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 $\phi=-\pi/4$

  • 2
    $\begingroup$ 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 $\frac{1}{\sqrt{2}}$. $\endgroup$ Jan 3 '20 at 19:12
  • $\begingroup$ Ok many thanks for try to solve but I make edit to brackets .please see the state again. @AlexeyKrugovets $\endgroup$
    – Ba. Taj
    Jan 4 '20 at 1:11
  • 1
    $\begingroup$ I don't see the difference in your edition. Pls look at the Martin Vesely comment $\endgroup$
    – Alex
    Jan 4 '20 at 5:32
  • $\begingroup$ which step do you not understand? $\endgroup$
    – Alex
    Jan 4 '20 at 11:52
  • $\begingroup$ in step 2 the $\cos(\pi/4)=\sin(\pi/4)$ replaced the $\frac{1}{\sqrt{2}}$ and the $e^{-i\pi/4}$ replaced $\frac{1-i}{\sqrt{2}}$ $\endgroup$
    – Alex
    Jan 5 '20 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 $\phi = -\frac{\pi}{4}$.


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