The generators of a ``Poincare-type'' group in momentum space

  • #1
redtree
292
13
TL;DR Summary
The mathematics of the Poincare group in position space X are well described. However, I have not found an analogous description of generators of an analogous ``Poincare-type'' group in momentum space K, where the boosts, rotations and translations involve k^u (as opposed to x^u in position space).
Can someone share a paper or chapter from a textbook if they know a good one?

I'm curious to see the explicit form of these matrices. In position space, the generators of boosts act on the rapidity, which can be related to velocity in X. Assuming the generators of boosts in K act on rapidity in K, what is the velocity in K related to the rapidity in K?
 
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  • #2
redtree said:
TL;DR Summary: The mathematics of the Poincare group in position space X are well described. However, I have not found an analogous description of generators of an analogous ``Poincare-type'' group in momentum space K, where the boosts, rotations and translations involve k^u (as opposed to x^u in position space).

I'm curious to see the explicit form of these matrices.
In position space, one typically uses differential operators to represent the Poincare generators. E.g., ##\partial_x## generates translations in the "x" direction. So I'm not sure why you want to use matrices (even though of course one can use 5x5 matrices for that purpose).

And what, precisely, do you mean by "Poincare-type" group? Do you mean simply representing the usual generators in a Fourier-transformed way, or do you mean (e.g.,) to include translations in momentum space (changing the momentum). If the former, ##\partial_x## goes over to ##k_x## (multiplied by some constant depending on your FT conventions).

redtree said:
In position space, the generators of boosts act on the rapidity, which can be related to velocity in X. Assuming the generators of boosts in K act on rapidity in K, what is the velocity in K related to the rapidity in K?
Velocity and rapidity are related by a tanh function. Wikipedia covers this.
 
  • #3
My reference is: Schwichtenberg, Jakob. Physics from symmetry. Springer, 2018.

In this text, matrices are utilized for the generators of the Poincare group and these matrices are 4x4 or 2x2.

In any case, the generators, particularly the boosts, are related to complex rotations over an angle ##\theta##, where ##\theta## often is denoted as ``rapidity'' (though not in this particular text). Rapidity in spacetime, i.e., X-space, can then be related to relative velocity. The Poincare group denotes the spacetime, i.e., X-space, isometries for translations, boosts and rotations.

The derivation of these generators with their Lie algebra is relatively simple. Importantly, the mathematics is not specific to spacetime in the sense that an analogous derivation for analogous generators preserving analogous isometries can be done for any space, such as a momentum space, i.e., K-space, or a generic space Z. The Poincare group specifically denotes these group actions in X-space. A analogous group structure in some other space is not a Poincare group, though it might be called a Poincare-type group in that other space. Both the Poincare group and analogous Poincare-type groups in other spaces can be utilized to represent the symmetries of basis transformations in the space.

Given an analogous derivation of a analogous group structure to the Poincare group but in K-space, the generators of "boosts" in K can be related to complex rotations in K over an angle, where this angle can be denoted as a "rapidity" in K, which then can be related to a velocity in K. Again, all of this is relatively straightforward mathematically.

Just as the Poincare group provides important mathematical structure to physics, it would seem that a "Poincare-type" group in K (or momentum space) might also provide insights, and I wonder if such a group structure has been described explicitly (and if so, where) and if such a structure has been utilized in ways that are analogous to the Poincare group in formulating physical theories.
 
  • #4
redtree said:
My reference is: Schwichtenberg, Jakob. Physics from symmetry. Springer, 2018.
In this text, matrices are utilized for the generators of the Poincare group and these matrices are 4x4 or 2x2.
Do you really mean the Poincare group (10 parameters describing rotations, boosts and translations) or are you interested in the Lorentz group (just the 6 parameters of rotations and boosts)?
Assuming the later, you might consult the 2002 article:
The generators of lorentz transformation in momentum space (Pengfei Zhang & Tunan Ruan)
Abstract: In the momentum space, the angular momentum operator and the boost vector operator, i.e. the generators for the Lorentz transformation of a particle with arbitrary spin and nonzero mass are discussed. Some new expressions are obtained in terms of the orbital and spin parts.
https://link.springer.com/article/10.1360/02ys9025
 
  • #5
What I really want is the Poincare-type group in K-space, which of course is trivial to construct from the Lorentz group in K-space.

The article you quote is interesting but suffers from the following problem:

Given a 4-momentum ##\textbf{p} \in K##:
$$\textbf{p} = \{E_0,p_1,p_2,p_3 \} = \{E_0, \vec{p} \}$$
where ##\vec{p} =\{ p_1,p_2,p_3\}##

The paper utilizes the following equivalence
$$\textbf{p} = \{\gamma_X M, \gamma_X \vec{v}_X M\}$$
where ##M## denotes mass, ##\vec{v}_X## denotes velocity in X-space, such that ##\vec{v}_X \in X \not\in K## and ##\gamma_X = (1 -\vec{v}_X^2)^{-1/2}##, such that ##\gamma_X \in X \not \in K##

From a mathematical perspective, this equivalence implies a mapping ##m: K \rightarrow X##, where
$$m(\textbf{p}): \{E_0, \vec{p} \} \rightarrow \{\gamma_X M, \gamma_X \vec{v}_X M\}$$
which for ##\vec{v}_X = 0## implies
$$m(\textbf{p}) \rightarrow \{m,0 \}$$

Of course, the problem should be clear in this context. Given a mapping ##m: K \rightarrow X##, this formulation of momentum, though a formulation of momentum, is no longer in K space but rather in X space.

I am hoping to see a formulation of a Lorentz-type and Poincare-type groups in K and not in X and thus this paper is not what I am hoping to find. I am specifically looking for formulations that do NOT utilize the mapping ##m##.
 
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  • #6
redtree said:
What I really want is the Poincare-type group in K-space, which of course is trivial to construct from the Lorentz group in K-space.
If it's trivial to construct, then why...

redtree said:
I am hoping to see a formulation of a Lorentz-type and Poincare-type groups in K and not in X and thus this paper is not what I am hoping to find. I am specifically looking for formulations that do NOT utilize the mapping ##m##.
...are you even asking this? Don't you already have the answer?

Basically you appear to me to be considering "K-space" as a Minkowski vector space in its own right, independent of any connection with anything else. If that's the case, the Minkowski vector space structure already gives you everything you need to know. What more do you want?
 
  • #7
I am considering K space as the Fourier conjugate space of X space, not a completely independent space, and I am requiring that dynamics in K space respect that Fourier conjugate relationship between X and K spaces.

My opinion is that this exercise is both trivial and useful, and I want to see if it has been done.

Ultimately, the reason for doing this is twofold: 1) Because Lorentz-type and Poincare-type groups in K would seem to be important symmetry groups in K space when considering basis transformations in K; 2) These symmetry groups can then be used to consider Lagrangian densities formulated in K, which of course is its own topic.
 
  • #8
redtree said:
I am considering K space as the Fourier conjugate space of X space
Then I don't understand your objections in post #5, since the mapping you object to is part of the Fourier conjugate relationship.
 
  • #9
How is the mapping ##m## part of the Fourier conjugate relationship? It does not respect the Fourier conjugate relationship between the spaces X and K. This is trivial to demonstrate.

Assuming ##\hbar =1 \rightarrow \textbf{p} = \textbf{k} = \{ k_0, \vec{k} \}##, where ##\vec{k} = \vec{p}##.

For an integral transform ##\mathscr{F}: X \rightarrow K##, i.e., the Fourier transform
$$f(x) \rightarrow \hat{f}(k): \hat{f}(k) = \int_{\mathbb{R}} f(x) e^{-2 \pi i k x} dx$$
with an inverse mapping ##\mathscr{F}^{-1}: K \rightarrow X##, i.e., the inverse Fourier transform
$$\hat{f}(k) \rightarrow f(x): f(x) = \int_{\mathbb{R}} \hat{f}(k) e^{+2 \pi i k x} dk$$

For ##\hat{f}(k) = \vec{k}##, with the mapping ##\mathscr{F}^{-1}: K \rightarrow X##
$$\mathscr{F} \left[ \vec{k} \right] = \frac{\delta'(\vec{x})}{2 \pi i}$$
while for ##\vec{k}##, with the mapping ##m: K \rightarrow X##
$$m\left[\vec{k} \right] = M \gamma_X \vec{v}_X$$

These are independent mappings.
 
  • #10
redtree said:
How is the mapping ##m## part of the Fourier conjugate relationship?
First of all, your Fourier transform in post #9 is wrong. The vector you are calling ##\mathbf{p}## in what you are calling K space (P space would be a better term) is, in Fourier transform terms, a delta function in K space--because it is a vector representing a single 4-momentum, i.e., a single point in K space. The Fourier transform of that is not a delta function in X space. It's the function ##e^{ipx}##.

Second, you didn't include the norm of ##mathbf{p}## in your analysis. That norm is ##M##. So there needs to be a factor of ##M## included.

Third, what is the thing in X space that you are calling ##m(\mathbf{p})##? It's not a vector; vectors in X space represent spacetime positions, i.e., events, i.e., points in X space--in Fourier transform terms, delta functions. And as above, ##\mathbf{p}## is a delta function in K space, whose Fourier transform in X space is ##e^{ipx}## (with a factor of ##M## put in in the appropriate place). What does this represent in X space? It represents a plane wave. And what is the ordinary 3-velocity of that wave in X space? It is what you are calling ##\vec{v}##. In other words, the thing in X space you are calling ##m(\mathbf{p})## is the Fourier transform of ##\mathbf{p}## to X space.
 
  • #11
PeterDonis said:
First of all, your Fourier transform in post #9 is wrong. The vector you are calling ##\mathbf{p}## in what you are calling K space (P space would be a better term) is, in Fourier transform terms, a delta function in K space--because it is a vector representing a single 4-momentum, i.e., a single point in K space. The Fourier transform of that is not a delta function in X space. It's the function ##e^{ipx}##.
For ##\hbar =1##, the choice of K space or P space is a matter of convention. I also choose to use the Fourier transform convention ##e^{2 \pi i k x}## instead of ##e^{i p x}##. I tried to be explicit about all this.

I apologize, but I don't know what you mean in saying my calculation is incorrect. This is not my calculation, I did it in Mathematica.

$$f(x) = \int_{\mathbb{R}} k e^{2 \pi i k x} dk = \frac{\delta'(x)}{2 \pi i}$$
where ##\delta'(x)## denotes the derivative of the delta function with respect to ##x##

This is the Mathematica code if you want to check:
FourierTransform[k, k, x, FourierParameters -> {0, +2*Pi}]

As to your second point, I was never calculating the norm of ##\textbf{p}##. I apologize again but don't see how the norm has relevance to the discussion.

I defined ##m## as a mapping and ##M## as mass. I was merely using a notation convention where one writes the mapping as a function acting on the variable, similar to ##f(x)## for ##f: X \rightarrow X##. If that notation makes it harder to understand my mathematical point then ignore it.

I would ask you to explain your point that momentum formulated in X space is not a vector. Are you saying that momentum formulated as a function of velocity in X is not in X space just because it does not represent a 4-position? What space is ##d \vec{x}/d x_0## in then, if not X space?
 
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  • #12
redtree said:
I don't know what you mean in saying my calculation is incorrect.
In momentum space, which is what you are calling K space, the object you are calling ##\mathbf{p}## is a point. The vector ##\mathbf{p}## is the vector from the origin to that point. But if you are going to Fourier transform it, then what you Fourier transform is a delta function in K space; that is what represents a specific value of the 4-momentum ##\mathbf{p}##.

redtree said:
This is not my calculation, I did it in Mathematica.
You don't need Mathematica to Fourier transform a delta function, or a complex exponential for that matter. You can do it by inspection. But you need to correctly understand what you are transforming.

redtree said:
As to your second point, I was never calculating the norm of ##\textbf{p}##.
You didn't calculate it, you just assumed it when you introduced the quantity ##M## into your equations. That's what ##M## is: the norm of ##\mathbf{p}##.

redtree said:
don't see how the norm has relevance to the discussion.
You could assume it to be ##1## if you like, just as you are assuming ##\hbar = 1##. But you didn't do that. That means you need to include it in your equations. I agree it's a minor point, though. The major issue is correctly understanding what you are transforming.

redtree said:
I defined ##m## as a mapping and ##M## as mass.
Yes, I know that. There is no confusion there and that choice of notation is not a problem.

redtree said:
I would ask you to explain your point that momentum formulated in X space is not a vector.
You can model 4-momentum as a tangent vector in X space if you want; then it's just a directional derivative along a particular curve in X space. But if you are going to Fourier transform between X space and K space, or vice versa, then that's not how you want to formulate momentum in X space.

As far as a Fourier transform is concerned, momentum in X space is the function ##e^{ipx}##, since that's the Fourier transform of the delta function ##\delta(p)## in K space. Or, if you want to be precise about exactly what kinds of products you are taking, it would be ##\delta(\mathbf{p})## in K space (where ##\mathbf{p}## tells you the particular point in K space that denotes the 4-momentum you are interested in), and its Fourier transform in X space would be ##e^{i \mathbf{p} \cdot \mathbf{x}}##, considered as a function over ##\mathbf{x}##, i.e., over the points in X space, with ##\mathbf{p} \cdot \mathbf{x}## being a dot product.
 
  • #13
redtree said:
What space is ##d \vec{x}/d x_0## in then, if not X space?
In the Euclidean 3-space that represents a spacelike 3-surface in X space, picked out by a constant value of the time coordinate ##t## in a chosen inertial frame. X space is a space of 4-vectors, not 3-vectors.

The 4-momentum in X space, considered as a directional derivative along some timelike curve, is ##M\ d\mathbf{x} / d\tau##, where ##\tau## is the affine parameter along the curve.
 
  • #14
Thank you for your response.

Why do you assume that ##\textbf{p} \in K## is a point relative to the origin, let alone one represented by a delta function? I have made neither of those assumptions. A vector may or may not include the origin as one of its endpoints. If I wanted to represent a delta function, I would have written ##\delta(\textbf{p})## or ##\delta(\vec{p})##.

The idea that momentum in momentum space must be represented by a delta function seems highly problematic. For example, it is inconsistent with the derivation of the uncertainty principle. (see https://en.wikipedia.org/wiki/Uncertainty_principle)

My point was to demonstrate that ##\mathscr{F}^{-1}## and ##m## are independent transformations by demonstrating how they act differently on the same mathematical object. That one is an integral transformation and the other a scaling transformation, where ##\hbar, M, \gamma_X## are all scalars, should really be enough to prove the point, but I wanted to demonstrate the mathematical fact explicitly.

The larger point is that I am looking for a published formulation of a Poincare-type group in K (but a Lorentz-type group in K would suffice) formulated without utilizing the scaling transformation represented by ##m##. If you know of any published sources, please let me know.
 
  • #15
redtree said:
Why do you assume that ##\textbf{p} \in K## is a point relative to the origin
It's not an assumption, it's what a vector in K space represents mathematically. Physically, it represents a state with a single definite 4-momentum.

redtree said:
let alone one represented by a delta function?
For purposes of doing a Fourier transform, a state with a single definite 4-momentum, in 4-momentum space, i.e., K space, is a delta function. That's how you describe a single point in K space when you want to Fourier transform.

redtree said:
I have made neither of those assumptions.
As above, they aren't assumptions, they're just correctly recognizing what you need to recognize in order to do correctly what you say you are trying to do.

redtree said:
A vector may or may not include the origin as one of its endpoints.
A vector in a vector space, if you insist on thinking of it as a little arrow with two endpoints, must have the origin as one of its endpoints.

The real fix for this issue is to stop thinking of a vector in a vector space as a little arrow with two endpoints. A vector space is a much more general concept, and even in special cases where you can sort of get away with the "arrow" metaphor, it can still cause confusion, as it is doing for you in this case.

redtree said:
The idea that momentum in momentum space must be represented by a delta function seems highly problematic. For example, it is inconsistent with the derivation of the uncertainty principle. (see https://en.wikipedia.org/wiki/Uncertainty_principle)
Um, what? Even leaving aside that Wikipedia is not a good primary source, the article you reference says:

"Mathematically, in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding orthonormal bases in Hilbert space are Fourier transforms of one another (i.e., position and momentum are conjugate variables)."

In particular, an eigenstate of momentum is a delta function in momentum space, and the Fourier transform of that, i.e., a complex exponential, in position space. Some sources get uneasy about this (and the Wikipedia article appears to share such uneasiness) because neither a delta function nor a complex exponential are, strictly speaking, in the Hilbert space of square integrable functions, but that is easily dealt with, for example by using the rigged Hilbert space formalism, as discussed in, e.g., Ballentine.

redtree said:
My point was to demonstrate that ##\mathscr{F}^{-1}## and ##m## are independent transformations by demonstrating how they act differently on the same mathematical object.
And your claimed demonstration is wrong, for reasons I have already given.

redtree said:
The larger point is that I am looking for a published formulation of a Poincare-type group in K (but a Lorentz-type group in K would suffice) formulated without utilizing the scaling transformation represented by ##m##.
And because of the various issues that I and others have pointed out, we still don't understand what you are actually looking for, or even whether what you are looking for makes any sense.
 
  • #16
I once found A. Cohen, An Introduction to Lie Theory of One-Parameter Groups, Baltimore 1911, by chance on the Internet. I wanted to read and understand the essential parts of it, for one because I'm interested in the history of science, and second I observed that physics hasn't really performed a transition of language, at least not fully. Many terms are still the same as they were when Cohen wrote that book.

I summarized the concepts in an insights article When Lie Groups Became Physics. The considered groups are far smaller than the ten-dimensional Poincaré group, however, the principles are comparable. So maybe, this helps.
 
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  • #17
fresh_42 said:
the ten-dimensional Poincaré group
Mathematically, this is a single group, but physically, it can have (at least) two distinct interpretations. We can think of it as the group of Killing vector fields on Minkowski spacetime, i.e., as the symmetry group of what the OP calls "X space". Or we can think of it as the symmetry group of what the OP calls "K space", and which is usually called "momentum space" in the literature (for example, QFT is most often done in momentum space since energy-momentum is typically what is measured in particle physics experiments). The latter interpretation appears to be what the OP is asking about.
 

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