# What’s the proper gate use for the following code? (Python)

I am trying to find a proper way to write a code that will give me the resulting quantum state $$i|10\rangle$$ using qiskit in Python (using Jupiter notebook). I think I have figured out that the following works:

from qiskit import QuantumCircuit, execute, Aer
from math import pi

qc = QuantumCircuit(2,2)

qc.x(1)
qc.crz(3*pi,1,0)


which (if I choose to print the state vector) gives me this result:

job = execute(qc, Aer.get_backend(‘statevector_simulator’),optimization_level=0)
current_quantum_state = job.result().get_statevector(qc)
print(current_quantum_state)
—————————————————————————————————————————————————————————————————————————————————————
Statevector([0.00000000e+00+0.j, 0.00000000e+00+0.j, -2.22044605e-16+1.j,
0.00000000e+00-0.j],
dims=(2,2))


Using an X gate first and then a CRZ gate I get the result I want, but I was told that we can get this through just one line (therefore one gate?) and not the last two lines I provided. I looked everywhere through my notes and the Qiskit manual online, but I can’t seem to find something that works and I was wondering if someone could direct me to the faster approach to this.

Note that $$Y|0\rangle = i|1\rangle$$ and so the circuit you are looking for is

• Oh! Then this would mean that qc.y(1) would suffice! Is that correct?
– Tita
Apr 13 at 15:55
• qc.y(0) since you want $|10\rangle$ and not $|01\rangle$. Apr 13 at 15:57
• I understand it better now, thank you very much!
– Tita
Apr 13 at 15:58
• However, in Python I’m still inclined to believe the code would be qc.y(1) because of the way it twists the qubits.
– Tita
Apr 13 at 16:18
• yeah. qiskit uses little endian convention so that makes sense quantumcomputing.stackexchange.com/questions/8244/… Apr 13 at 17:02