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.
[tex]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?[/tex]

 

[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.