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

In summary, the conversation discusses the validity of the statement f(\hat{x})=f(x) and its implications when acted upon by another function g(x). The notations are clarified and it is assumed that x is the eigenvalue of a position operator \hat x. It is mentioned that the statement may not always hold true, depending on the function f. The last question addresses the meaning of 1/\hat x and suggests considering the Hilbert space as being L^{2}(\mathbb{R},dx).
  • #1
zhaiyujia
6
0
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]?
 
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  • #2
Can you first clarify your notations, and also show work done and relations you know of etc.
 
  • #3
Who's "x" and who's [itex]\hat{x} [/itex] ?
 
  • #4
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?
 
  • #5
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.
 

1. What is the position operator, f(\hat{x})=f(x)?

The position operator, denoted as f(\hat{x})=f(x), is a mathematical representation of a physical quantity known as position. It is used in quantum mechanics to describe the position of a particle in a given system.

2. How does the position operator affect g(x)?

The position operator, f(\hat{x})=f(x), acts on a function g(x) by multiplying it with the position variable x. This results in a new function, f(g(x)), which describes the effect of the position operator on g(x).

3. What is the significance of the position operator in quantum mechanics?

In quantum mechanics, the position operator is a fundamental mathematical tool used to describe the position of a particle in a given system. It is also used to calculate the uncertainty in the position of a particle, which is a key concept in quantum mechanics.

4. How is the position operator related to the uncertainty principle?

The position operator is closely related to the uncertainty principle in quantum mechanics. According to the uncertainty principle, it is impossible to measure the exact position and momentum of a particle simultaneously. The position operator is used to calculate the uncertainty in the position of a particle.

5. Are there any limitations to the use of the position operator?

While the position operator is a useful tool in quantum mechanics, it has its limitations. It is not applicable to systems with multiple particles or continuous position measurements. Additionally, the position operator is only valid for position measurements and cannot be used for other physical quantities such as velocity or energy.

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