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What makes matrix mechanics matrix mechanics? |
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| Aug10-12, 10:02 AM | #18 |
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What makes matrix mechanics matrix mechanics?
The important point is that all pictures of time evolution are unitary equivalent by definition. This ensures that the physical meaning of the quantum theoretical formalism is independent of the choice of the picture. Neither state-reprsenting vectors nor eigenvectors of operators, representing observables, are measurable quantities but only the probabilities
[tex]P_{\psi}(a)=|\langle a| \psi \rangle|^2.[/tex] The time dependence of the mathematical quantities [itex]|\psi \rangle[/itex], [itex]\hat{A}[/itex] (and following from this [itex]| a \rangle[/itex] can be quite arbitrarily chosen. Usually it depends on the problem you want to solve, which choice is best. For the final result it doesn't matter at all. The wave function in the "a basis", [itex]\psi(a,t)=\langle a|\psi \rangle[/itex] is, by the way, a picture independent quantity. |
| Aug10-12, 10:03 AM | #19 |
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The Heisenberg picture and the Schrödinger picture have very little to do with the difference between matrix mechanics and wave mechanics. (Edit: "Wave mechanics" is what some people call the quantum theory of a single spin-0 particle influenced by a potential, i.e. the theory that's taught in introductory classes. I think it's essentially the same as Schrödinger's theory).
My understanding* is that Schrödinger was working with the semi-inner product space of square-integrable functions, while Heisenberg was working with an inner product space of "column vectors with infinitely many rows". (Yes, this seems rather ill-defined, but Heisenberg wasn't trying to do rigorous mathematics). *) I could be wrong, since I haven't read any of the original sources. I'm really just guessing based on what I know about Hilbert spaces and the stuff that's being taught in introductory QM classes. If you define two members of Schrödinger's inner product space to be equivalent if they differ only on a set of measure zero, and then consider the set of equivalence classes, you can easily give it the structure of an inner product space. This inner product space turns out to be a separable Hilbert space. The rows of Heisenberg's "∞×1 matrices" can be thought of as the components in some specific orthonormal basis, of a vector in a separable Hilbert space. This is the sense in which von Neumann's Hilbert space approach "unifies" the two original approaches by Heisenberg and Schrödinger. The two "pictures" on the other hand are only different opinions about whether the time dependent factors in an expression like ##\langle\alpha|e^{iHt}A e^{-iHt}|\alpha\rangle## should be considered part of the state vectors or part of the operators, i.e. should we say that ##|\alpha\rangle## is the state and ##e^{iHt}A e^{-iHt}## is the observable (Heisenberg picture) or should we say that ##e^{-iHt}|\alpha\rangle## is the state and ##A## the observable (Schrödinger)? |
| Aug10-12, 11:19 AM | #20 |
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| Aug10-12, 11:21 AM | #21 |
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As with so many times back in school or in research group meetings today, this is one of those questions that I was reticent to ask because I thought I'd look stupid but and very glad to find out that my lack of understanding (of this topic, at least) doesn't make me so.
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| Aug10-12, 11:25 AM | #22 |
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| Aug10-12, 11:28 AM | #23 |
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| Aug10-12, 11:41 AM | #24 |
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the mathematician ofcourse, tends to put hard problems in terms of problems that have been solved before. There is a combination of theories that goes behind why you can solve the wave equations in this manner, and going back to the beginning of what I was trying to say, it really boils down to the information you are able to supply or extract from your system experimentally.
If you really want to know it for sure, I recommend you take some lessons in partial differential equations, familiarise your self with the Sturm-Liouville equations and solutions, and then work with green's functions. here I supply the relevant page if you are in fact interested: http://en.wikipedia.org/wiki/Green's_function it so happens that green's functions solves any linear differential operator, and it so happens that matrices are linear mathematical objects (not by coincidence!). This is matrix mechanics but only if your Hamiltonian has a maximum of second order derivatives. any higher and you would need tensor algebra to solve the equations of motion. It has not a lot to do with matrices themselves but matrices so happen to take the shape of the tensors that we use, as the Hamiltonian is of second order in almost every case. the matrix algebra that we use simplifies things a lot of course. You will not be able to get far with solving just linear equations, but systems of linear equations, and to do so you need both the things you have learnt in schrodinger and the heisenberg picture. |
| Aug10-12, 11:55 AM | #25 |
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You've made a couple of allusions to what can be taken from experiments will determine which method you use; could you please focus on that aspect? What situations (that is, given what experimental values for what properties) would lead you to use "matrix mechanics" instead of "wave mechanics" and vice versa? |
| Aug10-12, 12:06 PM | #26 |
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i really really think its obvious. for example for the stern-gerlacht experiment, you know the state of the initial particle (spin) as you prepare them in a particular way, and you know the state will change because they will go through the potential, therefore state is time dependant and is given by solving the relevant matrix of equations, which takes into account 2 hamiltonians. In the case of the hydrogen molecules, you make the gross assumption that the system is in equilibrium, therefore the states are time independant, and thereby you can solve using linear differential equations.
in most cases you are sending particles through potentials and you already know their states are going to change. you work out what the states are going to be by prior experiments, and then you work out what the potential looks like by prior experiments, and then you put your particle through and solve the matrix of equations. |
| Aug10-12, 05:41 PM | #27 |
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| Aug10-12, 06:17 PM | #28 |
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| Aug10-12, 06:20 PM | #29 |
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| Aug10-12, 06:36 PM | #30 |
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(I have written another post just before your post. Maybe the wikipedia-article about matrix mechanics also helps.) |
| Aug12-12, 06:30 AM | #31 |
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ok there's one detail that I have ignored that may have been what you were looking for. wave mechanics failed to comply with the relativistic energy. at the time dirac solved this by introducing matrices that solved the wave equation if inserted, instead of just the single wave function. for details I recommend:http://en.wikipedia.org/wiki/Dirac_equation
the idea is that in the new tonsorial equation, the wave function must be a tensor itself, thereby introducing matrix mechanics to wave physics. |
| Aug12-12, 09:09 PM | #32 |
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