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Is [x,p]=i quantum mechanical?

  1. Apr 11, 2013 #1
    In CM, p is the generator of translations.

    On functions, exp[ i(-i) d/dx] is what translates.

    So the relationship [x,-id/dx]=i
    seems to just be some sort of group theoretic relation.

    Where did quantum mechanics come into this? Was it the fact that you represented the translation operator with the i in the exponential exp[ i(-i) d/dx] instead of writing it like exp[d/dx]? That is, a unitary operator operates on a state function f(x) to translate it to f(x+1)?

    I know in CM you can define an exponential map such that exp[operatator]x(0)=x(t), where x(t) is the particle trajectory. This looks suspiciously similar to exp[-iHt]f(0)=f(t) for the state function f(t), so I was wondering whether you can say that the difference between CM and QM has to do with the representation of the group elements that evolve time and also the objects on which the group elements act (in CM this object would be a function, and in QM this object would be a functional?), so the commutation relations are fundamental to the group but the different ways of representing the group distinguish CM from QM?
  2. jcsd
  3. Apr 12, 2013 #2


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    1) The momentum p is the generator of translations; this property follows already from the classical Poisson backet {x,p} = 1 and has nothing to do with quantum mechanics. You are right, there is a relation to group theory, namely the symplectic group Sp(2) acting on the phase space spanned by {x,p}.

    2) The translation into quantum mechanical language, so-called quantization, is a change in the formalism
    - introduce or construct a Hilbert space
    - translate variables on phase space into operators acting on this Hilbert space
    - translate Poisson brackets {x,p}=1 of canonically conjugate pairs (like x,p) into commutators [x,p]=i
    - translate the Hamilton function into an operator
    The two formalism look rather similar in the Heisenberg picture w/o introducing a representation (*) Here the classical e.o.m. derived from
    dx/dt = {x,H} and
    dp/dt = {p,H}
    are translated into Heisenberg e.o.m. for time dependent operators using [x,H] and [p,H]

    3) Translation of p into a differential operator x = -i d/dx is a second step, namely the choice of a special representation (*) for quantum mechanics. Here x is a number (!) acting on wave functions u(x) simply by multiplication. And p is a special differential operator. There are other possibilities, e.g. a representation where p is a number and x becomes a differential operator p = i d/dp acting on wave functions u(p). All representations must respect the fundamental canonical commutation relations.
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