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Can we model classical world using quantum laws?

  1. May 24, 2015 #1

    zonde

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    In QM there is no cloning theorem that says "we can't clone or erase quantum state". But in classical macro world we can copy information and erase it.
    If we should describe classical information as a very complex quantum state then we shouldn't be able to copy or erase it, right? And obviously we can do that.

    Any help with analysis in this direction?
     
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  3. May 24, 2015 #2

    atyy

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    I think it's because classically distinct nacroscopic states are orthogonal, for which the no cloning theorem fails.

    I'm not entirely convinced, but this is what Nielsen and Chuang suggest, as well as Zurek http://arxiv.org/abs/1412.5206 (see the section REPEATABILITY AND QUANTUM JUMPS on p2). This is consistent with ideas like the von Neumann projection postulate and decoherence, since those yield orthogonal states.
     
  4. May 25, 2015 #3

    Strilanc

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    The no cloning theorem stops you from making independent un-entangled copies, like an operation that takes any state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle) \otimes \left| 0 \right\rangle## and turns it into the state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle) \otimes (a \left| 0 \right\rangle + b \left| 1 \right\rangle) ##.

    The no cloning theorem does not stop you from making entangled copies, like the operation that takes any state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle) \otimes \left| 0 \right\rangle## and turns it into the state ##a \left| 00 \right\rangle + b \left| 11 \right\rangle##.

    Classical copies are possible because they correspond to entangled copies.

    (The reason the entangled copies don't act oddly, like bell pairs, is that there are massively redundant copies (instead of just 2). Also, thermodynamics scatters the copies all over the place, preventing you from collecting them all in practice.)
     
  5. May 25, 2015 #4

    atyy

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    @Strilanc: is my proposed answer essentially the same as yours, since that ##|0\rangle## and ##|1\rangle## must be orthogonal? Are they the same argument, with one being in Copenhagen, and the other being in MWI?
     
  6. May 25, 2015 #5

    Strilanc

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    Yes, it's the same argument but phrased differently.

    I was trying to be concrete about things that were and were not allowed, instead of giving the abstract reasons they weren't allowed. The shades of MWI comes from thinking about QM in terms of quantum computing. They end up sounding pretty similar because a controlled-not gate toggling an otherwise unused qubit acts exactly like a measurement.
     
  7. May 25, 2015 #6

    zonde

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    Thanks atty, Strilanc for your answers.
    I would like to ask question about no cloning theorem.
    If we have two states with the same eigenbasis and the same phase they are entangled states. But then we don't have meaningful definition what is independent clone of the state - such idea simply does not exist in QM. Right?
     
  8. May 25, 2015 #7

    Strilanc

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    I'm not sure what you're asking.

    Quantum states don't have a basis, they're represented with a basis. You can change the basis of representation easily, without changing anything about the underlying state.

    Observables have an associated basis, made up of eigenvectors, but the observable you choose doesn't affect the amount of entanglement present (though it can affect the correlations that you measure, of course).

    I gave a meaningful definition of an independent clone in my response. The state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle) \otimes (a \left| 0 \right\rangle + b \left| 1 \right\rangle)## has two unentangled independent copies of the state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle)##. No matter how you change the basis, you'll find you can represent this state as a tensor product of something with itself. It can always be written in the form ##(x \left| x \right\rangle + y \left| y \right\rangle)^{\otimes 2}##. (Well... any single-qubit basis. Obviously multiple qubit basises, like the bell basis, can result in arbitrary expressions.)
     
    Last edited: May 25, 2015
  9. May 25, 2015 #8

    zonde

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    Sorry for sloppy language.
    But let me use your definition to reformulate my question. In order to represent two states as product they can't be coherent. But now how we can talk about two identical states being not coherent (by themselves)?
     
  10. May 25, 2015 #9

    Strilanc

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    ... with the tensor product?

    Your confusion might be due to focusing too much on the word "identical", instead of the mathematical definitions. Identical in this context does not mean "will be observed as having the same value". It doesn't mean entangled. Identical in this context means "are described with the same state". It means cloned.

    Two cloned qubits will both be described as being in the state ##(a \left| 0 \right\rangle + b \left| 1 \right\rangle)##. They will have identical probability distributions for measuring any single-qubit observable you want to try, but you will gain no information about one by measuring the other (if you already knew the pure state).
     
  11. May 25, 2015 #10

    zonde

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    Yes, they shouldn't be entangled. But I corrected this when I reformulated my question.

    Okay, two wavefunctions are not necessarily entangled but can they evolve in time differently so that they have changing complex phase between them and be incoherent? If they can't then they can't be described by tensor product.
    Let's say if you let the two particle beams interact you will not observe interference when their joined wavefunction is described by tensor product. On the other hand if you observe interference their joined wavefunction can't be described by tensor product.
    Now if we would take two perfectly but classically identical lasers (no phase drift between them) they would produce interference. Do we have in this case something more than we would expect from cloned wavefunctions?
     
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