# Graph States for Quantum Secret Sharing

1. Dec 27, 2013

### limarodessa

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. Mar 12, 2014

### DonQubito

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].