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Quantum Computers, how do they work?

  1. May 14, 2012 #1
    Lately I've been hearing about latest advancements in quantum computers, such as the chinese group breaks distance record for teleporting qubits (As seen on phys.org, Unfortunately i cannot post that link because i am new). I was just wondering if anybody could clear up what a quantum computer is? And how does it work? All i know is that it uses quantum entanglement, and it stores information in qubits.
  2. jcsd
  3. May 15, 2012 #2
    There's different ways quantum computing can take place. Some revolve around the superposition of sub-atomic particles, some revolve around entanglement and/or teleportation of photons.
    Quantum computers use "qubits" as units of information, which is essentially every infinite number between 0 and 1, like an extension of the classical binary code. Particles on the sub-atomic scale don't really have a finite state or position, and instead occupy a multitude of probable positions and momentum simultaneously, and this property is called superposition.
    It's still being worked on, but it uses the vector states of these positions, since when you actually measure a particle, the path or momentum you see is defined and isn't a superposition, it's probability collapses to that of specific known values, and scientists are trying to work on a way to use this.
  4. May 16, 2012 #3

    Physics Monkey

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    You've got the basics already. Quantum information is stored in qubits while operations are performed by acting on the qubits with unitary "gates". For example, a very important gate is CNOT which acts on two qubits and flips the second qubit between 0 and 1 provided the first control qubit is 1. In equations, if x and y take values 0 and 1 then [itex] U_{CNOT} |x, y\rangle = |x, x+y \rangle[/itex] and the addition is binary i.e. mod 2. At this level the important differences between classical and quantum computation is that there are more kinds of quantum gates and quantum gates can act on superpositions e.g. [itex] U_{CNOT} (|00\rangle - |11 \rangle) = |00\rangle - |10\rangle . [/itex]

    However, a quantum computer is not just a massively parallel classical computer as is often claimed. There is a very simple reason why: whenever you measure the state of the computer you only get a single answer. Thus even if in some sense a quantum computer is "doing all calculations at once", you only get to see one answer at a time. The real power of a quantum computer comes from interface i.e. from the fact that amplitudes can be negative (or complex). The key is to use interference to your advantage so that a few measurements of the computer give you the answer you want. Thus quantum computers can solve some problems faster than classical computers but they are not all powerful computing machines.

    Finally, on a practical front, the crucial development of quantum error correction means that quantum computation is in principle feasible. We don't need any kind of crazy level of control to achieve the sort of superpositions and entanglement necessary to do quantum computation. In my opinion this makes the quantum computer a question of time only.

    Hope this helps.
  5. May 16, 2012 #4
    The only quantum computing thing I heard is it uses particle spin, and not as in "entanglement" but as in binary computing at nearly unrestricted speeds. (no heat/volts/durability issue)

    Quantum being a reference to the nano-tech, again not entanglement.

    Physics Monkey, is there a more easy to understand explanation for the "entanglement" computing you mention?

    btw this is all iirc
  6. May 21, 2012 #5
    So quantum computers use entanglement to perform a ton of operations simulataneously. N quibits holding the same state you would expect of 2^N bits, via entanglement. But can quantum computers be turing complete if they depend on linear operations on the quibits? Are there any nonlinear operations that can be performed on quibit strings without collapsing the state (voiding all the entangled information) and feeding into a classical stage?

    xor is linear. not is linear.

    or might be sort of kludged as a linear operation if you read everything over threshhold x as on, if you don't add too much signal to some bits and drown out others. But it would mess up subsequent operations.

    and appears nonlinear. if is nonlinear.

    edit: Oh wait! I get it! In order to perform irreversible/nonlinear operations, you would have to start with a buffer of uniformly zeroed quibits, or uniformly one'd quibits. Operations analogous to rotation can be performed on your working quibit array that amount to things like and-ing, if-ing, ect. But you would eventually run out of zero-buffer. Can you remove a buffer from a quantum system to refresh it without destroying your working buffer's state?
    Last edited: May 22, 2012
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