Position Operator: f(\hat{x})=f(x)? Effects on g(x)

[zhaiyujia]

is it true that: f(\hat{x})=f(x)?
What will happen if f(\hat{x})=\frac{\hat{x}}{\hat{x}+1} act on g(x)?

 

[malawi_glenn]

Can you first clarify your notations, and also show work done and relations you know of etc.

 

[dextercioby]

Who's "x" and who's \hat{x} ?

 

[CompuChip]

I assume that x is the eigenvalue of a position operator \hat x.

If f is a function that only depends on the operator \hat x, then the statement is true, as can be seen by expanding f(\hat x) in a series and acting it on a ket \left| x \right>.
In general this need not be true though, e.g.
f(\hat x, \hat p) = \hat x \hat p[/itex] will not give <i>x p</i>. <br /> <br /> For the last question, what is 1/\hat x supposed to mean?

 

[dextercioby]

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