I was reading a paper on a fault-tolerant quantum computation protocol based on tree cluster states. The author said that by applying a local Hadamard gate to two of the qubits in a 4 GHZ state, we create a three-qubit tree-cluster state where the central qubit is redundantly encoded in the two-qubit basis $|VV \rangle$, $|HH \rangle$.
Could someone explain this?
After applying hadamard gates on qubits 1 and 4
$$\frac{|0000\rangle+|1111\rangle}{\sqrt{2}}\to \frac{|+00+\rangle+|-11-\rangle}{\sqrt{2}}=\frac{|+HH+\rangle+|-VV-\rangle}{\sqrt{2}}$$
Now given a 3 qubit cluster state takes the form of $$\frac{|+0+\rangle +|-1-\rangle}{\sqrt{2}}$$
then in the case of applying the hadamards as shown, you've redundantly encoded $|0\rangle\to|00\rangle$ and $|1\rangle \to |11\rangle$.
Unless they mean redundantly encode it as bell states, which you can kind of do by applying the hadamard gates to the central two qubits, but you don't get $|00\rangle$ and $|11\rangle$ as the only values, you get bell states in the central qubits.
