# Help on applying a Hadamard gate and CNOT to two single q-bits

I am stuck on a few issues in this video. (Note: It is at the frame concerning this question.)

In it, from what I understand (which could be wrong) we first apply the Hadamard gate to a qbit in the $\lvert 0 \rangle$ state. Then on that we use a controlled not on the output value, together with another input from another qbit set at $\lvert 0 \rangle$. I don't understand how to work this out and how the $\lvert 00 \rangle + \lvert 10 \rangle$ is derived for the output of combination of the two gates.

Here is what I am thinking (neglecting the $1/\sqrt{2}$ multiplier):

$$\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \lvert 0 \rangle + \lvert 1 \rangle.$$

There however is another $\lvert 0 \rangle$ coming in from the second (bottom) wire and it interacts with the output from the Hadamard gate. I don't understand how to manipulate the values - have no idea if I can do something such as the following by distribution: $$(\lvert 0 \rangle + \lvert 1 \rangle) \lvert 0 \rangle = \lvert 0 \rangle \lvert 0 \rangle + \lvert 1 \rangle \lvert 0\rangle.$$

I have no idea why we can do this (if we can even do so), but this is the only way I can see that we could get $\lvert 0 \rangle \lvert 0 \rangle + \lvert 1\rangle \lvert 0 \rangle = \lvert 00 \rangle + \lvert 10 \rangle$, as the author of the video has on the particular time in the video.

Any clarification would be extremely appreciated.

Some notation: $|01 \rangle$ is the same as $|0\rangle |1 \rangle$. The first state is for the first qubit, the second state is for the second qubit.
Thus, the most general 2-qubit state would look something like $( a|0\rangle + b|1\rangle)(c|0\rangle + d|1\rangle)$, taking normalization into account of course.
Some basic linear algebra would then tell you that you can indeed use the distributive property, so $$(|0\rangle + |1\rangle) |0\rangle = |00\rangle + |10\rangle$$
If you want to go the long way to convince yourself that you can do these things, write down the Hadamard gate in the 2-qubit basis: You have your four basis states, $|00\rangle, |01\rangle, |10\rangle$ and $|11\rangle$ and you know how the Hadamard gate acts on the first qubit and that it does not change the second qubit, so you could write down the corresponding 4x4 matrix.
• Might be worth pointing out for the OP's sake that the implicit product here is the tensor product, $\otimes$, which does indeed distribute over addition. It might be "basic" for multiparticle QM, but I feel most pure linear algebra 101 courses for non-mathematicians won't cover it. – user10851 Jul 30 '13 at 1:54