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Hamiltonian in experiments

  1. Jul 2, 2004 #1
    Let say I have prepared two identical particle, both describable by a wavefunction Psi, whereby,
    Psi = a*1 + b*2, where, 1 and 2 are two stationary wavefunctions.

    If I perform an experiment to find out the systems' energy, this is equivalent to operating a Hamiltonian on Psi. Operating,
    HPsi = H(a*1 + b*2) = a*E1*1 + b*E2*2
    where, E1 is eigenvalue with eigenfunction 1,
    E2 = eigenvalue with eigenfunction 2.

    That means, I might get energy = E1 for the first particle from the experiment, and
    energy = E2 for the second particle.

    How can we get two different energy value E1 and E2 when I prepared both the particles exactly the same and both have the same wavefunction. So they must give me the same energy.
    Otherwise, where does the energy difference E1-E2 come from?"
  2. jcsd
  3. Jul 3, 2004 #2
    Not quite... If you perform an experiment to find out the system energy, you're
    measuring the energy. Therefore, you're projecting the state into the the corresponding eigenspace.

    You will measure E1 with probability |a|^2 and E2 with prob. |b|^2 and the final state will be either |1> or |2>.

    This comes from the way you prepared the state. The fact that the wave function is a superposition of both states means that you're not sure in which energy state it is. For example, you could have shone a laser on an atom trying to put it into an excited state, but you're not sure whether a photon was absorbed...

    I hope this will help you understand a little bit more of QM.
    Best regards
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