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Question about deriving TE Schrodinger WE

  1. Jul 20, 2014 #1
    To start, we have:
    Then, multiply both sides of energy equation by ψ to get:
    And replace [itex]-k^2[/itex] in the wave equation with [itex]-\frac{p^2}{hbar^2}[/itex]. Then plug into energy equation for [itex]p^2[/itex] to get:

    1. So why do we just assume the the wave function will be a complex exponential when it could be anything. If we used a sine and cosine wave function, this derivation wouldn't work, but yet, after you get the final equation, you can get a sine and cosine wave function out of it.

    2. This might be a dumb one, but why don't we just leave it at
    and call it a day, we have our function because now we could plug in our value for p and find the wave function.

    3. Also, since we used the De broglie equation for lambda, doesn't that mean that the De Broglie wavelength is really the wavelength of the particles wavefunction?
    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 20, 2014 #2


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    Actually the Schroedinger equation cannot be derived and is postulated as fundamental.

    But anyway, this derivation is a good way to guess it if you knew the wave function, but didn't know what equation it solved.

    The assumption of a complex exponential, when combined with the principle of superposition, says that an arbitrary wave is a sum of complex exponentials, where each exponential is a solution of the unknown wave equation.

    You already know the wave function, since that is what you guess at the start. The aim is to find out an equation for which the wave function is the solution. Basically, you already know the answer, the aim in this "derivation" is to guess the question. If you left it at that step, you wouldn't get the E and V terms of the equation, ie. you couldn't get the Schroedinger equation for systems like an electron in the hydrogen atom, where there is a potential energy due to the electrical attraction of the nucleus.

    BTW, this equation is only the time-independent Schroedinger equation. The full equation, which also tells us how the wave function changes with time is called the time-dependent Schroedinger equation.

    I think it is better to study the full Schroedinger equation first as an assumption, and treat this derivation as only a sort of way to guess the equation, not really to derive it. Maybe Schroedinger himself guessed the equation this way.
    Last edited: Jul 20, 2014
  4. Jul 20, 2014 #3
    Ok, I'm satisfied with that. So how did Schrodinger come up with it? Just a guess?

    Also I added a third question. Is the De Broglie wavelength the same as the wavelength of the particle's wave function?
  5. Jul 20, 2014 #4
    And to add on to question number 2, the time-dependent Schroedinger equation requires the time-independent to be solved specifically for [itex]E \psi[/itex]

    And to respond your third question, I'm not 100% sure because [itex]\displaystyle \lambda = \frac{h} {p} = \frac{h} {mv\gamma}[/itex] where [itex]\displaystyle \gamma = \frac{1} {\sqrt{1-\left(\frac{v} {c}\right)^2}}[/itex] so it certainly relates the mass of a particle to the De Broglie function. EDIT: Oh oops, sorry, I didn't understand your question properly.
    Last edited: Jul 20, 2014
  6. Jul 20, 2014 #5


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    I think it was an inspired guess. They knew that that particles were waves of some sort by that time, but they didn't know what equation governed these waves, and he guessed one of them. He didn't know the correct interpretation of the wave function even after getting the correct equation. The interpretation was discovered by Born, which is why we talk about the Born rule nowadays. Also the waves are really waves in an abstract space called Hilbert space. For one particle, the wave function can be considered a wave that travels in space, like a classical electromagnetic wave, but for many particles, this doesn't work, and the wave function for a system of many particles exists in a different space.

    Yes. The de Broglie wavelength is a concept that strictly applies only to a single free particle in a particular state (the initial state, which is a state of definite momentum). A general solution of the wave equation will be a superposition of the wave functions of different free particles with different de Broglie wavelengths.
    Last edited: Jul 20, 2014
  7. Jul 20, 2014 #6


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    You are using the relativistic de Broglie relations, which is correct. The de Broglie relations are consistent with special relativity.

    However, Schroedinger's equation is not consistent with special relativity.

    The correct equations consistent with special relativity are the Klein-Gordon equations and the Dirac equations. Actually, I think Schroedinger came up with the Klein-Gordon equation while looking for the Schroedinger equation. The Klein-Gordon and Dirac equation are not correct quantum equations for the wave functions of a fixed number of particles (we get negative probabilities), even though they are relativistic. However, their good relativistic features can be retained by using them as equations for quantum field theory, which can be roughly thought of as quantum mechanics in which the number of particles is not fixed.
    Last edited: Jul 20, 2014
  8. Jul 20, 2014 #7
    I thought that was what made QM different from classical mechanics- that negative probabilities were allowed?
  9. Jul 20, 2014 #8


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    Negative probabilities are not allowed in QM. The Wigner function, which is the quantum counterpart to the classical probability over canonically conjugate variables, can be negative. This is why the Wigner function is not a probability distribution. The probabilities of all observables calculated from the Wigner function are positive.
    Last edited: Jul 20, 2014
  10. Jul 20, 2014 #9


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    He made an analogy between mechanics and optics:

    • classical mechanics <--> geometrical optics (ray optics) and Fermat's principle.
    • quantum mechanics <--> wave optics and Huygens's principle.

    For some details:

  11. Jul 20, 2014 #10


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    That's not quite true.

    It can be derived from the Principle of Relatyivity (POR) and probability invariance (ie symmetry considerations). Specifically the probabilities of QM are frame invariant.

    See chapter 3 - Ballentine - QM - A Modern Development.

    Of course its replacing one assumption with another, but most recognise the POR, being a law about laws, is more fundamental than an instance of the law it applies to. Its actually downright weird that a general statement about laws actually implies specific instances of it. That's the power of symmetry.

    This is not the only area this happens eg see the derivation of Maxwell's equations from SR and Coulombs law:

  12. Jul 20, 2014 #11


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    Not quite.

    It was from analogies with the Hamilton Jacobi equation and waves:

    As the link above points out he made an error but got the correct answer anyway. The correct derivation via that method makes use of Feynman sum over histories approach.

    Like I said in my previous post the most fundamental derivation is from the POR as found in Ballentine.

    As Atty pointed out only in some instances eg a free particle.

  13. Jul 20, 2014 #12


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    You may be referring to the following link I often post:

    Its not allowing negative probabilities. As the lecture says 'Quantum mechanics is what you would inevitably come up with if you started from probability theory, and then said, let's try to generalize it so that the numbers we used to call "probabilities" can be negative numbers. As such, the theory could have been invented by mathematicians in the 19th century without any input from experiment. It wasn't, but it could have been.'

    Its making sense of probability theory with negative or even complex numbers - not that the probabilities themselves are negative - which would be a violation of the axioms of probability. The way it does that is to multiply the complex number by its conjugate which always gives a positive result - in fact its the square of its absolute value.

  14. Jul 20, 2014 #13
    I've tried to read that Scott Aaronson link a couple times now but for some reason can't ever make it through haha. I'll have to give it another shot. I'm glad the way I did it above isn't the official derivation because it felt a little fishy.

    Anyways, thanks for the help everyone.
  15. Oct 9, 2015 #14
    Hello there is p(momentum) operator or something else?And is k there some constant?
  16. Oct 9, 2015 #15
    Yes, p is the momentum and k is the wave number vector such that ##p = \hbar k##.
  17. Oct 9, 2015 #16
    Thanks:oldbiggrin:Your information helped me a lot!
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