Let's say we are given two probability values $p_1 > 0$ and $p_2 > 0$, with $p_1+p_2 \leq 1$ but not necessarily equals to $1$. We are asked to create a 2-qubit quantum circuit with state $|01\rangle$ probability as $p_1$ and state $|10\rangle$ probability as $p_2$. How to approach this type of problems? I think entanglement would help here but not sure how to come up with exact probabilities.

Also, is there a way to generalize this (e.g., build a n-qubit quantum circuit from given $k$ probabilities exactly for $k$ out of its $2^n$ superposition states)?

  • $\begingroup$ Do you mean your qubits are initialized in the all $|0\rangle$ state? $\endgroup$
    – Zhibo Yang
    May 6 at 22:29
  • $\begingroup$ Yes I assumed that pre-condition $\endgroup$ May 7 at 17:59

2 Answers 2


Qiskit has a function which initializes a circuit to a certain given state:


The paper that they use to implement the code apparently is:


A small example of code to implement the function is:

from qiskit.circuit import QuantumCircuit
from qiskit.circuit import QuantumRegister

nbqbits = 2 # Number of qubits

qr = QuantumRegister(nbqbits)
qc = QuantumCircuit(qr)

##In psi_0 you define the amplitudes of the state. These amplitudes are the square root
## of the probabilites you want to get for each state.

## Each state of the computational basis binary(i) get the amplitude psi_0[i], i.e:
## The state |00> get the amplitude psi_0[0]
## The state |01> get the amplitude psi_0[1]
## and so on.

psi_0 = [1, 2, 3, 4] # This state is not yet normalized

psi_0 = psi_0/np.linalg.norm(psi_0) # Normalization of the state

qc.initialize(psi_0) # Initialize function. Make sure that the vector you enter has 2**nbqbits entries.


You should get something like:

initialize state

If you decompose the circuit, you can get more information about what gates are inside:

d_c = qc.decompose().decompose().decompose().decompose()


and you would get:

decomposed initialize

  • $\begingroup$ Thanks, so in order to obtain the same result, we could have the initialization as psi_0 = [np.sqrt(0.1), np.sqrt(0.3), np.sqrt(0.6), 0] $\endgroup$ May 7 at 17:58

With qiskit I could solve the first part for a 2-qubit quantum circuit:

from math import asin, sqrt
from qiskit import QuantumCircuit, Aer, execute
from qiskit.visualization import plot_histogram

p1, p2 = 0.3, 0.6 # given probabilities

def prob_to_angle(prob):
   return 2*asin(sqrt(prob))

backend = Aer.get_backend('statevector_simulator')
qc = QuantumCircuit(2)

# Apply prior to qubit 0
qc.ry(prob_to_angle(p1), 0)

# Apply prior to qubit 1
qc.cry(prob_to_angle(p2/(1-p1)), 0, 1)

# execute the qc
results = execute(qc,backend).result().get_counts()

# plot the results

enter image description here


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