Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normalising wavefunctions

  1. Aug 5, 2015 #1
    ψ = Ae-kx2 ; A = normalisation constant
    For normalising,
    -infinf A2ψ°ψ dx = A2M (say) = 1
    so we put A = 1/√M
    My question is why we need 'A' ??

    The thing is either we find a particle or we do not and if we think of a simple waveform...'A' gives the amplitude part...so can we put it in this way - we are altering the amplitude of the wavefunction to make the probability of finding the particle 1..?? If it is so, how altering the amplitude can do it physically..(cannot form the mental picture clearly)..
    Last edited: Aug 5, 2015
  2. jcsd
  3. Aug 5, 2015 #2
    You made a math error. Double check your substitutions.
  4. Aug 6, 2015 #3
    Didn't do the integration at first so took the value to be 'M' .
    The integration gives A = √(2k/π) . So we still have to put this value to normalise the wavefunction ....right...??? Why ...??
  5. Aug 6, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper

    You don't alter the amplitude of a wavefunction physically in your experiment to make it representable as a probability. We have hypothesized that the modulus square of the expansion coefficient of a wavefunction in some basis represents the probability upon measurement to find the state in the basis state associated with that expansion coefficient. Therefore the sum of all modulus squared of the expansion coefficients must yield unity, and one can prove that if the representing basis states are orthonormal, the inner product of the system's wavefunction being expanded must be unity, in order to satisfy the assumption that the expansion coefficient represent a probability. What you do in you calculation is to adjust the coefficient A such that the ψ(x) can be interpreted as a probability.
  6. Aug 7, 2015 #5
    thanks...much clearer now...:smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook