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Translation operator

  1. Feb 21, 2013 #1
    [tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...=\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\frac{d^n}{dx^n}[/tex]
    Why this is translational operator?
    ##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)##
     
  2. jcsd
  3. Feb 21, 2013 #2

    tiny-tim

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    taylor expansion? :wink:
     
  4. Feb 21, 2013 #3

    G01

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    Consider alpha to be an infinitesimal translation. Expand [itex]f(x+\alpha )[/itex] for small [itex]\alpha[/itex] to first order.

    Do the same for the LHS of the equation and you should see that the equality is true for infinitesimal translations. We say that the operator [itex]\frac{d}{dx}[/itex] (technically [itex]\frac{d}{idx}[/itex]) is the 'generator' of the translation.

    EDIT: Beaten to the punch by TT!
     
  5. Feb 21, 2013 #4
    I have a problem with that. So
    [tex]f(x+\alpha)=f(x)+\alpha f'(x)+...[/tex]
    My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.
     
  6. Feb 21, 2013 #5
    I am not sure about your question. But that translation operator is a generic operator, which translate a function value at x to x+a.
     
  7. Feb 21, 2013 #6
    In Taylor series x is fixed, while in ##\frac{df}{dx}## ##x## isn't fixed. Well you suppose that is.
     
  8. Feb 21, 2013 #7

    tiny-tim

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    yes, but x here is a constant, and only α is the variable

    if you prefer, write f(xo + α) = f(xo) + α∂f/∂x|xo + … :wink:
     
  9. Feb 21, 2013 #8
    Ok but that is equal to
    [tex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}(\frac{df}{dx})_{x_0}[/tex]
    and how to expand now
    [tex]e^{\alpha\frac{d}{dx}}[/tex]
     
  10. Feb 21, 2013 #9

    tiny-tim

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    no, it's [itex]\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x_0}[/itex] :wink:
     
  11. Feb 22, 2013 #10
    I made a mistake. But I'm asking when you get that ##(\frac{d^n}{dx^n})_{x_0}##? Please answer my question if you know. In
    [tex]e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...[/tex]
    you never have ##x_0##.
     
  12. Feb 23, 2013 #11

    tiny-tim

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    [tex]\left(e^{\alpha\frac{d}{dx}}(f(x))\right)_{x_o}[/tex]
    [tex]= \left(\left(1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...\right)(f(x))\right)_{x_o}[/tex]
    [tex]= \sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\left(\frac{d^nf}{dx^n}\right)_{x_0}[/tex]
     
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