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Time Evolution of Wave Function

  1. Sep 12, 2011 #1
    I'm getting bogged down in what is probably a very basic subject and it's holding me back. I'm not really sure how to determine the wave function [itex]\psi[/itex](x,t) given a function [itex]\psi[/itex](x,t=0); and since this is pretty much the under-pinning of every homework problem I've seen so far it's a huge issue for me. Can anyone explain, at least generally, how I get from one to the other?
     
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  3. Sep 12, 2011 #2

    vela

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    The Schrodinger equation tells you how the state evolves over time. Your textbook should cover this.
     
  4. Sep 12, 2011 #3
    The book does, but I'm coming up short with applying it beyond the limited example in the text.

    So I understand (at least intellectually) for a particle in an infinite square well (between say 0 and a) I find the normalization constant and then the nth coefficient via integration from 0 to a. How do I go from this simple situation to the more general free particle of mass m? Is it simply a matter of adjusting my limits of integration?

    (I promise, I'm not trying to get something for nothing with this question)
     
  5. Sep 12, 2011 #4

    vela

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    Sorry, it's not clear at all to me what your question is. Could you please elaborate?
     
  6. Sep 12, 2011 #5
    Sorry, my own lack of clarity isn't helping...

    The general solution can be found as a superposition of eigenfunctions

    [itex]\psi[/itex](x,t)=[itex]\sum[/itex]cn[itex]\psi[/itex]ne-i/[itex]\hbar[/itex]Et

    how do I apply this to a given situation? If I start with [itex]\psi[/itex](x,0)=ei/h px , I should arrive at [itex]\psi[/itex](x,t)=ei/h (px-p2/2m t)

    how did I get from point a to point b?
     
  7. Sep 12, 2011 #6

    vela

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    The Hamiltonian for a free particle is [itex]\hat{H}=\hat{p}^2/2m[/itex]. Apply it to [itex]\psi(x,0)[/itex], which is an eigenstate of [itex]\hat{H}[/itex], to find E.
     
  8. Sep 13, 2011 #7
    Hmm, I agree with your explanation of Time Evolution of a Wave Function. I feel that any other explanation would be Shallow and Pedantic. Are you refering to a partical in a square well?
     
  9. Sep 13, 2011 #8
    You are already there. If the time-independent Hamiltonian is [itex]H(\hat{p},\hat{x})[/itex] and its eigenstates are [itex]\varphi_n[/itex] with energy [itex]E_n[/itex]. Then each will evolve in time by only acquiring a phase factor related to its energy [itex]\varphi_n \to \varphi_n e^{-it E_n}[/itex].

    So suppose you state start out at [itex]t=0[/itex] as some linear superposition of [itex]\varphi_n[/itex]: [itex]\psi (0) = \sum_{n} c_n(0) \varphi_n[/itex], at time t later. Each of the states evolves independently: [itex] \psi(t) = \sum_{n} c_n(0) e^{-i t E_{n}} \varphi_n[/itex]. Or another way of stating this, the coefficients of linear superposition change like this [itex]c_n(0) \to c_n (0) e^{-i E_n t}[/itex]

    The job is therefore to expand any general initial state into linear combination of energy eigenstates of the Hamiltonian.
     
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