# How do Hadamard and CNOT gates work on Qiskit SDK? Why is the output reversed?

Here is the code that I have been using on IBM Q Experience (should be the latest version of Qiskit). From my understanding it seems like the outputs of Hadamard and CNOT gates are reversed in a 2-qubit system in Qiskit: A Hadamard gate operating on state 0 actually acts on state 1 and vice versa, similarly with a CNOT gate. Is there something wrong with my understanding? The comments in my code summarize the output results I find surprising.

%matplotlib inline
# Importing standard Qiskit libraries
from qiskit import QuantumCircuit, execute, Aer, IBMQ
from qiskit.compiler import transpile, assemble
from qiskit.tools.jupyter import *
from qiskit.visualization import *
from iqx import *

from qiskit.quantum_info import Statevector

sv01 = Statevector.from_label('01')
print(sv01)
plot_state_qsphere(sv01.data)

h0circuit = QuantumCircuit(2)
h0circuit.h(0)
h0circuit.draw("mpl")

final01 = sv01.evolve(h0circuit)
print(final01)
plot_state_qsphere(final01.data)
#Shouldn't the output be state 1/sqrt(2) (|01> + |11>)  ????
#And why is the phase on the state |00> pi?
#Shouldn't the phase be pi for |01> instead on the qsphere (ambiguous, probably doesn't matter)?

h1circuit = QuantumCircuit(2)
h1circuit.h(1)
h1circuit.draw("mpl")

final01 = sv01.evolve(h1circuit)
print(final01)
plot_state_qsphere(final01.data)
#Shouldn't the output be state 1/sqrt(2) (|00> - |01>)  ????

#Conclusion of the above. An H gate on qubit 0 seems to apply to qubit 1. An H gate on qubit 1 seems to apply to qubit 0.

sv01 = Statevector.from_label('01')
print(sv01)
plot_state_qsphere(sv01.data)

c0xCircuit = QuantumCircuit(2)
c0xCircuit.cx(0,1)
c0xCircuit.draw('mpl')

final01 = sv01.evolve(c0xCircuit)
print(final01)
plot_state_qsphere(final01.data)
#Shouldn't this result be  |01> ?

c1xCircuit = QuantumCircuit(2)
c1xCircuit.cx(1,0)
c1xCircuit.draw('mpl')

final01 = sv01.evolve(c1xCircuit)
print(final01)
plot_state_qsphere(final01.data)
#Shouldn't this result be  |11> ?

#Conclusion of the above. The cnot (0,1) gate actually acts as a cnot (1,0) gate. The cnot (1,0) gate actually
#acts as a cnot (0,1) gate. The states 0 and 1 are flipped somehow.
$$$$
`

$$CX\ q_0, q_1 = I \otimes |0\rangle\langle0| + X \otimes |1\rangle\langle1| = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}$$