questionposter Posted April 19, 2012 Share Posted April 19, 2012 (edited) Oh. Usually "ionization energy" refers to the energy it takes to ionize a given electron from an atom. So I thought you were talking about something entirely different. I know nothing about electron capture. I know ionization energy isn't the right term, but like I said I don't know what the right term is, so I don't know what else to call it really. There's some kind of minimum energy required to excite an electron to the state where it's boundaries overlap a proton's enough to combine, but I just don't get why exactly it isn't enough on it's own. Edited April 19, 2012 by questionposter Link to comment Share on other sites More sharing options...

swansont Posted April 19, 2012 Share Posted April 19, 2012 There's some kind of minimum energy required to excite an electron to the state where it's boundaries overlap a proton's enough to combine There is? What do you mean it requires a "weak interaction"? What are particle colliders doing that a normal atoms doesn't? If anything, colliders add more energy, unless you mean the "weak force", in which case, what does the weak force have to do with particles combining? Wouldn't the strong force have more to do with that? No. Electrons don't interact via the strong force. So they aren't going to combine with a proton via that interaction. It would make sense if there was some kind of ionization energy for electron-proton, but so far I haven't found anything like that, and I also don't know specifically why that stops the particles from combining. Perhaps the quarks are held so tightly together that it takes massive amounts of kinetic energy to break them and then for an electron to combine with them? As I already stated, to combine, the electrons would have to interact with protons (i.e. one of the quarks) via the weak interaction. http://www.cyberphysics.pwp.blueyonder.co.uk/graphics/diagrams/Feynman/Feynmanelectroncapture.gif I was thinking because electrons are waves, they physically exist, and according to the math that describes it, they have physical oscillation. You talked about probabilities. The wave function is not a wave that physically exists. Link to comment Share on other sites More sharing options...

juanrga Posted April 19, 2012 Share Posted April 19, 2012 (edited) When I say oscillation patterns, I more or less mean the evolution of their wave functions over time, or how I can actually model the physical dimensional coordinates an electron is likely to occupy over time, and with all the quantinization you have the specific orbitals that they have, that's why nodal surfaces are generated. As has been said to you before, orbitals in atoms and molecules are solutions to stationary systems. The wave functions do not evolve over time in such systems. Physical probability is any real probability greater than 0. The probability of that two electrons are in the same quantum state is 0. That is real and a fundamental principle for understanding the physics and chemistry of matter! Your claim that electrons do not "came into contact with" protons is false. Core electrons in heavy atoms directly interact with protons (more correctly they interact with the quarks that a proton is made of). As already said in a previous post, the Hydrogen-like orbitals are based in the approximation of a point-like nucleus. When both the structure and the finite size of nucleus are taken into account it, we can derive corrections to Hydrogen-like orbitals for small r. What do you mean it requires a "weak interaction"? What are particle colliders doing that a normal atoms doesn't? If anything, colliders add more energy, unless you mean the "weak force", in which case, what does the weak force have to do with particles combining? Wouldn't the strong force have more to do with that? He used the correct term "weak interaction". "Weak force" is a misnomer. I was thinking because electrons are waves, they physically exist, and according to the math that describes it, they have physical oscillation. Electrons are particles, and their fundamental properties are studied in a branch of physics named particle physics. The belief that electrons are waves or behave as waves (electrons behave as quantum particles always) is a typical misconception that one finds in non-rigorous literature (science news, popular books...). For instance, the modern theoretical physics used in CERN does not longer work with the old "wave functions" but with Dirac kets in the momentum representation |p>. Edited April 19, 2012 by juanrga Link to comment Share on other sites More sharing options...

questionposter Posted April 19, 2012 Share Posted April 19, 2012 (edited) There is? No. Electrons don't interact via the strong force. So they aren't going to combine with a proton via that interaction. As I already stated, to combine, the electrons would have to interact with protons (i.e. one of the quarks) via the weak interaction. http://www.cyberphys...troncapture.gif Yes, and there's apparently a minimum energy required to force an electron and proton together for that interaction to take place, since as you said before that weak force interaction happens over a short distance, but I still don't understand what why the weak interaction allows things to combine in the first place and it normally it doesn't then if they overlap. If weak force carrying particles are emitted from a proton, and the electron comes into contact with that proton, why would weak bosons hit the electron? I could see how there's a small distance, but if it's actually touching the proton... Electrons are particles, and their fundamental properties are studied in a branch of physics named particle physics. The belief that electrons are waves or behave as waves (electrons behave as quantum particles always) is a typical misconception that one finds in non-rigorous literature (science news, popular books...). For instance, the modern theoretical physics used in CERN does not longer work with the old "wave functions" but with Dirac kets in the momentum representation |p>. Well not completely classical waves, but particles have many properties of waves even mathematically and I don't think they create the double slit experiment result by being more like a point particle, but I suppose they can be localized to packets and have vector states. Edited April 19, 2012 by questionposter Link to comment Share on other sites More sharing options...

mississippichem Posted April 19, 2012 Share Posted April 19, 2012 Well not completely classical waves, but particles have many properties of waves even mathematically and I don't think they create the double slit experiment result by being more like a point particle, but I suppose they can be localized to packets and have vector states. Juangra is saying that in modern QM we represent quantum states as abstract complex vectors on Hilbert spaces. It has little to do with "localization" of particles. Wave mechanics works fine but get's troublesome and cumbersome in more advanced scenarios. Link to comment Share on other sites More sharing options...

swansont Posted April 19, 2012 Share Posted April 19, 2012 Yes, and there's apparently a minimum energy required to force an electron and proton together for that interaction to take place, since as you said before that weak force interaction happens over a short distance, but I still don't understand what why the weak interaction allows things to combine in the first place and it normally it doesn't then if they overlap. If weak force carrying particles are emitted from a proton, and the electron comes into contact with that proton, why would weak bosons hit the electron? I could see how there's a small distance, but if it's actually touching the proton... "Apparently"? What is it? Electron capture is a spontaneous process, i.e. it takes place without any energy being added. Link to comment Share on other sites More sharing options...

juanrga Posted April 19, 2012 Share Posted April 19, 2012 (edited) Well not completely classical waves, but particles have many properties of waves even mathematically and I don't think they create the double slit experiment result by being more like a point particle, but I suppose they can be localized to packets and have vector states. I was not referring to "classical waves" but to waves. Quantum particles have properties of particles, not of waves (quantum or otherwise). The double slit experiment is the typical example given in that non-rigorous literature, which I alluded before. As has been explained to you the 'modern' [*] formulation of QM (e.g. that used in the CERN) uses Dirac ket formalism, kets as |p> are not wave functions. Moreover, any decent textbook in QFT (Mandl & Shaw, Weinberg...) explains why the solutions to the Klein & Gordon, and Dirac quantum equations cannot be interpreted as wave functions. The term wavefunction is still used in part of the literature for denoting the product <x|Psi>, but a wavefunction is not a particle (aka the converse "a particle is a wave" is incorrect), neither a property of a particle (aka "particles have many properties of waves" is not right). The properties of a particle (mass, energy, momentum, spin...) are obtained from the set of operators. Precisely particles are characterized (and discovered) by properties as mass and spin. An electron is a particle with mass m_e and spin 1/2. No 'wavelike' property is used to define what an electron is. Therefore an electron is a particle, not a wave. [*] Traced back to Dirac! Edited April 19, 2012 by juanrga Link to comment Share on other sites More sharing options...

questionposter Posted April 19, 2012 Share Posted April 19, 2012 (edited) "Apparently"? What is it? Electron capture is a spontaneous process, i.e. it takes place without any energy being added. Why would the government and investors spend billions of dollars to build a collider to make electrons go into a proton if it happens that naturally? It would have to happen pretty rarely on Earth, but still why doesn't it actually happen more often if the two particles actually come into contact with one another? I was not referring to "classical waves" but to waves. Quantum particles have properties of particles, not of waves (quantum or otherwise). The double slit experiment is the typical example given in that non-rigorous literature, which I alluded before. As has been explained to you the 'modern' [*] formulation of QM (e.g. that used in the CERN) uses Dirac ket formalism, kets as |p> are not wave functions. Moreover, any decent textbook in QFT (Mandl & Shaw, Weinberg...) explains why the solutions to the Klein & Gordon, and Dirac quantum equations cannot be interpreted as wave functions. The term wavefunction is still used in part of the literature for denoting the product <x|Psi>, but a wavefunction is not a particle (aka the converse "a particle is a wave" is incorrect), neither a property of a particle (aka "particles have many properties of waves" is not right). The properties of a particle (mass, energy, momentum, spin...) are obtained from the set of operators. Precisely particles are characterized (and discovered) by properties as mass and spin. An electron is a particle with mass m_e and spin 1/2. No 'wavelike' property is used to define what an electron is. Therefore an electron is a particle, not a wave. [*] Traced back to Dirac! So your saying the wave function is just to describe it's probability and not the particle itself, but then what is the actual mathematics for an actual particle? Why does it have that probability in the first place? And as I understand it, spin isn't a classical notion either. It has to be more complex than that, and I've run into multiple sources referring to wave-particle duality to describe the existence of particles. If an electron isn't a wave in any shape or form, why does it have the probability of one? It seems kind of weird to say a particle isn't anything like a wave when there are specific places it can't possibly exist that perfectly match the nodal surfaces predicted by wave mechanics. Edited April 19, 2012 by questionposter Link to comment Share on other sites More sharing options...

swansont Posted April 20, 2012 Share Posted April 20, 2012 Why would the government and investors spend billions of dollars to build a collider to make electrons go into a proton if it happens that naturally? It would have to happen pretty rarely on Earth, but still why doesn't it actually happen more often if the two particles actually come into contact with one another? The point of accelerating the particles (in general) is to probe higher energy collisions, such as would have been present in much hotter conditions present early on in the universe. They are not investigating the electron capture reaction. You collide protons with each other, possibly protons and antiprotons, electrons and positrons, higher-mass ions with each other, protons agains higher-mass targets, etc. I can't think of who collides protons with electrons — what lab does this? Link to comment Share on other sites More sharing options...

questionposter Posted April 20, 2012 Share Posted April 20, 2012 (edited) The point of accelerating the particles (in general) is to probe higher energy collisions, such as would have been present in much hotter conditions present early on in the universe. They are not investigating the electron capture reaction. You collide protons with each other, possibly protons and antiprotons, electrons and positrons, higher-mass ions with each other, protons agains higher-mass targets, etc. I can't think of who collides protons with electrons — what lab does this? So how often in a random object on Earth does it actually just have an electron spontaneously fall into the nucleus? Or is it only in radioactive materials? And still, why doesn't it happen more often if it doesn't require any added energy? Edited April 20, 2012 by questionposter Link to comment Share on other sites More sharing options...

swansont Posted April 20, 2012 Share Posted April 20, 2012 So how often in a random object on Earth does it actually just have an electron spontaneously fall into the nucleus? Or is it only in radioactive materials? And still, why doesn't it happen more often if it doesn't require any added energy? "Fall into the nucleus" paints a picture of Bohr orbits, and that's wrong. Electron capture happens in radioactive nuclei, by definition. The reaction won't happen in nuclei if forming the bound neutron doesn't release energy. It doesn't happen often because this is the weak interaction, which has a very short range and the cross-sections are really small. You didn't answer my question about what lab does e-p collisions? Link to comment Share on other sites More sharing options...

questionposter Posted April 20, 2012 Share Posted April 20, 2012 (edited) "Fall into the nucleus" paints a picture of Bohr orbits, and that's wrong. Electron capture happens in radioactive nuclei, by definition. The reaction won't happen in nuclei if forming the bound neutron doesn't release energy. It doesn't happen often because this is the weak interaction, which has a very short range and the cross-sections are really small. Ok, they only happen in radioactive materials, which I think have a lower binding energy per nucleon but still have a greater total amount of potential energy seeing as how it takes a supernova to form them and it would also take a higher amount of potential energy. With radioactive materials it seems to happen naturally, but otherwise not even degenerate matter seems to be able to do it, the electrons just get squeezed out, and I wouldn't be surprised if particle colliders can do it. And you still didn't answer my question of why it doesn't happen more often if the electrons in most atoms actually come into contact with protons. You didn't answer my question about what lab does e-p collisions? Actually, based on what I can find, I can't say any lab has done ANY particular combination of particles, all I can say for sure is they have done experiments involving protons, electrons, neutrons, positrons, etc. Edited April 20, 2012 by questionposter Link to comment Share on other sites More sharing options...

swansont Posted April 20, 2012 Share Posted April 20, 2012 And you still didn't answer my question of why it doesn't happen more often if the electrons in most atoms actually come into contact with protons. I guess I should have said something like "It doesn't happen often because this is the weak interaction, which has a very short range and the cross-sections are really small." Actually, based on what I can find, I can't say any lab has done ANY particular combination of particles, all I can say for sure is they have done experiments involving protons, electrons, neutrons, positrons, etc. Ah, I see. I guess I shouldn't be surprised at this. Link to comment Share on other sites More sharing options...

juanrga Posted April 20, 2012 Share Posted April 20, 2012 (edited) So your saying the wave function is just to describe it's probability I have not said such thing. And as I understand it, spin isn't a classical notion either. I never said that the spin of a quantum particle was a classical notion. It has to be more complex than that, and I've run into multiple sources referring to wave-particle duality to describe the existence of particles. If an electron isn't a wave in any shape or form, why does it have the probability of one? It seems kind of weird to say a particle isn't anything like a wave when there are specific places it can't possibly exist that perfectly match the nodal surfaces predicted by wave mechanics. I already explained why this is wrong. It was correct to discuss about a supposed "wave-particle duality" about 70 years ago, when quantum theory was in its infancy. It is no longer correct today within the modern theory of particle physics [yes, this discipline of physics is named particle physics not wave-particle physics or something as that]; although we can still find such discussions in very old references or in non-rigorous references (e.g. in some popular books written for general public). Edited April 20, 2012 by juanrga Link to comment Share on other sites More sharing options...

questionposter Posted April 20, 2012 Share Posted April 20, 2012 (edited) I have not said such thing. Well I don't see what else makes sense, since your not really explaining it. I never said that the spin of a quantum particle was a classical notion. Then why did you say it in support of particlism? I already explained why this is wrong. It was correct to discuss about a supposed "wave-particle duality" about 70 years ago, when quantum theory was in its infancy. It is no longer correct today within the modern theory of particle physics [yes, this discipline of physics is named particle physics not wave-particle physics or something as that]; although we can still find such discussions in very old references or in non-rigorous references (e.g. in some popular books written for general public). No, you didn't explain how it was wrong, all you said was "that" it is wrong, and I don't see how spin proves it wrong seeing as how you use spin with wave mechanics to determine probability. Furthermore, how do these "operators" account for the nodal surfaces and quantinization of energy, spin, momentum, etc? How do they explain it? I guess I should have said something like "It doesn't happen often because this is the weak interaction, which has a very short range and the cross-sections are really small." So it only occurs with the weak (force) interaction? Or it doesn't occur because the process itself is a weak interaction? Because otherwise I don't see many other explanation for why the contact doesn't trigger them combining if it only occurs with the weak interaction. Why doesn't it occur without a weak force interaction if they come into contact with each other? Edited April 20, 2012 by questionposter Link to comment Share on other sites More sharing options...

swansont Posted April 21, 2012 Share Posted April 21, 2012 So it only occurs with the weak (force) interaction? Or it doesn't occur because the process itself is a weak interaction? Because otherwise I don't see many other explanation for why the contact doesn't trigger them combining if it only occurs with the weak interaction. Why doesn't it occur without a weak force interaction if they come into contact with each other? It requires the weak interaction because it is a weak interaction. Gravity is negligible, the electrostatic interaction only binds the electron into an atomic orbital and electrons do not interact via the strong force. There's nothing left. Link to comment Share on other sites More sharing options...

questionposter Posted April 21, 2012 Share Posted April 21, 2012 (edited) It requires the weak interaction because it is a weak interaction. Gravity is negligible, the electrostatic interaction only binds the electron into an atomic orbital and electrons do not interact via the strong force. There's nothing left. So if they don't interact via the strong force, don't they have different oscillation? That or, doesn't it require energy for an electron to overcome the strong force? And anyway, why doesn't opposite charge cause them to interact if they come into contact anyway? So what if they don't interact via the strong force? Opposite charges still attract... Edited April 21, 2012 by questionposter Link to comment Share on other sites More sharing options...

juanrga Posted April 21, 2012 Share Posted April 21, 2012 (edited) Well I don't see what else makes sense, since your not really explaining it. My posts are rather precise, but you attribute to me stuff I have not said. Then why did you say it in support of particlism? Because particle spin (a quantum property) is one of the properties that define a quantum particle. As said before: The properties of a particle (mass, energy, momentum, spin...) are obtained from the set of operators. Precisely particles are characterized (and discovered) by properties as mass and spin. An electron is a particle with mass m_e and spin 1/2. No 'wavelike' property is used to define what an electron is. Therefore an electron is a particle, not a wave. If a given particle does not have mass m_{e} and spin 1/2, then it is not an electron but some other particle. No, you didn't explain how it was wrong, all you said was "that" it is wrong, and I don't see how spin proves it wrong seeing as how you use spin with wave mechanics to determine probability. Furthermore, how do these "operators" account for the nodal surfaces and quantinization of energy, spin, momentum, etc? How do they explain it? In the first place, I never said that spin proves wrong anything. You seem to be misreading me again. In the second place, those operators describe the observable properties of the particles. For instance, the mass operator gives the mass of the particle, the spin operator gives the spin of the particle and so on. In the third place, I already explained to you why your viewpoint is both wrong and outdated. For instance, I already explained to you that the modern formulation of quantum theory uses Dirac ket formalism and that the kets are not wavefunctions. I explained to you that wavefunctions are not used in experiments done at CERN, where new particles are discovered. Scattering events are interpreted using kets |p> in the momentum basis. I also said you that solutions to Dirac and Klein-Gordon equations cannot be interpreted as wavefunctions although, in the early years of quantum mechanics (about 1930), physicists believed the contrary. We are not living in the year 1930. Physics has advanced. I have even cited a standard modern textbook which will give you the technical details on why the solutions to those equations cannot be interpreted as wavefunctions. There are much more stuff which I am not saying here because lacking time and because you seem to be lacking the adequate background. For instance, originally Schrödinger worked with one-particle systems and then confounded the solution to his equation with some kind of wave somewhat as electromagnetic waves. However, it is now well-understood that for a multiparticle system, the 'wavefunction' is defined in a parametrized 3N space, which is not the four-dimensional space where electromagnetic waves are defined. Precisely the strong differences between both spaces generate well-known difficulties when trying to mix relativity and quantum mechanics. And this is related to why the attempt to interpret the solutions to Dirac and Klein & Gordon as wavefunctions fails. I am also not citing (and still less discussing) here recent works as [1] where it is shown that wavefunctions, or even Dirac kets |Psi>, cannot describe a large kind of quantum systems. If you want study the dissipative behaviour of an electron in an electron transfer reaction you cannot use wavefunctions neither Dirac kets. [1] PETROSKY, T; PRIGOGINE, I. The Liouville Space Extension of Quantum Mechanics. Adv. Chem. Phys 1997; Vol. 99, pp.1–120. Edited April 21, 2012 by juanrga 1 Link to comment Share on other sites More sharing options...

questionposter Posted April 21, 2012 Share Posted April 21, 2012 (edited) My posts are rather precise, but you attribute to me stuff I have not said. Because particle spin (a quantum property) is one of the properties that define a quantum particle. As said before: If a given particle does not have mass m_{e} and spin 1/2, then it is not an electron but some other particle. In the first place, I never said that spin proves wrong anything. You seem to be misreading me again. In the second place, those operators describe the observable properties of the particles. For instance, the mass operator gives the mass of the particle, the spin operator gives the spin of the particle and so on. In the third place, I already explained to you why your viewpoint is both wrong and outdated. For instance, I already explained to you that the modern formulation of quantum theory uses Dirac ket formalism and that the kets are not wavefunctions. I explained to you that wavefunctions are not used in experiments done at CERN, where new particles are discovered. Scattering events are interpreted using kets |p> in the momentum basis. I also said you that solutions to Dirac and Klein-Gordon equations cannot be interpreted as wavefunctions although, in the early years of quantum mechanics (about 1930), physicists believed the contrary. We are not living in the year 1930. Physics has advanced. I have even cited a standard modern textbook which will give you the technical details on why the solutions to those equations cannot be interpreted as wavefunctions. There are much more stuff which I am not saying here because lacking time and because you seem to be lacking the adequate background. For instance, originally Schrödinger worked with one-particle systems and then confounded the solution to his equation with some kind of wave somewhat as electromagnetic waves. However, it is now well-understood that for a multiparticle system, the 'wavefunction' is defined in a parametrized 3N space, which is not the four-dimensional space where electromagnetic waves are defined. Precisely the strong differences between both spaces generate well-known difficulties when trying to mix relativity and quantum mechanics. And this is related to why the attempt to interpret the solutions to Dirac and Klein & Gordon as wavefunctions fails. I am also not citing (and still less discussing) here recent works as [1] where it is shown that wavefunctions, or even Dirac kets |Psi>, cannot describe a large kind of quantum systems. If you want study the dissipative behaviour of an electron in an electron transfer reaction you cannot use wavefunctions neither Dirac kets. [1] PETROSKY, T; PRIGOGINE, I. The Liouville Space Extension of Quantum Mechanics. Adv. Chem. Phys 1997; Vol. 99, pp.1–120. I'm ok with a "quantum particle", but saying particles have nothing to do with waves makes no sense at this point seeing as how many aspects of them can be accurately described with wave mechanics, and these operators don't seem to explain why particles act how they act, it just seems to generate numbers according to what a mathematician or a programmer tells them, and nothing more. It doesn't explain why those numbers exist in the first place. It's like saying gravity isn't related to mass. Also, I did not see that book at the library. I don't think I ever said particles were "only" waves, but I just don't see how they aren't related to them. Edited April 21, 2012 by questionposter Link to comment Share on other sites More sharing options...

juanrga Posted April 22, 2012 Share Posted April 22, 2012 (edited) I'm ok with a "quantum particle", but saying particles have nothing to do with waves makes no sense at this point seeing as how many aspects of them can be accurately described with wave mechanics, and these operators don't seem to explain why particles act how they act, it just seems to generate numbers according to what a mathematician or a programmer tells them, and nothing more. It doesn't explain why those numbers exist in the first place. It's like saying gravity isn't related to mass. Also, I did not see that book at the library. Well, I think that I already explained why "wave mechanics" is a misnomer, there is none wave, which is the historical origin for this misnaming, and why "wave mechanics" gives only an approximated description (i.e. it is far from being fundamental). Regarding your other claims, an operator does not "generate numbers", but gives the observable (mass is not a number, energy is not a number...). When scientists write something as H |Psi> = E |Psi>, with H the Hamiltonian operator, the E is not a number but a physical quantity. I don't think I ever said particles were "only" waves, but I just don't see how they aren't related to them. I have not checked the entire thread, but in #19 you wrote that "electrons are waves". Even without the "only", what you said in #19 was not true. Edited April 22, 2012 by juanrga Link to comment Share on other sites More sharing options...

michel123456 Posted April 22, 2012 Share Posted April 22, 2012 (edited) (...)I already explained why this is wrong. It was correct to discuss about a supposed "wave-particle duality" about 70 years ago, when quantum theory was in its infancy. It is no longer correct today within the modern theory of particle physics [yes, this discipline of physics is named particle physics not wave-particle physics or something as that]; although we can still find such discussions in very old references or in non-rigorous references (e.g. in some popular books written for general public). So you say this article about wave-particle duality is obsolete? Edited April 22, 2012 by michel123456 Link to comment Share on other sites More sharing options...

questionposter Posted April 22, 2012 Share Posted April 22, 2012 (edited) Regarding your other claims, an operator does not "generate numbers", but gives the observable (mass is not a number, energy is not a number...). When scientists write something as H |Psi> = E |Psi>, with H the Hamiltonian operator, the E is not a number but a physical quantity. I guess a Hamiltonian operator can still account for a bit of it, but what about the exact matches of nodal surfaces both in atomic orbitals and in the double slit experiment that can be accurately described by wave mechanics? How does a Hamiltonian operator account for those without wave mechanics? Also, can't these operators be set equal to a wave function? I problem that I see with your operators is that they are nothing more than shortcuts, they could only have been created with previous knowledge, and that previous knowledge was wave mechanics. People found out from experiments that there can only be specific results that work, so instead of constantly trying to re-create those results from scratch, they simply created operators specifically designed for creating those results since the results an operator generates are the only possible results (if the correct information in put in), where-as with a sine wave there's millions of possible ways to combine them, but you can still create an accurate model of an atom using them if you do a ton of work. Edited April 22, 2012 by questionposter Link to comment Share on other sites More sharing options...

Xittenn Posted April 22, 2012 Share Posted April 22, 2012 For instance, I already explained to you that the modern formulation of quantum theory uses Dirac ket formalism and that the kets are not wavefunctions. I explained to you that wavefunctions are not used in experiments done at CERN, where new particles are discovered. Scattering events are interpreted using kets |p> in the momentum basis. I also said you that solutions to Dirac and Klein-Gordon equations cannot be interpreted as wavefunctions although, in the early years of quantum mechanics (about 1930), physicists believed the contrary. We are not living in the year 1930. Physics has advanced. I have even cited a standard modern textbook which will give you the technical details on why the solutions to those equations cannot be interpreted as wavefunctions. Thanks for pointing these things out juanrga! Link to comment Share on other sites More sharing options...

Royston Posted April 22, 2012 Share Posted April 22, 2012 (edited) I guess a Hamiltonian operator can still account for a bit of it, but what about the exact matches of nodal surfaces both in atomic orbitals and in the double slit experiment that can be accurately described by wave mechanics? How does a Hamiltonian operator account for those without wave mechanics? Also, can't these operators be set equal to a wave function? I think you could do with a few preliminaries, because you're not making any sense. This is the basics, (we'll stick with one dimension) but hopefully you'll get the idea. Operators An operator which is denoted with a hat i.e [math]\hat{O}[/math] transforms a function, say [math]f(x)[/math] to another function. So if the function is [math]f(x)=2x^2[/math] and the operator is [math]\frac {d}{dx}[/math] then [math]\hat{O}f(x) = \frac{d}{dx}2x^2 = 4x[/math] There's a good tutorial on differentiation here, if the above doesn't make sense to you. Eigenfunctions and eigenvalues Considering the role of the operator above, we can move on to eigenvalue equations. These have the relationship... [math]\hat{O}f(x) = \lambda f(x)[/math] where [math]\lambda[/math] is a complex constant. So you can see, in order for it to be an eigenvalue equation, the operator has to return the same function multiplied by a constant. The constant is the eigenvalue, and [math]f(x)[/math] is the eigenfunction. So for instance [math]\hat{O}f(x) = \frac{d^2}{dx^2}(sin(bx)) = \frac{d}{dx}(b\, cos(bx)) = -b^2 sin(bx)[/math]. In this case, [math]-b^2[/math] is the eigenvalue, and the function has not changed, so [math]sin(bx)[/math] is an eigenfunction of the operator. The de Broglie wavefunction The de Broglie relationship for an electron in an isolated system i.e free from any disturbances (forces) is [math]\lambda_{dB} = \frac{h}{p}[/math]. Where [math]\lambda_{dB}[/math] is the de Broglie wavelength, h is Planck's constant, and p is momentum. You've probably seen a wave equation before, such as [math]f(x,t) = A\,cos(kx-wt)[/math]. Where A is the amplitude of the wave, k is the wave number, [math]k = \frac{2\pi}{\lambda}[/math] and w is the angular frequency [math]w = 2\pi f[/math] (f is the frequency i.e 1/T where T is the period). The x in the equation, just denotes the direction the wave is travelling. Now, wave functions are complex, so we can write [math]\Psi_{dB}(x,t) = A[cos(kx-wt) + i\,sin(kx-wt)]=A\,e^{i(kx-wt)}[/math] (see Euler's formula) Remembering the de Broglie relationship, this implies that [math]k = \frac{2\pi\,p}{h} = \frac{p}{\hbar}[/math] where [math]\hbar = \frac{h}{2\pi}[/math] Also the photon energy is [math]E=\hbar w[/math] so [math]w = \frac{E}{\hbar}[/math] The Hamiltonian function You're probably familiar with Newtons equation [math]F=ma[/math]. This can be reformulated, by noting that F relates to the gradient of the potential energy function V(x). So for an object moving in the x direction, we have [math]F_x = -\frac{\partial V}{\partial x}[/math] Look up partial differentiation, if you're not sure about what the RHS means. Further, the derivative of velocity is acceleration, and p=mv (i.e momentum equals mass times velocity) so [math]ma_x = \frac{d(mv_x)}{dt} = \frac{dp_x}{dt}[/math] therefore we have... [math]\frac{dp_x}{dt} = -\frac{\partial V}{\partial x}[/math] Now, kinetic energy in terms of the momentum of a particle is [math]E_k = \frac{p^2}{2m}[/math] The Hamiltonian function, is the sum of the kinetic energy and potential energy of a system, with a particle of mass m, and potential energy V(x), so... [math]H = \frac{p^2_x}{2m} + V(x)[/math]. Now [math]\frac{p^2_x}{2m}[/math] is not explicitly dependant on x, so [math]\frac{\partial H}{\partial x} = \frac{\partial V}{\partial x}[/math] So, recalling the reformulated Newtonian equation, we have [math]\frac{dp_x}{dt} = -\frac{\partial H}{\partial x}[/math] The Hamiltonian operator Now we need to convert an observable (in this case the energy of the system) into an operator. Recalling the de Broglie wave function, if we partially differentiate it twice, we get (1) [math]\frac{\partial}{\partial x}\Psi (x,t) = ik\Psi(x,t)[/math] (2) [math]\frac{\partial^2}{\partial x^2}\Psi (x,t) = (ik)^2\Psi(x,t) = -k^2 \Psi(x,t)[/math] Now to obtain the eigenvalues for kinetic energy and momentum, we multiply eq (1) with [math]-i\hbar[/math] and eq (2) with [math]-\hbar^2 / 2m[/math] so (1.1) [math]-i\hbar\frac{\partial}{\partial x}\Psi (x,t) = \hbar k\Psi(x,t)[/math] (2.2) [math]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi (x,t) = -\frac{(\hbar k)^2}{2m} \Psi(x,t)[/math] So, [math]p_x \Longrightarrow \hat{p_x} = -i\hbar\frac{\partial}{\partial x}[/math] and [math] E_k \Longrightarrow \hat{E}_k = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}[/math] The arrow just indicates we're going from a classical variable to a quantum operator. Therefore the kinetic energy operator is simply [math]\frac{\hat p^2_x}{2m} = \frac{1}{2m} \left(-i\hbar \frac{\partial}{\partial x} \right)^2 = - \frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}[/math] So the Hamiltonian operator is (remembering the Hamiltonian function) [math]\hat{H} = \frac{\hat{p^2_x}}{2m} + \hat{V}(x) = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2} + V(x)[/math] The Schrodinger equation So finally, we're in a position to state the Schrodinger equation, which for a free particle could take the form [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = \hat{E}_k \Psi(x,t)[/math] But for a bound particle, i.e there is a potential energy term, we require that [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = (\hat{E}_k + \hat{V}(x)) \Psi(x,t)[/math] So using the kinetic energy operator, and the potential operator, this gives... [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x) \Psi (x,t)[/math] more compactly... [math]i \hbar \frac{\partial}{\partial t}\Psi (x,t) = \hat{H}\Psi(x,t)[/math] Hopefully, this should give you a better understanding of operators, the Hamiltonian and wave functions (it's just a basic treatment). I'm not sure what your maths level is, but you can always ask if there's anything that you find confusing. Edited April 22, 2012 by Royston 6 Link to comment Share on other sites More sharing options...

michel123456 Posted April 22, 2012 Share Posted April 22, 2012 So you say this article about wave-particle duality is obsolete? Interesting: This page was last modified on 22 April 2012 at 05:27. quoted from the linked wiki page. Link to comment Share on other sites More sharing options...

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