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limarodessa
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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
‘…The “encoded graph state” is $$\left| {{G_{{{\vec l}_{*2}}}}} \right\rangle = \mathop \otimes \limits_i Z_i^{{l_{i2}}}\left| G \right\rangle $$ . (3.9) …
Example 1. The three-qubit labeled graph state presented
in Fig. 1 is the encoded graph state
$$\left| {{G_{{{\vec l}_{*2}}}}} \right\rangle = Z_1^{{l_{12}}} \otimes Z_2^{{l_{22}}} \otimes Z_3^{{l_{32}}}\left( {\frac{{\left| {0 + + } \right\rangle + i\left| {1 - - } \right\rangle }}{{\sqrt 2 }}} \right)$$ , (3.13) …’
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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