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Confusion about preparing ensemble of particles and no-cloning theorem

  1. Jan 20, 2010 #1
    To get a distribution of some dynamic variable of a wavefunction, we actually need to prepare an ensemble of particles, in which all the particles have the same wavefunction, right?
    And no-cloning theorem states that it's impossible to copy an unknown quantum state.
    So is this a contradiction?
    I've figured that in order to prepare an ensemble we have to know the wavefunction, so it doesn't really fit the situation under which the no-cloning theorem applies. But I can't really give myself a clear and detailed reasoning to convince myself. Could you guys help me explain it?
     
  2. jcsd
  3. Jan 20, 2010 #2
    Maybe I was not specific enough, let's start with: what's the essential difference between preparing an ensemble and copying an unknown quantum state?
     
  4. Jan 22, 2010 #3
    Emm...still no one, that's strange, is there something wrong about my question, or the way I asked?
     
  5. Jan 22, 2010 #4

    f95toli

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    I am not sure I understand the question. An ensemble is something with statistical properties, usually described via a density matrix. There is no reason to assume that all the individual systems in the ensemble have the same wavefunction (in fact, one usually assumes that this is NOT the case).
    Making a measurement on an ensemble is usually equivalent to repeated measurements on a single system, i.e. the ergodic hypothesis.
     
  6. Jan 22, 2010 #5
    Well, seems I was under the false impression that all particles in an ensemble should have a same wavefunction, thanks for clarifying. But then I can't understand ensemble interpretation, I always thought to verify the Born's statistical interpretation, we need to prepare exactly the same wavefunctions.
     
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