Re: Quantum slip - Quantum conspiracy

From: Unruh (unruh-spam_at_physics.ubc.ca)
Date: 05/23/05


Date: 23 May 2005 16:52:44 GMT


"Matt Mahoney" <matmahoney@yahoo.com> writes:

>Unruh wrote:
>> "Matt Mahoney" <matmahoney@yahoo.com> writes:
>>
>> >1. put n qubits in a superposition of 2^n states
>>
>> n operations

>That is my question. I agree with your other points.

>Has there been any experimental verification that randomizing n qubits
>(for example, by measuring their spins perpendicular to the output
>detector angle) puts them in a superposition of 2^n states, rather than
>n superpositions of 2 states? Isn't randomizing a qubit a statistical
>process that is tied to the environment, which would therefore tie the
>qubits to each other?

They are NOT randomized. They are rotated. You start off with all the spins
in the single state with all spins pointing down say. Then you rotate each
spin so it points sideways. That is n operations -- a rotation of each
spin. But that state is a superposition of all possible spin up/down states
with equal probability.

>In current quantum computers with 5 or 8 qubits with NMR I/O (molecules
>in solution), the first step is implicit. How does this experimental

No, it is explicit in that you need to do the experiment 2^n times with
different sequences and then combine all those 2^n outcomes to get the one
outcome corresponding to having taken the intitial state to be all down
spins. In liquid state nmr, you cannot start out with the system in all
down spins-- the environmental interaction gives you only a system which is
an incoherent cojunction of all the 2^n possible states. (ie, each of the
molecues is in a different state) and you have to use a selected series to
pick out only those molecules which are in the desired state. This is what
sets the limit of about 10 qubits to the size of an NMR liquid state QC.

>setup distinguish between each molecule being in a superposition of 2^8
>states, and the case where 2^-8 of the molecules are in each state.
>NMR just averages the results from all the molecules, so it seems to me
>the two cases would be indistinguisable. But in the latter case, QC
>would break down around 20-25 qubits when you no longer have many
>molecules for each possible state.

>-- Matt Mahoney