Second Qubit Not Flipped in Hadamard Gate I'm very new to QM and Quantum Computing and I have a likely simple question, It may simply stem from my lack of knowledge of vector calculus.
We have a 2-qubit quantum state:
$$
\mid\psi\rangle = \alpha\mid00\rangle + \beta\mid01\rangle+\gamma\mid10\rangle+\delta\mid11\rangle
$$
Passed through the Hadamard Gate which is the following unitary matrix:
$$
H = \frac{1}{\sqrt2}\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix}
$$
I fail to understand how the outcome of passing, say, $\mid00\rangle$ through this would yield:
$$\frac{\mid00\rangle + \mid10\rangle}{\sqrt2}
$$
I've attempted to recreate this, but I don't know how to multiply different dimensional matricies. If someone wouldn't mid multiplying these out so I can understand a bit more of what's going on, I'd be very appreciative.
 A: The actual state of $|00\rangle$ is represented by 1×4 matrix, the matrix of Hadamard transformation is 2×2. You can perform matrix multiplication if, and only if the number of columns in first matrix is equal to the number of rows in the second matrix. 
Thus it is impossible to apply the Hadamard transformation described by your matrix.
This is because the matrix you've shown is meant to represent transformation on single qubit, and you are trying to apply it to two qubit system.
What you are probably looking for is how to apply the transformation to get the state of $(|00\rangle+|10\rangle)/\sqrt{2}$. You can do this by applying the transformation to the first qubit, an then you tensor multiply it with the second qubit to get the full state.
$$|A\rangle=H|0\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$$
$$|B\rangle=|0\rangle$$
$$|A\rangle\otimes |B\rangle=(|0\rangle+|1\rangle)/\sqrt{2}\otimes|0\rangle$$
$$|A\rangle\otimes|B\rangle=(|00\rangle+|10\rangle)/\sqrt{2}$$
Hopefully it will help you.
