So suppose I have the following Quantum Circuit:
A ---- |Control| -----|Hadamard|----
B ---- |xxxxxxx|------------------------
Which is a 2 input Controlled Gate (applying some gate of two choices to Qubit B, depending on the value of Qubit A) followed by a single Hadamard Gate acting on Qubit A.
Initially the Qubits are in states $$a_0\left| 0 \right> + a_1\left| 1 \right> $$ $$b_0\left| 0 \right> + b_1\left| 1 \right> $$ Respectively. So the combined system is in state
$$ a_0 b_0 \left| 00 \right> +a_0 b_1 \left| 01 \right> + a_1 b_0 \left| 10 \right> + a_1 b_1 \left| 11 \right>$$
At the beginning.
After the application of the controlled Gate, the combined superposition can easily be in a state that CANNOT be factored into a tensor product of two states. Any superposition of the form
$$ q_0 \left| 00 \right> +q_1 \left| 01 \right> + q_2 \left| 10 \right> + q_3 \left| 11 \right>$$
Where $q_0/q_1 \ne q_2/q_3$ for example couldn't be factored into a tensor product.
But now when we apply that Hadamard gate, it is applied onto a single Qubit. What is it doing then? Given that the "state" of a single qubit cannot be independently factored and considered, how does the hadamard gate now affect the state of system?
How this is different than:
Help on applying a Hadamard gate and CNOT to two single q-bits
In the linked question, we are dealing with a factorable state, that then is given a CNOT transform. That computation is obvious since the factorable state (post Hadamard) can be expressed as:
$$a_0\left| 0 \right> + a_1\left| 1 \right> $$ $$b_0\left| 0 \right> + b_1\left| 1 \right> $$
yielding superposition state:
$$ a_0 b_0 \left| 00 \right> +a_0 b_1 \left| 01 \right> + a_1 b_0 \left| 10 \right> + a_1 b_1 \left| 11 \right>$$
Which can now be easily computed with the $4 \times 4$ CNOT operator.
In our question we go the other way. WE start off with teh application of a $4 \times 4$ controlled operator to generate an entangled superposition
$$ q_0 \left| 00 \right> +q_1 \left| 01 \right> + q_2 \left| 10 \right> + q_3 \left| 11 \right>$$
And now am attempting to determine how the behavior of a gate acting on a SINGLE Qubit affects the whole superposition.
The link is irrelelvant here since our system is no longer factorable as a tensor product of independent superpositions.
What I'm asking can be summarized succinctly as: How can I write a single Qubit operator $O$ (given as a $2 \times 2$ matrix) as a multiQubit operator $O'$ (given as a $2^k \times 2^k$ matrix) that acts as the identity on all inputs except the first where it acts as $O$ traditionally would.
To that end, the question offers no hint of how to go about it.
Work so far
My intuition suggests given the system:
$$ q_0 \left| 00 \right> +q_1 \left| 01 \right> + q_2 \left| 10 \right> + q_3 \left| 11 \right>$$
We can artificially believe that the first qubit is in the state
$$ (q_0 + q_1) \left| 0 \right> + (q_2 + q_3) \left| 1 \right> $$
And that the entire superposition is in:
$$ (q_0 + q_1) \frac{q_0}{q_0 + q_1}\left| 00 \right> +(q_0 + q_1) \frac{q_1}{q_0 + q_1} \left| 01 \right> + (q_2 + q_3) \frac{q_2}{q_2 + q_3} \left| 10 \right> + (q_2 + q_3) \frac{q_3}{q_2 + q_3} \left| 11 \right>$$
So when we apply the Hadamard to the Qubit we send:
$$ (q_0 + q_1) \left| 0 \right> + (q_2 + q_3) \left| 1 \right> $$
To
$$ \frac{q_0 + q_1+q_2 + q_3}{\sqrt{2}} \left| 0 \right> + \frac{q_0 + q_1-q_2 - q_3}{\sqrt{2}} \left| 1 \right> $$
And thus the entire system now is in:
$$ \frac{q_0 + q_1+q_2 + q_3}{\sqrt{2}} \frac{q_0}{q_0 + q_1}\left| 00 \right> +\frac{q_0 + q_1+q_2 + q_3}{\sqrt{2}} \frac{q_1}{q_0 + q_1} \left| 01 \right> + \frac{q_0 + q_1-q_2 - q_3}{\sqrt{2}} \frac{q_2}{q_2 + q_3} \left| 10 \right> + \frac{q_0 + q_1-q_2 - q_3}{\sqrt{2}} \frac{q_3}{q_2 + q_3} \left| 11 \right>$$
But I'm not sure why I would rigorously believe this.
(Renormalize where necessary)
Answer
Given that the "state" of a single qubit cannot be independently factored and considered, how does the hadamard gate now affect the state of system?
To apply a single-qubit operation $M$ to an n-qubit system you hit the system with $I \otimes M \otimes I \otimes ... \otimes I$. The position of $M$ within that tensor product determines which qubit you hit.
You can also use simple equivalent rules, like these:
Group the states by the uninvolved qubits.
$|\psi\rangle = a|00⟩ + b|01⟩ + c|10⟩ + d|11⟩$
$= \Big(a|0⟩ + c|1⟩\Big)|0⟩ + \Big(b|0⟩ + d|1⟩\Big)|1⟩$
Apply the operation within each group.
$H_0 |\psi\rangle = \Big(H(a|0⟩ + c|1⟩)\Big)|0⟩ + \Big(H(b|0⟩ + d|1⟩)\Big)|1⟩$
$= \Big(\frac{a+c}{\sqrt 2}|0⟩ + \frac{a-c}{\sqrt 2}|1⟩\Big)|0⟩ + \Big(\frac{b+d}{\sqrt 2}|0⟩ + \frac{b-d}{\sqrt 2}|1⟩\Big)|1⟩$
Ungroup
$=\frac{a+c}{\sqrt 2}|00⟩ + \frac{b+d}{\sqrt 2}|01⟩ + \frac{a-c}{\sqrt 2}|10⟩ + \frac{b-d}{\sqrt 2}|11⟩$
Sometimes it makes sense to just keep the grouping around. For example, when the system is spread across two parties you might as well keep it laid out in a grid where columns are the Alice groups and rows are the Bob groups. Gives you a kind of "matrix of amplitudes", which is convenient because Alice's operations correspond to left-multiplying that matrix and Bob's operations correspond to right-transpose-multiplying it.
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