Qiskit: Initializing $n$ qubits with binary values ($0$s and $1$s)

Question

Is there a way in Qiskit to initialize $n$ qubits with binary values (0s and 1s)? For example, how can I initialize two qubits in the state $|11\rangle$? Here is a code snippet:

from qiskit import QuantumCircuit
import numpy as np

n = 2
circuit = QuantumCircuit(n, n)

# Seeking some sort of initialization routine like this
circuit.initializeQubits(initialState=np.ones(n), ...)   

# Define rest of the circuit
...

I am aware of the method in this tutorial, which is also referenced here and here. This method creates an arbitrary qubit state by defining an $N$ dimensional ($N = 2^n$) state vector of amplitudes. The problem with this method is it requires creating a state vector which is exponentially large. I'm trying to initialize the qubits by defining an $n$ dimensional binary vector, which for the above example would be [1, 1].

Answer 1

You can also create a Statevector, that can be directly initialized as follows:

from qiskit.quantum_info import Statevector
sv = Statevector.from_label('11')

You can use sv.evolve(qc) to apply an operator/circuit to the state, where qc is the operator/circuit. sv.data gives you the numpy array, containing the actual implementation of the state.

Check this for more details.

Answer 2

qiskit-terra 0.16 or lower

As answered, probably the most canonical way to do this is with Statevector.from_label and initialize.

Here is the full example:

from qiskit import *
from qiskit.quantum_info import Statevector

n = 2
qc = QuantumCircuit(n)

qc.initialize(Statevector.from_label('1'*n).data, range(n))
qc.draw()

     ┌──────────────────────┐
q_0: ┤0                     ├
     │  initialize(0,0,0,1) │
q_1: ┤1                     ├
     └──────────────────────┘

You could confirm the result like this:

qc.measure_all()
execute(qc, backend=BasicAer.get_backend('qasm_simulator')).result().get_counts()

qiskit-terra 0.17 or higher

This questions inspired a new way to initialize a qubits in the basis states of the Pauli eigenstates Z, X, Y (à la Statevector.from_label).

from qiskit import QuantumCircuit
from qiskit.extensions.quantum_initializer.initializer import Initialize

circuit = QuantumCircuit(6)
circuit.append(Initialize("10+-lr"), range(6))
circuit.draw()

     ┌──────────────────────────┐
q_0: ┤0                         ├
     │                          │
q_1: ┤1                         ├
     │                          │
q_2: ┤2                         ├
     │  initialize(1,0,+,-,l,r) │
q_3: ┤3                         ├
     │                          │
q_4: ┤4                         ├
     │                          │
q_5: ┤5                         ├
     └──────────────────────────┘

The decomposition of this gate is the following:

circuit.decompose().draw()

          ┌───┐ ┌───┐ 
q_0: ─|0>─┤ H ├─┤ S ├─
          ├───┤┌┴───┴┐
q_1: ─|0>─┤ H ├┤ SDG ├
          ├───┤└┬───┬┘
q_2: ─|0>─┤ X ├─┤ H ├─
          ├───┤ └───┘ 
q_3: ─|0>─┤ H ├───────
          └───┘       
q_4: ─|0>─────────────
          ┌───┐       
q_5: ─|0>─┤ X ├───────
          └───┘       

Answer 3

Qiskit assumes that initially each qubitt is set to the $|0\rangle$ state. So if you have $n$ qubits, the initial state is $|00..0\rangle$. If you want to flip the state of some specific qubits, you have to apply an $X$ gate to each of those specific qubits. For example, the following code sets the $|11\rangle$ state:

qr = QuantumRegister(2)
cr = ClassicalRegister(2)
circuit = QuantumCircuit(qr, cr, name='mycircuit')
circuit.x(qr[0])
circuit.x(qr[1])

Answer 4

may be you can define a function like below

from qiskit import QuantumCircuit
def qubitinitialize(n,n_init):
#n = number of qubits
#n_init = initial state of each qubit
    qc = QuantumCircuit(n)
    for i in n_init:
        if i == 1:
            qc.x(i)
return(qc)

This function will return a circuit with initial state as you provide in n_init

n = 4  #number of qubits 
n_init = [0,0,1,0] #a bit string of qubit initialization we need 
qc_n = qubitinitialize(n,n_init)
qc_n.draw('mpl')

just remember that 0,1,2,3 in the bit string will convert to q3,q2,q1,q0

Answer 5

Aer simulators will support what you're looking for very soon: https://github.com/Qiskit/qiskit-aer/pull/834