• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,079 questions , 2,229 unanswered
5,348 answers , 22,758 comments
1,470 users with positive rep
819 active unimported users
More ...

  How to set a "correct" integral measure in (nonstandard) quantum mechanics?

+ 2 like - 0 dislike


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).

asked May 27, 2021 in Theoretical Physics by andrehgomes (10 points) [ revision history ]
edited May 31, 2021 by andrehgomes

1 Answer

+ 2 like - 0 dislike

One does it the other way around, starting with the factor and deducing $x$.

Use $M=Q^*Q$ and the relation $\phi^*A\phi=\psi^*MA'\psi$ if $A'=Q^{-1}AQ$.

If $Q$ is a function of $p$, $\psi^*M\psi$ defines an inner product with a momentum dependent measure, and $A'$ is the appropriate version of $A$ in the new inner product. Now put $A=x$ and simplify $A'$ using the CCR to get the correct expression for $x$ in the new inner product. 

answered May 28, 2021 by Arnold Neumaier (15,787 points) [ revision history ]
edited May 28, 2021 by Arnold Neumaier
Most voted comments show all comments

@ArnoldNeumaier I believe your answer is perfectly well-suited for both one- and three-dimensional cases (or for arbitrary n-dimensions for all that matter). It seems to me the issue is with my definition of $X_i$ as in it I choose $f \partial_{p_i}(\dots)$ (and something similar for $g$) intead of $\partial_{p_i}(f\dots)$. This ambiguity is unphysical as it only results in some extra function of $p$ in the definition of $X_i$ which may be eliminated by a redefinition of $\gamma$ --- compare to (2) for context --- but it seemly changes in different ways what I define as $\gamma$ whether I'm considering the one- or the three-dimensional case, I haven't figured out the details yet, but I believe this is behind my confusion.

Since your formula for $x$ has two parameters and the metric has only one, it is clear  that you cannot reproduce the 2 parameter formula. Instead you get just the case $\beta'=0$, which is presumably the only reasonable choice.

One can call anything a position operator, but it is unlikely to lead to anything of interest if it is not naturally related to the canonical position operator.

Since by direct computation I verify $ \langle\phi|X_i|\psi\rangle = \langle\psi|x_i|\phi\rangle^* $ provided definitions (2), (3) and (4), I don't really understand why shouldn't it be possible to reproduce these results using canonical transformations as well...

@ArnoldNeumaier I finally found where my trouble was, and in advance many thanks for the discussion.

Canonical transformations in the context of my third comment above only relates $x_i$ symmetric under measure $d^3p/\mathcal{M}$ to $x'_i$ symmetric under $d^3p/\mathcal{M}^{1-\alpha}$, but imposes nothing to ensure these are indeed symmetric. It is only assumed this symmetricity is true. On the other hand, imposing $x_i$ to be symmetric under $d^3p/\mathcal{M}$ requires adding $\frac{i\hbar}{2}\mathcal{M} \partial_j [(1+\beta p^2)\delta_{ij}+\beta' p_i p_j)\mathcal{M}^{-1}]$ to the very definition of $x_i$. This extra piece together with the term proportional to $\alpha$ in (7) gives what we define as $\gamma p_i$ in (2) and for the choice $\mathcal{M}=1+(\beta+\beta')p^2$ we precisely get relation (4).

In general, you can add to $X$ any expression $X$ that is symmetric in your scalar product, and get another $X'$. For example you can add to the canonical solution (with $\beta'=0$) a central multiple of $Z$ with $Z_i=x_i-p_i(i\hbar-p\cdot x)$ and get a 2-parameter formula equivalent to yours. (Fix $\alpha,\delta$  and write $\beta=\delta-\beta'$, $\gamma=\beta'+\alpha\delta$.) This leaves a lot of freedom to go beyond canonicality - but the resulting operators mean nothing since what you can add is almost unrestricted!

Most recent comments show all comments

I'm probably missing something, but I couldn't get the correct expressions for the three-dimensional case... I do get $x_i$ as in the original post, but with incorrect relation between $\gamma$ and $\alpha$.

I believe my trouble is that maybe I should generalize your arguments to taking into account that the integration measure $M$ doesn't directly cancel $M_{ij}$ in $\hat{x}_i = i\hbar [(1+\beta p^2)\delta_{ij}+\beta'p_i p_j] \partial_{p_i} = i\hbar M_{ij} \partial_{p_i}$. Nevertheless, I cannot see how to do that due to the very scalar nature of $M=Q^*Q$.

@andrehgomes: In QM, the metric $M$ is necessarily a scalar. I don't know where your 3D formulacomes from, but unless there is a canonical transformation to the standard position operator, there is no right to call your $x$ a position operator. So maybe your original source had the relation between $\gamma$ and $\alpha$ wrong?

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights