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Graph States for Quantum Secret Sharing

  1. Dec 27, 2013 #1
    In Graph States for Quantum Secret Sharing on page 3 :

    I understand that $$\mathop \otimes \limits_i Z_i^{{l_{i2}}} = Z_1^{{l_{12}}} \otimes Z_2^{{l_{22}}} \otimes Z_3^{{l_{32}}}$$

    But I don’t understand why $$\left| G \right\rangle = \left( {\frac{{\left| {0 + + } \right\rangle + i\left| {1 - - } \right\rangle }}{{\sqrt 2 }}} \right)$$

    I ask explain to me – why it is so
     

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    Last edited: Dec 27, 2013
  2. jcsd
  3. Mar 12, 2014 #2
    It might just be a typo. From definition 3, [itex]\left( \left|0++\right>+i\left|1--\right> \right)/\sqrt{2} [\itex] is supposed to be the graph state [itex]\left| G \right>[\itex]. If I am correct in that assumption then [itex] \left|G\right> [\itex] cannot have any imaginary amplitudes. This is because you can create the graph by doing a series of controlled phases, one for each edge of the graph, to the state [itex]\left|+\right>^{\otimes 3}[\itex]. Controlled phases can only change the phase of a computational basis state by [itex]-1[\itex].
     
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