3
$\begingroup$

(Just so you know my background) I have taken a graduate course in quantum mechanics. I have also learned about information entropy in various places (statistical mechanics, information theory, dynamical systems).

It recently occurred to me that the HUP might have an interpretation purely in terms of entropy. I pondered this for a bit, but didn't get very. My intuition for why this seems plausible is that the HUP acts a bound on the amount of information one can retrieve from a system in a single measurement, and entropy is the instrument we use to measure information (or lack thereof) in a system.

Is there, in fact, a deep connection there? Or am I incorrectly identifying two completely distinct ideas since they both can be described loosely using the word "information"?

$\endgroup$
0

1 Answer 1

4
$\begingroup$

Congratulations! You stumbled upon entropic uncertainty relations (a good review here). They are a reformulation of the usual uncertainty principles using entropy instead of variance. The simplest and most famous one is probably the Maassen Uffink relation: let $\rho$ be a quantum state, let $A$ and $B$ be two observables and

$$ \mathcal M_O:\rho\mapsto\sum_i \langle o_i|\rho |o_i\rangle|o_i\rangle\langle o_i|$$

be the measurement channel for an observable $O$ where $|o_i\rangle$ denotes the eigenstates of $O$, then

$$ S(\mathcal M_A(\rho))+S(\mathcal M_B(\rho))\geq \log\frac 1c$$

where $S(\tau)=-\mathrm{Tr}(\tau\log\tau)$ is the Von Neumann entropy and $$ c=\max_{i,j}|\langle a_i|b_j\rangle|^2.$$

I think this corresponds to the intuition you're talking about: if you're able to guess really well the outcome of an $A$ measurement, then the classical probability distribution $\mathcal M_A(\rho)$ is very peaked around one value and hence has low entropy, this means that to satisfy the bound $\mathcal M_B(\rho)$ must have a high entropy, hence this distribution is closer to a uniform distribution and it's hard to guess the outcome before measuring.

Notice also that if $[A,B]=0$, then $A$ and $B$ share eigenvectors, hence $c=1$ and the bound is trivial, as you'd expect, i.e. both entropies can vanish at the same time.

Many interesting generalizations have been put forward (multipartite, continuous variable, with quantum memory and others) and you can find a lot of them in the review I mentioned, which also makes good arguments as to why you should care about entropy instead of variance, and also to what extent you can recover one kind of uncertainty relation from the other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.