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Thread: Why the i?

  1. #1 Why the i? 
    Forum Bachelors Degree
    Join Date
    Mar 2009
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    421
    Something that just occurred to me, that I probably should have figured out back when I was in college. Oh well, better late than never.

    As I said in a previous thread, to go from a classical system to a quantum system, simply promote observables to self-adjoint operators. For f,g two observables, define

    [f,g] = i{f,g}

    where {f,g} is the classically define Poisson bracket.

    So why the i? It's because we want an operator to correspond to the observable quantity {f,g}. A natural choice is [f,g], but the problem is that [f,g] is actually skew-adjoint:



    Meanwhile, if {f,g} is observable, it must be promoted to a self-adjoint operator. So if we let

    [f,g] = i{f,g}, then both sides are skew-adjoint, so the equation makes sense. We can easily pick the sign of i at this point by observing that p = (1/i) d/dx in position space and that {x,p} = 1, so that as operators [x,p] = i = i{x,p}.


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