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say that i have 2 qubits - 2 spin half fermions. my initial condition is |00> in the spin-wave function and some anti-symmetrical spacial wave function. I'm wondering about what happens when applying a NOT gate. The fermions total wave function still has to be inverted when swapping the particles. so does it mean that a NOT gate has to effect the spacial wave function? y.a. |
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First of all, you need to be careful what you mean by a NOT gate. If you have only one qubit, you could use the gate (typically called X) $$\begin{pmatrix}0 & 1\\1& 0\end{pmatrix},$$ which flips the two basis states of the qubit. Since your system here consists of only one electron, this won't affect the 'spatial wave function' very much. If you're talking about two or more qubits, you need to keep in mind that quantum operations are always done whilst (physically) identifying each of the qubits. If, for example, you'd have the following operation: $$|00\rangle \mapsto \frac{1}{\sqrt{2}}\left[|01\rangle - |11\rangle\right],$$ that means that the 'left' bit is mapped to 0 or 1, whilst the 'right' one is mapped to 1. You're therefore never really 'swapping particles'. In fact, if you were to do quantum computing on an entangled two-fermion system, (say: a deuterium atom), how would you probe the system, i.e., how would you experimentally distinguish between $|01\rangle$ and $|10\rangle$? What you could do, is swap the basis states of the two qubits, using a gate like $$\begin{pmatrix}1 & 0 & 0 & 0\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\0 & 0 & 0 & 1\end{pmatrix},$$ where the basis states are labeled $|00\rangle,\quad |10\rangle,\quad|01\rangle,\quad|11\rangle.$ [The matrix above is indeed unitary.] In fact, this quantum operation would leave the spin-symmetric states $|00\rangle, \quad |11\rangle$ alone. |
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It is not exactly clear which of the quantum "NOT gates" you meant - there are many quantum counterparts of the classical "NOT gate". But you clearly meant an operation that changes the antisymmetry with respect to the spins, assuming that the qubits are encoded in the spins. Such operations obviously exist. Well, if the spin-part of the wave function changes from an antisymmetric one to a symmetric one, and if the electrons fully carry the information about the qubits, then the spatial part of the wave function has to change from a symmetric to an antisymmetric one. That's because the total wave function is antisymmetric for identical fermions. That's not "difficult" in any sense because in a quantum computer, if the information is stored by the electrons, they're supposedly located at different places - or different states - so the position (or another observable) effectively makes them distinguishable particles: you may identify the individual electrons employed by the computer by their (discrete) position, by the "gate" where they belong etc. Their identical character only becomes important if you try to physically exchange the fermions - their positions - or if you're trying to squeeze them to the very same place. But these conditions don't hold for electrons used in a quantum computer so the fact that they're identical fermions is pretty much irrelevant. Obviously, the wave function of $N$ qubits in a quantum computer have to be allowed to be symmetric, antisymmetric, or having no symmetry, otherwise it wouldn't be $N$ independent qubits of information and they couldn't be used to calculate anything of value. |
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