Jeremy0922 Posted March 14, 2010 Posted March 14, 2010 (edited) The model of hydrogen atom that the electron moves in the center field produced by the proton is the inevitable result with pure classical theory, and the reduced mass of the electron must be introduced to correct the result. With QM (quantum mechanics), we accept and select this model which belongs to pure classical conception to build Schrödinger equation of it, but then deny the classical motion of the electron. I think that is self-contradictory. Since we have the perfect description about the particle by QM, so could build the model of hydrogen atom with it. Why don't we build the model of hydrogen atom independently by QM? more inforrnation see: http://www.scienceforums.net/forum/showthread.php?t=48176 Edited March 14, 2010 by Jeremy0922 modification
swansont Posted March 14, 2010 Posted March 14, 2010 The model of hydrogen atom that the electron moves in the center field produced by the proton is the inevitable result with pure classical theory, and the reduced mass of the electron must be introduced to correct the result. With QM (quantum mechanics), we accept and select this model which belongs to pure classical conception to build Schrödinger equation of it, but then deny the classical motion of the electron. I think that is self-contradictory.Since we have the perfect description about the particle by QM, so could build the model of hydrogen atom with it. Why don't we build the model of hydrogen atom independently by QM? more inforrnation see: http://www.scienceforums.net/forum/showthread.php?t=48176 We deny the classical motion of the electron because it does not behave that way. The Schrödinger equation did not come from the solution of the hydrogen atom, the solution for the hydrogen atom comes from solving the Schrödinger equation.
Jeremy0922 Posted March 15, 2010 Author Posted March 15, 2010 QM: Considering the motion of proton, the reduced mass of the electron replaces the mass of the electron to correct the result from Schrödinger equation, then the result is more closed to experiment observation. In this situation, what is the reference frame for the description of the motions of the electron and the proton? Is it the center of mass coordinate system of hydrogen atom???
swansont Posted March 15, 2010 Posted March 15, 2010 Yes. You need to have a spherically symmetric system to use the spherical coordinate system.
Jeremy0922 Posted March 15, 2010 Author Posted March 15, 2010 QM: The motions of the electron and the proton of hydrogen atom are described by their probability waves, so, the position of the center of mass of the two-particle system will be of probability. The description of physical phenomena in random reference frame can not be used to compare with experimental observation. So, by interpretation of probability wave about the motion of particle, it is impossible to obtain the model of hydrogen atom that is fit to compare to experimental data. The physical phenomena described by Schrödinger equation need to be reconsidered, and The conception of particle-wave duality of particle need to be reconsidered
swansont Posted March 15, 2010 Posted March 15, 2010 (edited) QM: The motions of the electron and the proton of hydrogen atom are described by their probability waves, so, the position of the center of mass of the two-particle system will be of probability. The description of physical phenomena in random reference frame can not be used to compare with experimental observation. It's not a random reference frame. The center of mass is stationary in the COM frame. By definition. The position of the center of mass will be a probability when measured by a fixed frame of reference, aka the lab frame. The conception of particle-wave duality of particle need to be reconsidered And how exactly does one explain away atoms interfering with each other, or diffracting? Edited March 15, 2010 by swansont Consecutive posts merged.
Jeremy0922 Posted March 15, 2010 Author Posted March 15, 2010 COM is random to lab We could try to select the resonance effect to explain
timo Posted March 15, 2010 Posted March 15, 2010 You could try to understand what a full solution of the hydrogen atom with all six spatial degrees of freedom looks like. I think that is better-invested time.
swansont Posted March 15, 2010 Posted March 15, 2010 COM is random to lab Yes. Why is this a problem? What error does this introduce, and can we possibly measure it?
Jeremy0922 Posted March 16, 2010 Author Posted March 16, 2010 (edited) Hi, timo. Thank you give me a good suggestion. the full solution of the two-particle system need six freedom to describe their motion at least as you say. If we want to known the releationship of some physical quantities between resonant vibrations of structure, such as the modal frequency, we could obtain the result by condition equation of the physical phenomena. Of course, that is not enough to describe the motions of the particles. If the linear spectrum from hydrgen atom could be understood as the resonance effect (modal response) of orbits of particles in hydrogen atom, the modal frequencies will be solved by resonance equation. the static Schrödinger equation of the hydrogen atom could be deduced from standing wave equation of reonant vibrations of ground orbits of particles by mathematical method (see my manuscript). Merged post follows: Consecutive posts mergedHi, swansont According to the pure probability wave interpretation, the center of mass coordinate system of hydrogen atom is random to lab, and the description of physical phenomena in random reference frame can not be used to compare with experimental observation. And, that interpretation can not give independently the model of hydrogen atom in which the electron is moving in the steady center field produced by the proton. Edited March 16, 2010 by Jeremy0922 Consecutive posts merged.
swansont Posted March 16, 2010 Posted March 16, 2010 Hi, swansont According to the pure probability wave interpretation, the center of mass coordinate system of hydrogen atom is random to lab, and the description of physical phenomena in random reference frame can not be used to compare with experimental observation. And, that interpretation can not give independently the model of hydrogen atom in which the electron is moving in the steady center field produced by the proton. You've said this twice now, but have not explained why it is true. You should be able to quantify the error it would introduce and show it is a problem. I think your premise is false. If I have a collection of atoms I can describe their collective behavior, even though they are moving randomly with respect to the lab frame. That, and the fact that QM actually agrees with experimental determination of the hydrogen atom. Arguing that they can't be compared is a tough sell.
Jeremy0922 Posted March 16, 2010 Author Posted March 16, 2010 Respected swanont: As you known, the interpretation of pure probability wave that is so-called "Copenhagen interpretation" does not equal to QM (quantum mechanics), and is only a interpreation of wave function from Schrödinger equation. Schrödinger equation provides the precices mathematical foundation for us to solve the problems of atomic, molecular and solid-state structure. Since it was arised by Bohr etc, "Copenhagen interpretation" was repelled stoutly by M. Planck founder of quantum, and other famous scientist including Schrödinger and Einstain did not accept it, because "Copenhagen interpretation" voilates the principle of science. Based on the analysis above, my viewpoint is that we can not independently obtain the steady model of hydrogen atom by "Copenhagen interpretation". "Copenhagen interpretation" is incorrect.
ajb Posted March 16, 2010 Posted March 16, 2010 The "Copenhagen interpretation" is an interpretation, it is related but independent of the mathematical formulation via functional analysis. Any interpretation is really a description of the mathematical constructions and calculations. I don't see how any interpretation can be correct or incorrect. It maybe useful in driving new ideas and re-formulations etc, but it is not on par with the real mathematical formulation.
swansont Posted March 16, 2010 Posted March 16, 2010 Your problem now appears to be with the Copenhagen interpretation of QM rather than with the Schrödinger equation. As ajb said, that's an interpretation. A tool for understanding, but not the theory itself. The bottom line is that the Hydrogen atom solution to the Schrödinger equation predicts answers that are confirmed by experiment. The vast part of QM is insanely successful in this way.
Jeremy0922 Posted March 18, 2010 Author Posted March 18, 2010 (edited) Thank you discuss with me. For physcis, any interpretation should depend on physical phenomena, mathematical tool is used to express the relationship between the physical quantities. For example, an object rotates on a circular orbit. we may select the motion equation to describe that physical phenomena. according to the motion equation we can get the position of the object at any time. With orbit equation, we can not get the position of the object at the specifitied time, but know it is on the obit. A mathematical equation possibly describes the relationship of some quantities for several kind of physical phenomenon, and the relationship between quantities of a physical phenomena might be describe by several mathematical equation. Thereby, we must be careful to explain a mathematical expression, and must consider the integrality of it. Merged post follows: Consecutive posts mergedBy my opinion, since Copenhagen interpretation fails to give an reliable model of hydrogen atom, the heart of quantum mechanics that is a blurred picture of probabilities need to be reconsidered. I don't disbelieve the outstanding working out the consequencs of Schrödinger equation, and trust that equation provides a precise mathematical foundation for description of the relationship between some physical quantities while the atomic, molecular and solid-state structures are changing. Resonace is a natural phenomenon of a object, and ables to change the strucure of it. When resonace takes place, the object will be in one modal of resonant vibrations by energy exchange with resonant wave. and the modal frequency is determined by the mechanical properties of the object. If the linear spectrum from hydrogen atom could be understood the effect caused by the modal response of its structure, then whether or not Schrödinger equation of hydrogen atom could be explained the modal equation of it?? Edited March 16, 2010 by Jeremy0922
swansont Posted March 18, 2010 Posted March 18, 2010 By my opinion, since Copenhagen interpretation fails to give an reliable model of hydrogen atom, the heart of quantum mechanics that is a blurred picture of probabilities need to be reconsidered. The Copenhagen interpretation is not a model, and it is not a theory. It is an interpretation of QM, which makes your statement a non-sequitur. QM itself (QED) gives a very, very reliable model of the hydrogen atom.
Jeremy0922 Posted March 19, 2010 Author Posted March 19, 2010 (edited) I have studied quatum theory including QM, quantum theory of atomic strucure (by J. Slater)etc. in college since 1984, and achieved doctor degree in atomic and molecular physics in 1994. By my knowledge, particle with wave duality of particle (de Broglie wave) is the most important conception of quantum theory for atomic and molecular world, and is described by probability wave (that is "Copenhagen interpretation"). That is the heart of QM, and the motion and other physical quantities of particle is understood with this conception. The problem of atomic structure is treated with central-field model (or approximation) as like that of hydrogen atom. I don't kown whether or not there is an other model about atomic structure. If not, our scientist have to face to the serious problem that we can not obtain central-field model of atomic structure independently by the conception of QM. So,we should improve the current QM, or find a new theory to settle it down. Edited March 19, 2010 by Jeremy0922
Mr Skeptic Posted March 19, 2010 Posted March 19, 2010 What if instead of the Copenhagen interpretation you use one of these?
swansont Posted March 19, 2010 Posted March 19, 2010 So,we should improve the current QM, or find a new theory to settle it down. Is there any physical evidence that shows that QM is wrong? QED is confirmed to an amazing precision. What, exactly, would you seek to improve?
ambros Posted March 20, 2010 Posted March 20, 2010 We deny the classical motion of the electron because it does not behave that way. It behaves like that whenever we can actually see it or measure it, in bubble chambers and electron beams for example. It is only when we can not "see" and properly measure it that we say to not know its motion, which is reasonable, but it is not sufficient argument to refute what is otherwise very usual and commonly observed kind of motion for electrons. As ajb said, that's an interpretation. A tool for understanding, but not the theory itself. The bottom line is that the Hydrogen atom solution to the Schrödinger equation predicts answers that are confirmed by experiment. The vast part of QM is insanely successful in this way. Schrödinger equation would work even if trajectories are continuous, you can use it to describe planetary orbitals too. I think that equation can even be used to predict weather, or wash clothes and cook dinner. Is there any physical evidence that shows that QM is wrong? QED is confirmed to an amazing precision. What, exactly, would you seek to improve? It would be hard to discredit QM in such way because it is statistical description of the database of many measurements, but that still does not prevent it to have incorrect interpretation of those results and observations, it also does not mean we are not oblivious to some error or whatever we have missed to discover so far. In any case, I do not think 'continuous trajectories' are fundamentally incompatible with QM. Q: If electrons are not continuously "sliding", then they must be moving by the means of "appear-disappear" kind of thing, which seem rather strange as electrons would need to be "nowhere" at certain points in time, so can this be explained and is this what QM thinks is the motion of electrons in atom orbitals?
swansont Posted March 20, 2010 Posted March 20, 2010 It behaves like that whenever we can actually see it or measure it, in bubble chambers and electron beams for example. It is only when we can not "see" and properly measure it that we say to not know its motion, which is reasonable, but it is not sufficient argument to refute what is otherwise very usual and commonly observed kind of motion for electrons. The context here is the hydrogen atom. The ground state has zero orbital angular momentum, yet the classical solution, the circular orbit, has nonzero angular momentum. The hydrogen atom does not obey the classical solution, so it must be discarded. Particles are also found not to obey classical solutions in other situations as well. It would be hard to discredit QM in such way because it is statistical description of the database of many measurements, but that still does not prevent it to have incorrect interpretation of those results and observations, it also does not mean we are not oblivious to some error or whatever we have missed to discover so far. In any case, I do not think 'continuous trajectories' are fundamentally incompatible with QM. If the interpretation is wrong (i.e. contradicts the data), then you get rid of it. But this seems to be more of a case of the interpretation being philosophically or metaphysically unsatisfying. Q: If electrons are not continuously "sliding", then they must be moving by the means of "appear-disappear" kind of thing, which seem rather strange as electrons would need to be "nowhere" at certain points in time, so can this be explained and is this what QM thinks is the motion of electrons in atom orbitals? QM does not predict the motion, per se. QM tells you you can't know it at the scale we are talking about — the classical solution gives a position uncertainty of the same order as the orbit diameter (which should make sense, since the ground state Bohr orbit was chosen to be one deBroglie wavelength in circumference)
Jeremy0922 Posted March 20, 2010 Author Posted March 20, 2010 swansont: Would you please tell me which experiment prove the ground state has zero orbital angular momentum.
swansont Posted March 21, 2010 Posted March 21, 2010 swansont:Would you please tell me which experiment prove the ground state has zero orbital angular momentum. Um, all of them? Spectroscopy only makes sense in light of the QM solutions. Specifically, the existence of only two hyperfine ground states indicates an absence of orbital angular momentum; any spectroscopy which resolves L+S coupling; the metastable nature of the 2S state (which doesn't even exist in the Bohr model) because it shows there is no ground state with zero angular momentum — it not be a forbidden decay. But you have a degree in atomic physics, so you already know this.
ambros Posted March 21, 2010 Posted March 21, 2010 The context here is the hydrogen atom. The ground state has zero orbital angular momentum' date=' yet the classical solution, the circular orbit, has nonzero angular momentum. The hydrogen atom does not obey the classical solution, so it must be discarded.[/quote'] "Classical solution" does imply angular momentum to be non zero, but it is not necessarily 'circular orbit', it's similar to planetary orbits if we model electric fields ONLY, but if you include magnetic interaction due to 'moving charge' and spin magnetic moment then the orbit is rather spherical and much more "irregular". Being unable to measure velocity or displacement of electrons in atom and unable to know motion and its angular momentum is different thing from actually measuring the value of zero. Is it not the postulate of QM that we can not measure any such thing, so how can we measure it to be zero? If we indeed measured that, instead of simply failing to measure anything, then that would mean electron has no velocity as derivative of position (as we know it), and/or it makes "sharp turns". What would be angular momentum if electron was "going left-right and making 180 degree turns? If the interpretation is wrong (i.e. contradicts the data), then you get rid of it. But this seems to be more of a case of the interpretation being philosophically or metaphysically unsatisfying. Well, that's quite a pickle actually. Remember Heliocentric system and Galileo, both theories described observation pretty well, yet they were completely opposite. Remember Dirac and Nobel Prize for positron, he was talking about "aether" and "holes" in it, which accurately modeled quite a few experimental observations and predicted positron, and though later *interpretation* changed, "Dirac sea" equation is still valid, aether or not. I agree it sounds impossible to have so much theory matching so many observations and yet have any fundamental misunderstandings about it, but apperantly it is possible. One more example, Broglie–Bohm theory seem to be making valid predictions just like mainstream QM, and it is using some sort of "classical trajectories". Specifically, the existence of only two hyperfine ground states indicates an absence of orbital angular momentum; any spectroscopy which resolves L+S coupling; the metastable nature of the 2S state (which doesn't even exist in the Bohr model) because it shows there is no ground state with zero angular momentum — it not be a forbidden decay. I have no idea really, but it does not sound to me as if any momentum is being measured there, indirectly perhaps. Zero angular momentum is very suspicious value to be measured, that's kind of value you also get for "out of range" and other 'unsuccessful' type of measurements.
swansont Posted March 21, 2010 Posted March 21, 2010 I have no idea really, but it does not sound to me as if any momentum is being measured there, indirectly perhaps. Zero angular momentum is very suspicious value to be measured, that's kind of value you also get for "out of range" and other 'unsuccessful' type of measurements. As with most quantum mechanical measurements, it is indirect. The allowable angular momentum sublevels for an electron are quantized, and limited to differences of h-bar. We see only two for the ground state; since we know that there are two spin orientation states (± 1/2), this implies no orbital angular momentum. Degeneracy is not masking the states, because you can split the degeneracy by adding an external magnetic field. Still only have two states. States with L=1 have four possibilities (±3/2, ±1/2) which you see in (excited, in the case of alkali atoms) D states; all of the alkali atoms easily show this pattern. These are not experiments where unsuccessful gives you the same result as successful (those are difficult experiments indeed). You see the spectroscopy signals for the existing states. Nothing suspicious about it.
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