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Measuring a qbit

  1. May 20, 2015 #1
    I am newly learning quantum computing and am confused about some concepts. Suppose your qbit is the electron of a hydrogen atom and its in the state α|0> + β|1> . As far as I can understand, this means that if you measure the qbit in |0>, |1> basis, you will get a ground state electron with the probability |α|2 and an exited one with the probability |β|2. So what does it mean when you say you are measuring the state in some other basis? say |+>, |-> basis. How do you physically do this? Thanks.
     
  2. jcsd
  3. May 20, 2015 #2

    naima

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    Gold Member

    It is easier to see that things when you measure a spin along a direction with a Stern and Gerlach. You get 0> or 1> You will get another orthogonal basis by rotating the SG apparatus.
     
  4. May 20, 2015 #3
    I kind of understood the thing about spin and how it is measured but I am trying to do understand the same thing for the H atom system. There has to be some similar measuring in a different basis thing for this as well right?
     
  5. May 20, 2015 #4

    Strilanc

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    Science Advisor

    Generally what you do is apply some operations that do a basis transform so the basis you want gets mapped to the computational basis, then measure, then undo the basis transform.

    For example, suppose you want to measure the X observable (instead of the Z observable that the computational basis corresponds to). The Hadamard operation happens to switch between those two basises, so you can use this circuit:

    Code (Text):

        |ψ> ──[H]──[Measure]──[H]──
     
    or this:

    Code (Text):

        ψ ──H──•──H── ψ [collapsed]
               │
        0 ─────X───── result
     
    If you don't want to figure out the basis transform, but have a gate whose operation is the observable's matrix U, and are able to do a controlled-U, then you can always use this circuit:

    Code (Text):

        ψ ─────U───── ψ [collapsed]
               │
        0 ──H──•──H── result
     
    (The X observable is unique in that *both* solutions use the Hadamard gate so you can reverse which gates go on which wire and it would still work.)

    Try measuring in the Y basis with this simulator.
     
    Last edited: May 20, 2015
  6. May 20, 2015 #5
    Thank you so much. It makes much more sense now. So basically, you apply a gate to the hydrogen atom and then you measure it in the standard basis. After the gate, the electron will be in a corresponding superposition and then when you measure, you will get outcomes depending on the new state and the new probability amplitudes(which depends on the gate you used). Did I get everything right? Thanks again.
     
  7. May 22, 2015 #6

    naima

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    Gold Member

    here u> and d> are spin of a particle along z.
    If you couple your atom with this particle, the state is g> d> + e> u>
    = g> (|+> + |->) + e> (|+> - |->) = (|g> + |e>)|+> + (|g> - |e>) |->
    So if you rotate your SG along x and measure its spin, your atom is projected on one of your Schrodinger cat's state.
     
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