2
$\begingroup$

I know how this can be proved using the quantum relative entropy. However, even with this proof, and am still confused about how this emerges.

Say I have a source that produces two states $\rho_1$ and $\rho_{2}$ with probability a half each, and both are mixed states, ie $S(\rho_i)>0$ for each of them. The dimensions of the Hilbert space is $2$.

How can $S(\rho)=H(p_{i})+\sum_{i}p_{i}S(\rho_{i})\le \log(d)$, given that $H(p_{i})=1$ and $S(\rho_i)>0$, given that $\log(d)=\log(2)=1?$

I am assuming that I am missing something obvious in the actual construction of $\rho$, in that something is bounding $H(p_{i})$ away from 1. I am assuming this has something to do with orthogonal supports, as $S(\rho) \le H(p)+\sum_{i}p_iS(\rho_{i})$ if they are not orthogonal.

$\endgroup$
4
  • $\begingroup$ The two states you are taking a mixture of are pure, so their entropy is zero. $\endgroup$
    – Rammus
    Aug 23 '20 at 21:40
  • $\begingroup$ As I stated in the question, $|\psi\rangle\langle\psi|$ and $|\phi\rangle\langle\phi|$ are mixed, not pure. Perhaps there is some confusion around the notation I used? $\endgroup$ Aug 25 '20 at 9:08
  • $\begingroup$ Yeah I used the incorrect notation. My bad. $\endgroup$ Aug 25 '20 at 12:00
  • $\begingroup$ Yeah I didn't realise what I had wrote until @Rammus pointed it out $\endgroup$ Aug 25 '20 at 14:53
3
$\begingroup$

I guess you meant to consider generic (not necessarily pure) states $\rho_1,\rho_2$ rather than $|\psi\rangle\langle\psi|$ and $\lvert\phi\rangle\langle \phi\rvert$, so I'll consider the slightly modified question "how is $$S(\rho)= H(p)+\sum_i p_i S(\rho_i)\le \log d$$ possible for a state $\rho=\sum_i p_i\rho_i$ with $H(p)=\log d$?"

The first thing to notice is that, in general, $S(\rho)\le H(p)+\sum_i p_i S(\rho_i)$. This is not an identity unless $\rho_i$ have orthogonal support. You then also have $S(\rho)\le\log d$. From these, you cannot imply $H(p)+\sum_i p_i S(\rho_i)\le \log d$, which is indeed untrue in general, as you observed.

When $\rho_i$ do have orthogonal support, then $S(\rho)= H(p)+\sum_i p_i S(\rho_i)$, and therefore you must also have $H(p)+\sum_i p_i S(\rho_i)\le \log d$. So then why don't we have a contradiction? Well, consider the case with $H(p)=\log d$. This means that you are dealing with a state $\rho$ that is a balanced mixture of $d$ different states with orthogonal support. The only way to get $d$ states with orthogonal support in a $d$-dimensional space is that they each have unit rank, i.e. they are all pure, and thus $S(\rho_i)=0$ for all $i$.

$\endgroup$
10
  • $\begingroup$ So if $\sum_{i}p_{i}\rho_{i}=\frac{1}{2}\rho_{1}+\frac{1}{2}\rho_{2}$, $H(p)\ne1$ unless the supports of $\rho$ are orthogonal? Does that mean that the supports of both $\rho_{1}$ and $\rho_{2}$ are orthogonal w.r.t themselves, or do they also have to be orthogonal to the supports of eachother as well? $\endgroup$ Aug 25 '20 at 12:04
  • $\begingroup$ yes, the entropy is not maximal if the supports are not orthogonal and $p_1=p_2=1/2$. I don't understand the second sentence, both statements look equivalent to me. $\text{supp}(\rho_1)$ and $\text{supp}(\rho_2)$ are two subspaces of the underlying space $\mathcal H$ which are orthogonal, that's it $\endgroup$
    – glS
    Aug 25 '20 at 12:38
  • $\begingroup$ I think I have misunderstood. If the supports of $\rho_{1}$ and $\rho_{2}$ are orthogonal, then they will be pure, so $S(\rho_{i}) =0$ and $H(p_{i})=1$. If they are not orthogonal, then I take it that the entropy will always be strictly less than $H(p_{i})+\sum_{i}p_{i}S(\rho_{i})$. $\endgroup$ Aug 25 '20 at 14:56
  • $\begingroup$ Am I to take it then that the only scenario wherein $S(\rho)=H(p_{i})+\sum_{i}p_{i}S(\rho_{i})$ and $S(\rho_{i})$ is when both the supports are orthogonal, but $\rho$ is not a balanced mixture of supports, ie all $\rho_{i} \ne \frac{1}{d}$? $\endgroup$ Aug 25 '20 at 15:07
  • $\begingroup$ @GaussStrife $d$ states with orthogonal support in a space of dimension $d$ must be pure (and thus have zero entropy), that's what I'm saying. $\endgroup$
    – glS
    Aug 25 '20 at 15:12

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.