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?


1 Answer 1


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 $\bigg| \dfrac{4}{5} \bigg|^2 = \dfrac{16}{25}$.

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.