4
$\begingroup$

I am a beginner at QC. I was going through the basics of multi-qubits I encountered a state $|2\rangle|3\rangle$.

I want clarification on the following points:

  1. Can I write $|2\rangle$ as $|10\rangle = |1\rangle|0\rangle$ always?
  2. If $|2\rangle = |10\rangle = |1\rangle|0\rangle$, Can I write $|2\rangle|3\rangle = |10\rangle|11\rangle = |1011\rangle$?

Assume $|0\rangle$, $|1\rangle$ in computational basis. $|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$\endgroup$
2
  • $\begingroup$ I've never seen decimal notation like $|2\rangle$; it is not clear whether it means 2-qubit state $|10\rangle$ or 3-qubit state $|010\rangle$ or other multiqubit state. $\endgroup$
    – kludg
    Dec 1, 2019 at 10:41
  • $\begingroup$ @kludg: It is generaly basis vector representing decimal number 2. You are right that it can be $|011\rangle$ as well, however, it should be clear from dimensionality of a problem. $\endgroup$ Dec 1, 2019 at 11:44

1 Answer 1

3
$\begingroup$

Your interpretation is right.

1) $|2\rangle$ is $|10\rangle$, or vector $\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$ in Hilbert space representing two q-bits system.

2)Yes, it is possible as well. The first two q-bits in in state $|10\rangle$ and last two q-bits in state $|11\rangle$, hence the state of all four q-bits is $|1011\rangle$, or $|11\rangle$ (eleven!) in decimal but I would recommend to avoid writing decimal numbers in this case as they can be confused with binary numbers easily.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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