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Suppose I have 01 and 11 in classical register 1 and 2 respectively in IBM quantum experience circuit. I want 01 + 11 = 00 mod 4. Can it be done?

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You can add two qubits firstly in quantum registers and then measure qubit with results and put it to classical register.

Adding two qubits modulo 4 can be do with this circuit:

Circuit

First $CNOT$ is fan-out and make a "copy" of qubit $q_0$ to $q_2$, second $CNOT$ realizes XOR function between $q_1$ and $q_2$. Eventually you have $(q_0 + q_1) \mod 4$ in $q_2$


EDIT: Expanded for two qubits adder modulo 4, based on comment by Adam Levine:

Here is a code implementing a modulo 4 adder for two qubits:

OPENQASM 2.0;
include "qelib1.inc";

qreg q[6];
creg c[6];

//first input q[0]q[1]
//second input q[2]q[3]
//output q[4]q[5]

//input
//q[0]q[1] = 01
id q[0];
x q[1];
//q[2]q[3] = 11
x q[2];
x q[3];

//first qubit:
//q[1] + q[3] - sum
cx q[1],q[5];
cx q[3],q[5];
//q[1], q[3] - carry
ccx q[1], q[3], q[4];

//second qubit:
//q[0] + q[2] - sum
cx q[0],q[4];
cx q[2],q[4];
//no carry, modulo 4 sum

$I$ and $X$ gates set an input (based on the comment), $CNOT$ gates acting on qubit $q_5$ return sum of last (lower) qubits of input numbers. Toffoli gate is used for calculation of carry form last qubits and next two $CNOT$ gates add together the carry and first (higher) qubits of input numbers. Carry from this sum is not performed as the adder is modulo 4. The result is in qubits $q_4$ and $q_5$ where higher qubit is $q_4$ and lower $q_5$.

Note: this is based on classical approach how to construct an adder. There is also quantum approach based on quantum Fourier transform. See details here: Addition on a Quantum Computer

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  • $\begingroup$ Can I have a circuit for adding 2 qubits: 01 + 11 = 00 mod 4? $\endgroup$ – Adam Levine Feb 24 at 6:32

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