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Expanding a translation operator

  1. Nov 17, 2012 #1
    I'm trying to understand the construction of the T(ε) operator and why it is equal to I-iεG/hbar.

    The textbook I'm using (Shankar) talks defines the translation operator with the phase factor:

    [itex]T(ε)\left|x\right\rangle=e^{i \epsilon g(x)/\hbar}\left|x+\epsilon\right\rangle[/itex]

    and translationational invariance

    [itex]\langleψ| H|ψ\rangle=\langle ψ_\epsilon| H|ψ_\epsilon\rangle[/itex]

    The book then says

    "To derive the conservation law that goes with the above equation, we must first construct the operator T(e) explicitly. Since ε=0 correspons to no translation, we may expand T(ε) to order (ε) as
    [itex]I-\frac{i ε}{\hbar} G[/itex]

    Why is this so? How can you find an equation for only T without it acting on anything?
  2. jcsd
  3. Nov 17, 2012 #2
    It's less an equation for T than a definition of G. There's some operator T(a), which depends on a, such that multiplying any state by T(a) translates that state by a distance a. Shankar assumes that we can expand the function T(a) as a power series in a. The zeroth order term must be the identity operator, because T(0) is the identity operator. The first order term Shankar calls (-i/hbar * G), which is just a definition of the operator G
    Last edited: Nov 17, 2012
  4. Nov 18, 2012 #3


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    T(ε) is an uniparametric group of unitary operators on a Hilbert space. Its generator G is a self-adjoint operator acting on the same space (dense everywhere subset of it). Mathematically, this is covered through Stone's theorem and its reverse.

    The + or - sign when linearizing the exponential is a convention for the so-called active vs passive view of symmetries. He chose - which IIRC stands for the passive view of looking at space translations.
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