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Why is the wave function not measurable alone?

  1. Apr 1, 2012 #1
    why is the wavefunction not measurable as it is, but is measurable when the square of the absolute value is taken?

    Thank you
  2. jcsd
  3. Apr 1, 2012 #2

    The wavefunction represents the probabilities of finding a particle in specific positions. So, let's say we have a particle that, upon observation, can express one of two traits, we'll call them Y and Z. Since it is in a superposition of both, we would have to write this as: [tex]\psi =\alpha (Y)+\beta (Z)[/tex]Where [itex] \alpha [/itex] and [itex] \beta [/itex] are known as probability amplitudes. So, we write that the probability of finding the particle in state Y is [itex]\left | \alpha \right |^{2}[/itex]

    A normalized wavefunction, [itex] \psi [/itex] is a probability amplitude. If we are talking about a position x, this is also a probability amplitude. So, to write the probability of a wavefunction collapsing to the position x we must express it as [tex]\left | \psi (x) \right |^{2}[/tex] See here for more: http://en.wikipedia.org/wiki/Probability_amplitude
    Hope I helped!
  4. Apr 2, 2012 #3
    I think that because of diffraction in QM,we must guest the wavefunction as probability amplitude but square of wavefunction is probability.Then the wavefunction is not measurable.
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