Star graph state

Each qubit is initialized in the state $∣ + ⟩$ : $\begin{align*} \text{Central qubit } a &: |+\rangle_a = \frac{1}{\sqrt{2}} (|0\rangle_a + |1\rangle_a), \\ \text{Peripheral qubits } b_i &: |+\rangle_{b_i} = \frac{1}{\sqrt{2}} (|0\rangle_{b_i} + |1\rangle_{b_i}) \text{ for } i = 1, \ldots, n. \end{align*}$ The controlled-Z operation between qubits $a$ and $b_{i}$ transforms their joint states as follows:

C Z_{a b_{i}} ∣0 ⟩_{a} ∣0 ⟩_{b_{i}} C Z_{a b_{i}} ∣0 ⟩_{a} ∣1 ⟩_{b_{i}} C Z_{a b_{i}} ∣1 ⟩_{a} ∣0 ⟩_{b_{i}} C Z_{a b_{i}} ∣1 ⟩_{a} ∣1 ⟩_{b_{i}} = ∣0 ⟩_{a} ∣0 ⟩_{b_{i}}, = ∣0 ⟩_{a} ∣1 ⟩_{b_{i}}, = ∣1 ⟩_{a} ∣0 ⟩_{b_{i}}, = - ∣1 ⟩_{a} ∣1 ⟩_{b_{i}} .

Initially, all qubits are in a tensor product state:

(\frac{1}{2} (∣0 ⟩_{a} + ∣1 ⟩_{a})) \otimes (\frac{1}{2} (∣0 ⟩_{b_{1}} + ∣1 ⟩_{b_{1}})) \otimes \dots \otimes (\frac{1}{2} (∣0 ⟩_{b_{n}} + ∣1 ⟩_{b_{n}}))

Apply CZ between a and each $b_{i}$ . The final state $∣ G ⟩$ is: $∣ G ⟩ = (\prod_{i = 1}^{n} C Z_{a b_{i}}) (\frac{1}{2 ^{n + 1}} (∣0 ⟩_{a} ⨂_{i = 1}^{n} (∣0 ⟩_{b_{i}} + ∣1 ⟩_{b_{i}}) + ∣1 ⟩_{a} ⨂_{i = 1}^{n} (∣0 ⟩_{b_{i}} - ∣1 ⟩_{b_{i}})))$ This can be expressed as:

∣ G ⟩ = \frac{1}{2 ^{n + 1}} (∣0 ⟩_{a} i = 1 ⨂ n (∣0 ⟩_{b_{i}} + ∣1 ⟩_{b_{i}}) + ∣1 ⟩_{a} i = 1 ⨂ n (∣0 ⟩_{b_{i}} - ∣1 ⟩_{b_{i}})) = \frac{1}{2} (∣ 0 ⟩_{a} ∣ + ⟩_{b_{i}}^{\otimes n} + ∣ 1 ⟩_{a} ∣ - ⟩_{b_{i}}^{\otimes n})

Meignant-et-al The choice of which vertex is the center is arbitrary, and can be changed by local operations such as two successive local complementations.

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Star graph state

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