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I have 4 states say $|00\rangle, |01\rangle,|10\rangle, |11\rangle$. I want to add the states in a manner such that

  1. $|a\rangle=|00\rangle\otimes|01\rangle\to |00\rangle\otimes|01\rangle$
    and
  2. $|b\rangle=|10\rangle\otimes|11\rangle\to |10\rangle\otimes|01\rangle$.

That is the sum operator where the result is saved in the second register, i.e $|x,y\rangle\to|x, \mathrm{sum}(x,y)\rangle$.

But there is a rule that:
if $i=1$ then compute case $a$
if $i=2$ then compute case $b$.

How do I implement this conditional statement in a quantum circuit?

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  • $\begingroup$ Can you clarify what you mean by "case $a$" and "case $b$"? Are the only possible inputs $|a\rangle$ and $|b\rangle$? Or do you mean something else? $\endgroup$ – NNN May 18 '19 at 18:02
  • $\begingroup$ alex since i have made some progress regarding this question, will it be okay if I modify the question so that it is more clear $\endgroup$ – Upstart May 18 '19 at 18:04
  • $\begingroup$ That’s fine, yes. $\endgroup$ – NNN May 18 '19 at 22:07
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I guess the easiest way to implement a conditional operator is to add a quantum register ($cond_0$) that will contain the conditional parameter and then have a succession of control-gates, controlled by $cond_0$.

In your case lets's call A, $Op A$ and B, $Op B$. $op_0$ and $op_1$ will be the registers involved in $Op A$ or $Op B$. Obviously the same reasoning works with operations on more than 2 qubits.

The conditional quantum circuit is then :

from qiskit import QuantumCircuit, QuantumRegister
import numpy as np
import math
from scipy.linalg import expm


A = np.random.random((4,4))
A = expm(A*A.T*1.j*math.pi/2) #"whatever unitary defines operation A"
B = np.random.random((4,4))
B = expm(B*B.T*1.j*math.pi/2) #"whatever unitary defines operation B"

#register acting as the conditional statement
qr1 = QuantumRegister(1, name='cond')

#registers involved in the operation
qr2 = QuantumRegister(2, name='op')

#Circuit
qc = QuantumCircuit(qr1, qr2)
gate=ex.UnitaryGate(A, label="Op A").control(1)
qc.append(gate, [0,1,2])
qc.x(0)
gate=ex.UnitaryGate(B, label="Op B").control(1)
qc.append(gate, [0,1,2])
qc.x(0)


print(qc.draw('text'))

The resulting circuit is:
enter image description here

As you can see if $cond_0$=$|0\rangle$ then $Op A$ is not applied but $cond_0$ is bit-flipped so $Op A$ is applied.
In the same way if $cond_0$=$|1\rangle$ then $Op A$ is applied but $cond_0$ is bit-flipped so $Op A$ is not applied.

The last bit-flip on $cond_0$ is done to restore the value of $cond_0$. (Potentially unnecessary)

Hope that helps.

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