Recently I've been learning about quantum mechanics through studying quantum computers. I understand how a unitary transformation can be used to effect the amplitudes of a single qubit; for example, the Hadamard gate will do this:
$$ H(\alpha|0\rangle + \beta|1\rangle) \rightarrow \frac{\alpha + \beta}{\sqrt 2}|0\rangle + \frac{\alpha - \beta}{\sqrt 2}|1\rangle $$
So that, when the qubit is measured, we will see it is in state $|1\rangle$ with a probability of $\left|\frac{\alpha - \beta}{\sqrt 2}\right|^2$. However, this seems a lot more complicated to me when you take into account that qubits' states can be entangled with one another. My question is, what exactly happens when you apply a transformation to a single qubit in a many qubit system?
This is easy when the state is separable. For example, if the state of a 3-qubit quantum register is this:
$$ (\alpha|0\rangle + \beta|1\rangle)\otimes(\gamma|0\rangle + \delta|1\rangle)\otimes(\epsilon|0\rangle + \eta|1\rangle) $$
then applying Hadamard to the second qubit will result in this:
$$ (\alpha|0\rangle + \beta|1\rangle)\otimes\left(\frac{\gamma + \delta}{\sqrt 2}|0\rangle + \frac{\gamma - \delta}{\sqrt 2}|1\rangle\right)\otimes(\epsilon|0\rangle + \eta|1\rangle) $$
This calculation doesn't work in general though, because you can't necessarily write down the global state as a tensor product of separate qubits to begin with. In a quantum register, all you really know at any given time is that you have a vector of amplitudes:
$$ \begin{bmatrix} \alpha_{000} & \alpha_{001} & \alpha_{010} & \alpha_{011} & \alpha_{100} & \alpha_{101} & \alpha_{110} & \alpha_{111} \end{bmatrix} $$
such that $\sum |\alpha_i|^2 = 1$. Mathematically, what is the result of applying a transformation to a single qubit? How would I calculate the final state with a pen and paper?
I have an idea (which I'm not sure about), that passing a qubit to a quantum gate means measuring it first, which would implicitly mean measuring any other qubits it's entangled with. For example, if I measured the second qubit and saw it was $|0\rangle$, then every amplitude contradicting that measurement would collapse to zero, and the new (renormalized) vector would look like this:
$$ \begin{bmatrix} \alpha^\prime_{000} & \alpha^\prime_{001} & 0 & 0 & \alpha^\prime_{100} & \alpha^\prime_{101} & 0 & 0 \end{bmatrix} $$
I don't know if I'm thinking about this wrongly, but I'm at least still missing something. I still can't construct the new vector like so:
$$ (\alpha|0\rangle + \beta|1\rangle)\otimes H(|0\rangle)\otimes(\epsilon|0\rangle + \eta|1\rangle) $$
because if the first and third qubits are entangled, then decomposing the state into a tensor product will still be impossible -- even after measuring the second qubit! So how does a quantum gate actually change the state of the system in a quantum circuit?