Is 0^0 indefinite in Von Neuman Entropy?

[anuttarasammyak]

TL;DR

Shannon and Von Neumann entropy assume x log_a x = 0 for limit of small x. Is it justified in mathematics ?

Hello. In Von Neuman Entropy
S(\rho):=-k_B \ \Sigma_j \ p_j \ \log_e \ p_j
It is assumed that for $p_j=0$
p_j \ \log_e p_j=0
Is it correct mathematics though $0^0$ is indefinite.

 

[mfb]

For positive real x you get $\displaystyle \lim_{x \to 0} x log(x) = 0$. It's plausible to define the expression to be zero at x=0. That way impossible states do not contribute to the entropy, as expected.

 

[anuttarasammyak]

Thanks. I would derive it. Let y=1/x
\lim_{x \rightarrow +0} \ x \log x= - \lim_{y \rightarrow +\infty} \frac{\log y}{y} = - \lim_{y \rightarrow +\infty} \frac{1}{y} = -0
At the last step I used L'Hopital's rule though I am not certain for this infinite limit. I should appreciate your advice.

 

[Office_Shredder]

I'm a bit confused on the first step you did, shouldn't you get $\log (1/y)$?

Edit: oh, that's where the minus sign is from. Looks fine

I think you can just leave it as x also.

$$ x\log(x) = \frac{\log(x)}{ 1/x}.$$

 

[anuttarasammyak]

Thanks. Yes, $\log \frac{1}{y} = - \log y$ , I made.
Following your way
\lim \frac{\log x}{1/x}=\lim \frac{1/x}{-1/x^2} = - \lim x = -0
Again I used L'hopital's rule.

 

[Office_Shredder]

That looks right to me!