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

  • $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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