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Operator Language?

  1. Mar 13, 2010 #1
    When linear operators A and B act on a function ψ(x), they don't always commute. A clear example is when operator B multiplies by x, while operator A takes the derivative with respect to x. Then
    ba139ec4de3277dec6bf746cf8fb941f.png

    which in operator language means that

    f54e3202f439d8acb22e6675932c7659.png

    To get the last equation they divided through by ψ but why is it true? I guess what I'm trying to say is that the second equation makes no sense => d/dx*x - x*d/dx doesn't always equal 1... so why do they say that?
     
  2. jcsd
  3. Mar 13, 2010 #2
    First, they don't divide it by [itex]\Psi[/itex]; they simply consider it the argument of the operator AB-BA. It's akin to denote a function by f only, instead of f(x), where the variable is explicit.

    Second, in this context, d/dx*x - x*d/dx doesn't always equal 1; in fact, it never equals 1: it equals the identity operator I, the one that satisfies [itex]I\Psi=\Psi[/itex].
     
  4. Mar 13, 2010 #3
    Thanks for stating it so clearly and concisely. Many issues were resolved.
     
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