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Question about schroedinger equation

  1. Jun 8, 2004 #1
    pliz help me how to solve the schroedinger equation from the time dependent to become the time independent?

    thank you
     
  2. jcsd
  3. Jun 8, 2004 #2
    You'll have to excuse any Latex errors I haven't quite worked out what Tex this board uses yet. anyway:
    [tex]
    $$
    \frac{{ - \hbar ^2 }}
    {{2m}}\frac{{\partial ^2 \Psi (x,t)}}
    {{\partial x^2 }} + V(x)\Psi (x,t) = i\hbar \frac{{\partial \Psi (x,t)}}
    {{\delta t}}
    $$

    \]

    [/tex]

    Assume seperable:
    [tex]$\Psi (x,t) = \psi (x)\phi (t)$
    [/tex]
    Substitute:
    [tex]$$
    $$
    \frac{{ - \hbar ^2 }}
    {{2m}}\frac{{\partial ^2 \psi (x)\phi (t)}}
    {{\partial x^2 }} + V(x)\Psi (x,t) = i\hbar \frac{{\partial \psi (x)\phi (t)}}
    {{\delta t}}
    $$

    [/tex]
    Take terms out of x derivate that are t-dependant, and vice-versa - also divide by [tex]$\psi (x)\phi (t)$
    [/tex]:
    [tex]$$
    {{ - \hbar ^2 } \over {2m}}{1 \over {\psi (x)}}{{\partial ^2 \psi (x)} \over {\partial x^2 }} + V(x) = i\hbar {1 \over {\phi (t)}}{{\partial \phi (t)} \over {\delta t}}
    $$
    [/tex]

    Set each side equal to a costant:
    [tex]$$
    \hbar i{1 \over {\phi (t)}}{{d\phi (t)} \over {dt}} = E
    $$
    [/tex]
    This yields:
    [tex]$$
    \phi = Ae^{ - i\omega t}
    $$
    [/tex]
    The other equation, becomes (upon rearranging):
    [tex]$\frac{{ - \hbar ^2 }}
    {{2m}}\frac{{d^2 \psi (x)}}
    {{dx^2 }} + V(x)\psi (X) = E\psi (x)$

    [/tex]
    There is a quick sketc. However, it is a worthwhile problem to do yourself - seperation of variables is a very useful technique!
     
    Last edited: Jun 9, 2004
  4. Jun 9, 2004 #3
    Basic Separation of Variables

    HI

    You have not mentioned the situation for which you want the solution. So I am assuming that you are interested in getting the standard one,

    [tex]
    \nabla^2 \psi + \frac{8\pi^{2}m}{h^{2}} (E-V)\psi = 0
    [/tex]

    Here, E is the total particle energy, V is the potential energy, psi is the wavefunction for the particle, m is its mass, h is the Planck Constant.

    Steps

    We wish to write the wave equation for a particle, so generally, it is

    [tex]
    \nabla^2 \psi = \frac{1}{v^{2}} \frac{\partial^2{\psi}}{\partial{t}^2}[/tex]

    This is the starting point for all our computations. Now, in order to separate the time dependent parts (containing t) and the time-independent parts (containing x,y,z) we perform a technique called Separation of Variables. It is more generally used to solve partial differential equations.

    We write the total wavefunction as a product of two wavefunctions--one dependent only on spatial coordinates (x,y,z) and not on time and the other dependent only on time and not on (x,y,z).

    [tex]
    \psi_{(xyzt)} = \psi_{(xyz)}g(t)
    [/tex]

    Now clearly g(t) must be solely time dependent, so for the sake of convenience, we usually write

    [tex]
    g(t) = g_{0}e^{2 \pi i \nu t} \qquad \mbox{where i = \sqrt{-1}} \qquad \mbox{g_{0} = const.}
    [/tex]

    Note that g(t) is the most general complex function depending only on time.

    Substitute the (assumed) expression for the total wavefunction (including g(t) as a product) in the wave equation. Note that the laplacian is a space operator only (contains only space coordinates) and so g(t) remains unaffected by it. After a little bit of algebraic manipulations which should now be clear, you get

    [tex]
    \nabla^{2}\psi_{(xyz)} = -\frac{4 \pi^{2}\nu^{2}}{v^{2}}\psi_{(xyz)}
    [/tex]

    If the particle analog of the expression,

    [tex]
    c = \lambda \nu
    [/tex]

    that is,

    [tex]
    v = \lambda \nu = \frac{h\nu}{p}
    [/tex]

    is substituted for v in the above wave equation, it becomes

    [tex]
    \nabla^{2}\psi_{(xyz)} = -\frac{4 \pi^{2}p^{2}}{h^{2}}\psi_{(xyz)}
    [/tex]

    The linear momentum of the particle p, is related to the kinetic energy T as

    [tex]
    T = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m}
    [/tex]

    Substituting for p into the equation and writing the kinetic energy T as (E-V) where E is the total energy of the particle and V is its potential energy, the wave equation becomes

    [tex]
    \nabla^2 \psi + \frac{8\pi^{2}m}{h^{2}} (E-V)\psi = 0
    [/tex]

    (This psi depends only on space coordinates, the (xyz) subscripts dropped.)

    This is the time independent form of the Schroedinger Wave Equation. The algorithm for obtaining it is stated thus:

    1. Consider the general wave equation in terms of the velocity of the particle.
    2. Write the total wavefunction as a product of two wavefunctions, one containing space coordinates only and the other containing time only. Simply to remove the time dependence.
    3. Use De-broglie's Relation to introduce particle character.
    4. Write T in terms of p.
    5. Replace T by E-V.
    6. Bingo!

    Hope that helps...

    Cheers
    Vivek
     
    Last edited: Jun 9, 2004
  5. Jun 9, 2004 #4
    thank you maverick280857
     
  6. Jun 9, 2004 #5
    thank you heardie
     
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