How to write the state $|ψ\rangle=|00\rangle+\sqrt{i}|01\rangle+(3+i)|11\rangle$ as a column vector?

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?

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?

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 $$\bigg| \dfrac{4}{5} \bigg|^2 = \dfrac{16}{25}$$.
• I recommend including a comment on the normalization of $|\psi\rangle$ in your answer (since the OP's vector is not normalized). Feb 17, 2022 at 4:10