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Bra-ket notation and qubit issue.

  1. May 7, 2013 #1
    I am having trouble understanding the following:

    Uf: |x>|y> → |x>|y [itex]\oplus[/itex]f(x)>

    [itex]\oplus[/itex] being a mod 2 operation (nand)? I suppose I don't understand how to read the "ket" states so well. As far as I understand we have that since x and y can be 0,1 only if |x=1>|y=1> then if f(x) = 1 then |y [itex]\oplus[/itex] f(x)> would come out to be → |1>|0>

    but the main part is that if we are dealing with a quantum computer then we can chose the input state to the a superposition of |0> and |1>. That if the second qubit is initially prepared in the state 1/[itex]\sqrt{2}[/itex](|0> - |1> then... The issue is with this equation in that how does |0> - |1> mean superposition? This is the part of the bra-ket notation that I don't understand
     
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  3. May 7, 2013 #2

    kith

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    What definition of superposition do you use then?

    If you want to apply your gate U to a superposition |x1 y1>+|x2 y2>, you can simply apply it to each term because of the linearity of U. So U(|x1 y1>+|x2 y2>) = U|x1 y1> + U|x2 y2>
     
    Last edited: May 7, 2013
  4. May 9, 2013 #3
    Thanks kith. Well as far as superposition goes - I am not quite sure of the definition he (Preskill) is using in his Notes: www.theory.caltech.edu/people/preskill/ph229/notes/book.ps‎ [Broken]

    I am guessing just some sort of overlap of states that is treated as a vector space. I say this as he talks about states being "ray's" in a Hilbert space. That these rays are vectors. He does say that every ray corresponds to a possible state so that given two spaces |θ>, |ψ> we can create a state a|θ> + b|ψ> by the superposition principle as he puts it (Pg. 38).

    Is it just that the states |0> and |1> are perpendicular and thus 1/sqrt(2) is the normalization? I.e. in 2-D Euclidean space we would have 1x + 0y being denoted as |0> and 0x + 1y as |1> hence the distance between them is sqrt((1-0)^2 + (0-1)^2) = sqrt(2)?

    He has something similar where |cat> = 1/sqrt(2) [itex]\cdot[/itex] |dead> + |alive> so since these are distinct (can't be both dead and alive) we have that they are perpendicular hence the normalization of 1/sqrt(2).

    I suppose I am looking for a better understanding of |0> + |1>. are these indeed considered perpendicular? But furthermore I am somehow looking at a probability density in |0> and |1> respectively? So I would be looking at some sort of superposition of two perpendicular waves?

    Thanks,

    Brian
     
    Last edited by a moderator: May 6, 2017
  5. May 9, 2013 #4

    kith

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    Yes.

    Yes. |0> and |1> are eigenstates of a certain self-adjoint operator which corresponds to a physical quantity (usually spin). Such eigenstates have the property of being perpendicular.

    Why a probability density? Do you know how probabilities are related to superpositions? Do you know the postulates of QM?

    Waves belong to infinite-dimensional spaces while our space is two-dimensional.
     
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