# Quantum embeddings¶

A quantum embedding represents classical data as quantum states in a Hilbert space via a quantum feature map. It takes a classical datapoint $$x$$ and translates it into a set of gate parameters in a quantum circuit, creating a quantum state $$| \psi_x \rangle$$. This process is a crucial part of designing quantum algorithms and affects their computational power—for more details, see [R1] and [R3].

Let’s consider classical input data consisting of $$M$$ examples, with $$N$$ features each,

$\mathcal{D}=\{x^{(1)}, \ldots, x^{(m)}, \ldots, x^{(M)}\},$

where $$x^{(m)}$$ is a $$N$$-dimensional vector for $$m=1,\ldots,M$$. To embed this data into $$n$$ quantum subsystems ($$n$$ qubits or $$n$$ qumodes for discrete- and continuous-variable quantum computing, respectively), we can use various embedding techniques, some of which are explained briefly below.

## Basis Embedding¶

Basis embedding associates each input with a computational basis state of a qubit system. Therefore, classical data has to be in the form of binary strings. The embedded quantum state is the bit-wise translation of a binary string to the corresponding states of the quantum subsystems. For example, $$x=1001$$ is represented by the 4-qubit quantum state $$| 1001 \rangle$$. Hence, one bit of classical information is represented by one quantum subsystem.

Let’s consider the classical dataset $$\mathcal{D}$$ mentioned above. For basis embedding, each example has to be a N-bit binary string; $$x^{(m)}=(b_1,\ldots,b_N)$$ with $$b_i \in \{0,1\}$$ for $$i=1,\ldots,N$$. Assuming all features are represented with unit binary precision (one bit), each input example $$x^{(m)}$$ can be directly mapped to the quantum state $$| x^{(m)}\rangle$$. This means that the number of quantum subsystems, $$\mathbf{n}$$ , must be at least equal to $$\mathbf{N}$$. An entire dataset can be represented in superpositions of computational basis states as

$| \mathcal{D} \rangle = \frac{1}{\sqrt{M}} \sum_{m=1}^{M} |x^{(m)} \rangle.$

For example, let’s say we have a classical dataset containing two examples $$x^{(1)}=01$$ and $$x^{(2)}=11$$. The corresponding basis encoding uses two qubits to represent $$| x^{(1)} \rangle=|01 \rangle$$ and $$| x^{(2)} \rangle=|11 \rangle$$, resulting in

$| \mathcal{D} \rangle = \frac{1}{\sqrt{2}}|01 \rangle + \frac{1}{\sqrt{2}} |11 \rangle.$

Note

For $$N$$ bits, there are $$2^N$$ possible basis states. Given $$M \ll 2^N$$, the basis embedding of $$\mathcal{D}$$ will be sparse.

## Amplitude Embedding¶

In the amplitude-embedding technique, data is encoded into the amplitudes of a quantum state. A normalized classical $$N$$-dimensional datapoint $$x$$ is represented by the amplitudes of a $$n$$-qubit quantum state $$| \psi_x \rangle$$ as

$| \psi_x \rangle = \sum_{i=1}^{N} x_i |i \rangle,$

where $$N=2^n$$, $$x_i$$ is the $$i$$-th element of $$x$$, and $$| i \rangle$$ is the $$i$$-th computational basis state. In this case, however, $$x_i$$ can have different numeric data types, e.g., integer or floating point. For example, let’s say we want to encode the four-dimensional floating-point array $$x=(1.0, 0.0, -5.5, 0.0)$$ using amplitude embedding. The first step is to normalize it, i.e., $$x_{norm}=\frac{1}{\sqrt{31.25}}(1.0, 0.0, -5.5, 0.0)$$. The corresponding amplitude encoding uses two qubits to represent $$x_{norm}$$ as

$| \psi_{x_{norm}} \rangle = \frac{1}{\sqrt{31.25}}\left[|00 \rangle - 5.5|10 \rangle\right].$

Let’s consider the classical dataset $$\mathcal{D}$$ mentioned above. Its amplitude embedding can be easily understood if we concatenate all the input examples $$x^{(m)}$$ together into one vector, i.e.,

$\alpha = C_{norm} \{ x^{(1)}_1, \ldots, x^{(1)}_N, x^{(2)}_1, \ldots, x^{(2)}_N, \ldots, x^{(M)}_1, \ldots, x^{(M)}_N \},$

where $$C_{norm}$$ is the normalization constant; this vector must be normalized $$|\alpha|^2=1$$. The input dataset can now be represented in the computational basis as

$| \mathcal{D} \rangle = \sum_{i=1}^{2^n} \alpha_i |i \rangle,$

where $$\alpha_i$$ are the elements of the amplitude vector $$\alpha$$ and $$| i \rangle$$ are the computational basis states. The number of amplitudes to be encoded is $$N \times M$$. As a system of $$n$$ qubits provides $$2^n$$ amplitudes, amplitude embedding requires $$\mathbf{n \geq \log_2({NM})}$$ qubits.

Note

If the total number of amplitudes to embed, i.e., $$N \times M$$, is less than $$2^n$$, non-informative constants can be padded to $$\alpha$$ [R1].

For example, if we have 3 examples with 2 features each, we have $$3\times 2= 6$$ amplitudes to embed. However, we have to use at least $$\lceil \log_2(6)\rceil = 3$$ qubits, with $$2^3=8$$ available states. We therefore have to concatenate $$2^3-6=2$$ constants at the end of $$\alpha$$.

Important

There are many other embedding techniques with similar embedding protocols. For example, squeezing embedding and displacement embedding are used with continuous-variable quantum computing models, where classical information is encoded in the squeezing and displacement operator parameters. Hamiltonian embedding uses an implicit technique by encoding information in the evolution of a quantum system [R1].