**Context**

Studying a paper exploring possible consequences of a fundamental length scale in Nature, particularly in the context of one-dimensional quantum mechanics [1], the authors argue for a modification of the $\hat{x}$ operator in momentum-space representation: $\hat{x}=i\hbar(1+\beta p^2)\frac{\partial}{\partial p}$, with $\beta$ a positive constant. The momentum operator is as usual in this case, $\hat{p}=p$. It is then discussed $\hat{x}$ is no longer hermitian, but still symmetric nevertheless, *i.e.*, $\langle\psi|\hat{x}|\phi\rangle=\langle\phi|\hat{x}|\psi\rangle^*$, as long as the scalar product is defined as

$$ \langle\psi|\phi\rangle= \int \frac{dp}{1+\beta p^2} \psi^*(p) \phi(p). \tag{1}$$

This is very reasonable because the factor $1/(1+\beta p^2)$ exactly cancels the factor $(1+\beta p^2)$ in $\hat{x}$ and one is left with verifying that $\hat{x}$ is symmetric by a simple integration by parts. In a sense, the extra factor on the definition above could be guessed without much effort.

On the other hand, in the context of three-dimensional quantum mechanics this is not so clear to me. For instance, in [2] the authors propose the following representation of $\hat{x}_i$:

$$ \hat{x}_i = i\hbar(1+\beta p^2)\frac{\partial}{\partial p_i} + i\hbar \beta' p_i p_j \frac{\partial}{\partial p_j} + i\hbar \gamma p_i, \tag{2} $$

with $p\equiv|\vec{p}|$ and where $\beta$, $\beta'$, and $\gamma$ are constants. Similar to the previous case, this operator is not hermitian, but it is symmetric under the definition

$$ \langle\psi|\phi\rangle= \int \frac{d^3p}{[1+(\beta+\beta')p^2]^{1-\alpha}} \psi^*(\vec{p}) \phi(\vec{p}), \tag{3}$$

where $\alpha$ depends on our choice of $\gamma$ as

$$ \alpha = \frac{\gamma-\beta'}{\beta+\beta'}. \tag{4}$$

I can derive this extra factor not by simple guess as before, but by pattern recognition after studying other simpler cases, some trial and error, and so on. At the end of the day, I cannot derive it from some basic understanding.

**The question**

How can we derive the extra factor on the definition of the scalar product not by somehow guessing it?

**My efforts**

I still don't have the mathematical background to be comfortable with the topic, but some research indicates me the extra factor in the integrals above are different choices of "integral measure". I found mathematical texts on that, but too technical to grasp useful understanding. Indication of any physical approach to this topic, or at least some not too technical but enough for basic application, would be really appreciated.

[1] A. Kempf, G. Mangano, and R. B. Mann, Phys. Rev. D **52**, 1108 (1995).

[2] R. Akhoury and Y.-P. Yao, Phys. Lett. B **572**, 37 (2003).