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Schrodinger equation dedution step?

  1. Mar 7, 2013 #1
    schrodinger.png

    My chemistry teacher showed this slide but didn't explained how the marked step in the dedution of the schrodinger equation. Please help! I really don't know where I can find the answer, it's the first time I see the Schrodinger equation!
     
  2. jcsd
  3. Mar 7, 2013 #2

    bhobba

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    It comes from replacing the E in the first equation by the energy operator. Check out:
    http://www.eng.fsu.edu/~dommelen/quantum/style_a/ops.html

    As to why - that's a very deep question. It has an answer, and a mathematically very beautiful one to do with symmetries, but for a first brush with QM its a bit too advanced. At this stage its simply best to accept it. If you want to check it out see Chapter 3 - Quantum Mechanics- A Modern Development by Ballentine - but for now I wouldn't worry about it.

    Thanks
    Bill
     
  4. Mar 7, 2013 #3

    jtbell

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    More precisely, the kinetic energy operator.

    If we accept that the momentum p corresponds to the operator

    $$\hat p = -i \hbar \frac {\partial}{\partial x} = -\frac{ih}{2\pi}\frac {\partial}{\partial x}$$

    then from p = mv and ##K = \frac{1}{2}mv^2##, we can get

    $$\hat K = \frac{\hat p^2}{2m} = - \frac{\hbar^2}{2m} \frac {\partial^2}{\partial x^2}
    =-\frac{h^2}{8 \pi^2 m} \frac {\partial^2}{\partial x^2}$$

    But why should we accept that particular form for ##\hat p##?

    If we accept that a particle with momentum p has a wave function of the form

    $$\psi = A e^{i(kx - \omega t)} = Ae^{i(\frac{2\pi}{\lambda}x - \omega t)}
    = Ae^{i(\frac{2\pi p}{h}x - \omega t)} = Ae^{i(\frac{p}{\hbar}x - \omega t)}$$

    then

    $$\hat p \psi = -i \hbar \frac {\partial}{\partial x} \left[ Ae^{i(\frac{p}{\hbar}x - \omega t)} \right]
    = p Ae^{i(\frac{p}{\hbar}x - \omega t)} = p \psi$$

    But why should we accept that a particle with momentum p has a wave function of that form? Basically because we observe wavelike aspects of particle beams (diffraction and interference) that agree with de Broglie's hypothesis that ##\lambda = h / p##; and that is the generic (complex) wave function. (for one spatial dimension, of course)

    And because the full-blown QM that we develop from this starting point (Schrödinger's equation etc.) makes predictions that agree with experiment, so far.
     
  5. Mar 9, 2013 #4
    Thanks! You guys are awesome but i still don't get it why the last equation justifies the linear momentum operator.

    why this

    [itex]\hat{p}\Psi=p\Psi[/itex]

    makes justifies the linear momentum operator?
     
  6. Mar 9, 2013 #5

    bhobba

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    Not 100% sure what your issue is but any wave function by means of a Fourier transform can be put in the form of an integral of functions Jtbell gave. By moving the operator under the integral sign it can be applied to any wave function.

    But like I said in my previous post the real justification lies in much deeper symmetry arguments that are bit more advanced than is desirable to give in a first brush with QM. My suggestion is at this stage don't worry too much about it but when you get a bit more advanced get a hold of Ballentine's book and read the first three chapters where the full detail and justification of this stuff is given.

    Thanks
    Bill
     
  7. Mar 9, 2013 #6

    jtbell

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    I think tsuwal is simply asking, why is ##\hat p \psi = p \psi## important? Or, why do we want the momentum operator to give us that result?

    My answer would be that as tsuwal gets further into QM, and learns more about things like operators and expectation values, he will see that it can be shown that if an operator ##\hat Q## represents some physical quantity, and if ##\psi## satisfies the equation ##\hat Q \psi = Q \psi##, where Q is some value of that quantity, then the quantum state represented by ##\psi## definitely (without any uncertainty) has the value Q for that quantity.

    Here, we're assuming that ##\psi = Ae^{i(\frac{p}{\hbar}x - \omega t)}## is the appropriate wave function for a definite momentum p, and using the condition ##\hat p \psi = p \psi## to deduce the form of ##\hat p##.

    About forty years ago, the professor in our second-year "introductory modern physics" course told us that "you can't really understand any particular piece of QM without first understanding some other piece of QM." But after you've seen most of the pieces, you'll begin to appreciate how they fit together.
     
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