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Physical Chemistry: Math

  1. Oct 11, 2007 #1
    hello all,

    I am doing some homework and need some verification on the math:

    [Px, Px^2]f= PxPx^2f - Px^2Pxf

    Px= -iħd/dx
    Px2= -i^2ħ^2d^2/dx^2

    -iħd/dx(-i^2ħ^2d^2/dx^2f) - (-i^2ħ^2d^2/dx^2)( -iħd/dxf)

    i^3ħ^3d/dx(d^2f/dx^2) - i^3ħ^3 d^2/dx^2(df/dx)

    the number following the "^" sign just means raised to that power. Would someone verify if this is correct please? thank you.
     
  2. jcsd
  3. Oct 13, 2007 #2

    Gokul43201

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    When you write Px and Px^2, do you mean [itex]P_x[/itex] and [itex]P_x^2[/itex]?

    1. If the above is true, then there's a very simple value for the commutator that can be got without having to write the momentum operator in the position representation ([itex]-i \hbar \partial/\partial x [/itex]). The last line of your calculation simplifies to this final answer.

    2. [tex]\frac{\partial}{ \partial x} \frac{\partial f}{ \partial x} = \frac{\partial ^2f}{ \partial x^2 } [/tex]

    and so on...
     
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