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Schrodinger's Equation

  1. Dec 26, 2004 #1
    Can Schrodinger's Equation be derived without a boundary condition?

    Particles according to quantum physics are only "partly localised", so does it mean that Schrodinger's equation can only be applied in a confined region of space?

    Also, from what I read from my text book, Schrodinger's Equation is applied to wave packets, because it has an "estimated" boundary of [tex]\Delta x[/tex] of large magnitude. If so, how can a simple harmonic quantum oscillator exist? An ideal simple harmonic motion is represented by pure sine or cosine waves, where [tex]\Delta x = \infty[/tex].
     
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  3. Jan 2, 2005 #2

    dextercioby

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    Schroedinger's equation cannot be derived.It is accepted as a postulate.U mean 'solved'.In that case,it depends on the typical problem it is applied.Generally we take into account wavefunctions defined on all R^{3n}.

    No,Schroedinger's equation can be applied for all space,for all conditions,for all possible cases.However,not every solution to this equation describes a possible quantum state of a system.

    Yes,it's applied to to wave packets,simply because de Broglie's plane momentum waves do not represent phyiscal states of a quantum particle as they cannot be normalized.

    It can exist,as all solution of the Schroedinger's equation can be normalized and hence describe possible quantum states.Pick one state of the QSHO and compute [tex] \Delta \hat{x} [/tex] and see whether it is infinite or not...

    Daniel.
     
  4. Jan 3, 2005 #3
    @dextercioby:

    I've seen you state a few times that Schrodinger's equation cna't be derived, but needs to be accepted as a postulate or axiom. I think that would be a bit strange, since what then would bring schrodinger to formulate it? Also, I've read a derivation of the equation following from Feynman's path integral formulation of QM, that, as far as I could tell, didn't take the Schrodinger equation as a known fact. But I'm quite the newbie on path integral QM, so I could be wrong.
     
  5. Jan 3, 2005 #4

    dextercioby

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    Traditional nonrelativistic QM has 3 versions/formulations:
    1.Formulation with vectors and operators,due to mostly to Paul Adrien Maurice Dirac and founded mathematically by John von Neumann.
    2.Formulation with operators due to John von Neumann.
    3.Formulation with path-integrals due to Richard P.Feynman.

    At conceptual level all these 3 formulations are equivalent.That means one implies the other and viceversa.In particular,the Schroedinger equation can be obtained from the treatment by Feynman and naturally viceversa.Actually Feynman took Schroedinger's eq.for granted and proved its complete equivalence with equations containing path-integrals.

    So,in the Schroedinger picture of Dirac/traditional formulation of nonrelativistic QM,the IV-th postulate contains the Schroedinger's eq.But in the Feynman formulation,this eq.can be derived through a more tricky procedure.This formulation due to Fynman is most commonly made in the Heisenberg picture,so one has to use the equivalence between the Heisenberg picture and the Schroedinger's one to get the equation.

    Anyway,the key-word is "equivalence".

    Daniel.
     
  6. Jan 3, 2005 #5

    jtbell

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    Schrödinger's inspiration (not a derivation in the rigorous mathematical sense) was to make an analogy between mechanics and optics. I have a PDF copy of Schrödinger's first English-language paper on the subject [Phys. Rev. 28, 1049 (1926)] following on his original German-language papers. He starts with the Hamiltonian action integral of classical mechanics:

    [tex]W=\int{(T-V)dt}[/tex]

    and considers the family of surfaces in space on which W is constant. It turns out that the possible trajectories of a particle in that system are perpendicular (normal) to those constant-W surfaces. He comments:

    A bit further on:

    He discusses the relationship between geometrical (ray) optics and wave optics, and then:

    Schrödinger associates Hamilton's W-function with the phase of his [itex]\psi[/itex]-function:

    [tex]\psi=A(x,y,z) \sin(W/\hbar)[/tex]

    He assumes that this has to satisfy the usual differential wave equation, with velocity

    [tex]u=E/\sqrt{2m(E-V)}[/tex]

    and on substituting these into the differential wave equation, out pops what we now know as the time-independent Schrödinger equation!
     
    Last edited: Feb 10, 2012
  7. Jan 3, 2005 #6

    dextercioby

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    To carry on from where Jtbell left off,i'll post Schroedinger's view on the (Hydrogen) atom.
    However,though it has been obtained from classical phyiscs,this equation is still the basis of quantum physics and it cannot be applied to macroscopic level/classical systems.

    Daniel.
     
  8. Mar 5, 2005 #7
    I never had a formal QM course but I'm jumping into it by myself. For the last four days I've tried to understand the "derivation" of Schrodinger's equation. As a matter of fact, as I now type this reply I have 5 books on my desk open in Schrodinger chapter. Of course I have not been able to fully understand why the "derivation" process was jumping from one integral to another. Now I see.
    I guess I should not feel bad if I don't understand its derivation as there is no real rigorous mathematical derivation but it is in part based on considerations. Right?
     
  9. Mar 6, 2005 #8

    dextercioby

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    Well,there are formulations of QM in which SE is not a postulate.Even in Dirac'ss formulation,if you postulate Heisenberg equation,you'll be able to prove SE...

    Daniel.
     
  10. Mar 6, 2005 #9
    I have the same question.:-)
     
  11. Mar 6, 2005 #10

    ZapperZ

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    You have a considerable misunderstanding of what a "Schrodinger Equation" is, and, I suspect, a lack of knowledge on the mathematics involved. The Schrodinger equation itself is INDEPENDENT of any boundary condition. It is, after all, simply a 2nd order differential equation. It is the SOLUTIONS to the Schrodinger equation that is dependent on the boundary conditions. The solutions are the ones tailored to whatever boundaries that exist.

    Does this, outright, answer your question?

    Zz.
     
  12. Mar 8, 2005 #11
    The SE is independent of any boundary condition. There is a simple way to "prove" the SE. Let's assume we have no problem with wavefunctions (why should we, right ?). Given the equation for a wavefunction (ie the exponential structure) we now that taking the second derivative with respect to position yields the momentum p squared : p² (with some constants). The first derivative with respect to time yields the energy E. Now we know that E = p²/2m so if you substitute p² and E with the derivatives of the wavefunctions, you get the SE

    regards
    marlon
     
  13. Mar 8, 2005 #12
    marlon, your derivation does nothing more than show:

    [tex] \psi(\vec {r}, t) = e^i^(^\vec {p} \cdot \vec{r} + Et) [/tex]

    Is a solution to a particular equation, a game we could play all night (it is like asking "for what equations is 2 a solution?").

    That is really a backwards derivation, and I don't see how it applies in any case other than that of a plane wave (at most, maybe you only meant one dimensional).

    If you want to see the quick and dirty derivation of SE:

    1) Assume there is a wave function which contains all of the observable information it is possible to know about a particle.

    Cosideration: In order to find this function we must construct it by giving it properties. That is, it should be constructed as to give us information about the system when we apply a relatively simple operator.

    2) Whatever the wave function is, I want to get momentum with a space derivative and energy with a time derivative, that would be simple.

    Since energy and momentum are related as they are, we have the shrodinger equation! Fortunately the uniqueness of solutions to a partial differential equation gaurantees our success.
     
    Last edited: Mar 8, 2005
  14. Mar 8, 2005 #13

    dextercioby

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    That's both fishy and logically incorrect,because you know where you to get (the SE) and you know that the result you're expecting (SE) is correct...

    Daniel.
     
  15. Mar 8, 2005 #14
    :rofl: :rofl: :rofl:

    I don't really think you got the point. However, since i don't want to play with you all night long, let us just drop it...ok ?

    marlon
     
  16. Mar 9, 2005 #15

    EL

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    well, it's not the first time this is discussed here... :tongue2:
     
  17. Mar 9, 2005 #16

    dextercioby

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    True,these subjects in Quantum Physics forum are kinda circular.The EPR gedankenexperiment is still the most debated.

    Daniel.
     
  18. Mar 10, 2005 #17
    That and Afshar also

    marlon
     
  19. Mar 10, 2005 #18
    Not that I know of. Its only possible by way of using a different set of postulate and then the Schroedinger's equation simply fall out of it.

    Pete
     
  20. Mar 10, 2005 #19
    No, once again: the Schrodinger equantion (SE) is independent of boundary conditions. The SE is a mere manifestation of E = p²/2m, translated into derivative operators working on a wavefunction. The solutions (ie the wavefunctions) are dependent on boundary conditions in order to assure their normalizability (so they are "physical" if you will) and their continuity at boundaries (like in the case of the potential-well)

    marlon
     
  21. Mar 10, 2005 #20

    dextercioby

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    Marlon,SE refers to the abstract equation which is postulated.You've been refering to (not only in this thread) is the simple equation found by Erwin Schrödinger in 1926.Dirac axiomatized and abstractized everything...

    Daniel.
     
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