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

Question

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).

Answer 1

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. 

Comments

@ArnoldNeumaier Thanks! It worked beautifully for the one-dimensional example above and I'll be checking it soon for the three-dimensional one. Very interesting that the answer revolves around a canonical transformation.

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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 $M=Q^*Q$.

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@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?

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(I'm not totally aware of all policies of physicsoverflow -- I'm new here -- but here it goes a looong comment.)

@ArnoldNeumaier Yes, $x$ is no longer connected to the standard position operator by a canonical transformation and we may not regard it as such indeed -- not in the usual sense at least. Equation (2) above may be written more generally as $X_i = f(p)x_i + g(p) p_i p_j x_j + i\hbar\gamma(p) p_i$ with $[x_i,p_j]=i\hbar\delta_{ij}$ as in standard QM but now it is the commutator $[X_i,p_j]=i\hbar(f\delta_{ij}+g p_i p_j]$ that defines our model (whether $X$ or $x$ is the physical observable would be a topic for another discussion indeed). Maybe I should've mentioned all of that in the original post... for all that matters, I'll take this context for granted in what follows.

Operator $X_i$ is not hermitian, but we can make it a symmetric operator defining the scalar product as Eq. (3) above. I checked the relation between $\gamma$ and $\alpha$ is correct by direct calculation to show $\langle\phi|X_i|\psi\rangle=\langle\psi|X_i|\phi\rangle^*$. So I tried to adapt your answer to derive (4) above. I started defining $X_i=i\hbar(f\delta_{ij}+g p_ip_j)\partial_{p_i}$ because of the commutator $[X_i,p_j]$ and also the scalar product as

$$ \langle\phi |X_i |\phi\rangle \equiv \int \frac{d^3p}{\mathcal{M}} \phi^* X_i \phi \tag{5}$$

with the factor $\mathcal{M}(p)$ to ensure $X_i$ is symmetric. Introducing the canonical transformation $Q(p)$ such that $\phi=Q\psi$ and $X'_i=Q^{-1}X_i Q$ we get

$$ \langle\phi |X_i |\phi\rangle = \int \frac{d^3p}{\mathcal{M}} M \psi^* X'_i \psi \equiv \langle\psi |X'_i |\psi\rangle_\alpha \tag{6}$$

where I set $M=Q^*Q\equiv\mathcal{M}^\alpha$ to defined the scalar product $\langle\psi |X'_i |\psi\rangle_\alpha$ with measure $d^3p/\mathcal{M}^{1-\alpha}$ which is to be compared to (3) in the original post.

To sum up, $X_i$ is symmetric under the  scalar product measure $d^3p/\mathcal{M}$ while $X'_i$ is symmetric using the measure $d^3p/\mathcal{M}^{1-\alpha}$, but both operators are connected by the canonical transformation $X'_i=Q^{-1}X_i Q$ with $Q=\mathcal{M}^{\alpha/2}$. Using the commutator $[X_i,p_j]$ I can solve for $X'_i$:

$$ X'_i = i\hbar(f\delta_{ij}+g p_ip_j)\partial_{p_i} + i\hbar \left[ \frac{\alpha \mathcal{M}'}{2p} \mathcal{M}^{-1}(f+gp^2) \right] p_i, \tag{7}$$

where $\mathcal{M}'\equiv\partial\mathcal{M}/\partial p$. The piece inside the square brackets is just $\gamma$ (compare to (2) in the original post), then

$$ \alpha = \frac{2\gamma p}{\mathcal{M}'} \frac{\mathcal{M}}{f+gp^2}. \tag{8}$$

Here's where things get curious: if we go to one-dimension, this is the correct relation between $\alpha$ and $\gamma$ (not shown in the original post); but it is just not correct for three-dimensional case as we notice it doesn't reproduce (4) in the particular case $f=1+\beta p^2$, $g=\beta'$ and $\mathcal{M} = 1+(\beta+\beta')p^2$. As a matter of fact, the correct version of (8) has $\gamma-g$ instead of $\gamma$. Despite re-doing the above calculations a variety of times, I may be messing with some computation, but if some of the steps above is unreasonable, that may the source of trouble.

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@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.

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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.

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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...

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@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).

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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!