Whenever we calculate entropy we make use, for example, $\log(P)$ for $P$ a projection defined for some arbitrary finite dimensional Hilbert space.

But for projection operators this is not well-defined by means of the functional calculus. My question is: although we are more interested in the terms $P \log(P)$ what is really happening is that we use this only for a symbol whenever we have projections and write $0\log(0)=0$ as the eigenvalue description using the functional calculus for the operator $P\log(P)$?

Is there a way to make $\log(P)$ a valid, well-defined, mathematical description?

  • $\begingroup$ Does this answer help? In particular, it turns out that checking whether $\exp (\log P) = P$ is sufficient to make sense of $\log P$. $\endgroup$ Jul 17, 2020 at 19:43
  • $\begingroup$ Do you have any example where a "bare" log(P) appears? It typically only appears as P log(P) and any textbook will tell you that the convention is that 0 log(0)=0. $\endgroup$ Jul 17, 2020 at 23:29
  • $\begingroup$ Let P= I - |0><0|, Plog(P) = (I-|0><0| )log(P) = log(P) - |0><0| log(P), if log(P) is well defined, by linearity of the trace we get tr(log(P)) - tr(|0><0|log(P)). Hence, by noticing the unexistence of such constructions it helps so that one does not falls into such considerations. Mistakes like this are fairly common, since we are applying heavy machinery we should be clear into what is and what is not allowed. Att. R.W. PS: This is why I think the question is interesting. $\endgroup$
    – R.W
    Jul 17, 2020 at 23:34
  • $\begingroup$ Well, no. This transformation is not allowed. You shall not take the difference of two infinities without good reason. Differently speaking: The function $X\log(X)$ is well-defined also at $0$. The function $\log X$ isn't. So your second "=" sign is incorrect. (P.S.: Please use @[username] to ping me. Otherwise I won't be notified.) $\endgroup$ Jul 17, 2020 at 23:37
  • $\begingroup$ Dear @NorbertSchuch thank you for pointing out my mistaken consideration. In fact, this was already clarified by the below answer. Att. R.W. $\endgroup$
    – R.W
    Jul 17, 2020 at 23:39

1 Answer 1


$P$ is a projection operator in the limiting case where $P$ represents a state that is completely known, i.e. a pure state, so the entropy is zero. As a limiting case, the valid and well-defined mathematical description is $$\lim_{P \rightarrow 0^+} P \log(P)=0.$$ This is still a bit sloppy. Since $P$ is a matrix, we're actually taking the trace and $P \rightarrow 0^+$ means as P goes to a representation of a pure state. It's a bit more clear to consider entropy in terms of the eigenvalues of $P$, $$S=\sum \lambda_i \log \lambda_i,$$ so that there is no ambiguity in interpreting $$\lim_{\lambda \rightarrow 0^+} \lambda \log(\lambda)=0,$$ for any given term in the sum.

In response to your comment, when $P$ is a projection operator, the expression $\log(P)$ is inherently undefined. You can see this by noting that projection operators are idempotent by definition, so $P^2=P$.

Consider the proposition that we can express $P$ in exponential form $P=e^x$ for some unknown $x$. The idempotentcy of $P$ requires $e^x=e^{2x}$, which tells us that $x$ is imaginary with magnitude $2 \pi n, \, n \in \mathbb{Z}$, over some $d$-dimensional basis.

Whatever appropriately normalized basis we pick, say $b_d$, the inevitable result is $e^{i2\pi n b_d}=I_d$. So we get back the identity instead of our projection operator, contradicting the proposition.

The is one of several ways to see that we can't define $P$ as an exponential, and that $\log(P)$ is hopelessly undefined.

  • $\begingroup$ Dear @Jonathan I understand that this is the situation for the entropy. But I am thinking if the operator $\log(P)$ is meaninful. The limit you are taking/considering is with respect to which notion of convergence? I do think that it has not fullfilled entirely the question. Att R.W. $\endgroup$
    – R.W
    Jul 17, 2020 at 18:05
  • 1
    $\begingroup$ @R.W I added some commentary on why $\log(P)$ is inherently undefined. Hopefully it's helpful. $\endgroup$ Jul 17, 2020 at 18:56
  • $\begingroup$ Dear @Jonathan, your answer is very satisfactory, I shall upvote it. I was thinking weather it is possible to make sense of log(P) in some way. Thank you for the pacience, I'll wait two days to accept it as the final answer. Att. R.W. $\endgroup$
    – R.W
    Jul 17, 2020 at 23:29

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