I've been reading pages such as this one: http://comp.uark.edu/~jgeabana/blochapps/bloch.html

Which talk about the Bloch sphere, but I've been unable to figure out how to plot states on the sphere due to both states being a complex number, thus resulting in 4 "coordinates".

Many sources say it's only the relative angle between the states that matters, but that seems odd to me, since there is an absolute $|0\rangle$ and $|1\rangle$ state, which makes it think that both imaginary parts of the complex values matter.

So, how would I plot things like these, or at least, how would I convert them to spherical coordinates for plotting on the bloch sphere?

$$1/\sqrt{2}(|0\rangle + |1\rangle)$$ $$1/\sqrt{2}(|0\rangle - |1\rangle)$$ $$1/2(\sqrt{3}|0\rangle + |1\rangle)$$

  • 1
    I dare you to pass the turing test Normal Human! Your response was too quick :P – Alan Wolfe Oct 16 '15 at 20:12
up vote 2 down vote accepted

Using equation (1) of your link, consider your first qubit:

$$\cos\theta/2 = 1/\sqrt 2, \quad e^{i\phi}\sin\theta/2 = 1/\sqrt 2$$

Hence--given the range of values of $\theta$ and $\phi$--we have $\theta = \pi/2$ and $\phi = 0$.

Proceed similarly for the other qubits.


An important thing to remember about qubits is that we don't care about overall phase. Hence if we have a qubit of the form

$$|\psi\rangle = e^{i\alpha}a|0\rangle + e^{i\beta}b|1\rangle = e^{i\alpha} \left( a|0\rangle + e^{i(\beta-\alpha)}b|1\rangle \right) $$

we find the Bloch representation of $a|0\rangle + e^{i(\beta-\alpha)}b|1\rangle $.

  • Thanks! Why is it that we don't care about overall phase? if the answer isn't too long for a comment hehe – Alan Wolfe Oct 16 '15 at 20:40
  • Because a multiple of a phase $e^{i\theta}$ of a coefficient doesn't affect the probability of being in a given state. It is however worth keeping track of the phase difference between the components so we don't dispense with phase altogether. – Simon S Oct 16 '15 at 20:44
  • oh got it. thanks! – Alan Wolfe Oct 16 '15 at 20:44

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