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For a bibpartite system $A,B$ with $A\in\mathcal{H}_A$ and $B\in\mathcal{H}_B$, the Schmidt decomposition of its state $| \psi \rangle_{AB}$ is : \begin{equation} | \psi \rangle_{AB}= \sum_{i=0}^{\min \{d_A, d_B\}} \lambda_i | u_i \rangle \otimes | v_i \rangle \end{equation} where :

  • $\lambda_i\ge 0$ are the Schmidt coefficients,
  • $\{| u_i \rangle\}_i$ is an orthonormal basis for Hilbert space $\mathcal{H}_A$,
  • $\{| v_i \rangle\}_i$ is an orthonormal basis for Hilbert space $\mathcal{H}_B$,
  • $d_A$ and $d_B$ are respectively the dimensions of Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$.

Then, for two qubits $A$ and $B$, we obviously have $d_A=d_B=2$ and : \begin{equation}\tag{1} | \psi \rangle_{AB}= \lambda_0 | u_0 \rangle \otimes | v_0 \rangle + \lambda_1 | u_1 \rangle \otimes | v_1 \rangle \end{equation}

A well known example of this is the EPR state : \begin{equation} | \psi \rangle_{AB}= \frac{1}{\sqrt{2}} [ | 00 \rangle + | 11 \rangle ] \end{equation}

Qubits are entangled provided that the Schmidt rank (number of $\lambda_i >0$) is $>1$. Otherwise they are separated. But then, Eq. 1 means that either qubits are separated or maximally entangled, because maximally entangled means that all Schmidt coefficients are non zero, and there are only two Schmidt coefficients.

Finally, this is the same whatever the number of qubits, because the dimension of each qubit Hilbert space is always $2$. So, is it true to say about any qubit register that if it is entangled it is necessarily maximally entangled ?

As an example, the GHZ state for 3 qubits $(A,B,C)$ : \begin{equation} | \psi \rangle_{AB}= \frac{1}{\sqrt{2}} \biggl[ | 000 \rangle + | 111 \rangle \biggr] \end{equation} is maximally entangled.

One can say that $AB$ is entangled with $C$, or $A$ is entangled with $BC$, but however $A$ and $B$ are not entangled :

\begin{split} \rho_{AB} &=\text{Tr}_{C} (\rho_{ABC})\\ &= \frac{1}{2}[| 0 \rangle \langle 0 |_{A} \otimes | 0 \rangle \langle 0 |_{B} \otimes \text{Tr}(| 0 \rangle \langle 0 |_{C}) + | 1 \rangle \langle 1 |_{A} \otimes | 1 \rangle \langle 1 |_{B} \otimes \text{Tr}(|1 \rangle \langle 1 |_{C})\\ &\quad\quad\quad+ | 0 \rangle \langle 1 |_{A} \otimes | 0 \rangle \langle 1 |_{B} \otimes \text{Tr}(| 0 \rangle \langle 1 |_{C}) + | 1 \rangle \langle 0 |_{A} \otimes | 1 \rangle \langle 0 |_{B} \otimes \text{Tr}(| 1 \rangle \langle 0 |_{C})]\\ &= \frac{1}{2}[| 0 \rangle \langle 0 |_{A} \otimes | 0 \rangle \langle 0 |_{B} + | 1 \rangle \langle 1 |_{A} \otimes | 1 \rangle \langle 1 |_{B}]= \frac{1}{2}(| 00 \rangle \langle 00 |_{AB} + | 11 \rangle \langle 11 |_{AB}) \end{split} The reduced density of $AB$ is mixed separated, meaning that $A$ and $B$ are not entangled. The same between $B$ and $C$ or $A$ and $C$.

Is it not contradictory to say that the GHZ state for $ABC$ is maximally entangled although that any two qubits between the three are not entangled ?

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    $\begingroup$ "Eq. 1 means that either qubits are separated or maximally entangled, because maximally entangled means that all Schmidt coefficients are non zero, and there are only two Schmidt coefficients." something like $\frac{1}{\sqrt3}(|00\rangle+\sqrt2|11\rangle$ is a trivial counterexample $\endgroup$
    – glS
    Jan 14 at 21:56
  • $\begingroup$ related quantumcomputing.stackexchange.com/a/34751/25170 $\endgroup$
    – FDGod
    Jan 15 at 0:34
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    $\begingroup$ You can define "maximal entanglement" with respect to any entanglement measure you want. Schmidt rank is exceptionally poor because it's not continuous, and has almost resolution to tell you anything about low dimensional bipartite states. People usually take other entanglement measures to define maximally entangled states. $\endgroup$
    – DaftWullie
    Jan 15 at 8:10
  • $\begingroup$ @glS : could you explain why your example is not maximally entangled ? what does it concretely change that Schmidt coefficients be equal or not ? $\endgroup$
    – deb2014
    Jan 15 at 9:11

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All Schmidt coefficients being nonzero does not imply maximal entanglement.

There's a few different things in your question that need to be clarified:

  1. "But then, Eq. 1 means that either qubits are separated or maximally entangled, because maximally entangled means that all Schmidt coefficients are non zero, and there are only two Schmidt coefficients."

This seems to be a misconception, that is not how "maximally entangled" is defined. One typical definition for maximal entanglement of two qubits is that both Schmidt coefficients are equal. This means $\lambda_0 = \lambda_1 = 1/\sqrt{2}$. Note that maximal entanglement does imply all Schmidt coefficients are nonzero in this case, but the opposite isn't true.

  1. "Is it not contradictory to say that the GHZ state for $ABC$ is maximally entangled although that any two qubits between the three are not entangled ?"

Leaving aside the issue that maximal entanglement is not defined as having only nonzero Schmidt coefficients, there isn't really an agreed upon definition for maximal entanglement once three or more subsystems are involved. An early demonstration of this is Ref [1], where they distinguish the kind of entanglement found in GHZ states from W states. They even address your question:

For instance, [the GHZ state] maximally violates Bell-type inequalities, the mutual information of measurement outcomesis maximal, it is maximally stable against (white) noise and one can locally obtain from a GHZ state with unit propability an EPR state shared between any two of the three parties.

These include a lot of useful notions of maximal entanglement, even if tracing out a single qubit leaves the rest of the system in a separable state.


[1] Phys. Rev. A 62, 062314 (2000) https://arxiv.org/abs/quant-ph/0005115

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