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Consider the two-qubit state $|ψ \rangle= 1|00\rangle +\sqrt i |01\rangle + (3+i)|11\rangle$. How can I write the state $|\psi\rangle$ as a column vector? I'm confused.

And what if I want to measure in the $Z$-basis, what are the probabilities of the states $|00\rangle, |01 \rangle, |10\rangle,$ and $|11\rangle$? When it's two-bits how do I deal with it?

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First note that $|00\rangle = |0\rangle \otimes |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix} $. Given that $|1\rangle= \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, can you do the rest?


To your second point:

Given a state $|\psi \rangle = \sum c_i |e_i\rangle $. Note that a quantum state is always normalized, i.e. unit norm, hence we have must have that $\sum |c_i|^2 = 1$.

With this in mind the probability that you observe the basis state $|e_i\rangle$ upon measurement is $|c_i|^2$. This is a postulate of quantum mechanic.

Example: Suppose we have $|\psi \rangle = \dfrac{3}{5}|0\rangle + \dfrac{4}{5}|1\rangle $. Note that $|0\rangle$ and $|1\rangle$ are the eigenvectors of the observable $Z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$. Then measuring $|\psi\rangle$ in the $Z$ basis will result in the state $|0\rangle$ with probability $\bigg| \dfrac{3}{5} \bigg|^2 = \dfrac{9}{25}$, and $|1\rangle$ with probability $\bigg| \dfrac{4}{5} \bigg|^2 = \dfrac{16}{25}$.

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    $\begingroup$ I recommend including a comment on the normalization of $|\psi\rangle$ in your answer (since the OP's vector is not normalized). $\endgroup$
    – jecado
    Feb 17, 2022 at 4:10
  • $\begingroup$ for my first point, yes I can do the rest but at the end do I add the 4*1 matrices together or tensor product? $\endgroup$
    – n22
    Feb 17, 2022 at 22:41
  • $\begingroup$ you would multiply them with their respective coeff and add them together. Make sure your state is normalize... $\endgroup$
    – KAJ226
    Feb 17, 2022 at 22:52

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