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 $|0\rangle$.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.