1
$\begingroup$

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

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

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 $\hat{x}_i$:

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

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

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

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

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

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).

$\endgroup$

1 Answer 1

2
+100
$\begingroup$

The term you're looking for is "weighted inner product", which is usually first taught in the context of Sturm-Liouville operators. The basic idea, in one dimension, is that given a positive function $w(x) \geq 0$, one can define an inner product via $$\langle\psi|\phi\rangle = \int dx \, w(x) \psi^*(x)\phi(x)\tag{1}$$ where $w(x)$ is the "weight" of the inner product. The generalisation to higher dimensions is obvious. Then one can, for example, define a notion of "orthogonality" with respect to this inner product. If $w(x)\equiv 1$, then $(1)$ is the "usual" inner product, but from a purely mathematical standpoint $w \neq 1$ is an equally good choice and may be useful if you consider, like here, nonstandard quantum mechanics.

To really see that this is nothing weird, consider for example spherical coordinates in three dimensions. Then we can write the inner product in cartesian coordinates as usual, $$\langle\psi|\phi\rangle = \int dx \,dy\,dz\, \psi^*(\vec{x})\phi(\vec{x})$$ However, we can also go to spherical coordinates, for which $$\langle\psi|\phi\rangle = \int dr \,d\theta\,d\varphi\, r^2 \sin\theta\, \psi^*(\vec{x})\phi(\vec{x})$$ Here a "weight" $r^2\sin\theta$ has appeared. So as you can see it is natural to introduce weighted inner products when changing coordinate systems. So, as remarked before, one could in principle also have a weight in cartesian coordinates. Sometimes the whole object $dx\,w(x)$ is referred to as "the measure", for example $dr \,d\theta\,d\varphi\, r^2 \sin\theta$ is the "integral measure" in spherical coordinates. However note that "measure theory" in mathematics, while related to this, is a very complicated subject that is not useful in this case.

What the authors of the papers are trying to do is to find an appropriate weighted inner product where their modified operator $\hat x$ is still symmetric. To do so, they define the weighted inner product $$\langle\psi|\phi\rangle = \int d^3 \vec{p} \, w(\vec{p}) \psi^*(\vec{p})\phi(\vec{p})\tag{2}$$ where $w(\vec{p})$ is an appropriate weight so that the required condition ($\hat{x}$ symmetric) is satisfied. Apart from this, the choice of $w$ is completely arbitrary. In order to find $w$, you write the condition that $\hat{x}$ is symmetric, i.e. $$\langle\psi|\hat{x}|\phi\rangle=\langle\phi|\hat{x}|\psi\rangle^*\tag{3}$$ for generic $w$ with the weighted inner product $(2)$. This will give you a condition for $w$, which you can solve for. It should be emphasised that any $w$ that makes $(3)$ true will work, you don't need "the most general" $w$ or any "unique" $w$. As such, to make your life easier, you may assume for example that $w(\vec{p}) = w(p^2)$. Mathematically this doesn't have to be the case, but it's simple, and if you can find a $w$ with this property, you might as well go with it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.