-2
$\begingroup$

I have task to calculate this quantum circuit enter image description here

Where psi is:

x1 = 5
x2 = 1
r = x1 * x1 + x2 * x2
a = np.sqrt(1 + 2*r)
psi = [0, 0, 0.5, 0, 0, -0.5, 0, 0, 1/(np.sqrt(2)*a), 0, x1/(np.sqrt(2)*a), x2/(np.sqrt(2)*a), x1/(np.sqrt(2)*a), x2/(np.sqrt(2)*a), 0, 0]

I am doing this type of code:

circ = QuantumCircuit(4)

circ.h(0)

meas = QuantumCircuit(4, 4)
meas.barrier(range(4))
# map the quantum measurement to the classical bits
meas.measure(range(4), range(4))

# The Qiskit circuit object supports composition using
# the addition operator.
qc = circ + meas
qc.initialize(psi, 4)
#drawing the circuit
backend_sim = Aer.get_backend('qasm_simulator')

# Execute the circuit on the qasm simulator.
# We've set the number of repeats of the circuit
# to be 1024, which is the default.
qc.draw('mpl')

job_sim = execute(qc, backend_sim, shots=1000)

# Grab the results from the job.
result_sim = job_sim.result()
counts = result_sim.get_counts(qc)
print(counts)

But getting this error:

CircuitError: 'Index out of range.'

If I change qc.initialize like this:

qc.initialize(psi, 3)

I get this:

QiskitError: 'Initialize parameter vector has 16 elements, therefore expects 4 qubits. However, 1 were provided.'
$\endgroup$
0
$\begingroup$

As stated here, for initialization, you should make it as a list. Hence for your case, since it only uses 4 qubits, you could write

qc.initialize(psi, [0,1,2,3])

However, your circuit seems to be a bit off: the resulting circuit of yours would be a Hadamard gate first, followed by a series of measurements, and then the initialization at the end.

To build the circuit as in your picture, this code will do:

qc = QuantumCircuit(4)

qc.initialize(psi, [0,1,2,3])

qc.h(0)

qc.measure_all()

The resulting circuit would be: enter image description here

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.