2
$\begingroup$

Let a quantum system be given with Hilbert space $\mathscr{H}$. Suppose it is described by the quantum state $\rho$, i.e., a density matrix. We can define the Von-Neumman entropy as $$S(\rho)=-\operatorname{Tr}\rho\log\rho.$$

It can be shown that $S(\rho)=0$ if and only if $\rho$ is pure, in other words, if exists $|\psi\rangle\in \mathscr{H}$ such that $$\rho=|\psi\rangle\langle \psi|.$$

Next, suppose that $|\psi\rangle$ is a pure state. It obviously encodes some information about the system in question.

For example, suppose the system is a non-relativistic spin 1/2 particle of mass $m$. In that case, a complete set of states is $|\mathbf{p},\sigma\rangle$ the eigenstates of momentum and spin. Then

  • If $|\psi\rangle = |\mathbf{p},\sigma\rangle$ we have the information that momentum is $\mathbf{p}$, spin is $\sigma$ and mass is $m$.

  • If $|\psi\rangle = c_+|\mathbf{p},1/2\rangle+c_-|\mathbf{p},-1/2\rangle$ we have information that momentum is $\mathbf{p}$, that mass is $m$ and we have some information about spin, though it has uncertainty associated with it.

  • If $|\psi\rangle = \int f(\mathbf{p})|\mathbf{p},\sigma\rangle d^3\mathbf{p}$ we have the same reasoning above. We have information that spin is $\sigma$, mass is $m$ and we have some information about momentum.

Now, if $S(\rho)$ quantifies the information, we would necessarily have $S(|\psi\rangle\langle \psi|)=0$. Still it seems wrong to me to say that "information on this state is zero". As I said, in all these above examples we have at least some information about the system.

For starters in all states we know the mass for sure. This is information already.

Secondly, if $\mathbf{p},\sigma$ are described by the spinor wavefunction $$\Psi(\mathbf{p})=\begin{pmatrix}\psi_+(\mathbf{p})\\ \psi_-(\mathbf{p})\end{pmatrix},$$

we have some information aboute $\mathbf{p}$ and $\sigma$ contained in $\Psi(\mathbf{p})$ that should be quantified wth some measure of uncertainty.

In particular $\Psi(\mathbf{p})_{i}=\delta(\mathbf{p}-\mathbf{q})\delta_{ij}$ should mean "maximum information" - we know momentum and spin. And we should have "minimum information" - a state with maximum uncertainty of these variables.

So by all above it seems obvious that pure states have information associated. The information seems to be associated to observables. And finaly, it seems that this information intuitively should be quantified by a measure ranging from "minimum information/maximum uncertainty" and "maximum information/minimum uncertainty".

Moreover, the Von-Neumman entropy seems not to capture this since it says such a state has zero information associated which seems to not be the case.

So how do we quantify this information contained in a pure state in the end?

$\endgroup$
3
  • $\begingroup$ Nice question I think, but you may need to rearrange parts: check entropy=0 is actually maximal negentropy - it is that which measures information. $\endgroup$ – isometry Oct 12 '18 at 5:34
  • 1
    $\begingroup$ From information theory point of view, information means uncertainty. So for a pure state, von Neumann entropy =0 means there is no uncertainty about its state. But this does not mean we can not extract information about its state. Just as in a classical system, a bit with a fixed value 1 has no information since there is no uncertainty about it. But you can still argue we have the information that the bit is in state 1. We have to be careful when using the word 'information'. $\endgroup$ – XXDD Oct 12 '18 at 14:26
  • $\begingroup$ So in the end when one talks about information one is not talkin about "information we currently have" but actually "information which is available to be found by experiment"? So that when we know it all, no more can be learned and information is zero? $\endgroup$ – Gold Oct 18 '18 at 13:23
1
$\begingroup$

Pure states carry indeed maximal information. In fact, they are maximal with respect to a partial ordering $\prec$ on states defined as follows.

Let $\rho,\omega$ be two states. We say that $\rho \prec \omega$ iff there exists $0\leq\lambda\leq 1$ such that $\rho\geq \lambda \omega$, where $\geq$ is the bigger or equal relation on operators (i.e. $A\geq 0$ iff $\langle\psi, A\psi\rangle \geq 0$ for all $\psi$).

Now, $\omega$ is an extremal, or pure state iff it is maximal w.r.t. $\prec$, in the sense that $\omega\prec\varpi$ iff $\varpi=\lambda\omega$, for some $0\leq \lambda\leq 1$.

It is not difficult to understand by this definition why pure states carry maximal information on the system, among the states. In fact, let $\rho$ be a mixed state. Then there exist two states $\varpi_1$ and $\varpi_2$, and $0<\lambda<1$ such that $$\rho=\lambda\varpi_1 +(1-\lambda)\varpi_2\; .$$ In fact, there exist a (possibly infinite) set of pure states $\Omega=\{\omega_i,i\in I\}$, and coefficients $\{0<\lambda_i<1,i\in I\}$ such that $$\rho=\sum_{i\in I} \lambda_i\omega_i\; .$$

Therefore, the information carried by the mixed state $\rho$ (its description of the system) is completely encoded in the information carried by the set $\Omega$ of pure states, and in general it is encoded by other states. On the other hand, a pure state carries maximal information since the information they carry cannot be encoded in the one carried by any other state. The set $P$ of pure states is therefore exhausting (by convex combinations) all the possible configurations of the system, and thus it contains in some sense all the possible knowledge about the system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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