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Homework Help: How to find eigenfunctions given the wave function?

  1. Nov 24, 2007 #1
    Hi all,

    How do we find the eigenfunctions if we are given the wavefunction? I have a wave function at time = 0 and it is of a *free* particle and I need to find the wave function at a later time t. I did :

    [tex]\Psi(x,t)=\Psi(x,0)*e^{-iHt/hbar}[/tex] then
    [tex]\Psi(x,t)=\sum_{n}(<\phi_{n}|\Psi(x,0)> |\phi_{n}> e^{-iE_{n}t/hbar})[/tex]

    I have [tex]\Psi(x,0)[/tex] so the only thing I need to know is [tex]\phi_{n}[/tex] which are the eigenfunctions and I have no idea how to do this. I solved Schroginger's equation and got [tex]\Psi(x)=Ae^{ikx}+Be^{-ikx}[/tex], where k[tex]^{2}[/tex]=[tex]\frac{2mE}{hbar^{2}}[/tex], and the particle is totally free.

    thanks in advance.
  2. jcsd
  3. Nov 25, 2007 #2
    The question seems incomplete to me. They should specify the wavefunction [itex]\psi (x,0)[/itex] if they are asking for eigenfuntions. The time dependent Schroedinger's equation permits solution of the initial value problem. All you need is:
    [tex]\psi (\vec r ,t) = e^{-i\omega_n t}\phi_n (\vec r)[/tex]

    If it is the eigenfunction they have asked then you already have your answer.
  4. Nov 25, 2007 #3
    I just don't understand the difference between the wavefunction and eigenfunction and how to get one from the other.
  5. Nov 26, 2007 #4


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    The 'physical' thing is the wave function: this describes the state of the system, in so far as it will give you the probability of measuring a certain value for certain quantities.

    An eigenfunction is a special wave function. Namely, an eigenfunction [itex]\psi[/itex] of an operator A is a function such that [itex]A\psi = \lambda\psi[/itex] for some (complex) number [itex]\lambda[/itex], which we call the eigenvalue. For example, if the wave function of a certain system is an eigenfunction of the Hamiltonian, measurement of the energy will always return the same definite energy, namely the eigenvalue. If the system is in an eigenfunction of some other (observable) operator, applying that operator (measuring the quantity) will always give the associated eigenvalue.

    Now a nice mathematical consequence is, that the eigenfunctions form -- what we technically call -- a complete set. This means that we can write any wave function as a linear combination of eigenfunctions. You can compare it to Fourier decomposition: we can write any function f(x) as a sum over sines and cosines with ever increasing frequencies. Only this time we don't make a sum over sines and cosines, but a sum over eigenfunctions.
  6. Nov 26, 2007 #5


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    What CompuChip says, in summary:
    Wave function is a linear combination of eigenfunctions. It could be composed of only one, or many.

    HOW to get eigenfunctions to different operators depends on how the operator "looks" like.

    But how to get eigenfunctions to a operator from an arbitrary wave function you just decompose it.
  7. Nov 29, 2007 #6
    CompuChip- thanks for the explaination. It was great!

    malawi_glenn (or CompuChip): How can I decompose the wavefunction if I don't know how the eigenfunctions of the certain operator look like?
  8. Nov 29, 2007 #7


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    Then you work them out first.

    The eigenfunction for a free particle, is a plane wave.
  9. Nov 29, 2007 #8


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    Formally it is possible to do it though.
    Suppose that we have a wavefunction [itex]\psi[/itex] and that [itex]\phi_n (n \in \mathbf{N})[/itex] is our set of eigenfunctions of the Hamiltonian (for simplicity, I'll assume that they can be indexed by integers n, but if n runs through a continuous set then the argument is similar except there are integrals over dn instead of sums over n).
    We have an inner product [itex]\langle \cdot | \cdot \rangle, which takes two wavefunctions and returns a number. The decomposition in eigenfunctions is then just
    [tex]\Psi = \sum_n c_n \phi_n[/tex]
    where the coefficients are given by
    [tex] c_n = \langle \phi_n | \Psi \rangle \phi_n[/tex]
    Now if the wave-function depends on time as in
    [tex]\Psi = \Psi_0 e^{-i H t / \hbar}[/tex]
    we do the same:
    [tex]\Psi = \sum_n \langle \phi_n | \Psi_0 e^{-i H t / \hbar}\rangle \phi_n[/tex]
    and we can show that if we let the exponent work to the right, we get
    [tex]\Psi = \sum_n \langle \phi_n | \Psi_0 e^{-i E_n t / \hbar} \rangle \phi_n
    = \sum_n \langle \phi_n | \Psi_0 \rangle \phi_n e^{-i H t / \hbar} \phi_n[/tex]
    where I took the exponent outside the inner product because it's now just a number, and the En are the eigenvalues
    [tex]H \phi_n = E_n \phi_n[/tex].

    Recently, I was sort of accidentally flipping through the first chapters of D.J. Griffiths book on Quantum Mechanics, and noticed that he explains all this really well in the first chapters (also, where the time dependence factor comes from, why this special time dependence is not a loss of generality, etc).
    If you ever want to do anything with QM again, whatever it is, I strongly recommend you buy this book as it covers almost everything you'd learn in a first (and in my case, even second) course in Quantum Mechanics and he explains it quite clearly. I myself consult this book from time to time now, though it has been a couple of years since I took QM :smile:
  10. Nov 29, 2007 #9
    Cool, thanks. I'll see if the book is available in the library. Thanks alot again
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