Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physical meaning of the 2 eigenfunctions of Free Particle

  1. Feb 11, 2008 #1

    cks

    User Avatar

    For Schrodinger's equation

    [tex] \frac{\d^2\psi}{dx^2} = - \frac{2mE}{\hbar^2}\psi [/tex]

    Solving to find that

    [tex] \psi = Aexp(ikx)+Bexp(-ikx) [/tex]

    I am curious about the physical meanings of the two terms of the solutions.

    In solving a free particle encountering a potential barrier, In the region before the encounter of the barrier, the solutions of the Shcrodinger equation is just the free particle equation above. My teacher says the term with the positive sign means it's a wave going towards the barrier, whereas the negative signs is the wave that reflect from it.

    Well, the wave function is just the solution of the Schrodinger's equation and how does my teacher derives the physical meaning from it?? I mean the exponential function has complex term, which are actually sinusoidal but doesn't tell us anything about the direction of going??
     
  2. jcsd
  3. Feb 11, 2008 #2

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That is because the operator of MOMENTUM, being P = hbar/i d/dx, gives you, applied to the first term:

    P {exp(ikx) } = hbar k exp(ikx), meaning that the first term is an eigenfunction of the momentum operator with eigenvalue hbar k.

    P {exp(- ikx) } = - hbar k exp(-ikx), meaning that the second term is an eigenfunction of teh momentum operator with eigenvalue - hbar k.

    So the first wavefunction represents also a state with momentum + hbar k,
    while the second wavefunction represents a state with momentum - hbar k.
     
  4. Feb 11, 2008 #3

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Although I wouldn't worry about it for your course, you should perhaps at least be aware that your solution for a free particle is not the true solution, since it isn't square integrable. You'll see why if you do more advance QM courses.
     
  5. Feb 12, 2008 #4

    cks

    User Avatar

    I see. The reason that a momentum operator acts on the eigenfunction produces a positive eigenvalue means its direction is towards the barrier.

    thanks .
     
  6. Feb 12, 2008 #5

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, it means that your momentum is a positive number. Depends then on how your axes are defined and where the barrier is of course...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Physical meaning of the 2 eigenfunctions of Free Particle
  1. The free particle (Replies: 6)

Loading...