Experimental setup

The model is based on a quantum experimental arrangement. We assume that the only possible measurement outcomes, as well as input states, of this system are either 0 or 1. They are the eigenstates of an observable $S$ of the system and can be written as ${\left\vert 0 \right\rangle}_k$ and ${\left\vert 1 \right\rangle}_k$, where the subscript $k$ represents the position of the quantum bit (qubit). We further assume that the qubit on the first input, the second input, as well as the output are represented by different eigenstates,^6.2 so we have in total 6 orthogonal eigenstates^6.3, which we claim forms a complete basis of a Hilbert space in which the state of affairs is to be discussed. A quantum computation is then to be understood as a trajectory in this abstract space. At the beginning of each quantum experiment, the initial state is prepared as a superposition of input eigenstates -- either $\left\vert 0 \right\rangle$ or $\left\vert 1 \right\rangle$ on their corresponding positions. For example, to compute $XOR(0,1)=1$, the (initial) input state is prepared as:

$${c_{01} \left\vert 0 \right\rangle}_1+{c_{12} \left\vert 1 \right\rangle}_2,$$ (6.1)

where the subscripts of an eigenket $\left\vert x \right\rangle$ represent the position of the qubit and the coefficient $c_{mn}$ (a complex number) is the component of the corresponding eigenstate $\left\vert m \right\rangle$ at position $n$. The system is then set free to evolve without external perturbation. Technically speaking, the initial state is subject to a unitary evolution $U$, which is determined by the specific arrangement of the experiment. $U$ can be expressed as:

$$U=e^{-i{{H} \over \hbar }t}$$ (6.2)

where $H$ is the Hamiltonian (the energy operator) of the system and $\hbar$ is the Planck constant divided by $2\pi$. (For simplicity and without losing generality, in the implementation we can take $\hbar$ simply as 1.) The linear superposition is straight-forward for a ``parallel'' configuration, where both inputs are fed into the quantum logic gate at the same time. It is also the case for a ``serial'' configuration, which is the norm in natural language processing/understanding. In the case of classical XOR, these two schemes are equivalent. This can be justified as follows: suppose the input qubits are fed into the system one after another, say first $\left\vert 0 \right\rangle_1$ then, after the a time delay $t$, $\left\vert 1 \right\rangle_2$, we should have the end state of the system as:

$$U\left( {a_{01}\left\vert 0 \right\rangle _1+U'a_{12}\left\vert 1 \right\rangle _2} \right),$$

where $U'$ is another unitary operator that can be written in the following general form:

$$U'=e^{-i{{H'} \over \hbar }t},$$

where $H'$ is the Hamiltonian of a preparation process. Since the classical XOR is a function that does not tell the difference between an input's temporal position (therefore is symmetric with respect to the input position), we should have the same outcome from the system if we put $\left\vert 1 \right\rangle_2$ first and, after the a time delay $t$, $\left\vert 0 \right\rangle_1$ afterwards. Namely

$$U\left( {a_{01}\left\vert 0 \right\rangle _1+U'a_{12}\left\ve... ...\right\rangle _1+a'_{12}\left\vert 1 \right\rangle _2} \right).$$

The best way to achieve this is to assume that the preparation process will not mix each eigenstate with others. That is, $H'$ has to be a diagonal matrix. In this case, we can write the input as shown in equation 6.1. Furthermore, the absolute square of the coefficient of a particular eigenstate relative to the sum of absolute squares of all components (or the square of input vector length) is the probability of finding the system in this particular eigenstate. So, assuming symmetry, we have:

$$\left\vert {c_{01}} \right\vert=\left\vert {c_{12}} \right\vert=\sqrt {1/2}.$$

The end-state or output of the experimental setup is a superposition of eigenstates:

$$c'_{01}\left\vert 0 \right\rangle _1+c'_{11}\left\vert 1 \rig... ...ight\rangle _{out}+c'_{1,out}\left\vert 1 \right\rangle _{out}.$$

Generally speaking, there can be ``residues'' of the input eigenstates in the end state. However, since we are only interested in the relative probability of the output qubit, we expect to measure the output states either as $\left\vert 0 \right\rangle_{out}$ or $\left\vert 1 \right\rangle_{out}$, with corresponding probability of

$$p_0={{\left\vert {c'_{0,out}} \right\vert^2} \over {\left\ver... ...{0,out}} \right\vert^2+\left\vert {c'_{1,out}} \right\vert^2}}.$$

In this experiment, we have a total of 36 real parameters (the free variables of the targeted unitary operator) which are to be found. This can be easily transferred to a standard minimization problem by defining a cost function:

$$C(\vec v)=\sum\limits_{i\in \{ 0,1\} } {\left( {p_i-t_i} \right)^2},$$ (6.3)

where $\vec v$ is the 36-dimensional parameter vector and $t_i$ is 1 if the corresponding classical symbol ($\mathbb{T}$ or $\mathbb{F}$) is present, 0 otherwise. A standard minimization procedure can then be used. We use the conjugate gradient method [41] (see Appendix C), starting with a small random initial vector^6.4.