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Complete set of stationary solutions

  1. Apr 19, 2013 #1
    In the infinite square well, the stationary states solutions form a complete set, and therefore I can write a function such as ( f(x) = -x^2 + x ) as an infinite sum of them,

    But this function is, clearly, not a solution of S.E., although it's written as a sum of solutions.

    Why is it not a solution? is it because the sum is infinite or what?
  2. jcsd
  3. Apr 19, 2013 #2
    The eigenfunctions of the infinite square well form a complete basis only for functions which have compact support within the range of the well. In other words, you can only make functions which are zero outside of the well (since all of the eigenfunctions will be zero outside of it, there's no way you can add them up to get a nonzero number). Since the function you gave is nonzero outside of the well, it can't be represented as the sum of square well eigenfunctions.
  4. Apr 19, 2013 #3


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    The function [itex]f:[0,1]->\mathbb R[/itex], [itex]f(x) = -x^2+x[/itex] can definitely be expanded in square well eigenfunctions, but it still isn't a solution to the Schrödinger equation.

    The answer to the question is that only because two functions are eigenfunctions of some operator, their sum doesn't need to be an eigenfunction anymore (unless they lie in the same eigenspace). I suggest you try it yourself with a simple matrix like [itex]M = \mathrm{diag} (1,2)[/itex]. However, you can use the function [itex]f[/itex] as initial condition at [itex]t=0[/itex] for the time-dependent Schrödinger equation and get a completely valid solution.
  5. Apr 19, 2013 #4
    Whoops, you're right, I misread what the question was asking.

    Sounds like the issue here is confusion between the time-independent and time-dependent Schrodinger equations. The former is not really a differential equation so much as a family of differential equations: [itex](-\frac{\hbar^2\nabla^2}{2m} + V(x))\psi(x) = E\psi(x)[/itex], for various values of [itex]E[/itex]. The square well solutions form a complete basis for describing functions over that range, but each function has a different eigenvalue, so adding together functions with different eigenvalues does not produce another solution to the TISE.

    On the other hand, the time-DEPENDENT equation is [itex]i\hbar\frac{\partial}{\partial t}\psi(x,t)=(-\frac{\hbar^2\nabla^2}{2m} + V(x))\psi(x,t)[/itex]. This is a single linear differential equation, not a family of equations, so you can add together any number of solutions to this, and the result will still be a solution. So you can take your parabola and decompose it into solutions of the TISE (this is basically a Fourier transform), and then add those all up and use them in the TDSE to determine the system's time evolution.
  6. Apr 20, 2013 #5


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    It's a simple linear algebra problem. Suppose you have an eigenvalue equation with two solutions (eigenvectors and -values; i=1,2)

    (H - Ei) ui = 0

    What about

    v = a1u1 + a2u2

    We apply H

    Hv = H (a1u1 + ...) = a1E1u1 + ...

    but for different E1 ≠ E2 this is no longer an eigenvector of H
    Last edited: Apr 20, 2013
  7. Apr 20, 2013 #6


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    The way it works is this: If [itex]\psi_n(x,t)[/itex] is a stationary state, that doesn't really mean that it is time-independent, it means that the time-dependency is very simple:

    [itex]\psi_n(x,t) = \psi_n(x,0) e^{- \omega_n t}[/itex]

    where [itex]\omega_n[/itex] is the characteristic frequency of state [itex]\psi_n[/itex].

    Now, if you express an arbitrary function as a combination of stationary states, you can do it at a single moment (say t=0):

    [itex]f(x) = \sum C_n \psi_n(x,0)[/itex]

    But this [itex]f(x)[/itex] is not a stationary state, because if you initially choose a wave function [itex]\Psi[/itex] with [itex]\Psi(x,0) = f(x)[/itex], and you wait a while, the wave function will change to:

    [itex]\Psi(x,t) = \sum C_n \psi_n(x,0)e^{- \omega_n t}[/itex]

    That's not stationary, except in the special case where all the [itex]\omega_n[/itex] have the same value.
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