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I need some help in multi-qubit controlled -Z rotation. Below is the qiskit code of triple controlled z rotation

def cccZ():                                      
qc = QuantumCircuit(4)
qc.cp(pi/4, 0, 3)            

qc.cx(0, 1)
qc.cp(-pi/4, 1, 3)

qc.cx(0, 1)
qc.cp(pi/4, 1, 3)

qc.cx(1, 2)
qc.cp(-pi/4, 2, 3)

qc.cx(0, 2)
qc.cp(pi/4, 2, 3)

qc.cx(1, 2)
qc.cp(-pi/4, 2, 3)

qc.cx(0, 2)
qc.cp(pi/4, 2, 3)


gate = qc.to_gate(label=' cccZ')
return gate

Please help me in modifying this code to hexa qubit z rotation? It would be really great if someone can explain its theory also. I am really struggling with random nature of quantum computing with no fixed pattern at all.

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The mostly flexible way to create gates beyond the ones included as methods is with the circuit library:

from qiskit import QuantumCircuit
from qiskit.circuit.library import ZGate

circuit = QuantumCircuit(7)
c6z = ZGate().control(6)
circuit.append(c6z, range(7))
circuit.draw()
q_0: ─■─
      │ 
q_1: ─■─
      │ 
q_2: ─■─
      │ 
q_3: ─■─
      │ 
q_4: ─■─
      │ 
q_5: ─■─
      │ 
q_6: ─■─

The decomposition of this gate is deep:

circuit.decompose().depth()
315

How to construct multi-controlled gates, in general

Any single-qubit gate can be arbitrarily-n-controlled with the method Gate.control. The parameter n sets the amount of controlled qubits. As a consequence, the resulting controlled-gate will be a n-plus-one-qubits gate.

from qiskit.circuit.library import <some>Gate
cN_gate = <some>Gate(<gate_params>).control(<n>)
print(cN_gate.num_qubits)  # n+1

You can append this custom gate to a circuit using the QuantumCircuit.append method:

circuit.append(cN_gate, [....., i])
                         \_n_/

The first $n$ parameters are the qubits in which you want to control. The last one (i) is the target qubit.

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  • $\begingroup$ Thank you @Luciano. Please provide some methodology or trick to understand these quantum gates and circuit? $\endgroup$ Aug 19 at 6:20
  • $\begingroup$ I generalized it, have a look $\endgroup$
    – luciano
    Aug 19 at 8:00
  • $\begingroup$ Thank you @Luciano $\endgroup$ Aug 19 at 14:56

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