Bill Hobba Posted February 25 Report Share Posted February 25 Like many people, I have struggled to understand what QM is and what it says. Recently, after many years I have formed the following answer. There is a reality out there, independent of us and amenable to rational analysis by the conscious mind. Some parts of that reality we directly interact with every day. Others, like Electric fields, are necessary for well-understood laws to hold (eg Wigner proved if there were no electric fields, then conservation of momentum would not hold in violation of Noether's Theorem.) This is everyday stuff. But we know things like electrons exist, and we cannot directly interact with them; all we can do is interact with them using other things and find out what happens. They are equally as real as all the other things - it is just to know about them; we must interact with them somehow. At rock bottom, QM is a theory about such interactions. It follows from a straightforward model of interactions and their outcomes. Suppose two systems interact, and the result is several possible outcomes. We imagine that, at least conceptually, these outcomes can be displayed as a number on a digital readout. Such is called an observation, but it is an interaction between two systems. You may think all I need to know is the number. But I will be a bit more general than this and allow different outcomes to have the same number. To model this, we write the number from the digital readout of the ith outcome in position i of a vector. We arrange all the possible outcomes as a square matrix with the numbers on the diagonal. Those who know some linear algebra recognise this as a linear operator in diagonal matrix form. To be as general as possible, this is logically equivalent to a hermitian matrix in an assumed complex vector space where the eigenvalues are the possible outcomes. Why complex? That is a profound mystery of QM - it needs a complex vector space. Those that have calculated eigenvalues and eigenvectors of operators know they often have complex eigenvectors - so from an applied math viewpoint, it is only natural. But just because something is natural mathematically does not mean nature must oblige. So we have the first Axiom of Quantum Mechanics: To every observation, there exists a hermitian operator from a complex vector space such that its eigenvalues are the possible outcomes of the observation. This is called the Observable of the observation. But we have seen there is nothing mystical or strange about it - it is just a common sense way to model observations. The only actual physical assumption is it is from a complex vector space. Believe it or not, that is all we need to develop Quantum Mechanics. This is because of a theorem called Gleason's Theorem, a simple proof of which has recently been found: https://www.arxiv-vanity.com/papers/quant-ph/9909073/ This leads to the second axiom of QM. The expected value of the outcome of any observable O, E(O), is E(O) = trace (OS), where S is a positive matrix of unit trace, called the state of a system. Believe it or not, this is all that is needed to derive QM. See Ballentine - Quantum Mechanics - A Modern Development. https://www.amazon.com/QUANTUM-MECHANICS-MODERN-DEVELOPMENT-2ND-dp-9814578576/dp/9814578576 Quote Link to comment Share on other sites More sharing options...

Boydstun Posted February 25 Report Share Posted February 25 Hear! Hear! Can you say anything, Bill, about how operator representations in QM meld into algebraic-variable representations in the classical regime? Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 25 Report Share Posted February 25 2 hours ago, Bill Hobba said: The only actual physical assumption is it is from a complex vector space. What do you have in mind when you say this? How can a “physical assumption” be that something is from an abstraction? [I am very familiar with complex numbers and vector “spaces” and their applications] Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 25 Author Report Share Posted February 25 22 minutes ago, StrictlyLogical said: How can a “physical assumption” be that something is from an abstraction? If we assume a complex space, linking the theory when developed further to physical concepts can be done. It's just one way of doing it, but we find the Poisson bracket from classical mechanics needs complex numbers: https://farside.ph.utexas.edu/teaching/qm/Quantum/node20.html Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 26 Report Share Posted February 26 7 hours ago, Bill Hobba said: If we assume a complex space, linking the theory when developed further to physical concepts can be done. It's just one way of doing it, but we find the Poisson bracket from classical mechanics needs complex numbers: https://farside.ph.utexas.edu/teaching/qm/Quantum/node20.html I see nothing here which gives rise to a “physical assumption”. The physical things and attributes (observables) are measured as possessing real magnitudes. How we calculate expected measurements involves complex numbers. These are distinguishable. What is the “physical assumption”? Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 26 Author Report Share Posted February 26 I think you are correct. It is not a physical assumption - it is a consistency assumption. If we did not allow complex numbers then the poisson bracket would not be an observable. Thanks for that. Thanks Bill Quote Link to comment Share on other sites More sharing options...

AlexL Posted February 26 Report Share Posted February 26 (edited) 21 hours ago, Bill Hobba said: I have struggled to understand what QM is and what it says. Recently, after many years I have formed the following answer [...] Intriguing... Just a question of detail: 21 hours ago, Bill Hobba said: Wigner proved if there were no electric fields, then conservation of momentum would not hold in violation of Noether's Theorem. Can you please elaborate - at any level you feel comfortable with? I am surprised because, in classical mechanics at least, the total 3-momentum of a system is conserved (in the absence of external forces/fields), whatever the internal interactions are, that is with or without the electromagnetic fields. And this holds also for the relativistic 4-momentum. In both cases, the conservation of momentum follows from the translation invariance, that is from the homogeneity of space (and time)... Did Wigner imply that without the existence of the e.m. interaction the translation invariance would break ??? That, IOW, the existence of the e.m. interaction can be deduced from the premise of translation invariance? Edited February 26 by AlexL Boydstun 1 Quote Link to comment Share on other sites More sharing options...

Boydstun Posted February 26 Report Share Posted February 26 (edited) Alex, ". . . conservation of [linear] momentum follows from the translation invariance, that is from the homogeneity of space (and time)." From the Hamiltonian formulation of classical mechanics (meaning non-relativistic), invariance under temporal translation implies conservation of energy, not conservation of momentum. Were you adding the parenthetical "and time" due to energy-momentum four-vector for the wider special relativity formulation of Hamiltonian mechanics? That is, due to meaning by "momentum" in the part I quoted the "relativistic 4-momentum"? The conservation of energy is implied also via the Bianchi identities in the Einstein Field Equation, as I recall. But this is not such a jolt as the jolt you mention concerning the Wigner demonstration concerning E-M field, as Bill described it, because of the equivalence of inertial and gravitational mass in GR. The Wigner showing is especially jolting to me because of Einstein's failure to find his much-hunted classical unified field theory. And in all of the derivation of E-M from SR by Rosser in his Classical Electromagnetism via Relativity, you have to have introduced the extra-mechanics concept electric charge. Edited February 26 by Boydstun Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 26 Report Share Posted February 26 (edited) 9 hours ago, Bill Hobba said: I think you are correct. It is not a physical assumption - it is a consistency assumption. If we did not allow complex numbers then the poisson bracket would not be an observable. Thanks for that. Thanks Bill What would happen if second person came along and thought one did not need complex numbers to solve something, but instead thought only that the formulas and equations had to be adjusted… in other words instead of the division of labour for our abstractions being using complex numbers in simple equations, using simple numbers in complex equations? What if a third person came along and said the distinctions of division of labour for our abstractions is really quite superficial and illusory and really either approaches are simply the same thing expressed in different ways. EDIT: and to the extent correct or corresponding to reality, refer to the same things in reality. Edited February 26 by StrictlyLogical Quote Link to comment Share on other sites More sharing options...

AlexL Posted February 26 Report Share Posted February 26 (edited) 22 minutes ago, Boydstun said: Were you adding the parenthetical "and time" due to energy-momentum four-vector for the wider special relativity formulation of Hamiltonian mechanics? That is, due to meaning by "momentum" in the part I quoted the "relativistic 4-momentum"? Yes, I added "and time" because in the previous paragraph I was mentioning both non-relativistic and relativistic case. You quoted me: 22 minutes ago, Boydstun said: ". . . conservation of [linear] momentum follows from the translation invariance, that is from the homogeneity of space (and time)." but the insertion "[linear was yours]". My two paragraphs: 1 hour ago, AlexL said: in classical mechanics at least, the total 3-momentum of a system is conserved (in the absence of external forces/fields), whatever the internal interactions are, that is with or without the electromagnetic fields. And this holds also for the relativistic 4-momentum. In both cases, the conservation of momentum follows from the translation invariance, that is from the homogeneity of space (and time)... are thus correct and consistent. I believe the misunderstanding is thus solved. The main point, the e.m. field being necessary for the conservation of momentum - I hope @Bill Hobbawill explain his thoughts. Edited February 26 by AlexL for clarity Quote Link to comment Share on other sites More sharing options...

Boydstun Posted February 26 Report Share Posted February 26 (edited) Alex, I inserted "linear" (with square brackets indicating the word was an addition by me to the quoted material) because I thought (am I wrong?) that conservation of angular momentum falls out under reorientations, unlike conservation of linear momentum which falls out under translations. Edited February 26 by Boydstun Quote Link to comment Share on other sites More sharing options...

AlexL Posted February 26 Report Share Posted February 26 (edited) 13 minutes ago, Boydstun said: Alex, I inserted "linear" (with square brackets indicating the word was an addition by me to the quoted material) because I thought (am I wrong?) that conservation of angular momentum falls out under reorientations, unlike conservation of linear momentum which falls out under translations. Your insert "[linear]" suggested to me, maybe mistakenly, classical mechanics (vs relativistic); I didn't perceive it as "vs. angular". PS: Standard terminology: conservation of angular momentum is the consequence of the isotropy of the space, of the invariance under rotations, of the Lagrangian, for example. Edited February 26 by AlexL Boydstun 1 Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 (edited) 13 hours ago, AlexL said: Can you please elaborate - at any level you feel comfortable with? For classical mechanics, you are, of course, talking Noether and are correct. But when you go to relativity, things get a bit more complicated. Take two charged particles. If the momentum of one changes, i.e. goes from stationary to some velocity by momentum conservation instantaneously, the other particle should move. But it doesn't - from relativity, it happens a bit later. For a time, momentum is not conserved - in violation of Noether. To rectify this, Wigner said you need to introduce the concept of a field as a holder for the missing momentum. Wigner did interesting work in field theory. I tried to find more details from Physics Forums where I am a mentor. Here is the reply I got: https://www.physicsforums.com/threads/wigners-theorem-that-all-fields-must-be-tensors.984076/ BTW, rules are strictly enforced on that forum, and one rule, with a bit of leeway (as determined by mentors like me), is you can't discuss philosophy. I won't go into why other than to say it was once allowed but got out of hand. Added later: The result that the field must exist is called the no interaction theroem. Thanks Bill Edited February 27 by Bill Hobba Added about no interaction theorem Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 11 hours ago, StrictlyLogical said: What would happen if second person came along and thought one did not need complex numbers to solve something, but instead thought only that the formulas and equations had to be adjusted… Some recent work would suggest that while they may be successful initially, they will eventually run into problems: https://arxiv.org/abs/2101.10873 Thanks Bill Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 27 Report Share Posted February 27 (edited) 10 hours ago, Bill Hobba said: Some recent work would suggest that while they may be successful initially, they will eventually run into problems: https://arxiv.org/abs/2101.10873 Thanks Bill A funny thing about how we tend to use language, and funnier when we are talking about physicists, “Quantum Mechanics” is sometimes interpreted as referring to what reality does, when it is far more accurate to say QM is something we do, which to the extent it corresponds with observables of what reality does do, is valid and useful. That paper is more about how we process what reality does, not what attributes and properties which are possessed by entities. The third person would point out that a complex number is nothing more than a complicated (not very) combination of real values. They are absolutely and always reducible to real values. We happen to call them phase and magnitude. But again this is mere characterization of the abstraction which is QM, merely interpretations of the abstractions as more or less complex … when in fact it is all the same and beside the point. The processing abstraction is not the referent to which its predictions are directed. Observe there is no absolute phase in the complex coefficients, and also observe that statistical in nature they are not strictly speaking possessed by any single entity, and of course are never observable properties possessed by any single entity. As such, any assumption about complex numbers, I put to you, is more of an assumption about our abstractions referring to physical reality than an assumption about physical reality itself. Edited February 27 by StrictlyLogical tadmjones 1 Quote Link to comment Share on other sites More sharing options...

AlexL Posted February 27 Report Share Posted February 27 (edited) 15 hours ago, Bill Hobba said: For classical mechanics, you are, of course, talking Noether and are correct. But when you go to relativity, things get a bit more complicated [...] Your initial assertion was: Quote Electric fields, are necessary for well-understood laws to hold (eg Wigner proved if there were no electric fields, then conservation of momentum would not hold in violation of Noether's Theorem.) The implication seem to be that in an isolated system (isolated from external influences), where the constituents do NOT interact electrically (why only electrically, btw?), the total momentum is not conserved. (You added that this is specific to the 4-momentum, that is to special relativity only, because in the Newtonian mechanics the total 3-momentum is conserved.) 1. Do you agree that THIS would be the implication? 2. Do you insist that this is also factually true, that is that the total momentum of a system of non-interacting relativistic constituents is NOT conserved? I believe there must be a misunderstanding somewhere, because the 4-momentum conservation of an isolated system is a well established observational fact – whatever the interaction between the constituents is, including when it is absent. From this observational fact it follows that any considerations which contradict it (Wigner theorem, non-interaction theorem) are either false or do not, in fact, contradict it (the fact of conservation, that is.) (Just in case: please note that I do not define “total momentum” as the sum of individual momenta.) Edited February 27 by AlexL for clarity Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 (edited) I really should have started with the no interaction theorem. This states particles are unable have any interaction if the principle of relativity is satisfied. Take two charged particles - we know they do interact in violation of the theorem. To circumvent this something else is added than just two particles. This something is a field and when one writes the Lagrangian of the system it contains not only the particles, but the field as well. You then derive the equations of motion and get Maxwells equations and the Lorentz Force Law. The reason the no interaction theorem is not violated is because you also have more than two particles - you have a field as well. In fact from Noether this field has energy and momentum. Because of that, while I suppose you could mount an argument, they are just mathematical abstractions; if they have energy, due to E=MC^2 the most reasonable assumption is the field are real. To me this is a very profound result. Interestingly using Coulomb's Law and relativity one can derive Maxwells Equations: http://richardhaskell.com/files/Special%20Relativity%20and%20Maxwells%20Equations.pdf Hence you get the standard Lagrangian and that the field introduced in the above as a mathematical abstraction must have energy and momentum. The argument works just as well for other fields such as gravity. Here is a conundrum to think about. Exactly the same argument in the link above can be used to derive similar equations for gravity. They are wrong - but why? Have a think and we can discuss it (hint - gravity gravitates and 'gravitational charge' is not a scalar but the stress energy tensor). Thanks Bill Edited February 27 by Bill Hobba Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 27 Report Share Posted February 27 (edited) 41 minutes ago, Bill Hobba said: I really should have started with the no interaction theorem. This states particles are unable have any interaction if the principle of relativity is satisfied. Take two charged particles - we know they do interact in violation of the theorem. To circumvent this something else is added than just two particles. This something is a field and when one writes the Lagrangian of the system it contains not only the particles, but the field as well. You then derive the equations of motion and get Maxwells equations and the Lorentz Force Law. The reason the no interaction theorem is not violated is because you also have more than two particles - you have a field as well. In fact from Noether this field has energy and momentum. Because of that, while I suppose you could mount an argument, they are just mathematical abstractions if they have energy. Due to E=MC^2 the most reasonable assumtion is the field are real. To me this is a very profound result. Interestingly using Coulomb's Law and relativity one can derive Maxwells Equations: http://richardhaskell.com/files/Special%20Relativity%20and%20Maxwells%20Equations.pdf Hence you get the standard Lagrangian and that the field introduced in the above simple as a mathematical abstraction must have energy and momentum. The argument works just as well for other fields such as gravity. Here is a conundrum to think about. Exactly the same argument in the link above can be used to derive similar equations for gravity. They are wrong - but why? Have a think and we can discuss it (hint - gravity gravitates and 'gravitational charge' is not a scalar but the stress energy tensor). Thanks Bill Please excuse me I am unfamiliar with the details of Noether’s theories re. fields. If the electric field of an electron has a separate energy which you imply has a mass, then when you expend energy to accelerate the mass of an electron do you need to expend energy to accelerate the electric field of the electron which you imply also has mass? In fact, when you accelerate an electron are you accelerating i.e. moving, its electric field as well? Are you moving two things are one thing... are there two things or one thing? As student of philosophy what is your understanding of an attribute or property of an entity as regards its nature and relationship to the entity? Some would say they certainly are real but are not independently real, they are only always “of” entity or entities. Edited February 27 by StrictlyLogical Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 See: https://arxiv.org/pdf/1601.03616 Thanks Bill Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 27 Report Share Posted February 27 (edited) @Bill Hobba I note a great many physicists completely dispense with philosophy and philosophical theory and the rigours which may be attained therewith… do you bring any philosophical scrutiny to the physics you cite or do you simply take them and the interpretations therein, as true? Edited February 27 by StrictlyLogical Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 Yes, but at this stage of my philosophical journey, just some simple philosophical considerations, eg reality exists and is not observer dependent as in some interpretations of QM. Even just analysing QM, so it is in that form, has taken me decades - what I wrote in the QM post is the result of many years of struggle. I find the going tough when confronted with more complex philosophical issues. Due to my background being in math, I will often resort to mathematical arguments. They are logically sound. But, great physicists can see the 'substance' behind the equations. That represents a deep philosophical insight. They do it intuitively, These are people like Feynman, who, on the surface, was very anti-philosophy but was deeply philosophical. It's just that this substance behind the equations came so naturally to him he saw no reason to take it any further formally. This can lead to confusion. For example, in arguments that QM is whacko, Feynman's path integral approach is often bought up and, correctly attacked, as irrational - how can particles take all these different paths simultaneously? They forget that it is just a heuristic suggested by writing the equation in a certain way. Particles do not take all paths at the same time. Such people have not progressed to even looking at the superficial substance behind the equations. Thanks Bill Quote Link to comment Share on other sites More sharing options...

StrictlyLogical Posted February 27 Report Share Posted February 27 30 minutes ago, Bill Hobba said: Yes, but at this stage of my philosophical journey, just some simple philosophical considerations, eg reality exists and is not observer dependent as in some interpretations of QM. Even just analysing QM, so it is in that form, has taken me decades - what I wrote in the QM post is the result of many years of struggle. I find the going tough when confronted with more complex philosophical issues. Due to my background being in math, I will often resort to mathematical arguments. They are logically sound. But, great physicists can see the 'substance' behind the equations. That represents a deep philosophical insight. They do it intuitively, These are people like Feynman, who, on the surface, was very anti-philosophy but was deeply philosophical. It's just that this substance behind the equations came so naturally to him he saw no reason to take it any further formally. This can lead to confusion. For example, in arguments that QM is whacko, Feynman's path integral approach is often bought up and, correctly attacked, as irrational - how can particles take all these different paths simultaneously? They forget that it is just a heuristic suggested by writing the equation in a certain way. Particles do not take all paths at the same time. Such people have not progressed to even looking at the superficial substance behind the equations. Thanks Bill How familiar are you with Objectivism as a philosophy? As a former academic of physics, I highly recommend it. Interpretations of physics and in particular of QM are interesting but until we have a complete understanding of the mechanics of measurement (rather than a formalism and assumptions) I think they will remain rather fanciful and mystical… I have found THAT is where the mind goes when it encounters something it does not understand… not merely an acknowledgement of the unsolved but a kind of ecstasy in the “mysterious”. I would suggest you hold onto as much of the solid foundations of thought as possible when you encounter the myriad flights of fancy of both your common and uncommon physicist. Be rigorous and disappointingly real about the distinctions between our abstractions and entities referenced by them. Good luck on your journey! Quote Link to comment Share on other sites More sharing options...

AlexL Posted February 27 Report Share Posted February 27 (edited) 2 hours ago, Bill Hobba said: I really should have started with the no interaction theorem. This states particles are unable have any interaction if the principle of relativity is satisfied. Take two charged particles [...] I was trying to put some order in our discussion, which is about a subject which is already complicated and doesn’t need a multiplication of issues. I asked two questions. Please answer them. Here are they, with some of their context : 4 hours ago, AlexL said: Quote Electric fields, are necessary for well-understood laws to hold (eg Wigner proved if there were no electric fields, then conservation of momentum would not hold in violation of Noether's Theorem.) The implication seem to be that in an isolated system (isolated from external influences), where the constituents do NOT interact electrically (why only electrically, btw?), the total momentum is not conserved. (You added that this is specific to the 4-momentum, that is to special relativity only, because in the Newtonian mechanics the total 3-momentum is conserved.) 1. Do you agree that THIS would be the implication? 2. Do you insist that this is also factually true, that is that the total momentum of a system of non-interacting relativistic constituents is NOT conserved? I believe there must be a misunderstanding somewhere, because the 4-momentum conservation of an isolated system is a well established observational fact – whatever the interaction between the constituents is, including when it is absent. Depending on your answer, we will see - if there is indeed a contradiction between what we see and the results you’ve mentioned – the Wigner Theorem and the no-interaction theorem, - or that the results you’ve mentioned do not, in fact, imply non-conservation PS: regarding the necessity to also account for the interaction fields - I already addressed this by specifying: "please note that I do not define “total momentum” as the sum of individual momenta." This implies that the total momentum I was talking about also includes the contribution from the fields. Edited February 27 by AlexL Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 30 minutes ago, StrictlyLogical said: How familiar are you with Objectivism as a philosophy? I read a lot when young, but my interests turned more to math/physics as I got older. Enrolling in the Masters naturally has made me much more interested. Thanks Bill Quote Link to comment Share on other sites More sharing options...

Bill Hobba Posted February 27 Author Report Share Posted February 27 4 hours ago, AlexL said: 1. Do you agree that THIS would be the implication? 2. Do you insist that this is also factually true, that is that the total momentum of a system of non-interacting relativistic constituents is NOT conserved? 1. They are conserved - relativity demands it. In fact, the world lines of each particle must be straight. For charged particles that is not the case - hence the difficulty. For momentum to be conserved the field is needed and the field has momentum. The particles and field have momentum conserved. 2. It is conserved as per point 1. Thanks Bill Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.