Representing Quantum Gates in Tensor Product Space

Question

I want to write the matrix form of a single or two qubit gate in the tensor product vector space of a many qubit system. Ill outline a simple example:

Both qubits, $q_0$ and $q_1$ start in the ground state, $|0 \rangle =\begin{pmatrix}1 \\ 0 \end{pmatrix}$. Then we apply the Hadamard gate, $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ on the $q_0$.

Here is my understanding:

The Hadamard gate on two qubit system only operates on $q_0$

$$ \hat{H}_0(q_0 \otimes q_1) = \hat{H}_0q_0 \otimes q_1 $$

$$\hat{H}_0(|0 \rangle \otimes |0 \rangle) = \hat{H}_0|0\rangle \otimes |0\rangle$$

$$\hat{H}_0 (\begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} )= \hat{H}_0 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

$$\hat{H}_0 \begin{pmatrix} 1 \\ 0 \\ 0 \\0 \end{pmatrix} = \frac{1}{\sqrt{2}} ( \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix})$$

$$\hat{H}_0 \begin{pmatrix} 1 \\ 0 \\ 0 \\0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$

I don't know exactly how to solve this but can give a guess.

$$\hat{H}_0 = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

How is $\hat{H}_0$ written in the tensor product space? Are there any resources that explain this well? Any help is appreciated!!

Answer 1

If you want the Hadamard gate to act on only one qubit, say only the first qubit, then the composite operator acting on the two-qubit state is given by $\hat{H} \otimes \mathbf{I}$ so that

$$(\hat{H} \otimes \mathbf{I})(|q_{0}\rangle \otimes |q_{1}\rangle=\hat{H}|q_{0}\rangle \otimes \mathbf{I}|q_{1}\rangle= \hat{H}|q_{0}\rangle \otimes |q_{1}\rangle$$

Here $\mathbf{I}$ is the unit operator or identity operator.

If you want the Hadamard gate to act on both the qubits, then the composite operator acting on both the qubits is given by $\hat{H} \otimes \hat{H}$ so that

$$(\hat{H} \otimes \hat{H})(|q_{0}\rangle \otimes |q_{1}\rangle=\hat{H}|q_{0}\rangle \otimes \hat{H}|q_{1}\rangle$$

The operators $\hat{H} \otimes \mathbf{I}$ and $\hat{H} \otimes \hat{H}$ can be expressed in the matrix form by using the rules of tensor product. See https://en.wikipedia.org/wiki/Tensor_product

Answer 2

First, let me fix a couple of misprints in your Hadamard gate definition, namely the absence of the factor $1/\sqrt{2}$ and the minus sign wrong position (take a look at this article): $$ H_0 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$ $H_0$ without $1/\sqrt{2}$ won't be a unitary matrix (a matrix that conserves a wavefunction norm). And the minus sign before the right bottom element guarantees not only antisymmetry of $H_0 |0\rangle$state, but also the Hermitian property $H_0^{\dagger} = H_0$. Thus, the conventional Hadamard gate used twice doesn't modify a given state: $H_0^2 = I$, where $I$ is a $2\times2$ identity matrix.

Now, for construction of the desired two-qubit gate, you need the same tensor product operation as you used for the vectors (see this):

$$ H_1 \equiv H_0\otimes I = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & 1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\\ 1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & -1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix} $$

where $H_1$ is a one-qubit Hadamard gate in the two-qubit space. The sense of the formula above is simple: applying $H_1$ you mix up the first qubit states and keep the second qubit state unchanged.

Indeed:

$$ H_1 \left(|0\rangle\otimes|0\rangle\right) = H_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle + |1\rangle\otimes|0\rangle\right) $$

If you wish to swap the gate action, i.e. change only the second qubit, you can write it in the same manner:

$$ H_2 \equiv I \otimes H_0 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} & 0 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\\ 0 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} & 1 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{pmatrix} $$

and

$$ H_2 \left(|0\rangle\otimes|0\rangle\right) = H_2 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle + |0\rangle\otimes|1\rangle\right) $$

Answer 3

To give you a general answer, everytime you have "no gate" on a qubit, this means that there is a identity operator described by unit matrix 2x2 (denote $I$). For example, you have $n$ qubits and want to apply a gate $U$ on $i$ th qubit. Then matrix form of this operation is $$ I_0 \otimes I_1 \otimes \dots\otimes I_{i-1} \otimes U_i \otimes I_{i+1} \dots \otimes I_{n-2} \otimes I_{n-1}. $$

Note that subscripts are indices of qubits going from 0 to $n-1$.