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The power of Qubits

  1. Oct 29, 2015 #1
    Hi,

    I'm having trouble understanding the power of qubits relating to quantum computers. I've read a number of times that the power comes from the fact that instead of simply holding an on or off state (1/0), they can hold both at the same time (superposition). However, when we measure them they 'decide' on a state.

    My question is, how can this third state of superposition provide a huge benefit over the 1 and 0 states of transistors, since once we use them (observe them) the states available to us is still only a 1 or a 0?

    Many thanks in advance http://www.thephysicsforum.com/images/smilies/smile.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 29, 2015 #2
  4. Oct 29, 2015 #3
    I'm looking at lecture 9 and there are many gaps. Maybe he's giving a slide show and the slides are not included.
     
  5. Oct 29, 2015 #4

    Strilanc

    User Avatar
    Science Advisor

    Quantum computers are not ternary computers. Qubits are not just bits with a third state.

    A qubit is a 2-level quantum system that can store states like ##a \left|0\right\rangle + b \left|1\right\rangle## where ##a^2 + b^2 = 1##.

    Put ##n## qubits together, and you get a ##2^n##-level quantum system. For example, 3 qubits can store states like ##a \left|000\right\rangle + b \left|001\right\rangle + c \left|010\right\rangle + d \left|011\right\rangle + e \left|100\right\rangle + f \left|101\right\rangle + g \left|110\right\rangle + h \left|111\right\rangle## where ##a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 = 1##.

    In other words, quantum computers can store a linear combination of the classical states. But the weights of the linear combination are not probabilities, which would have to satisfy ##a + b + ... + h = 1##, they are the square roots of probabilities and must satisfy ##a^2 + b^2 + ... + h^2 = 1##.

    Everything else flows from that square-root-of-probability thing. Operations correspond to complex orthonormal matrices. Destructive interference is possible. Copying doesn't quite work. Everything is reversible. Entanglement is a thing. Etc.
     
  6. Oct 29, 2015 #5
    Do you mean that you don't see any figures or formulas?
     
  7. Oct 29, 2015 #6
    I see

    ask me to
    exp
    the Bell inequality to them
    .

    BUT when I copy this from the lecture notes and paste here, it comes out correctly!

    ask me to explain the Bell inequality to them.

    So I can paste the entire lecture to a Physics Forums reply box and read it that way. Golly. Well, whatever works, works.
     
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