# Show how the Bell state arises from the circuit with Hadamard and CNOT, using matrix notation

I understand that starting with

,

we can get to $$\vert \Phi^+ \rangle$$. First, we start with $$\vert Q_1 \rangle \otimes \vert Q_2 \rangle = \vert 0 \rangle \otimes \vert 0 \rangle$$ and then applying $$H$$ on $$Q_1$$ which gives $$\left( \frac{1}{\sqrt{2}}\vert 0 \rangle + \frac{1}{\sqrt{2}}\vert 1 \rangle \right)\otimes \vert 0 \rangle$$. After $$t_1$$, we obtain $$\frac{1}{\sqrt{2}}\vert 00 \rangle + \frac{1}{\sqrt{2}}\vert 10 \rangle$$. Then, we apply the $$CNOT$$ gate to end up with $$\vert \Phi^+ \rangle = \frac{1}{\sqrt{2}}\vert 00 \rangle + \frac{1}{\sqrt{2}}\vert 11 \rangle.$$

I am wondering if there is a way to show all of this using the matrix representation of gates. I have tried the following way but I can't seem to get it properly:

We start with the state $$\vert 00 \rangle = \scriptstyle\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$$. Between $$t_0$$ and $$t_1$$, I applied $$H \otimes I = \begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0\\\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0\\ 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix},$$ since we are manipulating $$Q_1$$ and leaving $$Q_2$$ unchanged to obtain $$\scriptstyle\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\\0\end{bmatrix}$$. Lastly, we apply $$CNOT$$ which gives us $$\scriptstyle\begin{bmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\\0\end{bmatrix}.$$

I am really looking for the matrices required to achieve the desired result.

I think the problem of the calculations is in the $$H \otimes I$$, that should be equal to:

$$$$H \otimes I = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \cdot I & 1 \cdot I \\ 1 \cdot I & -1 \cdot I \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 &-1 & 0\\ 0 & 1 & 0 &-1 \end{pmatrix}$$$$

Then if we will apply this to $$|00\rangle$$ state we will obtain:

$$$$H \otimes I |00\rangle= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 &-1 & 0\\ 0 & 1 & 0 &-1 \end{pmatrix} \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\0\\1\\0 \end{pmatrix}$$$$

Now let's apply the CNOT gate:

$$$$CNOT \frac{1}{\sqrt{2}}\big(|00\rangle + |10\rangle\big)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1\\0\\1\\0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\0\\0\\1 \end{pmatrix} = \frac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle\big)$$$$

• My very silly mistake due to overworking. Thank you so much! – M. Al Jumaily Apr 12 '20 at 9:25