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Let $|\psi\rangle$ be an $n$ qubit quantum state on a line with Von Neumann entanglement entropy at most $r$ with respect to any bipartition of the qubits (does not have to be a contiguous bipartition). Now, consider a state

$$|\Psi\rangle = |\psi\rangle \otimes |\phi\rangle, $$

where $|\phi\rangle$ is a state on $\log n$ many qubits. Could we now put any bounds on the Von Neumann entanglement entropy of $|\Psi\rangle$ with respect to any bipartition? Can I say that it is $r + \log n $?

For contiguous bipartitions, I think it is easy to see that the entanglement is always bounded by $r$, when $r > \log n$. However, the problem is non-contiguous bipartition of qubits

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  • $\begingroup$ I seem to agree with your reasoning that the maximum Von Neumann entropy it can have is $r + \text{log}(n)$, since two systems $\psi$ and $\phi$ are by definition separated, the amount of Von Neumann entropy $\Psi$ can have should be the sum of maximum possible Von Neumann entropies of $\psi$ and $\phi$ which are $r$ and $\text{log}(n)$ respectively. So, the maximum possible entropy of entanglement for $\Psi$ should be $\max(r, \text{log}(n))$. $\endgroup$
    – FDGod
    Commented Sep 11, 2023 at 20:46
  • $\begingroup$ When there are $\log n$ qubits should the entropies be bounded by $\log log n$? $\endgroup$ Commented Sep 11, 2023 at 21:28

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Too long for a comment, partially an answer, but the question may need to be updated in response to this.

For $ n$ qubits, we expect the entanglement entropy to be upper bounded by $\log n$, so $\log n$ qubits, we expect the entanglement entropy to be upper bounded by $\log \log n$. Assuming OP intended that to be the question, we want to find the upper bound of $r+\log\log n$.

Let's try to find a counterexample. We can choose each state to have the same entanglement entropy by writing $$|\psi\rangle\propto |aa\rangle+|bb\rangle$$ and $$|\phi\rangle\propto |cd\rangle+|dc\rangle.$$ With this notation I am implying that $|a\rangle$, $|b\rangle$, etc. are multiqubit states. I'll choose something separable like $|a\rangle=|0\rangle^{\otimes n/2}$, $|b\rangle=|1\rangle^{\otimes n/2}$, $|c\rangle=|0\rangle^{\otimes m/2}$, $|d\rangle=|1\rangle^{\otimes m/2}$. We have entanglement entropies $r=\log 2$ and $\log 2$, regardless of the relative sizes of $n$ and $m$.

Putting these systems together and choosing another bipartition (middle two vs outside two) yields $$|\Psi\rangle\propto |ac\rangle\otimes |ad\rangle+|ad\rangle\otimes |ac\rangle+|bc\rangle\otimes |bd\rangle+|bd\rangle\otimes |bc\rangle.$$ Each of the terms is orthogonal to each other in my construction (the set $\{|ac\rangle,|ad\rangle,|bc\rangle,|bd\rangle\}$ is orthonormal for the first subsystem, similarly for the second) and so the entanglement entropy is $-\frac{1}{4}\sum_{i=1}^4 \log\frac{1}{4}=\log 4=r+\log 2$.

So if we choose $n=4$, $m=\log n=2$ in base 2, then the proposal is that the entanglement entropy is upper bounded by $r+\log m=\log 4$. OP's proposal holds, but the commenter's proposal of $\max(r,\log\log n)$ (with my edits...) is violated.

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  • $\begingroup$ Is there any general formula at play here, that we can derive? $\endgroup$
    – BlackHat18
    Commented Sep 12, 2023 at 3:27
  • $\begingroup$ @BlackHat18 probably! Write each state with its maximally entangled bipartition. The first will have $j$ nonzero Schmidt coefficients and the second will have $k$ (in my example $j=k=2$). Putting them together we get a superposition of $j\times k$ terms, so a new bipartition made from the four original partitions will have at most $jk$ nonzero Schmidt coefficients, each made from the product of the original Schmidt coefficients. Such as state's entanglement entropy is the sum of the entropies from the original two $\endgroup$ Commented Sep 12, 2023 at 13:36
  • $\begingroup$ Ah, I see.; My intuition was wrong! $\endgroup$
    – FDGod
    Commented Sep 14, 2023 at 2:38
  • $\begingroup$ I had another thought, What if we lay out qubits as vertices and entanglement as edges, then finding the maximally entangled bipartition is just like the maxcut problem? I may be really, really wrong. $\endgroup$
    – FDGod
    Commented Sep 14, 2023 at 2:43

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