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I am learning about entropy in quantum mechanics, and I am trying to develop some tools and intuition. One tool that I have found very helpful has been

$$\sum_{i=1}^k \lambda_i \mathbb{S}[\rho_i] \leq \mathbb{S}\left[\sum_{i=1}^k\lambda_i\rho_i\right] \leq \sum_{i=1}^k \lambda_i \mathbb{S}[\rho_i] + \left(-\sum_{i=1}^k \lambda_i \ln(\lambda_i)\right)$$

which shows how the entropy of a sum of density matrices is related to the entropies of the individual density matrices.

Is there an an extension of the above for the entanglement entropy of sums of wavefunctions? I have in mind a picture of spin chains. Let $\tilde{\mathbb{S}}[|\psi\rangle]$ be the half-chain entanglement entropy of the wavefunction $|\psi\rangle$. Then, is

$$\sum_{i=1}^k |c_i|^2 \mathbb{\tilde{S}}[|\psi_i\rangle] \leq \mathbb{\tilde{S}}\left[\sum_{i=1}^kc_i|\psi_i\rangle\right] \leq \sum_{i=1}^k |c_i|^2 \mathbb{\tilde{S}}[|\psi_i\rangle]+ \left(-\sum_{i=1}^k |c_i|^2 \ln(|c_i|^2)\right)$$

true? Could the inequality on the right side be made even stronger by using the overlap of $\langle \psi_i|\psi_j \rangle$?

EDIT: Norbert Schuch rightly points out that the left-hand side of my inequality is nonsense. A little embarassing on my part! Nevertheless, I am still curious about the right hand side. I will accept answers where the $|\psi_i\rangle$ are assumed orthogonal, with $\sum_i |c_i|^2 = 1$, although I am still curious about loosening that.

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  • $\begingroup$ Unless the psi_i are orthogonal, the c_i are not even normalized. Also, how many examples did you try? (That something like that must be wrong should already follow from the fact that you can go back and forth in such basis transformation, unlike for mixtures of states.) $\endgroup$
    – Norbert Schuch
    Commented Dec 14, 2020 at 23:29
  • $\begingroup$ @NorbertSchuch You are right to call me out! My attention was so focused on the RHS of the inequality that I failed to check the LHS, which suffers a number of obvious counterexamples like $1/\sqrt{2}(1/\sqrt{2}(|\downarrow\uparrow\rangle + |\uparrow\downarrow\rangle) + 1/\sqrt{2}(|\downarrow\uparrow\rangle - |\uparrow\downarrow\rangle)$. I will think more on the right hand side now. $\endgroup$
    – user196574
    Commented Dec 14, 2020 at 23:37
  • $\begingroup$ I think the whole idea suffers from the issue that $|c_i|^2$ is not a probability distribution unless your pure states are orthogonal. $\endgroup$
    – Norbert Schuch
    Commented Dec 14, 2020 at 23:42
  • $\begingroup$ @NorbertSchuch My dream was that if the states weren't orthogonal, using $|c_i|^2$ would give an overestimate for the right hand side in which I am most interested; that is, the cross terms would work against the entropy. For example, the state $|\psi\rangle = 1/\sqrt{3}(1/\sqrt{2}(|\uparrow\downarrow\rangle + |\uparrow\uparrow\rangle) + 1/\sqrt{2}(|\downarrow\uparrow\rangle + |\uparrow\uparrow\rangle))$ has $c_1 = c_2=1/\sqrt{3}$ corresponding to a RHS of $2/3\ln(3) > \ln(2)$, so the RHS is a valid, loose upper bound in that case. $\endgroup$
    – user196574
    Commented Dec 14, 2020 at 23:53
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    $\begingroup$ @NorbertSchuch Ultimately, my aim was to get a bound or intuition I could use in this question. There, it is straightforward to show that $S^+$ increases the entropy of a product state in the $S^z$-basis by at most $\ln(2)$, and so I was trying to get bounds I could use for sums of $S^z$-basis product states. $\endgroup$
    – user196574
    Commented Dec 15, 2020 at 0:03

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