By definition, a quantum state can be expressed as $$|\psi\rangle = a |0\rangle+b |1\rangle.$$ Here, $a, b\in\mathbb{C}$ and $|a|^2 + |b|^2 = 1$.

Now, I would like to take $a = \frac{1}{\sqrt{2}}(1 + i)$ and $b = 0$. Next, I would like to position it on the Bloch sphere. But the formula for a quantum state for the Bloch spere is $$|\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle.$$ Since $\theta\in[0, \pi]\subset\mathbb{R}$, the values of the $\sin(\theta/2)$ are real too, so I fail to see how we can obtain a complex value of $\frac{1}{\sqrt{2}}(1 + i)$ as a coefficient of $|0\rangle$.

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    $\begingroup$ You can multiply the $\psi$ state to any phase $e^{i\alpha}$ where $\alpha$ is real. These introduce the same states. In fact, quantum states are from a Hilbert space divided by an equivalence relation defined by this: $a$ and $b$ are in relation if and only if there exists a real number $\alpha$ such that $a=e^{i\alpha} b$. $\endgroup$ – heaven-of-intensity Jul 19 '16 at 13:26
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    $\begingroup$ So, my state is equivalent to $|0\rangle$ and it sould be depicted as the north pole of the Bloch sphere, right? $\endgroup$ – Igor Deruga Jul 19 '16 at 13:37
  • $\begingroup$ @KNP: Hey, I beleive I got enlightened by your comment. Can you put it as the answer to my question, please? $\endgroup$ – Igor Deruga Jul 19 '16 at 13:40
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    $\begingroup$ South pole cause $\theta=\pi$ is the answer for $\psi=0$. Someone just did it! $\endgroup$ – heaven-of-intensity Jul 19 '16 at 13:41
  • $\begingroup$ I am sorry, I think an error slipped into my formula, it should be $\cos$ that goes with $|0\rangle$, because $|0\rangle$ is usually depicted at the north pole. $\endgroup$ – Igor Deruga Jul 19 '16 at 13:49

Quantum states are rays in the Hilbert space. i.e. $e^{i\alpha}|\psi\rangle$ is the same as $|\psi\rangle$. The set of all $e^{i\alpha}|\psi\rangle$'s with $\alpha\in[0,2\pi)$ is said to form a ray in the Hilbert space. A point on the Bloch sphere represents a ray and not a state.

If you want you can think of the point $(\theta,\phi)$ on the Bloch sphere as representing the set of all vectors $|\psi\rangle = e^{i\alpha}(\cos(\frac{\theta}{2})|0\rangle+e^{i\phi}\sin(\frac{\theta}{2})|1\rangle)$.

If you want to understand how we get away working with states instead of rays(i.e. set of states differing by a phase), you can read up on projective representations from Weinberg's The Quantum Theory of Fields Volume I Chapter 3.

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    $\begingroup$ For 1 qubit, the proper geometrical structure is given by the first Hopf fibration as $S^{3}\rightarrow S^{2}$ with a $S^{1}$ fibre. The Bloch sphere is just the base space and all states on the same fibre are projected to the same base space point. $\endgroup$ – XXDD Jul 19 '16 at 15:04
  • $\begingroup$ Right, agreed. For every point in the Bloch sphere, is a circle. And this indeed gives a three sphere. Thanks for the exact mathematical statement. $\endgroup$ – BoundaryGraviton Jul 19 '16 at 16:01

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