What does the "monogamy" principle imply? At a superficial level, it seems to say that a particle can be entangled with at most one other particle.

However, I keep reading that several particles are entangled. For example:

  1. Quantum entanglement can reach into the past. Here they describe an experiment with two entangled pairs, $(A_1, A_2)$ and $(B_1, B_2)$. Victor is given $A_2, B_2$, Alice is given $A_1$ and Bob is given $B_1$. All three are separated by a large distance. If Victor entangles $A_2, B_2$ then he has entangled $A_1$ and $B_1$ as well. This somehow implies that each particle can be entangled with more than 1.

  2. 3,000 atoms entangled in bizarre state. Scientists somehow entangled $\approx$ 3000 atoms.

Both imply that entanglement is not monogamous. So what does the "monogamy" principle actually imply?


To my knowledge, Monogamy means that If particle A and B are in a maximally entangled state, for example one of the Bell states, then particle A or B can not have entanglement with the third particle C. This can be extend to more particle's systems. For example, for GHZ states, the entanglement between all possible bipartitions like A-BC are equal to one (by using normalized negativity), so A cannot have entanglement with any other particle D. This is true for any other two bi partitions.

In the material below I define the Bell states, the GHZ state and normalized negativity.

Bell state: Bell states are maximally entangled states for two particles. your particles can be Qubits (the Hilbert space of each one is two dimensional space), or they can be Qudits (the Hilbert space of each one is a d dimensional space). Here you can see Bell states for two qubits: \begin{eqnarray} \vert\phi_{\pm}^{AB}\rangle &=&\frac{1}{\sqrt{2}}(\vert 0^{A},0^{B}\rangle\pm\vert 1^{A},1^{B}\rangle)\:,\nonumber\\ \vert\Psi_{\pm}^{AB}\rangle &=&\frac{1}{\sqrt{2}}(\vert 0^{A},1^{B}\rangle\pm\vert 1^{A},0^{B}\rangle)\:. \end{eqnarray}

GHZ state: GHZ states, or Greenberger-Horne-Zeilinger states, are states with at least 3 particles. There exist certain kind of entanglement between their bipartitions. The GHZ state of M particles is defined as

\begin{equation} \vert\Psi_{GHZ}\rangle_{M} =\frac{1}{\sqrt{2}}\left(\vert 0\rangle^{\otimes M} +\vert 1\rangle^{\otimes M}\right). \end{equation}

For example, if M=3,

\begin{equation} \vert\Psi_{GHZ}\rangle_{3} =\frac{1}{\sqrt{2}}\left(\vert 000\rangle +\vert 111\rangle\right). \end{equation}

Here the entanglement between any of $\vert\Psi_{GHZ}\rangle_{3}$ bipartitions is equal to 1 and if we trace about one of the qubits there will be no entanglement between the remaining two qubits.

Normalized Negativity: The Peres-Horodecki criterion is one of the measures of multipartite entanglement, which is called Negativity and can be defined as

\begin{equation} N(\rho)=\frac{\Vert\rho^{T_{B}}\Vert_{1}-1}{d-1}, \end{equation}

where $ \rho^{T_{B}} $ is partial transpose of state $ \rho $ with respect to partition $ ``B" $ in a $ d\otimes d^{'} $-dimensional quantum system $ (d\leq d^{'}) $, and $ \Vert \cdot \Vert $ is the trace norm. The equivalent definition of negativity is based on negative eigenvalues of $ \rho^{T_{B}} $ \begin{equation*} N(\rho)=\frac{(\sum_{\lambda_{i}}|\lambda_{i}|)-1}{d-1} \end{equation*}

  • $\begingroup$ This seems good except that you should give more information so that someone without a background in quantum information understands. What does a Bell state and GHZ state look like? What is a normalized negativity measure? etc. $\endgroup$ – Rococo Nov 12 '15 at 23:47
  • $\begingroup$ Thank you so much for your comment. I will put this information in my answer. $\endgroup$ – Najmeh Nov 13 '15 at 5:08
  • $\begingroup$ Just to understand your example with GHZ, here you take $A$ as your first two qubits, $B$ as the third one, and you say that $A$ and $B$ are maximally entangled, thus $B$ cannot be entangled with any other system $C$. But I would like to check something : here you take $C$ as any of the first two qubits. The little thing that confuse me is that $C$ is thus "inside" $A$. It is not a problem for the definition ? It still works to justify ? (and also you talked about particle but the definition is valid for any quantum states right ?) $\endgroup$ – StarBucK Oct 26 '18 at 12:41

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