6

There’s not exactly a standard way of doing it. The first thing you have to do is make the transformation unitary. The way that’s guaranteed to work is to introduce 2 extra qubits. However that’s not necessary in this case. Instead, the circuit is very simple: controlled not controlled from y targeting z, followed by controlled not controlled from x and ...


4

While talking about knowing the position exactly is a nice theoretical ideal, in practice, you cannot do that. You'll really be asking: "In which 'bin' of width $\delta x$ where $x$ spans from $x_{\min}$ to $x_{\max}$ is the particle confined to?". This means that there's $(x_{\max}-x_{\min})/\delta x$ bins, and so you basically need $$ \log_2\left((x_{\max}-...


3

For a classical function $f:\{0,1\}^n\rightarrow\{0,1\}^m$, there is always a reversibile implementation (and hence corresponding unitary implementation) on $n+m$ bits. There could be one on as few as $n$ bits depending on the function. For example, when you say that controlled not is the reversible implementation of not (it corresponds to the upper bound), ...


2

Yes, when you run an algorithm, often specified as a sequence of gates, you get different results according to some probability distribution. Assuming we are talking about an error-free computation (this is where, in this field, we use the terminology "noise" which is distinct from the "noise" of "signal to noise ratio"), then most algorithms that you will ...


2

You can found some interesting approaches to decomposing gates also here: https://arxiv.org/abs/quant-ph/9503016 (Elementary gates for quantum computation).


2

I found paper Quantum Circuits for Isometries to be a useful reference on the topic. It describes several methods for decomposing multi-qubit unitaries into CNOT gates and qubit unitaries and also gives several references to earlier related works. There is also a Mathematica package that implements algorithms described in the mentioned paper. If you don't ...


2

The procedure I use e.g. in this answer works in the general case as well. The minimum number of ancillae that you need depends on how much "non-injective" the function is. By this I mean that, given a function $f$, the property that matters is the number of elements in the preimages of $f$: $|f^{-1}(y)|$, where $f^{-1}(y)\equiv\{x : f(x)=y\}$. More ...


1

A good way to start is to have a look at the truth table of the function: $$\begin{array}{ccccc} x & y & z & \text{out1} & \text{out2} \\\hline 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & ...


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