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Quantum conditions for position and momentum operators

  1. Jun 18, 2015 #1

    I'm currently making my way through the book "Quantum Field Theory of Point Particles and Strings" and on page 13 they talk are talking about quantization of the classical versions momentum and position. The first part to quantizing these is turning them into operators. The books goes on to say that "to complete the quantization, we must specify quantum conditions for the operators X and P, and that X and P must satisfy [X,P]=i, .

    Can anyone give me any insight into what this commutation relationship is saying and why it is the way it is?

  2. jcsd
  3. Jun 18, 2015 #2


    User Avatar
    Gold Member

    It basically says that, unlike their classical counterparts, quantum position and momentum will be incompatible observables, i.e., measurement of one quantity destroys previous information about the other, this is one of the key elements which are unique to QM. One of von Neumann's axioms of QM is precisely that incompatible observables are modeled by non-commuting operators.

    So, that's the content of the rule of 'canonical quantization of classical observables'. In classical mechanics, all observables are compatible, they all belong to the commutative C*-algebra of functions in phase space (the pointwise product of functions is used here). Linear functionals on this algebra give you states. GNS constructions of this give measures on Boolean algebras (of commuting projectors on a Hilbert space), i.e., classical probability. On the other hand, we have the classical Poisson bracket for these classical observables, {x,p}=1. But, in classical mechanics, this bracket doesn't interfere in the probability structure of the theory, characterized always by the commutative algebra.

    In QM, the fundamental C*-algebra which controls the type of probability theory is non-commutative, and thus GNS constructions give measures on non-distributive lattices of projectors, which gives then a non-classical probability theory, related to all the usual QM elements like incompatible observables. The thing that somehow dictates the way in which the C*-algebra doesn't commute is the classical Poisson bracket; in the Hilbert space of the GNS construction, the composition of operators is the image of the C*-algebra product, we make it non-commutative by adopting the canonital quantization recipe [x,p]=i id_H.

    Well, you can simply postulate the canonical quantization recipe as a physical principle. The justification is simply that it seems to work.

    In some cases, it can be derived from more fundamental assumptions related to symmetry principles. For example, via the Imprimitivity Condition, which encodes the homogeneity of space.
  4. Jun 18, 2015 #3


    Staff: Mentor

    If you want to see a deeper reason read Chapter 3 of Ballentine - Quantum Mechanics - A Modern Development. It explains why the commutation relations are in general like that. As far as QFT is concerned the model is you consider a field as a large number of blobs that obey the rules of classical mechanics, which is meant they are described by a Lagrangian, where its value is like position and has a conjugate momentum. Its formally the same as basic QM, and you can run through the same kind of symmetry arguments as per Ballentine giving the same commutation relations.

  5. Jun 23, 2015 #4
    Awesome answers! Thanks so much!
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