What do operations on single Qubits of Unfactorable Superpositions Do?

Question

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 1

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.

This has a very direct answer:

$$O' = O\otimes \underbrace{I_2 \otimes \cdots \otimes I_2}_{k-1\text{ factors}}$$

It's easy enough to see how you might come up with this for the two-qubit case. Let's use the basis $(\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle)$, which is handy because these basis states are not entangled; they can be factored into a state for qubit 1 and a state for qubit 2. For example: $$\lvert 00\rangle = \lvert 0\rangle_1 \otimes \lvert 0\rangle_2$$ or alternatively $$\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ 0\end{pmatrix}_1 \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}_2$$ The operator $O'$ should act like $$ O'\lvert 00\rangle = O'\bigl(\lvert 0\rangle_1\otimes\lvert 0\rangle_2\bigr) = \bigl(O\lvert 0\rangle_1\bigr)\otimes\bigl(\lvert 0\rangle_2\bigr)$$ or $$\begin{align} O'\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} &= \left[O\begin{pmatrix}1 \\ 0\end{pmatrix}_1\right] \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}_2 \\ &= \left[\begin{pmatrix}O_{11} & O_{12} \\ O_{21} & O_{22}\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}_1\right] \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}_2 \\ &= \begin{pmatrix}O_{11} \\ O_{21}\end{pmatrix}\otimes\begin{pmatrix}1 \\ 0\end{pmatrix}_2 \\ &= \begin{pmatrix}O_{11} \\ 0 \\ O_{21} \\ 0\end{pmatrix} \end{align}$$ Going through a similar procedure for each of the other three basis states, you find $$\begin{align} O'\begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix} &= \begin{pmatrix}0 \\ O_{11} \\ 0 \\ O_{21}\end{pmatrix} & O'\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix} &= \begin{pmatrix}O_{12} \\ 0 \\ O_{22} \\ 0\end{pmatrix} & O'\begin{pmatrix}0 \\ 0 \\ 0 \\ 1\end{pmatrix} &= \begin{pmatrix}0 \\ O_{12} \\ 0 \\ O_{22}\end{pmatrix} \end{align}$$ Putting these four columns together, you get the full matrix expression for $O'$: $$O' = \begin{pmatrix}O_{11} & 0 & O_{12} & 0 \\ 0 & O_{11} & 0 & O_{12} \\ O_{21} & 0 & O_{22} & 0 \\ 0 & O_{21} & 0 & O_{22}\end{pmatrix}$$ (check that this matrix reproduces the results derived above). This is, of course, just the tensor product of $O$ with the $2\times 2$ identity matrix, $O\otimes I_2$.

You could do the same calculation for three gates, or more, but it becomes increasingly tedious.

Answer 2

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:

1. Group the states by the uninvolved qubits.

   $|\psi\rangle = a|00⟩ + b|01⟩ + c|10⟩ + d|11⟩$

   $= \Big(a|0⟩ + c|1⟩\Big)|0⟩ + \Big(b|0⟩ + d|1⟩\Big)|1⟩$

2. Apply the operation within each group.

   $H_0 |\psi\rangle = \Big(H(a|0⟩ + c|1⟩)\Big)|0⟩ + \Big(H(b|0⟩ + d|1⟩)\Big)|1⟩$

   $= \Big(\frac{a+c}{\sqrt 2}|0⟩ + \frac{a-c}{\sqrt 2}|1⟩\Big)|0⟩ + \Big(\frac{b+d}{\sqrt 2}|0⟩ + \frac{b-d}{\sqrt 2}|1⟩\Big)|1⟩$

3. 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.