In last time there is a lot of questions how to find $\theta$ and $\phi$ for particular state on Bloch sphere. I think that it would be useful to solve one example to stop stream of very similar questions.

So my question is how to find $\theta$ and $\phi$ on Bloch sphere? Please demonstrate calculation on this quantum state: $$ \left| \varphi \right>=\frac{1+i}{\sqrt{3}} \left| 0 \right> + {\sqrt{\frac{1}{3}}} \left| 1\right> $$

Note to moderators and other users: please do not mark this question as duplicity but rather use it for closing similar questions on Bloch sphere coordinates calculation. It seems that explanation on particular example is asked rather than theoretical method how to solve such problems.


1 Answer 1


Generally, a quantum state can be expressed in this form:

$$ |\varphi\rangle = \cos\frac{\theta}{2}|0\rangle + \mathrm{e}^{i\phi}\sin\frac{\theta}{2}|1\rangle $$ Where $\theta$ and $\phi$ are coordinates on Bloch sphere.

Regarding the particular state in question, we firstly have to get rid of complex amplitude before $|0\rangle$ to have only real number here. We can do that by multiplying whole state by so-called global phase. This multiplication does not change the state as two states which differ in global phase are identical. You can for example check probabilities of $|0\rangle$ and $|1\rangle$ after multiplication. They remain same (for the state in question probability of measuring $|0\rangle$ and $|1\rangle$ in z-basis is $\frac{2}{3}$ and $\frac{1}{3}$, respectively).

Mathematically, the global phase is a complex number with absolute value 1.

In our paritucar case I multiplied $|\varphi\rangle$ with $\frac{1-i}{\sqrt{2}} = \mathrm{e}^{-\frac{\pi}{4}}$ (hence global phase is $-\frac{\pi}{4}$).

The result is

$$ \left| \varphi \right>={\sqrt{\frac{2}{3}}} \left| 0 \right> + \frac{1-i}{\sqrt{2}}\frac{1}{\sqrt{3}} \left| 1\right> $$

Since $\frac{1-i}{\sqrt{2}} = \mathrm{e}^{-i\frac{\pi}{4}}$, apparently $\phi = -\frac{\pi}{4}$.

Theta can be calculated from $\cos\frac{\theta}{2} = \sqrt{\frac{2}{3}}$. Hence

$$ \theta = 2\arccos\sqrt{\frac{2}{3}} = 1.2310. $$

We can verify $\theta$ with sine

$$ \theta = 2\arcsin\sqrt{\frac{1}{3}} = 1.2310. $$

Conclusion: $\theta = 1.2310$ and $\phi = -\frac{\pi}{4}$.

Another numerical example on Bloch sphere coordinates

  • 1
    $\begingroup$ Thanks for the detailed explanation ،I have an opinion, I think the angle of theta is 70.56° $\endgroup$
    – Ba. Taj
    Jan 8, 2020 at 21:12
  • $\begingroup$ @BasharTaj: I think that $\theta = 70^{o}31'52''$ since $180\cdot1.2310/\pi = 70.531 = 70+\frac{31}{60}+\frac{52}{3600}$. $\endgroup$ Jan 8, 2020 at 21:32

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