# How can I find the $\theta$ and $\phi$ values of a qubit on the Bloch sphere?

If I have the following state:

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

How can I find the $$\theta$$ and $$\phi$$ values of this qubit on the Bloch sphere?

See below: $$\left| \varphi \right>=\frac{1}{\sqrt{\sqrt{2}}}\left(\left(\frac{1+i}{\sqrt{2}} \right)\left| 0 \right> + \left| 1\right>\right) = \frac{1+i}{\sqrt{\sqrt{2}}}\left(\frac{1}{\sqrt{2}}\left| 0 \right> + \frac{1}{\sqrt{2}}\frac{1-i}{\sqrt{2}} \left| 1\right>\right) = \frac{1+i}{\sqrt{\sqrt{2}}}\left(\cos(\pi/4)\left| 0 \right> + \sin(\pi/4) e^{-i\pi/4} \left| 1\right>\right) = C\left(\cos(\theta/2)\left| 0 \right> + \sin(\theta/2) e^{i\phi} \left| 1\right>\right)$$

Therefore $$\theta = \pi/2$$ and $$\phi=-\pi/4$$

• Just note that state $|\phi\rangle$ is not normalized as $|a|^2+|b|^2 = \sqrt{2}$. Probably the fraction before left bracket should have been $\frac{1}{\sqrt{2}}$. – Martin Vesely Jan 3 at 19:12
• Ok many thanks for try to solve but I make edit to brackets .please see the state again. @AlexeyKrugovets – Ba. Taj Jan 4 at 1:11
• I don't see the difference in your edition. Pls look at the Martin Vesely comment – Alexey Krugovets Jan 4 at 5:32
• which step do you not understand? – Alexey Krugovets Jan 4 at 11:52
• in step 2 the $\cos(\pi/4)=\sin(\pi/4)$ replaced the $\frac{1}{\sqrt{2}}$ and the $e^{-i\pi/4}$ replaced $\frac{1-i}{\sqrt{2}}$ – Alexey Krugovets Jan 5 at 6:36

In last time there is a lot of questions how to find $$\theta$$ and $$\phi$$ for this particular state on Bloch sphere: $$\left| \varphi \right>=\frac{1+i}{\sqrt{3}} \left| 0 \right> + {\sqrt{\frac{1}{3}}} \left| 1\right>$$

I will try to demonstrate how to do so in more details in comparison with previous answer.

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 $$\phi = -\frac{\pi}{4}$$.