Hi @Manu_Chaudhary, thanks for the question, and thank you @Kamal_Mohamed for your input!
One way of thinking about these two questions that might help is to visualize these states and transformations on the Bloch sphere. In general, remember that the state of a qubit can be represented mathematically by (Equation (2) in Codebook Node I.6
|\psi \rangle = \cos \Big(\frac{\theta}{2} \Big) |0 \rangle + \sin \Big( \frac{\theta}{2}\Big) e^{i\phi} |1 \rangle
where we follow the convention for the angles shown here.
The state of qubit is then represented by a vector on the Bloch sphere. For example, the state |0 \rangle (which corresponds to \theta = 0 in the expression above) lies at the ‘north pole’ of the Bloch sphere.
To start, you can see the effect of a Hadamard gate on the |0 \rangle or |1 \rangle states geometrically by identifying the corresponding states |+ \rangle = H |0 \rangle, and |- \rangle = H |1 \rangle on the second picture on the textbook side (right-hand side) of the Codebook Node I.6.
Likewise, following the third picture on the textbook of the Codebook Node I.6 you can see the effect of each rotation gate: RX rotates the vector about the x-axis, RY rotates the vector about the y-axis and RZ rotates the vector about the z-axis.
For example, one way of going from the state |0 \rangle to the state |1 \rangle is to apply an RY gate. You can try to do this same rotation (getting from the |0 \rangle to the state |1 \rangle) using an RX gate. Try it out just by visualizing the rotations (don’t use matrices). Can you get from the |0 \rangle to |1 \rangle using an RZ gate?
Finally, for the global phase, you can refer to Codebook Node I.5 for details, but basically, a global phase would be something like
|\psi' \rangle = e^{i \delta} \Big[ \cos \Big(\frac{\theta}{2} \Big) |0 \rangle + \sin \Big( \frac{\theta}{2}\Big) e^{i\phi} |1 \rangle \Big]
You can see - or try convincing yourself - that the extra phase e^{i \delta} (this is a global phase) has no effect on the outcome of measuring the state of a qubit. In other words, whether your qubit is in state |\psi\rangle or |\psi' \rangle, the probability of measuring the state of the qubit (either |0 \rangle or |1\rangle) is the same for both.
Therefore, when you build a quantum circuit, it does not matter if you are off by a global phase since that does not change the outcome of any measurement. For example, if you are trying to go from |0\rangle to the state
\frac{1}{\sqrt{2}} \Big( |0\rangle + e^{5i \pi/4}|1 \rangle \Big)
it does not matter if your sequence of gates gets you to the state
i \frac{1}{\sqrt{2}} \Big( |0\rangle + e^{5i \pi/4}|1 \rangle \Big)
(with a global phase of i).
Hope that helps!
