I'm trying to understand a computation in my physics script.
To describe the Deltadistribution $\delta(x) $ correctly we would need the formalism of distributions, but one can also much less rigorously describe $\delta(x)$ as the function which is $\infty$ if $x = 0$ and $0$ everyhwere else. Thus an approximation of the Deltadistribution would be:
$\delta_\epsilon(H(x)-E) = \frac{1}{2\epsilon} \theta(|H(x) - E| < \epsilon)$ where $\theta(x)$ is the Heaviside function ($0$ if $x < 0$, $1$ if $x >0$)
In the script they write: $\delta_\epsilon log \delta_\epsilon = - log(2 \epsilon) \delta_\epsilon$
If we write that out:
$\delta_\epsilon log \delta_\epsilon = \delta_\epsilon log (\frac{1}{2\epsilon} \theta(|H(x) - E| < \epsilon)) = -\delta_\epsilon log(2\epsilon) + \delta_\epsilon \theta(|H(x) - E| < \epsilon)$ which would imply that $log(\theta(|H(x) - E| < \epsilon)) = 0$ if the above result holds true.
I have no Idea why that holds though.
Hope someone can explain.
Cheers!
Let $\psi_\epsilon=\delta_\epsilon\log\delta_\epsilon$ and $a=1/(2\epsilon)$. Recall that $0\log0=0$. If $\delta_\epsilon(t)=0$, then $\psi_\epsilon(t)=0$. Otherwise, $\delta_\epsilon(t)=a$ hence $\psi_\epsilon(t)=a\log a$. In both cases, $\psi_\epsilon(t)=\log a\cdot\delta_\epsilon(t)$, hence $\psi_\epsilon=\log a\cdot\delta_\epsilon$.