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Experimental Support of Shrodinger Equation

  1. Oct 16, 2014 #1
    Hi,

    The reason I'm posting this is because I'm trying to understand how to interpret the wave function of a particle. I'm trying to decide which interpretations of quantum mechanics are just speculation, and which ones are consistent with experimentation.

    I know that quantum objects such as electrons exhibit wave-like properties, but how do we know these observed properties are consistent with the Shrondinger equation? Is there any particular experiment which shows that the probability distribution of a quantum particle is consistent with the Shrodinger equation? Or any other types of experiments which support the Shrodinger equation?

    For example, can you construct an approximately infinite well and observe the appropriate energies and expectation values predicted by the Shrodinger equation? Etc etc.
     
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  3. Oct 16, 2014 #2

    atyy

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    Evidence for the non-relativistic Schroedinger equation comes from spectra of atoms and molecules, and properties of solids like their ability to conduct electricity.

    http://en.wikipedia.org/wiki/Hydrogen_spectral_series (atomic spectra)
    http://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2007/lecture-notes/lecture35.pdf (builds on the simple harmonic potential)
    http://en.wikipedia.org/wiki/BCS_theory (electrical conductivity)

    In relativistic situations, the Schroedinger equation must be generalized to quantum field theory. For example, a detail of the hydrogen spectrum that is not predicted by the non-relativistic Schroedinger equation is the Lamb shift. Quantum field theory gets this right.

    http://en.wikipedia.org/wiki/Lamb_shift
     
  4. Oct 16, 2014 #3

    Simon Bridge

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    ... in addition: all interpretations are speculation and all are consistent with experimentation.
    There's a good thread about this somewhere, but I keep losing it.
     
  5. Oct 17, 2014 #4

    f95toli

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    Also, a more "direct" demonstration would be quantum well type structures that are used in e.g. AlGaAs-GaAs laser diodes (i..e the type of diode you have in home electronics); there structures are designed by varying the amount of aluminium which raises/lowers the potential barriers for the electrons. This way you can create an arbitrarily shaped potential landscape and you get the properties of that landscape (e.g. the wavelength of the generated light) by solving the Schroedinger equation. In some cases you can actually solve the equations analytically and get very close to what is seen experimentally.
    There are many, many other examples. Nearly all modern electronics uses components where the designers that to solve the SE at one point or another to model their behavior. Hence there are literally billions of components out there that wouldn't work unless the SE was correct.

    The point I am trying to make is that the Schroedinger equation is not an "exotic" equation despite what you might think after reading pop-sci books, it is a fundamental equation in science and engineering..
     
  6. Oct 17, 2014 #5
    Why do people say that "a particle's wave function 'collapses' when a measurement is made"?

    If you have a simple system with a bound particle, the bound particle will have some wave function before the measurement is made (which tells us where we might find the particle). Then, once we've made the measurement, why would the particle's wave function be any different? It's still under the same potential. Maybe it will gain or lose some energy from the act of measuring, but disregarding that, if you were to solve the Shrodinger equation again immediately after the measurement, the wave function should turn out to be the exact same right? Even at the moment of measurement, wouldn't the solution to the Shrodinger equation still yield the same wave function?

    Is it because when you know the position of a particle this gives you a new "initial condition" to solve the Shrodinger equation for? The problem with this is that we don't actually measure the wave function, we just measure the position. So how could we ever get an "initial wave function" via measurement?
     
  7. Oct 17, 2014 #6
    This happens because the process of making a measurement involves obtaining a single result for a given value. If a particle's position wavefunction exists over a continuum (that is, when the particle does not have a single definite position), then measuring the particle's position requires interfering with it in such a way that it must assume a definite position. From the values of the wavefunction, you can calculate the probability that the position it assumes will be in a certain range.

    In the classic double-slit experiment, the measurement involves setting up a system where you can measure which slit an electron passed through (or just building a system where you could determine which slit an electron passed through - whether you look at the data yourself doesn't make a difference). In order to determine the position of an electron, one might track the strength of its electric field at an observation point next to the slits - in this case, the charged particle(s) you use to measure the field strength exert a pull on the electron disrupting its motion. Since the magnitude of the effect used to make the observation is relatively large, the wavefunction distribution becomes insignificant, and particle-like behavior becomes dominant.
     
  8. Oct 17, 2014 #7

    bhobba

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    If it happens is interpretation dependant.

    They probably simply haven't thought the issue through and/or don't know a lot about the various interpretations out there. They simply regurgitate what beginner texts say. Its the same with the wave-particle duality.

    Ballentine's text carefully explains exactly what's going on, but not all texts take that level of care - and even his text isn't perfect saying that Copenhagen requires collapse which is unreasonable physically. Its only unreasonable if you consider the wave-function is real - most versions of Copenhagen have it as simply as an aid to calculation.

    The issue is subtle though:
    http://arxiv.org/pdf/1111.3328v3.pdf

    But note, and people sometimes forget it with regard to that important theorem:
    'The argument depends on few assumptions. One is that a system has a real physical state" - not necessarily completely described by quantum theory, but objective and independent of the observer. This assumption only needs to hold for systems that are isolated, and not entangled with other systems. Nonetheless, this assumption, or some part of it, would be denied by instrumentalist approaches to quantum theory, wherein the quantum state is merely a calculational tool for making predictions concerning macroscopic measurement outcomes.'

    Thanks
    Bill
     
  9. Oct 17, 2014 #8

    Simon Bridge

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    If I roll a couple of die and hide the result - you would describe the possible outcomes by a specific probability distribution.
    If you then measure the result - by looking at it - you would describe the possible outcomes by quite a different probability distribution.

    If I ask you how tall a person is who you've never seen, you would reply in terms of the mean and standard deviation human heights ... but once you've seen that person, the probability distribution is quite different, and after taking a measurement with a ruler, different again - and also different if you use more or less accurate devices.

    These are easy to understand.

    Before and after a measurement the possible outcomes are different - in the above case because your state of knowledge of the system has changed. In the QM case, the wave-function still reflects the possible outcomes of a measurement. The act of measuring something affects the possible outcomes quite directly by interacting with the system.

    The slit situation is a good example of this:
    Shine a beam of light at a small hole in a screen: the photons can have a wide range of possible positions in the plane just before the hole, but just after the hole the possibilities are very curtailed simply because the screen has intercepted all the light that did not go through the hole. The screen acts to measure the position of the photons in that plane.

    When you learn QM, the word "measurement" gets used without talking abut the exact method used to make the measurement.
    As you advance you will learn about the classic experiments and how the measurements were made and so how those measurements affect the outcomes turns out to be consistent with what you are learning now.
     
  10. Oct 18, 2014 #9
    My understanding is that as soon as your system of a particle in a potential interacts with something else (such as interacting with your position measurement apparatus), you can no longer use Schrodinger equation of this system as if it is still an isolated one. If you do, you get erroneous results.

    The full correct treatment is to analyze the wavefunction of the bigger system now, one which includes measurement apparatus. In this bigger picture, you will have a wavefunction which is a superposition of many possible states where measurement apparatus detected the particle in every possible location.
     
  11. Oct 18, 2014 #10

    Doug Huffman

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    I've been watching Leonard Susskind's lectures, and am impressed with his steady delivery. I was also impressed with his brief rant about the volume of QM nonsense crap in the popular literature. The rant may have even been delivered during the Schrödinger equation lecture.
     
  12. Oct 18, 2014 #11

    mfb

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    The name is Schrödinger - or Schroedinger if writing "ö" is too complicated with your keyboard. Not Shrondinger, Shrodinger, Schrodinger or whatever else came up here.

    That depends on your measurement. If you can include the measurement process in the system and use the equation for the combined evolution of the system, then everything is fine.

    And if you do not stop doing that at arbitrary points, you get the many-worlds interpretation of quantum mechanics.
     
  13. Oct 18, 2014 #12

    bhobba

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    That's true. If you observe one part of an entangled system it looks exactly the same as a mixed state (called an improper mixed state) and you get apparent collapse. Many books explain this, but at the beginner level Susskind certainly does:
    https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465036678

    That's the view of dechohence.

    The modern version of collapse is the so called problem of outcomes - which is - why do we get any outcomes at all or, more technically, how does the improper mixed state become a proper one.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  14. Oct 18, 2014 #13

    bhobba

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    I think most people who have studied QM as the real deal from good physics books get very upset at populist crap such as What The Bleep Do We Know Anyway. I know I certainly do.

    Thanks
    Bill
     
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