# ket statevector of a 2-quibit system

Given the following code:

from qiskit import QuantumCircuit
from qiskit.quantum_info import Statevector

qc = QuantumCircuit(2)

# This calculates what the state vector of our qubits would be
# after passing through the circuit 'qc'
ket = Statevector(qc)

# The code below writes down the state vector.
# Since it's the last line in the cell, the cell will display it as output
ket.draw()


when executed, will output:

Statevector([1.+0.j, 0.+0.j, 0.+0.j, 0.+0.j] \n dims=(2, 2))

which is expected result, as per my understanding, as $$|00\rangle = [1 0 0 0 ]$$

But when we apply $$x$$-gate on qubit 0 as in:

qc.x(0)

ket = Statevector(qc)
ket.draw()


the output is:

Statevector([0.+0.j, 1.+0.j, 0.+0.j, 0.+0.j]\n dims=(2, 2))

which is not correct, as per my understanding, because applying the $$x$$-gate on first qubit will make the multi-qubit state $$|10\rangle = [0 0 1 0]$$, but the output of the above is $$[0 1 0 0]$$.

Also, here in the outputs dims=(2, 2)), shouldn't it be dims=(4, 1))?
The dims function on statevector is not giving you the dimensions of the vector. It is giving you a list of the dimensions of the individual quantum systems. Here you have two qubits (i.e. each of dimension 2).