I Dimensions of quantum cell automata's state space

[Jaime_mc2]

In the paper

C. S. Lent and P. D. Tougaw, "A device architecture for computing with quantum dots," in Proceedings of the IEEE, vol. 85, no. 4, pp. 541-557, April 1997, doi: 10.1109/5.573

about quantum dots, it is stated that the basis vectors in the space of quantum states for a single cell (four quantum dots) are of the form $$ |\phi_1\rangle = \left|\begin{array}{cccc}0&0&0&1 \\ 0&0&0&1\end{array}\right> \quad\cdots\quad |\phi_{16}\rangle = \left|\begin{array}{cccc}1&0&0&0 \\ 1&0&0&0\end{array}\right> $$ where the columns are related to the dot in which there is an electron, and the rows tell the projection of the spin (first row meaning that the spin points upwards). Therefore, the authors state that there are $16$ different basis states and that the dimension of the state space for $N$ cells is $16^N$.

However, I don't see why they only take into account states in which the electrons have opposite spin projections, and they are ignoring basis states like $$ \left|\begin{array}{cccc}1&1&0&0 \\ 0&0&0&0\end{array}\right> $$

Of course, because of the exclusion principle, the unique possibility for having two electrons in the same dot is that they have opposite spins, like in state $|\phi_1\rangle$, but I don't see why there should be such a restriction for two electrons being in different dots. If we take into account these extra basis states, the dimension of the state space for a single cell would be $$ \dfrac{8!}{2! \cdot 6!} = 28\ , $$ so we have $28^N$ for $N$ cells.

Why aren't these states taken into account?

 

[gentzen]

Of course, because of the exclusion principle, the unique possibility for having two electrons in the same dot is that they have opposite spins, like in state $|\phi_1\rangle$, but I don't see why there should be such a restriction for two electrons being in different dots.

You certainly "want" your basis states to have nearly identical energy levels (and being close to "some" ground state). Otherwise their relative phase will spin (much) faster than your "classical" control can handle (and if you are not close to "some" ground state, you risk that your qubits will decay to "some" ground state before you can perform useful computations on them).

If we take into account these extra basis states, the dimension of the state space for a single cell would be $$ \dfrac{8!}{2! \cdot 6!} = 28\ , $$ so we have $28^N$ for $N$ cells.

Why aren't these states taken into account?

Because then you would have different types of basis states, and those different types would almost certainly sit at completely different energy levels.