# Norm of position operator of quantum mechanics

Let the position operator be $$\begin{array}{cccc} \hat{x}: & L_2[a,b] & \to & L_2[a,b] \\ & f(x) & \mapsto & xf(x) \end{array}\ .$$

We can prove that this operator is bounded: $$||\hat{x}f(x)||_2 = ||xf(x)||_2 = \sqrt{\int_a^b |xf(x)|^2\ dx} = \sqrt{\int_a^b |x|^2|f(x)|^2\ dx}\ ,$$ now define $$M = \max\{|a|,|b|\}$$ to bound that integral, and $$||\hat{x}f(x)||_2 \leq \sqrt{M^2 \int_a^b |f(x)|^2\ dx} = M ||f(x)||_2\ ,$$ proving that $$\hat{x}$$ is a bounded operator.

However, this also makes a restriction on the norm of the operator, such that $$||\hat{x}|| \leq \max\{|a|,|b|\}\ .$$ But I don't know how to calculate the norm from here. I know that three expressions can be used, $$||\hat{x}|| = \sup_{||f(x)||\leq 1} ||\hat{x}f(x)|| = \sup_{||f(x)|| = 1} ||\hat{x}f(x)|| = \sup_{||f(x)||\neq 0} \dfrac{||\hat{x}f(x)||}{||f(x)||}\ ,$$ so I tried finding a $$g(x)$$ such that $$||g(x)|| = 1$$, and use it as a lower bound: $$||\hat{x}|| = \sup_{||f(x)|| = 1} ||\hat{x}f(x)|| \geq ||\hat{x}g(x)||\ ,$$ but I didn't have too much success on this.

I would appreciate any help on this problem.

Let $$f_n=\sqrt{n}1_{[b-1/n,b]}$$ and note that for $$n\geq\frac{1}{b-a}$$, $$f_n\in L^2([a,b])$$ and $$\|f_n\|_{L^2}=1$$. Furthermore, we have $$\int_a^b x^2f_n(x)^2\textrm{d}x=n\int_{b-1/n}^b x^2\textrm{d}x=\frac{n}{3}(b^3-(b-1/n)^3)=\frac{n}{3}\left( \frac{1}{n^3}-3\frac{b}{n^2}+3\frac{b^2}{n}\right),$$ which converges to $$b^2$$ as $$n\to\infty$$. Accordingly, $$\|\hat{x}\|\geq |b|.$$ Similarly, we can get $$\|\hat{x}\|\geq |a|$$ and thus, you get that $$\|\hat{x}\|=\max\{|a|,|b|\}$$.
• Yes, $1_A(x)$ is the function which is $1$ if $x\in A$ and $0$ else. It's called the indicator function or characteristic function of $A$. Jan 7, 2020 at 11:31