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We know the four Bell states are the maximal entangled states for two-qubit states, and we know if a state cannot be written as the tensor product by its subsets, then it is a entangled state, so is there a definition of maximal entangled state?

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  • $\begingroup$ are you asking specifically about maximal entanglement for multipartite states? $\endgroup$
    – glS
    Commented Sep 22, 2021 at 14:13
  • $\begingroup$ Yes, some description about maximal entanglement for multipartite states. $\endgroup$
    – karry
    Commented Sep 22, 2021 at 14:15

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A pure state is said to be maximally entangled if the Von Neumann Entropy $S(\rho_{A})$, where $\rho_{A}=Tr_{B}(\rho_{AB})$ is the maximum value. ie $S(\rho_{A})=log(d)$ where $d$ is the dimension of the subsystem $A$

Edit: Just gonna add in here that, in the case of multipartite states, the entropy of the marginals in any bipartition of said state should be log(d), where $d$ in this case is the minimum over the dimensions of the systems involved in said bipartition.

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  • $\begingroup$ What if three-qubit system (that is , multipartite system)? $\endgroup$
    – karry
    Commented Sep 22, 2021 at 13:56
  • $\begingroup$ You don't necessarily need subsystem B to be a single subsystem. What I am taking here are biparitions. A tripartite state would be maximally entangled if, upon tracing out any of the other 2 subsystems, the entropy of the remaining state in the subsystem that is left is log(d) $\endgroup$ Commented Sep 22, 2021 at 13:59
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    $\begingroup$ you might want to specify that this applies to pure states. Otherwise, the bipartite state $I\otimes I/dd'$ (with $d,d'$ dimensions of two spaces) also has "maximal entropy". $\endgroup$
    – glS
    Commented Sep 22, 2021 at 14:15
  • $\begingroup$ This is a good catch. I guess I just assumed that was what they were referring to given the only mentioned state was the bell states. $\endgroup$ Commented Sep 22, 2021 at 14:16

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