# Representing qubit swap using linear algebra

I want to write matrix representation of qubit swap algorithm, but I seem to be stuck. Here is the circuit I am trying to calculate using linear algebra:

Initially $$q_0 = |0\rangle$$ or $$\begin{pmatrix}1 & 0\end{pmatrix}^T$$ while $$q_1 = |1\rangle$$ or $$\begin{pmatrix}0 & 1\end{pmatrix}^T$$ and at the end of measurement I should be getting opposite outcomes. I know that to apply first $$CNOT$$ gate I should first do Kronecker product on the two vectors and than multiply it by $$CNOT$$ matrix, but I cant figure out what needs to be done next, particularly how to apply second $$CNOT$$ to my quantum state (control bit $$q_1$$, target bit $$q_0$$).

Any help would be appreciated.

Here, what you need to do is to understand writing CNOT gate based on the control qubit.

1. Your first CNOT gate has qubit 1 as control and qubit 2 as target. So, what this means is the second qubit will not be flipped until qubit 1 is set to zero. I am going to use computational basis for this $$CNOT_1\left|00\right> = \left|00\right>$$, $$CNOT_1\left|01\right> = \left|01\right>$$, $$CNOT_1\left|10\right> = \left|11\right>$$,$$CNOT_1\left|11\right> = \left|10\right>$$.

$$CNOT_1 = \left[\begin{matrix}1 & 0 &0 &0\\ 0 & 1&0&0\\0&0&0&1\\0&0&1&0\end{matrix}\right]$$.

1. Your second CNOT gate has qubit 1 as target and qubit 2 as control. Thus, until qubit 2 is set to 1, qubit 1 will not flipped. Again, using the computational basis $$CNOT_2\left|00\right> = \left|00\right>$$, $$CNOT_2\left|01\right> = \left|11\right>$$, $$CNOT_2\left|10\right> = \left|10\right>$$,$$CNOT_2\left|11\right> = \left|01\right>$$.

$$CNOT_2 = \left[\begin{matrix}1 & 0 &0 &0\\0&0&0&1\\0&0&1&0\\ 0 & 1&0&0\end{matrix}\right]$$.

1. Your third CNOT gate is $$CNOT_1$$.

What you can do to get the SWAP gate is apply these CNOT gates in the order they were applied.

The Swap gate is represented by the matrix $$\begin{bmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$

-- Example from book Quantum Computing: An Applied Approach by Jack Hidary

• I am aware of that, I just want to see how I can derive correct results by using CNOT operations as shown in the circuit. – Ach113 Dec 24 '19 at 17:47

I would add that there is also a Hadamard gate applied on a second qubit, so you will get states $$|00\rangle$$ and $$|10\rangle$$, both with a probability 50 %.

Matrix representatation of your circuit is

$$$$CNOT_1CNOT_2CNOT_1(I\otimes H)$$$$

To implement only swap gate, you should remove the Hadamard gate.

Note that I used notation $$CNOT_1$$ and $$CNOT_2$$ introduced by Purva Thakre.

• True, I forgot to remove H gate from the circuit. It did not really play any role in this problem of mine – Ach113 Dec 25 '19 at 7:27