What's the matrix representation of the CSWAP?

Question

I don't know how to represent the matrix format of the CSWAP gate in the circuit:

Despite reviewing some material about CSWAP from Qiskit CSWAP, I am still unable to understand the concept. I am seeking to obtain the 16x16 matrix format in order to calculate the quantum state vector after the CSWAP.

In addition, it can be noted that q2 and q3 are fully entangled (I am unsure if this information is relevant).

Any help would be greatly appreciated, thank you!

Answer 1

To get the $CSWAP$ matrix representation in Qiskit, you can use the Operator class defined in the qiskit.quantum_info module:

from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator
from qiskit.visualization import array_to_latex

qc = QuantumCircuit(4)
qc.cswap(0, 1, 3)

array_to_latex(Operator(qc), max_size=16)

If you want to do all the actual maths, you should start from the following: $$ CSWAP_{0 \rightarrow 1, 3} = I \otimes I \otimes I \otimes |0 \rangle \langle 0| + SWAP_{1, 3} \otimes |1 \rangle \langle 1| $$ The formula above basically means: "if qubit $q_0$ (right-most) is in state $|0\rangle$, don't do anything; if it is in state $|1\rangle$, swap qubit $q_1$ and $q_3$". The $SWAP_{i,j}$ gate can then be decomposed in 3 controlled-not operations as $CX_{i,j} \cdot CX_{j,i} \cdot CX_{i,j}$. So, in this specific case, you have to compute: $$ CX_{1, 3} = I \otimes I \otimes |0 \rangle \langle 0| + X \otimes I \otimes |1 \rangle \langle 1| $$ $$ CX_{3, 1} = |0 \rangle \langle 0| \otimes I \otimes I + |1 \rangle \langle 1| \otimes I \otimes X $$

Finally, if you put all together and perform the calculations, you will get the same 16x16 unitary matrix returned by Qiskit.

Answer 2

2-qubit gates can be simulated effectively, much faster than potentially large matrix multiplies. They can be applied basically with linear complexity over the size of the full state (Python example here).

A Swap(x, y) gate can be implemented as:

  Cnot(x, y)
  Cnot(y, x)
  Cnot(x, y)

Correspondingly, a controlled Swap gate would be:

  CCnot(x, y)
  CCnot(y, x)
  CCnot(x, y)

So the final question is - how to implement this CCnot gate with two qubit gates? The answer is - with the so-called Sleater-Weinfurter construction: