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According to An introduction to quantum machine learning (Schuld, Sinayskiy & Petruccione, 2014), Seth Lloyd et al. say in their paper: Quantum algorithms for supervised and unsupervised machine learning that classical information can be encoded into the norm of a quantum state $\langle x|x \rangle = |\vec{x}|^{-1}\vec{x}$. I'm not sure I understand their notation.

Let's take a simple example. Say I want to store this array: $V = \{3,2,1,2,3,3,5,4\}$ of size $2^{3}$ in the state of an $3$-qubit quantum system.

I can represent the state of an $3$-qubit system as:

$|\psi\rangle = a_1|000\rangle + a_2|001\rangle + a_3|010\rangle + a_4|011\rangle + a_5|100\rangle + a_6|101\rangle + a_7|110\rangle + a_8|111\rangle$ (using standard basis) where $a_i\in \Bbb C \ \forall \ 1 \leq i\leq 8$.

I could represent $V$ as a vector $\vec{V} = 3 \hat{x}_1 + 2 \hat{x}_2 +... + 4 \hat{x}_8$ where $\{\hat{x}_1,\hat{x}_2,...,\hat{x}_8\}$ forms an orthonormal basis in $\Bbb R^{8}$, and write the standard Euclidean norm for it as $|\vec{V}|=\sqrt{3^2+2^2+...+4^2}$.

After this, I'm confused as to how I'd get the coefficients $a_1,a_2,..,a_8$. Should I just assign $3$ to $a_1$, $2$ to $a_2$ and so on?

** But, then again**:

Consider the vector $N=2^{n}$ dimensional complex vector $\vec{v}$ with components $\{v_i=|v_i|e^{i\phi_i}\}$. Assume that $\{|v_i|,\phi_i\}$ are stored as floating point numbers in quantum random access memory. Constructing the $\log_2 N$ qubit quantum state $|v\rangle = |\vec{v}|^{-1/2}\vec{v}$ then takes $\mathcal{O}(\log_2 N)$ steps as long as the sub-norms are also given in the qRAM in which case any state can be constructed in $\mathcal{O}(\log N)$ steps.

**Firstly**, I don't understand their notion of a $2^n$ dimensional *complex* vector. If each of the components of their classical data array has two floating point numbers, wouldn't encoding that into a $n$-qubit quantum state be equivalent to storing a $2\times 2^{n}$ size classical array in a $n$-qubit system? Yes, I do know that $a_1,a_2,..,a_{2^n}$ are complex numbers having both magnitude and direction, and hence can store $2\times 2^{n}$ amount of classical information. But they don't mention anywhere how they will convert classical data (say in form of a $2\times 2^{n}$ array) into that form. Moreover, there seems to be a restriction that phase of a complex number $a_i$ can only range from $-\pi$ to $+\pi$.

**Secondly**, let us assume that the initial data array we wanted to store in our quantum system was actually $V=\{\{3,\phi_1\},\{2,\phi_2\},...,\{4,\phi_8\}\}$.

If they define $|v\rangle$ as $|\vec{v}|^{-1/2}\vec{v}$ then $|V\rangle$ in our example would look something like $(\sqrt{3^2+2^2+...+4^2})^{-1/2}(|3e^{i\phi_1}||000\rangle + |2e^{i\phi_2}||001\rangle + ... + |4e^{i\phi_8}||111\rangle)$. But then we're losing all the information about the phases $\phi_i$, isn't it? So what was the use of starting with a *complex* vector (having both a phase and magnitude) in the first place, when we're losing that information when converting to $|V\rangle$ anyway? Or are we writing supposed to consider $|V\rangle$ as $(\sqrt{3^2+2^2+...+4^2})^{-1/2}(3e^{i\phi_1}|000\rangle + 2e^{i\phi_2}|001\rangle + ... + 4e^{i\phi_8}|111\rangle)$?

It would be really helpful if someone could explain where I am going wrong using some concrete examples regarding storage of classical data in an $n$-qubit system.

1The Schuld

et al.paper was written quite early in the age of "quantum machine learning", and I have never had a deep enough interest in quantum machine machine learning (yet) to spend too much time learning it. So I won't try to to answer the question, but one thing I can contribute is to answer your confusion about therestrictionfor the complex phase to be between $-\pi$ and $\pi$. This range of $-\pi$ to $\pi$ is actually not a "restriction" because it covers all possible mathematical phases that can ever exist. $-\pi$ to $\pi$ means -180 to 180 degrees, which is a full circle. – user1271772 – 2018-05-28T00:55:23.993Anything beyond the range of $-\pi$ to $\pi$ is like saying 370 degrees, which is a complete circle plus another 10 degrees. So 370 degrees is equivalent to 10 degrees, and likewise for anything outside the range of $-\pi$ to $\pi$. – user1271772 – 2018-05-28T00:56:49.090