IBM quantum circuit - order of tensor product for equivalent matrix

Question

I'm trying to understand how to apply tensor products on 3-qubit systems (or well at least 2 qubits). Let's take a basic example:

where $$\lvert \psi \rangle = \lvert q_2q_1q_0\rangle $$ with $q_2$ being the most significant bit and $q_0$ the least significant (matching the above schematic for the circuit).

H matrix (for a single qubit): $$H = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$

I understand that I need to use the identity matrix and tensor product in order to format the Hadamard matrix (and any other in the circuit) to be applied to 3 qubits. What I do not understand is how, generally, I determine the order of the tensor product? In order words, with the above description, should I do:

$$M=(H\otimes I)\otimes I = \frac{1}{\sqrt{2}} \begin{bmatrix} I & 0 & I & 0\\ 0 & I & 0 & I\\ I & 0 & -I & 0\\ 0 & I & 0 & -I\\ \end{bmatrix}$$

or

$$M=(I\otimes I)\otimes H = \begin{bmatrix} H & 0 & 0 & 0\\ 0 & H & 0 & 0\\ 0 & 0 & H & 0\\ 0 & 0 & 0 & H\\ \end{bmatrix}$$

More generally, how should I think about why the correct one is the right one? What is the logic I need to understand with regards to the "extension" of my gate's matrices and in which order I need to apply the Identity matrices in the tensor product?

Answer 1

Borrowing from Lenny's Paperback, you can value a product state as follows: \begin{eqnarray} |\psi_1\rangle &=& a|0\rangle + b|1\rangle\\ |\psi_1\rangle &=& c|0\rangle + d|1\rangle\\ |\psi_3\rangle &=& |\psi_1\rangle \otimes \,|\psi_2\rangle \\ |\psi_3\rangle &=& ( a|0\rangle + b|1\rangle) \otimes ( c|0\rangle + d|1\rangle) \\ |\psi_3\rangle &=& ac |0\rangle \otimes |0\rangle + ad |0\rangle \otimes |1\rangle + bc |1\rangle \otimes |0\rangle + bd|1\rangle \otimes |1\rangle \end{eqnarray}

Using this logic into the operators:

\begin{eqnarray} O &=& H \otimes I \otimes I \\ |\psi\rangle &=& |q_1\rangle \otimes |q_2\rangle \otimes|q_3\rangle \\ O|\psi\rangle &=& \Big (H \otimes I \otimes I \Big) |q_1\rangle \otimes |q_2\rangle \otimes|q_3\rangle \\ O|\psi\rangle &=& (H|q_1\rangle)\otimes ( I|q_2\rangle ) \otimes ( I|q_3\rangle ) \end{eqnarray}

And if you want the other operator: \begin{eqnarray} O' &=& I \otimes I \otimes H \\ |\psi\rangle &=& |q_1\rangle \otimes |q_2\rangle \otimes|q_3\rangle \\ O|\psi\rangle &=& \Big (I \otimes I \otimes H \Big) |q_1\rangle \otimes |q_2\rangle \otimes|q_3\rangle \\ O|\psi\rangle &=& (I |q_1\rangle)\otimes ( I|q_2\rangle ) \otimes ( H|q_3\rangle ) \end{eqnarray}

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