Conditioned gates with multiple classical bits

Question

Consider a circuit where you have 6 qubits and apply some gates. Then, you measure qubits 1 , 2 , 3 with results $c_1 , c_2 , c_3$. Next you apply $X^{c_1}$ on the 4-th qubit, $X^{c_1 + c_2}$ on the 5-th qubit and $X^{c_1 + c_2 + c_3}$ on the 6-th qubit. How can I write this circuit in Qiskit? I can't find a way to sum the results of the classical register.

Answer 1

You can use bit_xor function from the newly added classical expressions module

from qiskit.circuit import QuantumCircuit, QuantumRegister, ClassicalRegister
from qiskit.circuit.classical import expr

qr = QuantumRegister(6, 'q')
c1 = ClassicalRegister(1, 'c1')
c2 = ClassicalRegister(1, 'c2')
c3 = ClassicalRegister(1, 'c3')
out = ClassicalRegister(3, 'o')
circ = QuantumCircuit(qr, c1, c2, c3, out)

# Sample operations for demonstration purposes
circ.h([0, 1, 2])

circ.measure(qr[0], c1[0])
circ.measure(qr[1], c2[0])
circ.measure(qr[2], c3[0])

with circ.if_test((c1, 1)):
    circ.x(qr[3])

with circ.if_test(expr.equal(expr.bit_xor(c1, c2), 1)):
    circ.x(qr[4])

with circ.if_test(expr.equal(expr.bit_xor(expr.bit_xor(c1, c2), c3), 1)):
    circ.x(qr[5])

circ.barrier()
circ.measure(qr[3:6], out)

circ.draw('mpl', style='textbook')

The circuit should look like:

Answer 2

This can be done quite easily. For instance, take the case where you have two classical bits $c_1, c_2$ and would like to implement $X^{c_1+c_2}$ on a qubit $q$. Clearly, applying $X^{c_1}$ followed by $X^{c_2}$ on $q$ applies $X$ on $q$ only if exactly one of $c_1$ and $c_2$ equals $1$. If both $c_1$ and $c_2$ are $0$ then $X$ is never applied, and if both bits are $1$, then $X$ is applied twice on $q$, which is $X^2 = I$.

Similarly, if you have three classical bits and would like to apply $X^{c_1+c_2+c_3}$ on $q$, then applying $X^{c_1}, X^{c_2}$ and $X^{c_3}$ serially, gives you the desired result of $X^{c_1+c_2+c_3}$ on $q$. As a check, see that if exactly one of $c_1, c_2, c_3$ equals $1$, then $X$ is implemented, and if exactly two of the bits are $1$, then $X$ is applied twice on $q$, which is $X^2=I$. Finally, if all three bits are $1$, then $X$ is applied thrice on $q$, which is $X^3 = X$ as required.

Note that this will not just hold for $X$ gate but for any unitary gate $U$.