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O Plus symbol

  1. Dec 10, 2009 #1
    "O Plus" symbol

    I keep running across this symbol in a introduction to quantum computing text and I can't figure out what it stands for. If anyone could help me out I would appreciate it. The symbol is [tex] \oplus [/tex].
     
  2. jcsd
  3. Dec 10, 2009 #2

    Hurkyl

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    Re: "O Plus" symbol

    An example of usage would help.

    I've seen that symbol used primarily in two ways.

    The first way is the "direct sum" of two vector spaces, or similar objects. If you have a vector space V and a vector space W, then [itex]V \oplus W[/itex] is the vector space of ordered pairs whose first element is in V and whose second element is in W. With chosen bases, you have [itex]\mathbf{C}^m \oplus \mathbf{C}^n \cong \mathbf{C}^{m+n}[/itex].


    The other way is binary exclusive or a.k.a. addition modulo two. The commutative, associative operation whose values are given by the table
    [tex]
    \begin{array}{c|cc}
    & 0 & 1 \\
    \hline
    0 & 0 & 1 \\
    1 & 1 & 0
    \end{array}
    [/tex]
     
  4. Dec 11, 2009 #3
    Re: "O Plus" symbol

    Here's two examples of usage

    1.
    (Note irreversible NAND gate is [itex] (a, b) \rightarrow ¬(a \Lambda b) [/itex]
    We can actually construct a reversible version
    of the NAND gate that preserves all the information about the input: For
    example, the (Toffoli) gate

    [tex] (a, b, c) \rightarrow (a, b, c \oplus a \Lambda b) [/tex]

    is a reversible 3-bit gate that flips the third bit if the first two both take
    the value 1 and does nothing otherwise.

    2.
    Now suppose we have a quantum black box that computes f(x). Of course
    f(x) might not be invertible, while the action of our quantum computer is
    unitary and must be invertible, so we’ll need a transformation Uf that takes
    two qubits to two:

    [tex]U_f : \mid x \rangle \mid y \rangle \rightarrow
    \mid x \rangle \mid y \oplus f(x)\rangle [/tex]

    (This machine flips the second qubit if f acting on the first qubit is 1, and
    doesn’t do anything if f acting on the first qubit is 0.) We can determine if
    f(x) is constant or balanced by using the quantum black box twice.
     
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