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 Quantum Physics Quantum Physics Help Forum Jul 16th 2015, 02:49 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 Is this Shrodinger Equation derivation a it fishy? [Edit: Title should have read "Is this Shrodinger Equation derivation a bit fishy?" ] DrPhysicsA, on youtube, does great short mathematical derivations that I don't need 5 post graduate degrees to follow. But I kind of got to thinking about this: (Its only 7 minutes long). However, I have some questions about this: 1. Is P = h / lambda true for all waves (ie water waves) or only electromagnetic waves? 2. If Psi(x,t) is a probability wave then how can a probability wave have a momentum? 3. If Psi(x,t) = exp(i(kx-omega*t)) then how can this even be massaged into bell shaped curves or curves with damped oscillations at the edges? For instance with the double slit experiment the probability distribution at the screen doesn't extend to infinity (or not without significant dampening as it extends to positive or negative infinity). 4. If its a superposition of Psi(x,t) eigenfunctions then how can you even produce a double slit interference pattern with linear combinations of Psi(x,t) even with different frequencies. 5. A fourier series is always going to be periodic but the Psi(x,t) functions are not necessarily so. I am starting to wonder if QM is "glorified curve fitting"? How is it that QM can be ontological with so much hand waving? Your thoughts? Regards KiwiHeretic   Jul 16th 2015, 04:13 PM #2 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 Infinite square well The other problem I foresee is in the case of an infinite square well where the potential U(x) is zero within the well and is infinite at the edges (say abs(x) > L). In this case the waves are standing waves and so can the waves of a standing wave hae any momentum at all? If E = p^2/2m + U(x) then wouldn't that imply that E = U(x) for a standing wave? How then can you have multiple discrete energy levels based on the number of nodes of the standing wave?   Jul 16th 2015, 05:20 PM #3 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,858 This won't be in order. 2. The S-equation derivation does tend to be a bit "dodgy" but in the end it is simply a statement of conservation of energy. We start with a Mathematical object called the wavefunction. The wavefunction contains all of the information about the system. We employ "operators" such as the momentum operator, (hbar/i)*(d/dx), to get information from the wavefunction. The wavefunction itself is not a probability distribution, the quantity |psi(x)|^2 tells you the probability that the particle is located between x and x + dx. The wavefunction gives information about physical properties, it does not contain those properties physically. Another way to derive the S-equation is to use Dirac's "bra-ket" notation. To my mind it is a better derivation using a lot of Linear Algebra directly (vectors spaces as opposed to what I would call "function spaces.") It gives exactly the same S-equation, but with more transparency. 1. p = h/lambda is only correct for particles with mass. 3. Plane waves do not fit a "bell curve." Plane waves are somewhat exotic creatures in the sense that they are not localized in space at all and carry only a single value of momentum. Not very "Heisenberg" of them. Plane waves are an idealization and strictly speaking don't exist. On the other hand they make good basis states to represent properties of wavefunctions. 4. The same way you do it with water waves. I'm not sure what your question is. 5. If you are doing a Fourier series analysis then you expect to get a periodic function out of it...That's what the analysis does. You can, however, analyse a non-periodic function by limiting the extent of your function over a finite domain. QM is not "glorified curve fitting," though that can be the case if you are trying to analyze a wavefunction's properties before putting the problem into the S-equation. For example the S-equation is a second order linear partial differential equation. We often apply boundary conditions by making physical assumptions about the wavefunction. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.   Jul 16th 2015, 05:45 PM #4 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 Yeah, I stand corrected regarding the probability of the wave being psi(x,t)^2 However, if psi(x,t) itself is not physical I am trying to see what justification there is in saying psi(x,t) = exp(i(kx-omega*t)) and then saying k = p / h_bar where p is momentum. How can you add momentum into a wave function that is not physical and its square is a probability density function and therefore has no units? Just squaring something doesn't remove its units. Thank you for your advice on the Dirac derivation being a bit more transparent. Will look into that. [ Note: Yeah, I know I'm not quite correct with my naive squaring of psi(x,t) as I know you have to multiply by its complex conjugate. ] Last edited by kiwiheretic; Jul 16th 2015 at 05:47 PM. 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