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.

  • $\begingroup$ The two states you are taking a mixture of are pure, so their entropy is zero. $\endgroup$
    – Rammus
    Commented Aug 23, 2020 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? $\endgroup$ Commented Aug 25, 2020 at 9:08
  • $\begingroup$ Yeah I used the incorrect notation. My bad. $\endgroup$ Commented Aug 25, 2020 at 12:00
  • $\begingroup$ Yeah I didn't realise what I had wrote until @Rammus pointed it out $\endgroup$ Commented Aug 25, 2020 at 14:53

1 Answer 1


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 $i$.

  • $\begingroup$ 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? $\endgroup$ Commented Aug 25, 2020 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
    Commented Aug 25, 2020 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 $H(p_{i})+\sum_{i}p_{i}S(\rho_{i})$. $\endgroup$ Commented Aug 25, 2020 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 $\rho_{i} \ne \frac{1}{d}$? $\endgroup$ Commented Aug 25, 2020 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
    Commented Aug 25, 2020 at 15:12

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