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I'm not a physicist, but rather a control (feedback) systems engineer eager to understand more than just a cursory explanation of quantum mechanics. The StackExchange has been an excellent forum for helping me piece together facts that are not always expressed well in the textbooks or other online information, and I have great appreciation for whoever conceived and continues to maintain its existence. Thank you. But on to my questions..

I've read that one can take the mathematics of Schrodinger's wave equation and Heisenberg's matrix equations and show that they are 'equivalent'. In control systems we model physical (macro)systems by either a set of differential equations or as a state space system, and indeed can show equivalence between these models whether linear or nonlinear.

  1. Specifically between Schrodinger and Heisenberg mathematics, are they each modeling the same quantum wave mechanics, and is this equivalence shown in the same manner that I show equivalence in the physical systems modeling I've described above (differential equations and state space systems)? If not then what is it that is shown to be equivalent between these two systems of mathematics?

Since I am a control systems engineer I'm familiar with linear systems theory, state space systems, the concept of observability, and the construction of the observability matrix from a linear state space model. In linear systems theory the observability matrix is derived from the system matrix (holds information on the eigenvalues) and the output coupling matrix ( a matrix that couples the system states to measureable outputs). The observability matrix basically determines what states within the system can be measured (observed).

  1. Does the Heisenberg description of quantum mechanics also lead to such an observability matrix (or some equivalent) explaining why only probability amplitude can be observed (and not the complex phases)?

In controls we can also take the system models and express them in terms of a Hamiltonian, and change the basis of a system description using matrix rotations. But we typically don't use 'Bra' and 'ket' notation - < and >.

  1. So are control systems engineers and quantum physicists basically using the same mathematical tools to describe system models, but perhaps using slightly different terminology?
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I would like to begin by your last question.

I am of the opinion that this is correct. I am not versed in systems engineering but I see the equivalence in the mathematical treatments. And QM is very much about the mathematics, is more a descriptive body of knowledge than it is an explanatory. And proof of this is the many interpretations it has, all of them valid within its mathematical context, which describes very well the experimental results.

So this means to me that QM strength is the mathematical description and accuracy of results with reality, but is weak in its comprehension of the physical world, and hence its mathematical formulation is its main feature. Since this is equivalent to systems engineering's math in several aspects (in the cases they are mathematically equivalent) then there is no distinction, like there is no distinction when the same statistics is used in Statistical Mechanics and demographics analysis: conceptually is the same.

On to question two.

No, Heisenberg description cannot explain why we only observe probability amplitudes. It can obtain the observed probability amplitudes under its assumptions, but in its fundamentals it accepts the fact that all observables are Real-valued quantities.

Finally, to your first question, perhaps the hardest to answer.

They are equivalent in the results they get: both are consistent with each other but completely different formulations.

Schrodinger formulation describes the system with a complex valued function, which depends on time and the relevant independent variables, like coordinates or momenta of the particles for example. And here measurables are equivalent to operators, which do not depend on time. But obtaining their values involves the time-dependent wave-function and the operators, which can yield time dependent or time independent values, always real numbers.

Heisenberg formulation describes systems in terms of matrices which depend on time and are the quantum equivalents of the classical magnitudes. The problem here is the determination of the matrix elements, which means knowing all the periodicity values of the system.

Obtaining the matrix elements means knowing all the information of the quantum system, whereas in the Schrodinger picture, all the relevant information is accessible through an operation on the wave function.

In summary,

...H describes the motion of a particle as with momenta and coordinates which are here matrices that otherwise behave pretty much like the classical concepts and follow the same relations...

...while S describes the system with a wave function, with which the measured quantities can be obtained by operating on it and sort of "folding it onto itself"....

They are different in their treatment, but consistent on their results. They describe the same systems, which

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  • $\begingroup$ Thanks for addressing my question - it's been a long time. After asking the question and not receiving any immediate replies I figured that a person knowing the details of quantum physics and control systems engineering may be a rare breed. Your answer somewhat answers my questions, except perhaps for 3.) In controls engineering we can model dynamic systems using differential equations and these can often be 'converted' into a state space (matrix) representation. By writing as a state space system, the dynamics and output coupling components of the system are more easily recogninized. $\endgroup$ – docscience Sep 18 '15 at 17:35
  • $\begingroup$ ... and my thoughts were that the output coupling has some parallel to observation/measurement and the 'collapse' of the wave function. I'm still not seeing if there is any connection of QM descriptions to the concepts in systems enegineering. $\endgroup$ – docscience Sep 18 '15 at 17:37
  • $\begingroup$ Well then I think I didn't understand your question. I thought you saw the math similarity and were wondering about the similarity in underlying physics. $\endgroup$ – rmhleo Sep 18 '15 at 18:04

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