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I have been trying to solve a puzzle (not homework) in which I need to derive a quantum circuit from given a superposition, $|\psi\rangle$, where

$$ |00\rangle: 20\%\\ |10\rangle: 40\%\\ |11\rangle: 40\%\\ $$

and I must generate an output of

$$ |00\rangle: 20\%\\ |11\rangle: 80\%\\ $$

I started by writing $|\psi\rangle$ out as a ratios of the input probabilities wich somehow has to get to the output via a transformation C $$ {\Huge C}\begin{bmatrix}1 \\\ 0 \\\ 2\\\ 2 \end{bmatrix}=\begin{bmatrix}1 \\\ 0 \\\ 0\\\ 4 \end{bmatrix} $$

It was then simple enough to determine that what was actually happening was $${\Huge C}=\begin{bmatrix}1 & 0 & 0 & 0\\\ 0 & 0 & 0 & 0\\\ 0 & 0 & 0 & 0\\\ 0 & 0 & 1 & 1\end{bmatrix}$$

My question is then, how can I derive which primitive quantum logic gates should be used to compose this operation, C? Even if it is intuitive to some people, I cannot recognise it and would appreciate a step-by-step way to break it down into regular gates. Ideally the solution would be able to generalise for other situations where I need to derive what logic gates are used to produce certain results.

If there is a problem with the formulation of the question or terminology used, feel free to suggest an edit. I am quite new to this.

Edit 1

Thanks for the comments. I have been working through the terminology in what you all have said and I can provide some amended information (though I can't personally see how it has a bearing on the solution).

The puzzle actually has a way of getting the amplitudes, yet none of them have a complex component. So if I got his right, the input state vector looks like this:

$$ \begin{bmatrix}\sqrt{0.2} + 0i \\\ 0 \\\ \sqrt{0.4} + 0i\\\ \sqrt{0.4} + 0i \end{bmatrix} $$

I also have no idea how to represent the "desired" vector. I do now understand why whatever C is needs to be unitary, but I don't know how to get a C at all now.

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    $\begingroup$ C isn't unitary so it's not a quantum operation. $\endgroup$ Mar 9, 2022 at 17:09
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    $\begingroup$ State vector should be amplitudes, not probabilities, no? And these may be complex-valued. $\endgroup$
    – jjgoings
    Mar 9, 2022 at 17:50
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    $\begingroup$ Your general approach of inverting $C$ is not bad but your matrix $C$ is not reversible/unitary. See here. Also your wavefunction $|\psi\rangle$ is composed of probabilities, but it needs to be amplitudes/roots of probabilities. $\endgroup$ Mar 9, 2022 at 18:11
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    $\begingroup$ controlled-Hadamard controlled off the first qubit targeting the second qubit. $$\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1/\sqrt{2} & 1/\sqrt{2} \\ 0 & 0 & 1/\sqrt{2} & -1/\sqrt{2} \end{array}\right)$$. $\endgroup$
    – DaftWullie
    Mar 10, 2022 at 15:24

1 Answer 1

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A quantum system is not specified by probabilities, but probability amplitudes. I'm going to assume that what you want to do is the conversion $$ \frac{1}{\sqrt{5}}\left(\begin{array}{c} 1 \\ 0 \\ \sqrt{2} \\ \sqrt{2} \end{array}\right)\rightarrow\frac{1}{\sqrt{5}}\left(\begin{array}{c} 1 \\ 0 \\ 0 \\2 \end{array}\right) $$ The way that I look at this, it says that the amplitudes on the states where the first qubit is in $|0\rangle$ just stay the same. I only have to change things if the first qubit is in the $|1\rangle$ state. Hence, we're looking for a unitary controlled-$U$ where $$ U\frac{1}{\sqrt{5}}\left(\begin{array}{c} \sqrt{2} \\ \sqrt{2} \end{array}\right)=\frac{1}{\sqrt{5}}\left(\begin{array}{c} 0 \\ 2 \end{array}\right), $$ subject to the constraint that $U$ is unitary, meaning $UU^\dagger=I$ and $U^\dagger U=I$. You can construct such a matrix very easily. In particular, multiply both sides of the previous equation by $U^\dagger$. We have $$ U^\dagger\left(\begin{array}{c} 0 \\ 1 \end{array}\right)=U^\dagger U\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ 1 \end{array}\right). $$ Now, let's temporarily write out $$ U^\dagger=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right). $$ Hence, $$ U^\dagger\left(\begin{array}{c} 0 \\ 1 \end{array}\right)=\left(\begin{array}{c} c \\ d \end{array}\right). $$ Thus, we can easily identify the elements $c$ and $d$ as both being $\frac{1}{\sqrt{2}}$. We have identified the second column of $U^\dagger$.

So, if we can identify the first column, i.e. the values $$ U^\dagger\left(\begin{array}{c} 1 \\ 0 \end{array}\right), $$ we're done. Note, however, that $$ \left(U^\dagger\left(\begin{array}{c} 0 \\ 1 \end{array}\right)\right)^\dagger U^\dagger\left(\begin{array}{c} 1 \\ 0 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \end{array}\right)UU^\dagger\left(\begin{array}{c} 1 \\ 0 \end{array}\right)=\left(\begin{array}{cc} 1 & 0 \end{array}\right)\left(\begin{array}{c} 1 \\ 0 \end{array}\right)=0 $$ but it is also $$ =\left(\begin{array}{cc} a^\star & b^\star \end{array}\right)\left(\begin{array}{c} c \\ d \end{array}\right)=a^\star c+b^\star d. $$ Said another way, the first column must be orthogonal in order to give unitarity. Also, each column must have length 1. There's some freedom in the choice - we can have any $a=e^{i\phi}$ with $b=-e^{i\phi}$, but a natural one would be $\phi=0$, thus giving $$ U^\dagger=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right)=\sqrt{Y} $$ and hence $$ U=\sqrt{Y}^\dagger. $$ Thus, the overall operation you're looking for is controlled-$\sqrt{Y}^\dagger$, which is written as $$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right) $$

This can subsequently be decomposed into more elementary gates if necessary, but since you don't specify what you consider to be your building blocks, I can't be more specific.

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  • $\begingroup$ I've been trying to understand your third and fourth paragraph. Would you be willing to tease out the reasoning/terminology a bit? I understand what those things mean on their own but I don't think I'm making the right connections. $\endgroup$
    – NeRoboto
    Mar 15, 2022 at 10:28
  • $\begingroup$ You mean the bit that goes from the action of $U$ on one particular vector to producing the matrix form of $U$? $\endgroup$
    – DaftWullie
    Mar 15, 2022 at 11:01
  • $\begingroup$ Yes. Particularly where you introduce U dagger. Like, for instance, you say "that's the second column of U dagger". I'm not sure which vector you mean (and I'm also assuming you mean a vector), because neither of the vectors are represented in U's matrix. $\endgroup$
    – NeRoboto
    Mar 15, 2022 at 20:40

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