How Does Qubit Orientation on the Bloch Sphere Indicate Quantum States?

[*FaerieLight*]

There is something that I don't quite understand in relation to the Bloch Sphere representation of qubits. I've read that any vector on the sphere is a superposition of two basic states, like spin up and spin down, denoted by |1> and |0>.
So does this mean that if the vector is at z=0 (pointing towards the equator), then it has neither spin-up nor spin-down components?
If the vector is pointing up 45 degrees above the equator towards |1>, then how is this represented by equations as a superposition of |1> and |0> ?
Also, if a qubit is not affected by external noise, then does it remain a static unprecessing vector on the Bloch Sphere?

 

[DrClaude]

So does this mean that if the vector is at z=0 (pointing towards the equator), then it has neither spin-up nor spin-down components?

No, it means that it is an equal superposition of spin-up and spin down,
$$
\frac{1}{\sqrt{2}} \left[ |0 \rangle + e^{i \phi} | 1 \rangle \right]
$$

If the vector is pointing up 45 degrees above the equator towards |1>, then how is this represented by equations as a superposition of |1> and |0> ?

The general equation is
$$
\cos \left( \frac{\theta}{2} \right) |0 \rangle + \sin \left( \frac{\theta}{2} \right) e^{i \phi} | 1 \rangle
$$
so $\theta = 135^\circ = 3\pi/4$ results in
$$
\frac{\sqrt{2 - \sqrt{2}}}{2} |0 \rangle + \frac{\sqrt{2 + \sqrt{2}}}{2} e^{i \phi} | 1 \rangle
$$
so a superposition that is more $|1 \rangle$ than $|0 \rangle$.

Also, if a qubit is not affected by external noise, then does it remain a static unprecessing vector on the Bloch Sphere?

Only if the two states have the same energy. This is not true in most cases, so one will see a precession as the angle $\phi$ will be time-dependent.

 

[Paintjunkie]

https://pdfs.semanticscholar.org/1627/1c6fce3564ee4e2882c6729a7167714da2c5.pdf
This helped me