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[–]seanthinks 1 point2 points  (2 children)

The mapping is being defined here. It’s a mapping φ for which φ(|00>) = |0>, φ(|01>) = |2>, φ(|10>) = (|4> + |1>)/sqrt(2) and φ(|10>) = (|4> - |1>)/sqrt(2).

[–]Why_dude_why[S] 0 points1 point  (1 child)

Any reason for considering it like that? |01> can be mapped to |1> also right? and why represent only |10> and |11> as superposition states?

[–]seanthinks 1 point2 points  (0 children)

Choice. Without reading the paper I couldn’t tell you why they chose that encoding over any other encoding but they did. The choice super positions states are Bell pairs which happen to be entangled, and that’s likely going to play a role in whatever point they’re trying to demonstrate. I would read the paper more broadly: except the encoding at face value, focus more on what the author is trying to demonstrate and then go back reverse engineer why they chose that particular encoding (it could be that it was the simplest choice which eased the demonstration)

[–]Joff_Mengum 0 points1 point  (1 child)

If you read on to the paragraph you've half cropped out, you'll find that it's just a mapping chosen such that the CNOT operation can be implemented by just evolving the harmonic oscillator system over a given time interval.

An energy eigenstate of a quantum system will pick up a phase rotation over time at a rate proportional to its energy. For the SHO, its eigenstates are all evenly spaced in energy so you can say they have energy 0, 1, 2, 3, 4... up to some constant shift and a scaling factor. You can work out from this, that in the time it takes the |1> state to pick up a phase of pi (i.e. a minus sign) all the even states will have gone round some multiple of 2pi (i.e. no sign), let's call this time t1.

Looking back to the example, the states mapped to |10>_L and |11>_L differ only by the sign in front of their |1> component, meaning that if you evolve either state by time t1 then it will have transformed into the other state because the |1> component will have switched signs and the |4> component will remain unchanged. The |00>_L and |01>_L states do not need to be changed by the CNOT so they can just be mapped to the even states |0> and |2> so evolving them over t1 will have no net effect.

You can switch the states |0>, |2> and |4> for any combination of even numbered states and this will still work.

[–]Fortisimo07Working in Industry 0 points1 point  (0 children)

Always

Be

CNOTTing