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[Q]Conserative theorem and parity

  1. Nov 16, 2008 #1

    I encountered three conserative theorem in textbook

    One of conserative theorem is involving my question.

    Liboff defined displacement operator as [itex] \hat{D}(\varsigma) = e^(\frac{i\varsigma\hat{p}_{x}}{\hbar}f(x) = f(x + \varsigma ) [/itex] but is it right?

    If system is displaced from x to [itex] x + \varsigma [/itex], function f should be displaced

    such as [itex] f(x - \varsigma) [/itex], it is function displacement from x to [itex] x + \varsigma [/itex]

    Why liboff defined displacement operator in that way?
  2. jcsd
  3. Nov 16, 2008 #2


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    You will notice that p is the momentum operator, and I hope you know this - (linear) momentum is the GENERATOR of translations.

    More than Liboff define the displacement operator as this, have you found any other suggestion?

    You can check this theorem by writing the momentum operator as the derivative operator, and using exponential of operator and finally taylor-expand this:

    [tex]\exp (i\zeta (-i\hbar\frac{d}{dx})/\hbar)f(x) \approx (1+\zeta \frac{d}{dx})f(x) = f(x) + \zeta f'(x) \approx f(x+\zeta) [/tex] (becomes more exact if keeping higher orders)
  4. Nov 16, 2008 #3
    I agree your calculation it is very insparational.

    Buy how could you explane my question? [itex] f(x + \varsigma ) [/itex] is translating f to left by [itex] \varsigma [/itex].

    I interprete [itex] \hat{D}(\varsigma) [/itex] is translating physics system to right by[itex] \varsigma [/itex]

    Is my interpretation wrong?
  5. Nov 16, 2008 #4


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    now this is not so stringent as my first post.

    One can look at translations in two ways,

    i) system is stationary and coordinate system is shifted.

    ii) coordinate system is stationary and system is moved.

    The operator moves the system from point x to point x + \zeta. So your interpretation is correct.
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