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Odd question

  1. Jan 19, 2010 #1
    friends,now, this may sound strange , but still......
    in my text while solving one dimension problems like barriers, step potential etc i noticed that they use a free particle eigenfunction coming from left ,. now why that is so did? i mean , if i need to describe a moving particle , will i not require time dependent scroedinger equation?
    yet, as far as i now, these are eigenfunctions of a time independent equation ie. momentum eigenfunctions.
    also, such eigenfunctions have actually infinite spread in posision space. so we always have a definite probability beyond the barrier?
    any help will be highly appreciated.
  2. jcsd
  3. Jan 19, 2010 #2
    In quantum mechanics we consider a particle to be a wave packet wich has finite width in both coordinate and momentum space. It must satisfy the time dependent Schrodinger equation:

    \hat{H}\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)
    [/tex] --- (1)

    In order to describe the wave packet propagation we should use Fourier method.
    (We can do that because of Schrodinger equation linearity)

    Let's represent the wave function as following expansion:

    \Psi(x,t) = \sum_k C_k \psi_k(x)\phi_k(t)
    [/tex] --- (2)

    Substitution of (2) into (1) yields

    \sum_k C_k \psi_k(x)\phi_k(t)
    \left[\frac{\hat{H}\psi_k(x)}{\psi_k(x)} - i\hbar\frac{\phi_k'(t)}{\phi_k(t)} \right] = 0
    [/tex] --- (3)

    In order to fit this equation for all [tex](x,t)[/tex] the following system should be satisfied for any [tex]k[/tex]:

    \hat{H}\psi_k(x) = E_k\psi_k(x).
    [/tex] --- (4a)

    i\hbar\frac{\partial}{\partial t}\phi_k(t) = E_k\phi_k(t);
    [/tex] --- (4b)

    (That's right only for time independent Hamiltonian)

    (4a) is well-known steady-state Schrodinger equation. (4b) produces

    \phi_k(t) = \exp\left(-i\frac{E_k}{\hbar}t\right).
    [/tex] --- (5)

    Thus the procedure of describing a moving particle is following:
    1) Calculate [tex]E_k[/tex] and [tex]\psi_k(x)[/tex] which are eigenvalues and eigenvectors of the Hamiltonian.
    2) Construct the eigenfunction expansion for [tex]\Psi(x,0)[/tex] (the initial condition). At this stage we will get coefficients [tex]C_k[/tex].
    3) Calculate [tex]\Psi(x,t)[/tex] using (2), (5) and [tex]C_k[/tex].

    If only transmission and reflection coefficients are required there is no need to watch the whole wave packet. It's enough to consider it's average energy and the corresponding wave function. This function behaves like free particle wave function far from the barrier where the potential is constant.
  4. Jan 19, 2010 #3


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    In deed, the situation is more subtle.
    The reason, that we can speak at all about the scattering of a particle being free in the infinite past and future from some finite range potential lies in the fact that a wavepacket will spread out more and more in time so that its overlapp with the scattering potential will tend to zero ultimately. This should be laid out in any book on scattering theory (e.g. Roger G. Newton, Scattering theory of waves and particles, Springer) or advanced QM (Ballentine?).
  5. Jan 20, 2010 #4
    thanks for the help,friends.

    ok, i asked this to my course instructor , who said that the calculasion of reflection and transmission coefficients yields the same values if i took steady functions and also if i took it as a gaussian wave packet, but it must have a sharp graph in momentum space ie. peaked at a specificated momentum.
    well, in a book , i found that procedure but one thing appeared strange to me. while constructing the propagator to evolve the gaussian incident packet, the book is using only eigenfunctions which satisfy E>barrier potential but not those for which E<V, as a reason it says that those eigenfunctions will not give a momentum close to incident momentum in momentum space. but i cannot understand why? is this just anticipation / can be understood beforehand???
  6. Jan 20, 2010 #5


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    Science Advisor

    Is this a semi-infinite barrier or a step?
  7. Jan 20, 2010 #6
    it is a step. like has value V for positive x and 0 for negative x.
    am i clear now, DRDU?
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