Understanding the effect of a quantum gate on an entangled system 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?
 A: If you know how to apply a linear operation to separable states and you want to extend it to an entangled state, then first you express your state in terms of separable states,
$$|\psi⟩ = \sum_n |a_n⟩\otimes|b_n⟩\otimes \cdots \otimes |c_n⟩,$$
and then you apply your operation term by term,
$$H|\psi⟩ = \sum_n |a_n⟩\otimes(H|b_n⟩)\otimes \cdots \otimes |c_n⟩,$$
which must hold by linearity. Since you know what the $H|b_n⟩$ are, you're done.
In general, an arbitrary entangled state can indeed be expressed as a column vector with a bunch of entries in a given tensor basis. In that case, if you want a matrix representation of a single-qubit operator, you need to use a Kronecker product of its matrix in the relevant sub-basis with a bunch of identity matrices on the left and on the right to make it address the correct part of the tensor product.
A: Reading a bit further on this stackexchange, I found this answer which showed me what I was looking for through its use of notation. Specifically, they write a Hadamard transformation on the first of two qubits as $H_A \otimes I_B |0_A0_B\rangle$, which is exactly what I needed to see!
By the same logic, in my example of applying $H$ to the second qubit of a 3-qubit system, the overall transformation applied to the state vector would be $I \otimes H \otimes I =$
$$
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\otimes
\frac{1}{\sqrt 2}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\otimes
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
In general, applying a unitary transformation, $U$, to the $n$th qubit in an $N$-qubit system would have the effect of applying the following transformation to the system: 
$$
\bigotimes_{i \in \mathbb Z_N}
\begin{cases}
U & i = n \\
I & \text{otherwise}
\end{cases}
$$
Essentially, a quantum gate can be thought of as the tensor product of all the operations being done to the individual qubits in the circuit - including the identity transformations on the channels not passing through the gate.
EDIT: Just to clear the misconception in the second half of the question: quantum gates do not measure their inputs. That whole thing about collapsing the amplitudes is wrong.
