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I'm learning about different possibilities to measure entanglement right now, and I'm struggling with the geometric measure. My question is probably quite trivial, but I can just not comprehend how the following example was calculated.

So, what I know is that if I have an $n$-partice pure state, in general: $$\lvert{\psi}\rangle = \sum_{p_1,p_2,\dots p_n} C_{p_1 \dots p_n}\lvert{p_n}^{(1)}\rangle \otimes \ \lvert{p_2}^{(2)}\rangle \otimes \dots \lvert{p_n}^{(n)}\rangle $$ Then I have to find the closest seperable pure state $$\lvert\phi\rangle.$$ The measure is then $$ \Lambda = \max_{\phi} \lvert\langle\phi|\psi\rangle\rvert$$

In the given example, there is the GHZ-State: $$ \lvert \text{GHZ}\rangle= \frac{1}{\sqrt{3}} ( \lvert001\rangle + \lvert010\rangle + \lvert100\rangle) $$ And the solution should be $\Lambda=2/3$. But I do not know, how this is calculated. The result implies (so I think) that the relevant $\lvert\phi\rangle$, which creates the maximum value, is a linear combination of 3 states of which 2 are also in the GHZ-state. But a state like: $$ \lvert\phi\rangle = \frac{1}{\sqrt{3}} ( \lvert001\rangle + \lvert010\rangle + \lvert000\rangle ) $$ Is not separable in factors of the 3 separate Hilbert-spaces. What do I miss? Or which thing did I misunderstood?

Edit: relevant link to slides : http://insti.physics.sunysb.edu/conf/simons-qcomputation/talks/wei.pdf Definition on Slide 10, Example on Slide 14)

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  • $\begingroup$ Although a good question in my opinion, you should consider to edit it to use \lvert and \rangle for dirac noation (see the help page on notation) as this will make it much easier to read your question. $\endgroup$ – Sebastian Riese May 31 '15 at 21:10
  • $\begingroup$ Where did you read this question and answer? $\endgroup$ – innisfree May 31 '15 at 21:34
  • $\begingroup$ As example here: arxiv.org/pdf/quant-ph/0212030v1.pdf Page 2, bottom right. I also found presentation-slides from the same author with the same example and result, but I cannot find the link to this anymore. $\endgroup$ – Djoser42 May 31 '15 at 21:47
  • $\begingroup$ Hmm, that paper isn't particularly lucid, is it! I think my answer was incorrect, I've removed it $\endgroup$ – innisfree May 31 '15 at 22:31
  • $\begingroup$ @Djoser42 Hi I am interested in learning about multipartite entanglement, do you have any other recommendations for basic notes which you found useful in outlining ideas of multipartite entanglement? $\endgroup$ – Moses Apr 11 '18 at 8:21
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This is a sketch of the calculation, take:

$$ \lvert\phi\rangle = \lvert\phi_1\rangle \otimes \lvert\phi_2\rangle \otimes \lvert\phi_2\rangle $$

The most general form of each component is

$$ \lvert\phi_i\rangle = \frac{1}{\sqrt{1+u_i^2+v_i^2}}\begin{bmatrix} u_i+iv_i \\ 1 \end{bmatrix}$$

where, $i=1,2,3$ and $u_i$ and $v_i$ are real

Writing the (square of the) inner product $\Lambda$ with and differentiating it with respect to the $u_i$'s and the $v_i$'s we obtain 6 equations, each couple of equations of the $u_i$'s and $v_i$'s can be subtracted to obtain linear equations whose final result implies that:

$$u_1=u_2=u_3 \doteq u $$

and

$$v_1=v_2=v_3 \doteq v $$

substituting thse results in the expression for $\Lambda$, we get

$$\Lambda = \frac{\sqrt{3r}}{(\sqrt{1+r})^3}$$

where $ r = u^2+v^2$

This expression attains its maximum at $r = \frac{1}{2}$ with $\Lambda = \frac{2}{3}$

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