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To put it straight - are qubits fermions, bosons or else?

For example, the Bell states that are frequently used in quantum computations have different symmetry (00 + 11 is symmetric, 10 - 01 is antisymmetric). Indistinguishable physical particles in a real world are either bosons (symmetric) or fermions (antisymmetric), not both.

Should we take into account permutation symmetry while discussing multiple qubit states?

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    $\begingroup$ I have a feeling that your question is more about physics than computer science. If this is the case then please let me suggest this site to ask on : physics.stackexchange.com . $\endgroup$ – Mohammad Alaggan Jun 15 '11 at 7:56
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    $\begingroup$ @M.Alaggan - It is hard to say where quantum computing ends and quantum mechanic begins. Let the question remain here for a day or two. If not answered here it may be transferred to physics. $\endgroup$ – kludg Jun 15 '11 at 8:17
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    $\begingroup$ @M. Alaggan: I would think this is a perfectly reasonable question for this site. $\endgroup$ – Joe Fitzsimons Jun 15 '11 at 22:33
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    $\begingroup$ The standard definition of qubit assumes that the particles are distinguishable (otherwise you don't get a $\mathbb{C}^{2^n}$ dimensional Hilbert space from $n$ qubits), so the question of whether they are bosons or fermions is irrelevant. There are no fundamental physics obstacles to implementing quantum computers using either bosons or fermions as qubits. $\endgroup$ – Peter Shor Jun 18 '11 at 10:21
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In general, a state of $n$ qubits is a state of $n$ distingushable particles. Eg. $|01\rangle$ means that the first particle is in the state $|0\rangle$, and the second is in the state $|1\rangle$.

How can be elementary particles distinguishable? Place two particles in different places. Eg. two electrons are in the positions $x_1$ and $x_2$, respectively. (Or two photons in two consecutive pulses, one in each.)

However, you can make the qubits indistinguishable by imposing a permutation symmetry, i.e. symmetrization and antisymmetrization.

And $n$ photons in the same place gives 'automatically' symmetrized state.

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  • $\begingroup$ Thank you. Can two qubits remain distinguishable after applying C-NOT gate? Quantum computing is effective if only the particles are entangled, is it possible for the entangled particles to be distinguishable? $\endgroup$ – kludg Jun 15 '11 at 9:02
  • $\begingroup$ @Serg: C-NOT at $|01\rangle$ gives the same state, so yes, they can. Entanglement of distinguishable particles is perfectly possible. For a clear example consider an entanglement state of a photon and an electron, eg. $\propto |0_e 0_p\rangle+|1_e 1_p\rangle$. $\endgroup$ – Piotr Migdal Jun 15 '11 at 10:50
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    $\begingroup$ One way to think about it is that an electron has both position and spin attributes. The overall wavefunction is antisymmetric, but if the positions are essentially classical, then they can label distinguishable particles whose only feature is their spin. $\endgroup$ – Aram Harrow Jun 15 '11 at 13:16
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    $\begingroup$ The same thing is possible classically. You can place n indistinguishable balls into (say) 2n bins in (2n choose n) ways. But if you say that the first ball has to be in bin 1 or 2, the second has to be in bin 3 or 4, etc. Then there are $2^n$ possibilities, corresponding to the state of n bits. $\endgroup$ – Aram Harrow Jun 15 '11 at 13:22
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    $\begingroup$ To follow up on Aram's comment, you can do exactly the same thing with bosons (photons for example), or systems which mix fermions and bosons. $\endgroup$ – Joe Fitzsimons Jun 15 '11 at 17:33

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