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Normalizing a wavefunction

  1. Feb 19, 2014 #1
    I was reviewing the infinite square well, using D.J. Griffiths, and came across this small point of confusion. The time-independent solution is shown to be Asin(kx), where the constant A is determined by normalization. Then, in assembling the complete (time dependent) solution, he writes that the most general solution "is a linear combination of stationary states", where each stationary state is assigned a coefficient c_n. Griffiths then explains that |c_n|^2 is the probability of observing that state, and "the sum of all these probabilities should be 1."

    My confusion is this: if c_n is the probability of observing each state, why do we use the normalization requirement to find A? Isn't this in effect normalizing twice?
  2. jcsd
  3. Feb 19, 2014 #2


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    You only normalize once to find A. Once that's done, c_n gives the probability without any further normalization.
  4. Feb 19, 2014 #3
    Thank you, that makes sense!
  5. Feb 19, 2014 #4
    It is quiet convenient to work in orthonormal set of basis. Therefore you must normalize each element of basis set such that
    \int_0^l f_n(x)f_m(x)\,dx = \delta_{n m}
    where [itex] f_n(x) = A_n\sin(n\pi x/l)[/itex].

    The total statefunction (wavefunction) can be written as a linear combination of the elements of basis set.
    \Psi(x,t) = \sum_{n=0}^\infty c_n(t)f_n(x).
    Now quantum mechanics says the total probability probability should always be unity, so
    \int_0^l \Psi(x,t)\Psi(x,t)\,dx = 1,
    which implies
    \sum_{n=0}^\infty c_n^2 = 1.
    (note: I have assumed real valued wave functions. For complex case, replace with complex conjugate wherever necessary)
    Normalisation of the basis set just makes the mathematics easy (it is not a necessary requirement), but the normalisation of total statefunction is a constraint of quantum mechanics)
    Last edited: Feb 19, 2014
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