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. $|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

  • $\begingroup$ 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


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.


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