studiot 2240 Posted April 26, 2018 Share Posted April 26, 2018 (edited) 9 hours ago, Rob McEachern said: 1) You stated you were only concerned with solutions "for a single particle translating freely in space": a potential, that has any effect whatsoever on a particle, implies that the particle is not free. 2) is not even a function of time, so it does not represent anything "translating" anywhere. 1) I don't think I said I was only concerned with solution for a single particle translating freely in space. I think I said this was the simplest solution. It is the simplest because of the second part of your statement. 2) Thank you for bringing to my attention that I missed a (very important) variable. Here is the correct equation, including Energy. [math]\Psi = A\sin \sqrt {\left( {\frac{{8{\pi ^2}mE}}{{{h^2}}}} \right)} [/math] Bearing in mind that the wave function itself, psi, does not translate - that is the activity of the 'particle', The question then arises, How can the energy of that particle change if it is (your words) not subject to any forces or (my words) not subject to any potential? Which brings me back to my question you have not answered What does your input have to do with a particular interpretation of QM as that is the OP? @swansont - any chance you could put a pointer on my previous incorrect post as I can't now alter it? Edited April 26, 2018 by studiot Link to post Share on other sites

Rob McEachern 14 Posted April 27, 2018 Share Posted April 27, 2018 15 hours ago, studiot said: How can the energy of that particle change It does not change in your solution. That solution was derived by assuming the potential is zero, and E is a constant. You can find a simple derivation of that solution, in the last answer given on this page: https://physics.stackexchange.com/questions/347996/when-solving-the-schrodinger-equation-how-do-we-know-what-functions-to-use-exp Just set the potential V_{0} = 0 and solve for "k". But note that now, you have left out the "x" in the solution. It should also be pointed out, that the very first step in this "derivation", assumed that the derivative of "k" with respect to x is zero. Think about the consequences of that assumption. 15 hours ago, studiot said: What does your input have to do with a particular interpretation of QM Everything. Because it demonstrates that Schrödinger's equation is just the expression for the Fourier transform of a wave function, transformed into a differential equation. Consequently, all the properties associated with wave functions (such as the uncertainty principle, superposition, entanglement, vacuum fluctuations, the Born rule etc.), turn out to be purely mathematical properties of Fourier transforms and have nothing to do with physics per se. In other words, these properties are properties of the mathematical language being used to describe physical phenomenon - not properties of the phenomenon being described. Consequently, attempting to interpret any of these properties as a physical phenomenon, rather than as mere side-effects of a specific mathematical technique (Fourier transforms) being employed, is a pitfall, that has confused the physics community for 90 years. 17 hours ago, swansont said: What about a constant positive potential everywhere? Would that have an impact on the particle being free? See: https://books.google.com/books?id=hEHCAgAAQBAJ&pg=PA232&lpg=PA232&dq="In+any+region+where+V+is+constant,+the+solution+of+the+wave+equation+is"&source=bl&ots=n0Kf89ulEE&sig=b-_mFvKHojT3NTPKOew-0CugX-g&hl=en&sa=X&ved=0ahUKEwiG2cTrvdnaAhWjrVkKHTgiCDoQ6AEIKjAA#v=onepage&q="In any region where V is constant%2C the solution of the wave equation is"&f=false Link to post Share on other sites

studiot 2240 Posted April 27, 2018 Share Posted April 27, 2018 4 hours ago, Rob McEachern said: It does not change in your solution. That solution was derived by assuming the potential is zero, and E is a constant. You can find a simple derivation of that solution, in the last answer given on this page: https://physics.stackexchange.com/questions/347996/when-solving-the-schrodinger-equation-how-do-we-know-what-functions-to-use-exp Just set the potential V_{0} = 0 and solve for "k". But note that now, you have left out the "x" in the solution. It should also be pointed out, that the very first step in this "derivation", assumed that the derivative of "k" with respect to x is zero. Think about the consequences of that assumption. So are you suggesting there is a solution to the circumstances presented whereby the energy can change? 4 hours ago, Rob McEachern said: Everything. Because it demonstrates that Schrödinger's equation is just the expression for the Fourier transform of a wave function, transformed into a differential equation. Consequently, all the properties associated with wave functions (such as the uncertainty principle, superposition, entanglement, vacuum fluctuations, the Born rule etc.), turn out to be purely mathematical properties of Fourier transforms and have nothing to do with physics per se. In other words, these properties are properties of the mathematical language being used to describe physical phenomenon - not properties of the phenomenon being described. Consequently, attempting to interpret any of these properties as a physical phenomenon, rather than as mere side-effects of a specific mathematical technique (Fourier transforms) being employed, is a pitfall, that has confused the physics community for 90 years. At last, thank you for answering the OP question. Can I take it your answer is No, Copenhagen is inappropriate? You could have said that a long time ago. I would certainly agree with you that many of physical phenomena are also described or inherent in the maths. But all this does is demonstrate what a jolly good model that particular piece of mathematics is. (and therefore why we use it). As to Mathematics itself, yes every differential equation can be remodelled as an integral equation - I have several books on the subject. But I would also council two things "Beware of all or nothing statements" They have a habit of tripping themselves up. "Beware of saying that it is a consequence of the maths and nothing more" There is more to the mathematical modelling of physical processes than that. Take 'centrifugal force' for example. This is physically evident from some points of view (models) but not others. Link to post Share on other sites

swansont 7597 Posted April 27, 2018 Share Posted April 27, 2018 9 hours ago, Rob McEachern said: I See: https://books.google.com/books?id=hEHCAgAAQBAJ&pg=PA232&lpg=PA232&dq="In+any+region+where+V+is+constant,+the+solution+of+the+wave+equation+is"&source=bl&ots=n0Kf89ulEE&sig=b-_mFvKHojT3NTPKOew-0CugX-g&hl=en&sa=X&ved=0ahUKEwiG2cTrvdnaAhWjrVkKHTgiCDoQ6AEIKjAA#v=onepage&q="In any region where V is constant%2C the solution of the wave equation is"&f=false That's for a potential well, i.e. V < 0, but for V > 0, how does that imply the particle is bound? Link to post Share on other sites

Rob McEachern 14 Posted April 27, 2018 Share Posted April 27, 2018 2 hours ago, swansont said: That's for a potential well There can be no well, if the potential is constant everywhere; wells require more than one level of potential. Where there is a well, there can be no free particle. 4 hours ago, studiot said: So are you suggesting there is a solution to the circumstances presented whereby the energy can change? Nothing can change in the solution you gave, because, in the derivation of that solution, "k" was unwittingly assumed to be independent of both time and position. Link to post Share on other sites

Endy0816 457 Posted April 27, 2018 Share Posted April 27, 2018 (edited) In the book, there's a figure of a square potential well above. Looks like talking about a constant V that is also less than zero. Edited April 27, 2018 by Endy0816 Link to post Share on other sites

studiot 2240 Posted April 27, 2018 Share Posted April 27, 2018 (edited) 2 hours ago, Rob McEachern said: Nothing can change in the solution you gave, because, in the derivation of that solution, "k" was unwittingly assumed to be independent of both time and position. You have said this twice, but that doesn't make it either true or an explanation/amswer to my question. I don't know is that is because you are entirely missing the point, or what. Potentials act over space not time. But you mentioned forces, not potentials and forces change the energy of a body by doing work on it. I do not need a fancy differential or integral equation to know that if no work is done on a free body the energy of a body does not change. Edited April 27, 2018 by studiot Link to post Share on other sites

swansont 7597 Posted April 27, 2018 Share Posted April 27, 2018 11 hours ago, Rob McEachern said: There can be no well, if the potential is constant everywhere; wells require more than one level of potential. Where there is a well, there can be no free particle. Maybe you could explain that to the person who linked to the example, and also answer my question. Link to post Share on other sites

Rob McEachern 14 Posted April 28, 2018 Share Posted April 28, 2018 23 hours ago, Endy0816 said: In the book, there's a figure of a square potential well above. Looks like talking about a constant V that is also less than zero. In the book, there is a two-step process to evaluate a square potential: (1) solve the equation when the potential is constant everywhere (Swansont's question), which is done in the first paragraph, and then (2) evaluate what happens when a second level is introduced. But, if there is no second level (as in the question that was asked), then all the rest of the book, after the first paragraph is irrelevant: There is no second level that is less than the first. Potential wells exist when there is a relative change in potential. 20 hours ago, studiot said: You have said this twice, but that doesn't make it either true The partial derivative of kx with respect to x, depends upon both the derivative of k and the derivative of x, with respect to x. The derivation of your solution assumed the derivative of k with respect to x is identically zero - there is no dependency on position. It also assumed that the derivative with respect to time is also identically zero - there is no dependency on time. Hence, k is being assumed to be a constant, right from the start; it depends on neither of the two variables (position and time) in the equation. I fail to see why you find that to be an interesting result; all it says is that a static sine-wave exists. How does that revelation enlighten your knowledge of reality? Link to post Share on other sites

studiot 2240 Posted April 28, 2018 Share Posted April 28, 2018 40 minutes ago, Rob McEachern said: In the book, there is a two-step process to evaluate a square potential: (1) solve the equation when the potential is constant everywhere (Swansont's question), which is done in the first paragraph, and then (2) evaluate what happens when a second level is introduced. But, if there is no second level (as in the question that was asked), then all the rest of the book, after the first paragraph is irrelevant: There is no second level that is less than the first. Potential wells exist when there is a relative change in potential. The partial derivative of kx with respect to x, depends upon both the derivative of k and the derivative of x, with respect to x. The derivation of your solution assumed the derivative of k with respect to x is identically zero - there is no dependency on position. It also assumed that the derivative with respect to time is also identically zero - there is no dependency on time. Hence, k is being assumed to be a constant, right from the start; it depends on neither of the two variables (position and time) in the equation. I fail to see why you find that to be an interesting result; all it says is that a static sine-wave exists. How does that revelation enlighten your knowledge of reality? This is still avoiding the point I made / question I asked which was How is it possible for a particle to change its energy if no force acts on it? Link to post Share on other sites

Endy0816 457 Posted April 28, 2018 Share Posted April 28, 2018 1 hour ago, Rob McEachern said: In the book, there is a two-step process to evaluate a square potential: (1) solve the equation when the potential is constant everywhere (Swansont's question), which is done in the first paragraph, and then (2) evaluate what happens when a second level is introduced. But, if there is no second level (as in the question that was asked), then all the rest of the book, after the first paragraph is irrelevant: There is no second level that is less than the first. Potential wells exist when there is a relative change Really doesn't sound correct based on the material. Link to post Share on other sites

Rob McEachern 14 Posted April 28, 2018 Share Posted April 28, 2018 10 hours ago, studiot said: How is it possible for a particle to change its energy if no force acts on it? It is not possible. The solution that you gave and claimed; "Which I think is more useful as you can plug numbers into it." Is not a solution to Schrödinger's equation, if the energy changes. 9 hours ago, Endy0816 said: Really doesn't sound correct based on the material. You will find the same thing in any other text book on the subject. The author of the text cited, was David Bohm - a rather highly regarded figure, in quantum theory. Link to post Share on other sites

Endy0816 457 Posted April 29, 2018 Share Posted April 29, 2018 Could you be cherry picking something that you think supports your position? Link to post Share on other sites

swansont 7597 Posted April 29, 2018 Share Posted April 29, 2018 On April 27, 2018 at 8:24 AM, Rob McEachern said: There can be no well, if the potential is constant everywhere; wells require more than one level of potential. Glad we cleared that up. "a potential, that has any effect whatsoever on a particle, implies that the particle is not free." is incorrect. In fact, the solution you provided shows that if E > V, the particle is free Link to post Share on other sites

Rob McEachern 14 Posted April 29, 2018 Share Posted April 29, 2018 1 hour ago, swansont said: "In fact, the solution you provided shows that if E > V, the particle is free Of course - because the potential has no effect, thus, the particle is free. Here is a quote from Merzbacher, "Quantum Mechanics", second edition, pages 80-81: 1 hour ago, Endy0816 said: Could you be cherry picking something that you think supports your position? . Link to post Share on other sites

swansont 7597 Posted April 29, 2018 Share Posted April 29, 2018 10 hours ago, Rob McEachern said: Of course - because the potential has no effect, thus, the particle is free. Here is a quote from Merzbacher, "Quantum Mechanics", second edition, pages 80-81: . It's not that the potential has no effect (if V changes the momentum is different, and thus the wavelength changes). The expectation value of finding the particle in some region will be different). It's that the particle is not "bound" to any particular area. Link to post Share on other sites

Rob McEachern 14 Posted April 30, 2018 Share Posted April 30, 2018 10 hours ago, swansont said: It's not that the potential has no effect Yes it is. E= (Kinetic energy + Potential energy) = (K+V). So (E - V) = K+V-V = K; the potential exactly cancels out in the special case, where V is a constant, regardless of what the value of the constant is. Note also that, given the expression for p, p^{2}/2m = (E-V) = K = mv^{2}/2, just as one would expect, for a free particle with only kinetic energy. The momentum cannot change if the particle is free. Link to post Share on other sites

swansont 7597 Posted April 30, 2018 Share Posted April 30, 2018 10 hours ago, Rob McEachern said: Yes it is. E= (Kinetic energy + Potential energy) = (K+V). So (E - V) = K+V-V = K; the potential exactly cancels out in the special case, where V is a constant, regardless of what the value of the constant is. Note also that, given the expression for p, p^{2}/2m = (E-V) = K = mv^{2}/2, just as one would expect, for a free particle with only kinetic energy. The momentum cannot change if the particle is free. So if I send an electron with some KE into a region at -500 volts, relative to that region, it will not change its KE, or its momentum? Quick, go tell the people working with particle accelerators that their devices don't work. The reason that's ridiculous is that since K = E-V, if you change V and E is constant, then K must change. Which should be obvious. Link to post Share on other sites

Rob McEachern 14 Posted May 1, 2018 Share Posted May 1, 2018 14 hours ago, swansont said: if you change V and E is constant, then K must change. Which should be obvious. It should be even more obvious, that if you change V, V will no longer be constant, as stipulated in the text book quotes, as the defining property of a free particle. Link to post Share on other sites

swansont 7597 Posted May 1, 2018 Share Posted May 1, 2018 9 hours ago, Rob McEachern said: It should be even more obvious, that if you change V, V will no longer be constant, as stipulated in the text book quotes, as the defining property of a free particle. That's not what it said. It said it was the trivial case of a free particle. Link to post Share on other sites

Anthony Appleyard 0 Posted May 19, 2018 Share Posted May 19, 2018 It seems to me that something "being noticed" is the sum of various neuro-electrical activity in the observer's eyes and brain, and is a part of the sum of all near and remote electrical and other effects that opening the box has. As soon as the alpha-particle flies and is detected, the effect and its wave equation have spread far further and over far more atoms than the sort of distance that quantum mechanics tends to usually extend. Link to post Share on other sites

Alex Caledin 2 Posted April 7, 2020 Share Posted April 7, 2020 (edited) Is any thought experiment with observable interference of Schrodinger's Cat(s) ever possible? (Probably, the answer is No, - and then that's why the Cat paradox just shows that we are agnostic about when the collapse occurs.) Edited April 7, 2020 by Alex Caledin Link to post Share on other sites

dimreepr 1101 Posted April 7, 2020 Share Posted April 7, 2020 (edited) How many light bulbs does it take to change a physicist? 🤒 Edited April 7, 2020 by dimreepr 1 Link to post Share on other sites

Mordred 1372 Posted April 8, 2020 Share Posted April 8, 2020 13 hours ago, dimreepr said: How many light bulbs does it take to change a physicist? 🤒 The formula would be be too complex to explain to the average layman lmao 1 Link to post Share on other sites

swansont 7597 Posted April 8, 2020 Share Posted April 8, 2020 20 hours ago, Alex Caledin said: Is any thought experiment with observable interference of Schrodinger's Cat(s) ever possible? (Probably, the answer is No, - and then that's why the Cat paradox just shows that we are agnostic about when the collapse occurs.) What would this interference look like? The point of the example was to show that applying QM to macroscopic systems leads to absurdities. Link to post Share on other sites

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