- Why the outcome after the controlled-phase and Hadamrad is equal to:
$$\alpha |++\rangle + \beta |--\rangle = (|0\rangle \otimes H|\psi\rangle+|1\rangle \otimes XH|\psi\rangle)/\sqrt{2}$$
Let us label the qubits $A$ and $B$ for clarity. The initial state for circuit (11) is
$$
|\psi_A \rangle \otimes |+_B\rangle = \left( \alpha |0_A\rangle + \beta |1_A\rangle \right) \otimes \frac{1}{\sqrt{2}}\left( |0_B\rangle + |1_B\rangle\right)
$$
which expands as
$$
|\psi_A \rangle \otimes |+_B\rangle = \frac{\alpha}{\sqrt{2}} |0_A 0_B \rangle + \frac{\alpha}{\sqrt{2}} |0_A 1_B \rangle + \frac{\beta}{\sqrt{2}} |1_A 0_B \rangle + \frac{\beta}{\sqrt{2}} |1_A 1_B \rangle
$$
The reason we need to expand is because the controlled-phase gate is defined on the basis states $|0_A 0_B \rangle$, $|1_A 0_B \rangle$, $|1_A 0_B \rangle$, $|1_A 1_B \rangle$ as $|xy\rangle \rightarrow (-1)^{xy}|xy\rangle$. So its action on the initial state produces
$$
|\Phi_{AB}\rangle = \frac{\alpha}{\sqrt{2}} |0_A 0_B \rangle + \frac{\alpha}{\sqrt{2}} |0_A 1_B \rangle + \frac{\beta}{\sqrt{2}} |1_A 0_B \rangle - \frac{\beta}{\sqrt{2}} |1_A 1_B \rangle = \\
= \alpha \;|0_A +_B \rangle + \beta \;|1_A -_B \rangle
$$
The subsequent Hadamard gate on qubit $A$ acts as $H_A |0_A\rangle = |+_A\rangle$ and $H_A |1_A\rangle = |-_A\rangle$ to give the output state
$$
|\Psi_{AB}\rangle = H_A|\Phi_{AB}\rangle = \alpha \;|+_A +_B \rangle + \beta \;|-_A -_B \rangle
$$
What we need now is to express $|\Psi_{AB}\rangle$ in a way that allows us to see what happens to the state of qubit $B$ after a measurement on qubit $A$ in the canonical basis $|0_A\rangle$, $|1_A\rangle$. That is, we need to rewrite
$$
|\Psi_{AB}\rangle = |0_A \rangle \otimes |\sigma_B \rangle + |1_A \rangle \otimes |\omega_B \rangle
$$
A little algebra gives us:
$$
|\Psi_{AB} \rangle = \frac{\alpha}{\sqrt{2}} \left( |0_A\rangle + |1_A\rangle \right) \otimes |+_B \rangle + \frac{\beta}{\sqrt{2}} \left( |0_A\rangle - |1_A\rangle \right) \otimes |-_B \rangle = \\
= |0_A\rangle \otimes \frac{1}{\sqrt{2}} \left( \alpha\; |+_B\rangle + \beta \;|-_B\rangle \right) + |1_A\rangle \otimes \frac{1}{\sqrt{2}} \left( \alpha\; |+_B\rangle - \beta\; |-_B\rangle \right) = |0_A \rangle \otimes |\sigma_B \rangle + |1_A \rangle \otimes |\omega_B\rangle
$$
But what we prefer is some relationship between the states $|\sigma_B\rangle$, $|\omega_B\rangle$ of qubit $B$ and the initial state $|\psi\rangle$ of qubit $A$. For this, recall that a Hadamard gate on qubit $B$ acts as $H_B |0_B\rangle = |+_B\rangle$ and $H_B |1_B\rangle = |-_B\rangle$. We can use this to rewrite $|\sigma_B\rangle$ as
$$
|\sigma_B \rangle = \frac{1}{\sqrt{2}} \left( \alpha \;|+_B\rangle + \beta \;|-_B\rangle \right) = \frac{1}{\sqrt{2}} \left( \alpha \; H_B|0_B\rangle + \beta \;H_B|1_B\rangle \right) = \frac{1}{\sqrt{2}} H_B |\psi_B \rangle
$$
As for $|\omega_B\rangle = \frac{1}{\sqrt{2}} \left( \alpha\; |+_B\rangle - \beta\; |-_B\rangle \right)$, let us remove the negative sign by recalling that states $|\pm\rangle$ are eigenstates of gate $X$, such that $X_B |+_B\rangle = |+_B\rangle$ and $X_B |-_B\rangle = -|-_B\rangle$. This means that we have
$$
|\omega_B\rangle = \frac{1}{\sqrt{2}} \left( \alpha\; |+_B\rangle - \beta\; |-_B\rangle \right) = X_B \frac{1}{\sqrt{2}} \left( \alpha\; |+_B\rangle + \beta\; |-_B\rangle \right) = X_B |\sigma_B \rangle = \frac{1}{\sqrt{2}} X_B H_B |\psi_B\rangle
$$
and the final form of the output state is
$$
|\Psi_{AB} \rangle = \frac{1}{\sqrt{2}}\left( |0_A \rangle \otimes H_B |\psi_B \rangle + |1_A \rangle \otimes X_B H_B |\psi_B\rangle\right)
$$
- Why isn't $X^m H|\psi \rangle$ the output of the first qubit?
Actually the output state before measurement on qubit $A$ is symmetrical wrt to qubits $A$ and $B$, so we can rewrite it as
$$
|\Psi_{AB} \rangle = \frac{1}{\sqrt{2}}\left( H_A |\psi_A \rangle \otimes |0_B \rangle + X_A H_A |\psi_A\rangle \otimes |1_B \rangle \right)
$$
But this form cannot tell us much about the state of $B$ after the measurement on $A$. The way it presents the information is useless.
- The cluster state: Should work similarly. Try breaking everything down in elementary steps until you get used to the patterns.
