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Energy of measurement

  1. Mar 1, 2010 #1
    Hi all,

    Consider a simple harmonic oscillator in the ground state. Measuring its energy yields the ground state energy with 100% probability. Now by measuring its position (thereby changing the wavefunction) and next measuring the energy again, there is now a nonzero probability of finding the particle in an excited state.

    (I've taken this example from
    http://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics#Example )

    If the particle has a probability of being found in a higher energy state, does this mean that this energy was somehow 'supplied' to the particle by measuring (i.e. sending photons to it to observe where it is), or just that the particle always had this energy, but by being prepared in the groundstate there was no way of measuring it?
     
  2. jcsd
  3. Mar 1, 2010 #2
    Hi.
    Measuring the position makes energy uncertain, I think.
    From uncertainty relation, delta x=0 causes delta p=infinity thus delta p^2=2m delta kinetic energy=infinity.
    Regards.
     
    Last edited: Mar 1, 2010
  4. Mar 2, 2010 #3
    Yes. To measure the position as accurately as possible, one must use the photon (let's take photon as example, maybe one can use other method to determine the position) with high frequce. As a result, the energy the measurement take into the oscillator is
    naturely high.
     
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