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Dynamical Variables As Operators

  1. Aug 20, 2014 #1
    In Quantum Mechanics, why do the dynamical variables become operators? What is the justification or motivation, if any exist?
  2. jcsd
  3. Aug 20, 2014 #2


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    It is an axiom that observables are represented by operators, and the motivation for choosing that and the other axioms of QM is that the mathematical results that follow from these axioms accurately describe the world around us.

    Although the math is more formal and less accessible to the average layperson, it's really no different than deciding to model the force of gravity as an inverse-square force, or at an even more natural level, accepting Euclid's axiom that in a plane, through a point not on a given straight line, at most one line can be drawn that never meets the given line. Accept these postulates and your mathematical results will accurately describe how the world works, and that's useful, so we're motivated to accept them.

    Of course, in practice these axioms didn't just magically appear. Just as Euclid and Newton did, we observe how the world works and then go looking for axioms that lead to mathematical theories that match what we observe. So it's no coincidence that we end up with axioms that "work" - we've rejected all the others.
  4. Aug 20, 2014 #3


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    Its got to do with the axioms of QM.

    Check out a post I did a while ago - see post 137

    In that approach you start with a basic axiom:
    An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

    Von Neuman Measurements are defined as POVM's that are disjoint - also called resolutions of the identity.

    Now there is a very nice theorem called the spectral theorem that links resolutions of the identity and Hermitian operators so there is a 1-1 correspondence between the two.

    Thus, given a Von-Neumann measurement, Ei, you can associate a real number yi with each outcome and get a Hermitian operator O=∑ yi Ei. These are the observables of a quantum observation.

  5. Aug 26, 2014 #4
    As the others have pointed out, the notion of an operator is axiomatic to quantum mechanics. This is not very satisfing though. The axiomaticness can be motivated though. Consider the equation for a plane wave in one dimension.

    [itex] \psi(x,t) = e^{i (kx + \omega t)} [/itex]

    Now I would like to define a differential operator whose eigenvalue is the momentum of my plane wave. By the deBroglie hypothesis ... I know that the momentum of my wave is [itex] \hbar k [/itex]. Then by inspection I can generate the momentum "as an eigenvalue" by.

    [itex] -i\hbar \partial_x \psi(x,t) = \hbar k \psi(x,t) [/itex]

    Thus the "momentum operator" [itex] -i\hbar \partial_x [/itex] acting on my plane wave [itex] \psi(x,t) [/itex] returns the product of the momentum of my plane wave and my plane wave [itex] \hbar k \psi(x,t) [/itex]. Note that the only thing "quantum" about this is the invocation of the deBroglie relation. The rest is simple calculus.

    Based on this, I can brazenly hypothesize for any thing (particle/field) I can define a differential operator which gives me the momentum of that thing times that thing. The argument is easily extended to other observable things like energy, angular momentum etc.
  6. Aug 27, 2014 #5
    kspace: is this argument easily extended to all other observables, because all dynamical variables can be written in terms of momentum?
    Last edited: Aug 27, 2014
  7. Aug 27, 2014 #6
    Excellent question !!

    Well. we not entirely ready for a complete extension ... but the momentum operator is a great start. consider the angular momentum which is the cross product of the position vector and the linear momentum vector. Lets try to work out the anglar momentum operator (now in 3D). Classically the angular momentum is ...

    [itex] L = r \times p [/itex]

    Recall that the 3D linear momentum operator is (generalizing our plane wave result to 3D):

    [itex] \hat{p} = -i\hbar \nabla [/itex]

    We would like to construct the angular momentum operator simply by substitution for [itex]r [/itex] and [itex] p [/itex] for their respective operators. We dont have a position operator yet though (though it is more or less trivial in this case since we are working in "position basis").

    [itex] \hat{r} = r [/itex]

    Now we are ready to go ! We can just write down the QM angular momentum operator.

    [itex] L = -i\hbar ( r \times \nabla ) [/itex]

    At this point you would be correct in saying that any classical observable can be obtained by extension of the position and momentum operators. How come ?? Well, from classical mechanics (Hamiltons fromulation) you might remember that momentum and position are conjugate variables. Knowledge of position and momentum is enough to determine the equation of motion of the system !! There is nothing stopping us from writing our classical equations of motion in terms of operators now. Generally the observable we are most interested in is the total energy (kinetic + potential).

    [itex] \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{r}) [/itex]

    Here [itex] V(\hat{r}) [/itex] (the potential energy) is of course just a function of the position of the particle. (think about [itex] V(\hat{r}) [/itex] being a polynomial in [itex] \hat{r} [/itex] for example) Already by substituion we have half of the schrodinger equation. Again the only "quantum" part of this entire line of reasoning is the invocation of the deBroglie relation way back when I was talking about the plane wave example.

    However, there are some truly "quantum" effects which dont have very good classical analogues, the best example being spin. Its pretty obvious that a particles spin (or internal angular momentum) cant be written in terms of its linear momentum and position. So some extensions are needed to handle these cases.

    There are a lot of very deep things going on here which can be discussed in a very abstract and mathematical way. I dont think this is helpful though for describing the basic ideas in terms of physical things that we are familiar with (like momentum). All we want to do (as beginners in QM) is to try to generalize Hamiltons formulation in a way which is consistent with the deBroglie relation.

    Hope this helps !!
  8. Aug 29, 2014 #7

    I would like to point out that the operator aproach is not unique to quantum mechanics. One can find operator description of classical mechanics (more like classical statistical mechanics), and actually that description is almost as old as quantum mechanics

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