SX operator and superposition

Question

I am running some tests using the probabilities we get from statevector to assert values in qiskit. For instance, with two qubits and a hadamard gate on the first one we have:

circuit = QuantumCircuit(2)
circuit.h(0)
trace_out_qubits = [1]
partial_trace = qi.partial_trace(qi.Statevector(circuit), trace_out_qubits)
print(partial_trace.probabilities())
plot_bloch_multivector(partial_trace)

The output of the print statement is [0.5 0.5] and the bloch sphere looks like this:

If we replace the Hadamard gate with the SX gate, we have:

circuit = QuantumCircuit(2)
circuit.sx(0)
trace_out_qubits = [1]
partial_trace = qi.partial_trace(qi.Statevector(circuit), trace_out_qubits)
print(partial_trace.probabilities())
plot_bloch_multivector(partial_trace)

The output is also [0.5 0.5], and the bloch sphere looks like this:

So, it looks like the sx acts similarly to the hadamard gate but with a rotation of pi/2, putting the |0⟩ in superposition.

My doubts:

1. Are the SX and Hadamard gate equivalent when it comes to superposition?
2. Can we use the SX gate together with the CX gate to entangle two qubits?
3. In case the answers to these two questions are positive, what are all the other operators that put qubits in superposition? Can any operator that puts the vector on the plane XY be a "superposition" operator?

Thank you,

Answer 1

1. if you mean whether they have the same impact on entangled states, the answer is generally a straight no, because the operators themselves are different ($√X \not= H$), as you can see from your own plots.

2. $CX$ is already an entangling gate (meaning that it is able to entangle states). Its combination with $√X$ can generate a maximally entangled state, similarly to what happens when applying $CX\cdot (H\otimes \mathbb{1})$ applied to $|00\rangle$. Nevertheless, the entangled state generated by $CX\cdot (√X\otimes \mathbb{1}) \cdot |00\rangle$ is different.

I hope answers 1. and 2. will help you understanding better your doubts in question 3.