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Quantum Computing

  1. May 18, 2010 #1
    Hi All - My name is Dan Sandberg and I am studying physical chemistry at the Univ. of Connecticut in the USA.

    I recently heard a talk on quantum computing and, at the time, began to picture a quantum computer which employed a photoactive protein. I decided to attempt to see this idea to fruition and present it to the department. However I can not seem to get a strong grasp on the intricacies of the quantum computer and I am stumbling with some of the jargon. I was hoping someone could assist me with answering some perhaps basic questions.

    For example,

    In 1993, Seth Lloyn, Prof. of quantum mechanical engineering at MIT, published a paper on "A Potentially Realizable Quantum Computer". In this paper he states, "the only logical operations that necessarily require dissipation are irreversible ones". What does he mean by "dissipation"?
     
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  3. May 18, 2010 #2

    f95toli

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    Dissipation basically means that the system is losing energy; or -more generally- is "leaking" (in an optical system dissipation could happen due to e.g. imperfect mirrors resulting in a loss of photons).
    A good example would be an LCR circuit where (ideally) the inductor and capacitor are lossless but energy is lost in the resistor (because electrical energy is converted into heat).
     
  4. May 19, 2010 #3
    I understand and I sort of assumed that dissipation meant energy dissipation. What I'm still struggling with still is 1) why the only logical operations that necessarily require dissipation are irreversible ones (Landauer's result) which I can look into further on my own but more importantly 2) why dissipation is required for error correction to quantum computers?
     
  5. May 19, 2010 #4

    f95toli

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    The answer to 1) goes back to the problem of Maxwell's demon. The basic point is that you need to dissipate energy (or -more accurately- increase the entropy by ln(2) ) to "forget" one bit of information (this is a result of information theory, it has nothing to do with quantum computing as such). Hence, an irreversible computer (which -by definition- is "throwing away" information) must be dissipative. A reversible computer could -in principle- be made dissipationsless (although an actual implementation of a computer will of course always dissipate energy).
    Note that reversible computing had been along for a long time when quantum computing came along (although classical reversible computers are not very useful). Hence, many of these results have nothing to do with quantum mechanics.

    I can unfortunately not help you with your second question.
     
  6. May 19, 2010 #5
    The answer to quesiton 1 is incredibly helpful and may actually help me answer question 2. I will look into information theory is more detail to see if I can form an answer to question 2.
     
  7. May 20, 2010 #6
    I'm not sure that it is. I seem to remember reading something about that, unfortunately I can't remember where. I'll have a look.

    I remember when I was learning about bit flip correction, I think I found a way to correct it with 3 toffoli gates without measurements. At least, I'm pretty sure it was separable. I don't know if it's possible with phase errors as well, could be.
     
  8. May 20, 2010 #7
    It was in Seth Lloyds paper on "A Realizable Quantum Computer" published in the 90s. I don't have the article in front of me but I know in the first paragraph it states the requirement for dissipation to achieve error correction.
     
  9. May 21, 2010 #8
    Here is the circuit I had. At least 2 of a, b and c are 0, and [itex]\bar{a}[/itex] means not a etc.

    I found the paper you mentioned:
    In the circuit I posted, instead of having a cost in terms of energy, it costs you 2 ancilla qubits of a known state, they end up in unknown states. So although the state after the error is unknown, so are the states of the 2 ancilla qubits at the end, so it doesn't require dissipation.

    Also, this circuit is reversible, but it doesn't do error correction when it runs backwards. It does initialization! The two unknown bottom qubits are transformed into 0, at the cost of the pure state [itex]\alpha|000\rangle+\beta|111\rangle[/itex].
     

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  10. May 24, 2010 #9
    Wow thats brilliant and will probly take me awhile to fully understand.

    I have another question regarding Dr. Lloyd's paper (A Potentially Realizable Quantum Computer): He states that the excitation pulses to induce bits into their states are given by a [tex]\pi[/tex] pulse where

    [tex]\pi[/tex]=[tex]\bar{h}[/tex]-1[tex]\int[/tex][tex]\vec{\mu}[/tex][tex]\bullet[/tex][tex]\hat{e}[/tex][tex]\epsilon[/tex](t)dt


    (super and sub scripts seem to be a little messed up, sorry)

    where mu is the induced dipole moment between ground and excited states, e is the polarization vector and epsilon is the magnitude of the pulse envelope at time t. To be honest, I'm so confused by this statement I don't even know what to ask. How does this define the pi pulse? If pi is the unknown (the pulse) then how do we know epsilon? Any assistance is greatly appreciated.

    I'm sure this is a foolish question but i'm a little out of my comfort zone with this project.
     
    Last edited: May 24, 2010
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