# Beginners doubt: calculating probability of one qubit based on other in a 2 qubit system

This is going to look like homework because it is from a self-inflected Coursera course. I have already found the answer by trial and error but I want to clear my logic. (Coz I'm not getting sleep otherwise and I can't ask the question in Coursera without revealing too much and ruining the quiz for others.)

Instead of doing the long math here I used the online calculator: here’s a link:

Once again online calculator to help : Link:

The result comes out as 3/4 but the expected answer is set as 1/2. Can anyone help? I might be missing something very simple.

### Edit

The complete solution:

\begin{aligned}|\phi| = {(1+ \sqrt{3}i) \over 4\sqrt{2}}(|+\rangle +|-\rangle)|0\rangle+{(1- \sqrt{3}i)\over 4\sqrt{2}}(|+\rangle + |-\rangle)|1\rangle \\ + {(1- \sqrt{3}i) \over 4\sqrt{2}}(|+\rangle-|-\rangle)|0\rangle+{(1+ \sqrt{3}i) \over 4\sqrt{2}}(|+\rangle - |-\rangle)|1\rangle\end{aligned}

By dropping the $$|+\rangle$$ combinations we are left with

\begin{aligned}|\phi| = {(1+ \sqrt{3}i) \over 4\sqrt{2}}|-\rangle|0\rangle-{(1- \sqrt{3}i) \over 4\sqrt{2}}\rangle|1\rangle+{(1- \sqrt{3}i) \over 4\sqrt{2}}|-\rangle|0\rangle-{(1+ \sqrt{3}i) \over 4\sqrt{2}}|-\rangle|1\rangle\end{aligned}

This is where I was wrong to not solve it even further.

$$|\phi| = ({(1+ \sqrt{3}i) \over 4\sqrt{2}}-{(1- \sqrt{3}i) \over 4\sqrt{2}})|-\rangle|0\rangle+({(1- \sqrt{3}i) \over 4\sqrt{2}}-{(1+ \sqrt{3}i) \over 4\sqrt{2}})|-\rangle|1\rangle$$

$$|\phi| = {\sqrt{3}i \over 2\sqrt{2}}|-\rangle|0\rangle-{\sqrt{3}i \over 2\sqrt{2}}|-\rangle|1\rangle$$

Next, we normalize this.

$$|\phi| = {({\sqrt{3}i \over 2\sqrt{2}}|-\rangle|0\rangle-{\sqrt{3}i \over 2\sqrt{2}}|-\rangle|1\rangle) \over \sqrt {{\left|\sqrt{3}i \over 2\sqrt{2}\right|^2}+{\left|\sqrt{3}i \over 2\sqrt{2}\right|^2}}}$$

$$|\phi| = {2\sqrt{3} \over 3} . {\sqrt{3}i \over 2\sqrt{2}}|-\rangle|0\rangle-{2\sqrt{3} \over 3}.{\sqrt{3}i \over 2\sqrt{2}}|-\rangle|1\rangle)$$

$$|\phi| = {1 \over \sqrt{2}}|-\rangle|0\rangle-{1 \over \sqrt{2}}|-\rangle|1\rangle)$$

Hence the probably of second qubit: $$P(|0\rangle)$$ when the first qubit is $$|-\rangle$$ is :

$$|\phi| = {\left ({1 \over \sqrt{2}}\right )}^2 = \frac{1}{2}$$

• Hey @S4nd33p! While the image format provided works, it's very preferable to use LaTeX in the future! If you have time, I'd recommend updating the question to be formatted in LaTeX. – C. Kang Nov 20 '19 at 6:03
• Thanks for the suggestion, i'll work on it. It does feel a bit stupid not being able to copy-paste the code. In the mean time, ctrl+/- to zoom in/out may help. – S4nd33p Nov 20 '19 at 6:18

Recognize that your misstep lies in the properties of the amplitude. Generally, for some $$a, b$$ components and $$x, y$$ separation:
$$|a + bi|^2 \neq |x + yi|^2 + |(a - x) + (b - y)i|^2$$
There is no long math, and you need not online calculators. When you simply omit $$|+\rangle$$ state of the first qubit after the measurement, the resulting 2-qubit state $$|\Phi\rangle$$ is unnormalized. You just need to factor out the state $$|-\rangle$$ of the first qubit and normalize the remaining state of the second qubit before obtaining the final answer.