# Find coordinates $\theta$ and $\phi$ on the Bloch sphere of a given qubit state [duplicate]

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.

• I'm sorry how is this not a duplicate of, e.g., quantumcomputing.stackexchange.com/q/9404/55? Could you not just add an answer to that question? – glS Jan 9 at 0:29
• also, you voted to close as duplicate this identical question and then asked it yourself? That doesn't make a lot of sense – glS Jan 9 at 0:35
• @glS I guess Martin wants this to be the canonical thread and mark those other questions as duplicates of this (even retroactively). I'm not sure if this is necessary, but well, the community's votes will decide. – Sanchayan Dutta Jan 9 at 2:08
• @glS: Yes, your are right. Firstly I considered the questions on Bloch sphere duplict, however, then I realized that users look for particular example with numbers. Therefore, I changed my mind. Sorry, if this looked confusing at first glance. Thanks for the link, however, I think that my example is more detailed. But I included the link to my answer as the second example. Overall, I think now there is enough material on Bloch sphere examples to refer to in new questions. – Martin Vesely Jan 9 at 5:49
• @SanchayanDutta: Thanks, that was my intention. – Martin Vesely Jan 9 at 5:50

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

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