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Calculating Cn for general wave solutionbasic Quantum physics

  1. Sep 17, 2006 #1
    This is for problem 2.5 in griffiths text

    basically they say that the wave function as Capital Si (x,0) = A (si1(x) + si2(x), where si1 and si2 are equal to the first two stationary states of an infinite square well.

    i need to get Cn..my normalization constant is sqrt(2)/2..and i'm using the same formula that they use to get it in example 2.2

    so I use equation 2.37

    Cn = sqrt (2/a) * int (sin (n*pi*x/a) * Si (x,0) dx, 0, a)

    into this equation i insert (Sqrt(2)/2) * si (x,0) which was given in the initial question as

    si (x,0) = A(si1(x) + si2(x)) where si1 and si2 are the solutions to the first two stationary states of the infinite square well

    these should be si1 = sqrt (2/a) * sin (pi*x/a) and
    si2 = sqrt (2/a) * sin (2*pi*x/a)


    anyways, i'm wondering if I"m doing it write up to here..because in the next step i'm not sure how to simplify the equation to integrate easily so i just plug it into the calculator.

    this gives me a function of n as is used in example 2.2 to get Cn...however, my equation has sin n*pi multiplied by the rest of the equation..the problem is that sin n*pi is always equal to zero for any integer n. Therefore i have no Cn.

    Obviously i'm doing something major wrong. So i was hoping to get some help on this. I've solved problem 36 ok since it doesn't require much.

    2.6 i haven't really tried yet since squaring the initial wave function looks like a beast since it looks like a pain to square (assuming you turn e^itheta into cos theta + isin theta), though i will get to it soon hoepfully..

    2.37 i also tried though..and normalizing it was ok (sqrt(16/5)), but i again was unable to get Cn..my calculator when giving the integral solution again leaves me with sin npi as a major factor..so i wonder if i'm plugging things in wrong.

    Anyways, getting Cn is a major factor in these problems so i could really use some help:\
     
  2. jcsd
  3. Sep 17, 2006 #2
    What do you know about eigenstates of the infinite square well? In particular, there is a nice property that makes this problem trivial. Try writing the product of wave function out, as
    [tex] \Psi^{\ast}\Psi = (\psi_1 +\psi_2)^{\ast}(\psi_1+\psi_2) [/tex]
    You should find this simplifies the calculation since the wave functions are orthogonal, which is what I'm referring to above. Another way to write this (though you probably haven't seen this yet) is in Dirac notation.
    [tex] \langle \Psi \mid \Psi \rangle = ( \langle 1 \mid + \langle 2 \mid)( \mid 1 \rangle + \mid 2 \rangle ) [/tex]
    where the bra's and ket's denoted with 1 and 2 are the corresponding stationary states of the infinite square well. Using this notation the problem is very easy to solve.
     
    Last edited: Sep 17, 2006
  4. Sep 17, 2006 #3
    my teacher told me that psi(x, 0) is fine with A = 1/sqrt(2)

    what is then needed is to write the time-dependent solution
    Psi(x, t) as A ( psi1(x) exp(-iE1 t/ hbar) + psi2(x) exp(-i E2 t /hbar) )

    using trigonometric identities you can then work out the
    probability density from the |Psi(x,t)|^2 and see how it oscillates with
    time.
    the integrals involved in c then only require trigonometric identities.


    however, this still doesn't tell me why i couldn't derive psi (x, t) from using the formula to get Cn.

    anyways, between your hint and the teachers i don't think doing the problem will be difficult now, though i'm still confused why what i was trying didn't work
     
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