# A question about momentum integrals and lengths

• I
• "Don't panic!"
In summary, in chapter 6 of Matthew Schwartz's book "Quantum Field Theory and the Standard Model", the author introduces the Lorentz invariant phase space measure (LIPS) and uses it to derive the differential cross-section for a ##2\rightarrow n## interaction. As part of the derivation, he notes that ##\int\frac{dp}{2\pi}=\frac{1}{L}## and ##V\times\int\frac{d^{3}p}{(2\pi)^{3}}=1##. The author also discusses the mathematical justification for this result and provides an analogy to explain it. Furthermore, the author explains the role of this result in dealing with singularities in cross section
"Don't panic!"
I've been making my way through Matthew Schwartz's QFT book "Quantum Field Theory and the Standard Model". In chapter 6 he derives the differential cross-section for a ##2\rightarrow n## interaction. As part of the derivation, he introduces the Lorentz invariant phase space measure (LIPS), and notes that $$\int\frac{dp}{2\pi}=\frac{1}{L}$$ which he uses to find that ##V\times\int\frac{d^{3}p}{(2\pi)^{3}}=1##.
Now, I can see how this is the case from dimensional analysis, since ##[p]=\frac{1}{\text{Length}}##, however, is there a way to show mathematically (at least at a physicists level of rigour) that this is true?

Can't you just consider each direction independently. Then multiply the three equations?

Jilang said:
Can't you just consider each direction independently. Then multiply the three equations?

It's not the ##\int\frac{d^{3}p}{(2\pi)^{3}}=\frac{1}{V}## that confuses me, I get that this follows from the one-dimensional case. What I'm not sure about is the one-dimensional case itself, i.e. ##\int\frac{dp}{2\pi}=\frac{1}{L}##. Can one derive this mathematically?

First, an analogy that might be somewhat less abstract: suppose a system consists of 1 kg masses that are connected in a line by light, rigid rods that each have length A. What is the linear mass density of the system?

bhobba
George Jones said:
First, an analogy that might be somewhat less abstract: suppose a system consists of 1 kg masses that are connected in a line by light, rigid rods that each have length A. What is the linear mass density of the system?

The linear mass density of the system would be ##\frac{1}{A}##, right?!

It's about the density of states for free particles. In order to avoid the ##\delta##-function singularity you need to make the space (for the cross section also time) finite. For simplicity choose a cube with length ##L##. In order to have a properly defined self-adjoint momentum operator you must make the boundary conditions on the mode functions periodic, i.e., ##\psi(\vec{x}+L \vec{e}_j)=\psi(\vec{x})## for ##j \in \{1,2,3 \}## and the ##\vec{e}_j## a Cartesian basis. Now the momentum eigenstates are the plane waves ##u_{\vec{p}}(\vec{x}) \propto \exp(\mathrm{i} \vec{p} \cdot \vec{x})## with ##\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3##.

Now make the volume pretty large and consider a volume ##\Delta^3 \vec{p}## in momentum space much larger than ##(2 \pi/L)^3##. Then you obviously have ##\frac{L^3}{(2 \pi)^3} \Delta^3 \vec{p}## momentum eigenmodes in this box. Now taking ##L \rightarrow \infty## you can write the sum over momentum states in terms of an integral, using the density of states just derived, which gives the rule
$$\sum_{\vec{p}} \rightarrow \frac{V}{(2 \pi)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}.$$
Such a substitution occurs pretty often in QFT, e.g., it's a shortcut to get the rule what to do with the S-matrix elements squared in cross section formulae, containing a ##\delta## function, when the usual asymptotic free plane-wave states are used: You just make the volume finite in the above described sense. Then the ##\delta## function becomes a Kronecker ##\delta## for momentum conservation, and you have no trouble anymore to square it. Now you want densities so you divide by ##V=L^3## and then take the limit. Of course the single Kronecker ##\delta## left in the formula then becomes a ##\delta## distribution in the infinite-volume limit. In the same way you deal with the time integral and energy conservation. For a more physical treatment you have to use true asymptotic free states, i.e., "wave packets" that are square integrable in the asymptotic free states and then take the limit to sharper and sharper peaked states in momentum space, until in the limit you get plane waves. Also in this approach finally one ##\delta## function is gone in the cross section formula, and you are left with only one energy-momentum conserving ##\delta## function as it must be. The finite-volume approach is only a bit more convenient to argue about (although the wave-packet approach is very illuminating; it can be found in the textbook by Peskin and Schroeder, where they discuss cross sections).

stevendaryl and bhobba
vanhees71 said:
Now the momentum eigenstates are the plane waves u⃗p(⃗x)∝exp(i⃗p⋅⃗x)u_{\vec{p}}(\vec{x}) \propto \exp(\mathrm{i} \vec{p} \cdot \vec{x}) with ⃗p∈2πLZ3\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3.

Does the condition ##p_{j}=\frac{2\pi\,n_{j}}{L}## (##j=\lbrace 1,2,3\rbrace## and ##n_{j}\in\mathbb{Z}##) follow from the periodicity of the mode functions, i.e. $$e^{i\mathbf{p}\cdot\left(\mathbf{x}+L\mathbf{e}_{j}\right)}= e^{i\mathbf{p}\cdot\mathbf{x}}\quad\Rightarrow\quad e^{i\left(\mathbf{p}\cdot\mathbf{e}_{j}\right)L}=1 \quad\Rightarrow\quad \left(\mathbf{p}\cdot\mathbf{e}_{j}\right)L=p_{j}\,L=2\pi\,n_{j}\quad\Rightarrow\quad p_{j}=\frac{2\pi\,n_{j}}{L}$$

vanhees71 said:
Now make the volume pretty large and consider a volume Δ3⃗p\Delta^3 \vec{p} in momentum space much larger than (2π/L)3(2 \pi/L)^3. Then you obviously have L3(2π)3Δ3⃗p\frac{L^3}{(2 \pi)^3} \Delta^3 \vec{p} momentum eigenmodes in this box.

I'm assuming this is because each eigenmode occupies a volume of ##\left(\frac{2\pi}{L}\right)^{3}## in momentum space, and so the number of momentum eigenmodes in a box of size ##\Delta^{3}\mathbf{p}## is simply ##\frac{\Delta^{3}\mathbf{p}}{\left(\frac{2\pi}{L}\right)^{3}}=\frac{L^{3}}{(2\pi)^{3}}\Delta^{3}\mathbf{p}##?!

vanhees71 said:
Now taking L→∞L \rightarrow \infty you can write the sum over momentum states in terms of an integral, using the density of states just derived, which gives the rule
∑⃗p→V(2π)3∫R3d3⃗p.​
\sum_{\vec{p}} \rightarrow \frac{V}{(2 \pi)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}.

In Matthew Schwartz's book he states that $$\int\frac{d^{3}p}{(2\pi)^{3}}=\frac{1}{V}$$ does this come from normalizing the sum such that $$\frac{L^{3}}{(2\pi)^{3}}\sum_{\mathbf{p}}\Delta^{3}\mathbf{p}=1\quad\to\quad V\int\frac{d^{3}p}{(2\pi)^{3}}=1$$ in the limit as ##L\to\infty##?!

I'm a bit in doubt, what's meant by the statement that
$$\int \mathrm{d}^3 \vec{p}$$
should be. Usually this means you integrate over the entire ##\mathbb{R}^3##, and this gives of course infinity. I'd rather use the arguments given in my previous post that the limit ##L \rightarrow \infty## makes
$$\sum_{\vec{p}} \rightarrow \frac{V}{(2 \pi)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}$$
in the sense of an operator equation acting on functions of ##\vec{p}##.

vanhees71 said:
I'm a bit in doubt, what's meant by the statement that
$$\int \mathrm{d}^3 \vec{p}$$
should be. Usually this means you integrate over the entire ##\mathbb{R}^3##, and this gives of course infinity. I'd rather use the arguments given in my previous post that the limit ##L \rightarrow \infty## makes
$$\sum_{\vec{p}} \rightarrow \frac{V}{(2 \pi)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}$$
in the sense of an operator equation acting on functions of ##\vec{p}##.

I find it confusing why he claims that ##\int\frac{dp}{2\pi}=\frac{1}{L}## - is he simply making the assumption that one is integrating over a finite region and then formally takes the limit?!

Also, is what I put in the first two parts of my last post (#7) correct at all?

Although I like Schwartz's book very much, I must admit, I've never checked his treatment of the cross section. I must say this is pretty inaccurate. I've no clue, what he means with this obviously nonsensical formulae you quote in your posting #1. I strongly recommend to read about it this in another book, because it's a pretty subtle subject and crucial for the understanding. Of course, the formulae are correct when you restrict the integration region in the spacetime domain to a finite four-volume (e.g., a 4D cubic box), and then also these formulae are correct.

I think, it's good to start with the most simple case of non-relativistic potential scattering theory. There a very thorough discussion is given in the classic textbook by Messiah.

For QFT I think of all introductory books, Peskin and Schroeder give the most careful analysis for the cross-section formula given S-matrix elements. These authors use the wave-packet approach, which is the most physical one.

The shortcut via the quantization spacetime volume is given in many if not most textbooks, e.g., in Weinberg vol. I. You also find it in my QFT manuscript:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf , Sect. 6.4

vanhees71 said:
Although I like Schwartz's book very much, I must admit, I've never checked his treatment of the cross section. I must say this is pretty inaccurate. I've no clue, what he means with this obviously nonsensical formulae you quote in your posting #1. I strongly recommend to read about it this in another book, because it's a pretty subtle subject and crucial for the understanding. Of course, the formulae are correct when you restrict the integration region in the spacetime domain to a finite four-volume (e.g., a 4D cubic box), and then also these formulae are correct.

I think, it's good to start with the most simple case of non-relativistic potential scattering theory. There a very thorough discussion is given in the classic textbook by Messiah.

For QFT I think of all introductory books, Peskin and Schroeder give the most careful analysis for the cross-section formula given S-matrix elements. These authors use the wave-packet approach, which is the most physical one.

The shortcut via the quantization spacetime volume is given in many if not most textbooks, e.g., in Weinberg vol. I. You also find it in my QFT manuscript:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf , Sect. 6.4

## 1. What is a momentum integral and how is it related to length?

A momentum integral is a mathematical tool used to calculate the total momentum of a system. It is related to length because it involves integrating the product of velocity and distance, which is a measure of displacement or length.

## 2. How is a momentum integral used in physics?

In physics, momentum integrals are used to analyze the motion of objects and systems. They are often used in calculations involving forces, energy, and collisions.

## 3. Can a momentum integral be negative?

Yes, a momentum integral can be negative. This can occur if the velocity and distance have opposite directions, resulting in a negative value for the integral.

## 4. What is the difference between a momentum integral and a momentum derivative?

A momentum integral is the sum of all the momentum values over a given length or time interval, while a momentum derivative is the rate of change of momentum with respect to time. In other words, a momentum integral gives the total momentum, while a momentum derivative gives the momentum at a specific moment in time.

## 5. How does the length of an object affect its momentum integral?

The length of an object does not directly affect its momentum integral. However, the length of an object can affect its velocity and displacement, which are both factors in calculating the momentum integral. A longer object may have a greater momentum integral if it has a higher velocity or travels a greater distance.

• Quantum Physics
Replies
24
Views
952
• Quantum Physics
Replies
1
Views
480
• Quantum Physics
Replies
15
Views
2K
• Quantum Physics
Replies
11
Views
2K
• Quantum Physics
Replies
14
Views
4K
• Quantum Physics
Replies
1
Views
1K
• Quantum Physics
Replies
1
Views
2K
• Quantum Physics
Replies
5
Views
1K
• Quantum Physics
Replies
4
Views
2K
• Quantum Physics
Replies
2
Views
2K