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)-\int dq f(\frac{-\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)\right]=\\ &=\partial_\epsilon\left(\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)\right) \end{align} $$

With $F(\epsilon)=\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})$. Now $\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)$ is a symmetric function that vanishes outside the width defined by $\alpha$, but its integral could be finite. One could therefore try expressing it as a delta-function. Thus the final for $\lim_{\beta \to 0}I(\alpha,\beta,\epsilon)$ expression could be some derivativeproduct of anti-symmetric function and a deltasymmetric function. If your $\epsilon>0$ thenSo the end result is still zero in the limit $\alpha\to 0$.

You could also explore what happens in the limit $\alpha=a\cdot\beta$ for finite $a$. If you stick to Gaussians, as definitions of delta functions, you should be able to evaluate the integrals. I believe you will still get zero.

You could try to tackle it by regularizing two delta-s. Let them be some gaussians of vanishing width $\alpha$, $\beta$. Then

$$ \begin{align} \int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta(x-y)\delta(\epsilon-|x-y|)\to I(\alpha,\beta)\\ I(\alpha,\beta,\epsilon)=\int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta_\alpha(x-y)\partial_y\delta_\beta(\epsilon-|x-y|) \end{align} $$

Replace coordinates to $q=x+y$ and $p=x-y$ and $\partial_y\delta_\beta(\epsilon-|x-y|)=\partial_\epsilon\delta_\beta(\epsilon-|x-y|)\,\text{sign}_\beta(x-y)$:

$$ \begin{align} 2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \\ &=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \end{align} $$

In the limit $\alpha \ll \beta,\epsilon$:

$$ \begin{align} \lim_{\alpha \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dq\, f(\frac{q}{2},-\frac{q}{2})\delta_\beta(\epsilon)\text{sign}_\beta(0) \\ &=0 \end{align} $$

In the limit $\beta \ll \alpha,\epsilon$:

$$ \begin{align} \lim_{\beta \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\left[\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)-\int dq f(\frac{-\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)\right]=\\ &=\partial_\epsilon\left(\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)\right) \end{align} $$

With $F(\epsilon)=\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})$. Now $\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)$ is a symmetric function that vanishes outside the width defined by $\alpha$, but its integral could be finite. One could therefore try expressing it as a delta-function. Thus the final for $\lim_{\beta \to 0}I(\alpha,\beta,\epsilon)$ expression could be some derivative of a delta function. If your $\epsilon>0$ then the end result is still zero in the limit $\alpha\to 0$.

You could also explore what happens in the limit $\alpha=a\cdot\beta$ for finite $a$. If you stick to Gaussians, as definitions of delta functions, you should be able to evaluate the integrals. I believe you will still get zero.

You could try to tackle it by regularizing two delta-s. Let them be some gaussians of vanishing width $\alpha$, $\beta$. Then

$$ \begin{align} \int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta(x-y)\delta(\epsilon-|x-y|)\to I(\alpha,\beta)\\ I(\alpha,\beta,\epsilon)=\int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta_\alpha(x-y)\partial_y\delta_\beta(\epsilon-|x-y|) \end{align} $$

Replace coordinates to $q=x+y$ and $p=x-y$ and $\partial_y\delta_\beta(\epsilon-|x-y|)=\partial_\epsilon\delta_\beta(\epsilon-|x-y|)\,\text{sign}_\beta(x-y)$:

$$ \begin{align} 2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \\ &=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \end{align} $$

In the limit $\alpha \ll \beta,\epsilon$:

$$ \begin{align} \lim_{\alpha \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dq\, f(\frac{q}{2},-\frac{q}{2})\delta_\beta(\epsilon)\text{sign}_\beta(0) \\ &=0 \end{align} $$

In the limit $\beta \ll \alpha,\epsilon$:

$$ \begin{align} \lim_{\beta \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\left[\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)-\int dq f(\frac{-\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)\right]=\\ &=\partial_\epsilon\left(\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)\right) \end{align} $$

With $F(\epsilon)=\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})$. Now $\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)$ is a product of anti-symmetric function and a symmetric function. So the result is zero.

You could also explore what happens in the limit $\alpha=a\cdot\beta$ for finite $a$. If you stick to Gaussians, as definitions of delta functions, you should be able to evaluate the integrals. I believe you will still get zero.

Source Link
Cryo
Cryo
  • 918
  • 4
  • 14

You could try to tackle it by regularizing two delta-s. Let them be some gaussians of vanishing width $\alpha$, $\beta$. Then

$$ \begin{align} \int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta(x-y)\delta(\epsilon-|x-y|)\to I(\alpha,\beta)\\ I(\alpha,\beta,\epsilon)=\int_{-\infty}^{+\infty} dx\int_{-\infty}^{+\infty} dy\, f(x,y)\delta_\alpha(x-y)\partial_y\delta_\beta(\epsilon-|x-y|) \end{align} $$

Replace coordinates to $q=x+y$ and $p=x-y$ and $\partial_y\delta_\beta(\epsilon-|x-y|)=\partial_\epsilon\delta_\beta(\epsilon-|x-y|)\,\text{sign}_\beta(x-y)$:

$$ \begin{align} 2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \\ &=\partial_\epsilon\int_{-\infty}^{+\infty} dp\int_{-\infty}^{+\infty} dq\, f(\frac{p+q}{2},\frac{p-q}{2})\delta_\alpha(p)\delta_\beta(\epsilon-|p|)\text{sign}_\beta(p) \end{align} $$

In the limit $\alpha \ll \beta,\epsilon$:

$$ \begin{align} \lim_{\alpha \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\int_{-\infty}^{+\infty} dq\, f(\frac{q}{2},-\frac{q}{2})\delta_\beta(\epsilon)\text{sign}_\beta(0) \\ &=0 \end{align} $$

In the limit $\beta \ll \alpha,\epsilon$:

$$ \begin{align} \lim_{\beta \to 0}2I(\alpha,\beta,\epsilon)&=\partial_\epsilon\left[\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)-\int dq f(\frac{-\epsilon+q}{2},\frac{-\epsilon-q}{2})\delta_{\alpha}\left(\epsilon\right)\right]=\\ &=\partial_\epsilon\left(\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)\right) \end{align} $$

With $F(\epsilon)=\int dq f(\frac{\epsilon+q}{2},\frac{-\epsilon-q}{2})$. Now $\left[F(\epsilon)-F(-\epsilon)\right]\delta_{\alpha}\left(\epsilon\right)$ is a symmetric function that vanishes outside the width defined by $\alpha$, but its integral could be finite. One could therefore try expressing it as a delta-function. Thus the final for $\lim_{\beta \to 0}I(\alpha,\beta,\epsilon)$ expression could be some derivative of a delta function. If your $\epsilon>0$ then the end result is still zero in the limit $\alpha\to 0$.

You could also explore what happens in the limit $\alpha=a\cdot\beta$ for finite $a$. If you stick to Gaussians, as definitions of delta functions, you should be able to evaluate the integrals. I believe you will still get zero.