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Error in Griffiths QM: completeness of stationary states

  1. Aug 15, 2013 #1
    According to Griffiths QM book, after he derived the stationary state solutions to the Schrodinger equation for a particle in an infinite potential well, which are just functions of sine, he claims that these stationary solutions are orthogonal and complete.

    I agree that they are orthogonal (since sin(nx) and sin(mx) are orthogonal for n!=m), but I definately disagree that they are complete (meaning that ANY function f(x), odd or even, can be written in terms of these stationary states). For example, you definatelly cannot write f(x)=cos(x) using only a basis of sine functions.

    Is there something I am missing here???



    Thanks!
     
    Last edited: Aug 15, 2013
  2. jcsd
  3. Aug 15, 2013 #2

    jtbell

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    Staff: Mentor

    Therefore the well has its left side at x = 0, correct?

    If the left side of the well is at x = 0, can cos(kx) be a valid solution?
     
  4. Aug 15, 2013 #3
    nope! cos cannot be a solution. I realized that maybe I am being too picky, but Griffiths' exact words are "The functions ψ_n (x) [where ψ_n (x) are the solutions to the infinite potential well] have some interesting properties...4. They are complete, in the sense that any other function f(x) can be expressed as a linear combination of them"

    He goes on to say that this is Dirichlet's theorem.

    By "any other function", I took it to literally mean any other function, not any other function that could possibly represent the wavefunction of a particle in infinite potential well.
    Having studied Fourier analysis in detail, this is definately not how I have seen the word "complete" and Dirichlet's theorem defined...
     
  5. Aug 15, 2013 #4
    Well Griffiths is being a little loose and he admits it when he comes to his discussion of the completeness of stationary states.

    He is solving the infinite square well in the region (0, a). If you look at a good reference on Fourier series (For an intro QM level, I would recommend Mary Boas' "Mathematical Methods in the Physical Sciences", chapter 7 in the 3rd edition), you'll see that for functions which are periodic on an interval (0,a), the functions sin(nπx/a) form a basis for functions which are odd on that interval, while the functions cos(nπx/a) form a basis for even functions on that interval, and putting them together you can see that the functions e(inπx/a) form a complete basis for arbitrary functions (remembering all these functions must have period a).

    The reason he forgets about the cos(nπx/a) terms is because the boundary condition ψ(x=0)=0 ensures that the solutions to the Schrodinger equation must be odd. Don't worry that the wavefunction doesn't really "continue" outside the well--just imagine that the solution repeats itself with period a, while respecting the boundary condition. [This is allowed because the potential is infinite outside the well.] So the sine functions he mentions form a complete basis for solutions to the Schrodinger equation since they must be odd with period a.

    As a general recipe, you can represent pretty much any function in terms of sinusoidal functions with arbitrary period
    f(x) = ∫ c(k) e(ikx) dk = ∫a(k) cos(kx) + b(k) sin(kx) dk

    But if we limit to functions with period a, then we don't need any of the contributions from k≠nπ/a
    thus if f(x)=f(x+a) for all x
    then f(x) = Ʃn an cos(nπx/a) + bnsin(nπx/a)

    For odd functions, you get no cos contributions, while for even functions, you get no sin contributions.

    Edit: The font that PF appears in on my computer makes the n's look very similar to the π's. Sorry for that.
     
    Last edited: Aug 15, 2013
  6. Aug 16, 2013 #5

    tom.stoer

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    For a well-defined statement one has to specify the space (including boundary conditions) on which the Hamiltonian is self-adjoint and for which completeness shall hold
     
  7. Aug 16, 2013 #6

    Demystifier

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    Inside the potential well, you can make a Fourier series out of sin(nx) functions, equal to cos(x) EVERYWHERE in the well except at the boundary.
     
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