Is the mathematical expression between the CNOT gates arranged in this way a tensor product or a concatenated product?

Question

Is the mathematical expression between the CNOT gates arranged in this way a tensor product or a concatenated product? That is \begin{matrix} \text{CNO}{{\text{T}}^{\otimes n}} & or & \prod\limits_{n}{\text{CNOT}} \\ \end{matrix}?

Answer 1

$CNOT^{\otimes n}$ is a matrix of size $4^n$ and corresponds to the following circuit :

$\prod_{i=1}^{n}{CNOT}$ is a matrix of size $2$ and corresponds to the following circuit :

The circuit you describe is more something like :

$\prod_{i=1}^{n}{I_{2^{i-1}}\otimes CNOT \otimes I_{2^{n-i}}}$

where $I_n$ is the identity matrix of size $n$

edit : explaination

You apply successively $CNOT$ gates on different qubits. In the circuit you present, when you apply a $CNOT$ gate on 2 qubits, you do not perform any transformation on other qubits, this is equivalent to perform an identity transformation (which does nothing).

The first $CNOT$ gate is applied on the two first qubits and an identity transformation is applied on all other qubits (n-1 qubits). The full transformation is then :

$CNOT \otimes I^{\otimes n-1} = CNOT \otimes I_{2^{n-1}}$

The second $CNOT$ gate is applied on the 2nd and 3rd qubits and an identity transformation is applied on the first qubit and all the qubits after the third. The full transformation is then :

$I \otimes CNOT \otimes I^{\otimes n-2} = I \otimes CNOT \otimes I_{2^{n-2}}$

And so on, if you apply a $CNOT$ gate on the qubits $i$ and $i+1$, you will apply an identity on all the qubits before the $i^{th}$ and all the qubits after the $i+1^{th}$, the transformation will be :

$I^{\otimes i-1} \otimes CNOT \otimes I^{\otimes n-i} = I_{2^{i-1}}\otimes CNOT \otimes I_{2^{n-i}}$

The overall circuit is the succession of all this transformations for i going from 1 to n and is therefore described by :

$\prod_{i=1}^{n}{I^{\otimes i-1}\otimes CNOT \otimes I^{\otimes n-i}} = \prod_{i=1}^{n}{I_{2^{i-1}}\otimes CNOT \otimes I_{2^{n-i}}}$