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Position operator

  1. Oct 9, 2007 #1
    is it true that: [tex]f(\hat{x})=f(x)[/tex]?
    What will happen if [tex]f(\hat{x})=\frac{\hat{x}}{\hat{x}+1}[/tex] act on [tex]g(x)[/tex]?
     
  2. jcsd
  3. Oct 10, 2007 #2

    malawi_glenn

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    Can you first clarify your notations, and also show work done and relations you know of etc.
     
  4. Oct 11, 2007 #3

    dextercioby

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    Who's "x" and who's [itex]\hat{x} [/itex] ?
     
  5. Oct 11, 2007 #4

    CompuChip

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    I assume that x is the eigenvalue of a position operator [itex]\hat x[/itex].

    If f is a function that only depends on the operator [itex]\hat x[/itex], then the statement is true, as can be seen by expanding [itex]f(\hat x)[/itex] in a series and acting it on a ket [itex]\left| x \right>[/itex].
    In general this need not be true though, e.g.
    [tex]f(\hat x, \hat p) = \hat x \hat p[/itex] will not give x p.

    For the last question, what is [itex]1/\hat x[/itex] supposed to mean?
     
  6. Oct 11, 2007 #5

    dextercioby

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    If [itex] \hat{x} [/itex] is the position operator in QM, then you might as well consider the Hilbert space as being [itex] L^{2}(\mathbb{R},dx) [/itex] and you will find that the Schwartz space [itex] S(\mathbb{R}) [/itex] is not only a domain for essential selfadjointness of [itex] \hat{x} [/itex], but also a domain for any polynomial function of the operator "[itex] \hat{x} [/itex]". Now, series expansions of operators is a tricky business (due to convergence issues) and now i'm too tired to go there.
     
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