# What happens when I input two qubits starting at state $|00\rangle$ into a hadamard gate?

I was watching the following video https://www.youtube.com/watch?v=IrbJYsep45E
And around 3 minutes in, they do an example computation with two qubits. The qubits started off in the state $$|00\rangle$$, and then go through what I assume is a Hadamard gate which puts them in a superposition. Why is it a $$50\%$$ chance of being $$|01\rangle$$, and a fifty percent chance of being $$|10\rangle$$? Why is it not a quarter chance of being $$|00\rangle$$, $$|10\rangle$$, $$|01\rangle$$, and $$|11\rangle$$?

• It's fair to wonder why they describe quantum gates that way, but they don't actually suggest that it's a Hadamard gate that they're applying, or even any gate in particular; they just say "a quantum gate". They then follow it up with another seemingly arbitrary and anonymous "quantum gate". In fairness, that's better than many treatments I've seen in that they're using real math with actual irrational numbers, but they aren't doing an especially good job at explaining themselves at that stage. – Niel de Beaudrap Oct 19 '18 at 23:34

It is not two Hadamard gates.
You are 100% correct that two Hadamrd gates would put you in an equal superposition of 00, 01, 10, 11, meaning 25% chance of getting any of those.

Let's construct her gate based on the input $$|00\rangle$$ and the output $$\frac{1}{\sqrt{2}}\left(|10\rangle +|01\rangle\right)$$.

The final state is $$|\Psi^+\rangle$$ from this question.

So the state can be constructed with this circuit:

where the middle gate is a CNOT.

• You're right this isn't unitary, let me fix it. The reason I considered a symmetric matrix was because I was thinking of Hamiltonians instead of Unitaries. – user1271772 Oct 19 '18 at 23:38
• @NieldeBeaudrap, I fixed it. – user1271772 Oct 19 '18 at 23:49
• do you know what gates she used for the second computation @user121772 – John Jepson Oct 20 '18 at 16:32
• @JohnJepson: Yes I do, would you like to ask that as a question? I would be happy to answer it. – user1271772 Oct 21 '18 at 1:21

There is a slight misconception in your question. You can't apply a Hadamard gate on $$|00\rangle$$. Because the basis state $$|00\rangle$$ is actually a tensor product of a $$2\times 1$$ dimensional basis state $$|0\rangle$$ used twice. This results in $$|00\rangle = |0\rangle \otimes |0\rangle = \begin{bmatrix}1 \\0\\0\\0\end{bmatrix}$$. Now you must take the tensor product of Hadamard gate(which is $$2\times 2$$ dimensional) too in order to apply it in this state. Like so: $$H \otimes H (|0\rangle \otimes |0\rangle)$$ $$= H|0\rangle \otimes H|0\rangle$$ $$=\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ $$=\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$$. Now obviously $$4 \times (\frac{1}{2})^2 = 1$$, as you have correctly inferred.

• This doesn't answer the question though. – user1271772 Oct 21 '18 at 17:24