Field theory plane wave volume integral

[center o bass]

Hi! I'm used to integrating over infinite spaces when working with QFT so far, but in an exercise I stumbled across a statement that

\int_V d^3x e^{-i \vec x \cdot (\vec p - \vec p')} = V \delta_{\vec p \vec p'}

It is clear that this is okay when p = p', but it does not seem to make sense when this is not the case. I would however agree that it's approximately true when V is large. Is this what is meant by this statement? Is it a large V approximation?

 

[zje2009]

I guess so.

 

[Bill_K]

The integral of e^ipx is zero if you integrate over a complete cycle, e.g from x = 0 to x = 2π, or more generally n complete cycles. Usually what they have in mind is that V represents a cube whose sides are length L = 2nπ.

 

[vanhees71]

To make this a bit more clear: Usually you start the quantization problem of free fields within a finite volume in order to avoid trouble with the continuous single-particle momentum spectrum when dealing with the whole \mathbb{R}^3. This is an example for a regularization you will have to deal with a lot in quantum field theory.

The most simple way is to use a cubic box with length L and periodic boundary conditions, i.e., you assume (e.g., for a scalar field) \phi(\vec{x})=\phi(\vec{x}+L \vec{e}_i) with i \in \{1,2,3\}.

Then the plane-wave mode functions are given by
u_{\vec{p}}(\vec{x})=\frac{1}{\sqrt{V}} \exp(\mathrm{i} \vec{p} \cdot \vec{x})
with V=L^3. These functions are orthonormal in the Hilbert space L^2(V) since due to the boundary conditions the momenta can only take the discrete values
\vec{p}=\frac{2 \pi}{L} \vec{n}, \quad \vec{n} \in \mathbb{Z}^3.
This means
\int_{V} \mathrm{d}^3 \vec{x} \; u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x}) = \frac{1}{V} \int_V \mathrm{d}^3 \vec{x} \exp[\mathrm{i} \vec{x}(\vec{p}-\vec{p}')]=\delta_{\vec{p},\vec{p}'}.
Here, \delta_{\vec{p},\vec{p}'} is a usual Kronecker symbol which is 1 if \vec{p}=\vec{p}' and 0 otherwise, with the momenta running over the above given discrete values.

If you take the limit L \rightarrow \infty the momenta become continuous, i.e., \vec{p} \in \mathbb{R}^3, and you have to normalize the mode functions "as a \delta distribution", i.e., you have to set
u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}),
so that
\int_{\mathrm{R}^3} \mathrm{d}^3 \vec{x} \; u_{\vec{p'}}^*(\vec{x}) u_{\vec{p}}(\vec{x}) = \frac{1}{(2 \pi)^3} \int_{\mathrm{R}^3} \mathrm{d}^3 \vec{x} \; \exp[\mathrm{i} \vec{x} \cdot (\vec{p}-\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').
Of course plane waves are no normalizable states but generalized states in the sense of distributions!

 

[center o bass]

To make this a bit more clear: Usually you start the quantization problem of free fields within a finite volume in order to avoid trouble with the continuous single-particle momentum spectrum when dealing with the whole \mathbb{R}^3. This is an example for a regularization you will have to deal with a lot in quantum field theory.

The most simple way is to use a cubic box with length L and periodic boundary conditions, i.e., you assume (e.g., for a scalar field) \phi(\vec{x})=\phi(\vec{x}+L \vec{e}_i) with i \in \{1,2,3\}.

Then the plane-wave mode functions are given by
u_{\vec{p}}(\vec{x})=\frac{1}{\sqrt{V}} \exp(\mathrm{i} \vec{p} \cdot \vec{x})
with V=L^3. These functions are orthonormal in the Hilbert space L^2(V) since due to the boundary conditions the momenta can only take the discrete values
\vec{p}=\frac{2 \pi}{L} \vec{n}, \quad \vec{n} \in \mathbb{Z}^3.
This means
\int_{V} \mathrm{d}^3 \vec{x} \; u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x}) = \frac{1}{V} \int_V \mathrm{d}^3 \vec{x} \exp[\mathrm{i} \vec{x}(\vec{p}-\vec{p}')]=\delta_{\vec{p},\vec{p}'}.
Here, \delta_{\vec{p},\vec{p}'} is a usual Kronecker symbol which is 1 if \vec{p}=\vec{p}' and 0 otherwise, with the momenta running over the above given discrete values.

If you take the limit L \rightarrow \infty the momenta become continuous, i.e., \vec{p} \in \mathbb{R}^3, and you have to normalize the mode functions "as a \delta distribution", i.e., you have to set
u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}),
so that
\int_{\mathrm{R}^3} \mathrm{d}^3 \vec{x} \; u_{\vec{p'}}^*(\vec{x}) u_{\vec{p}}(\vec{x}) = \frac{1}{(2 \pi)^3} \int_{\mathrm{R}^3} \mathrm{d}^3 \vec{x} \; \exp[\mathrm{i} \vec{x} \cdot (\vec{p}-\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').
Of course plane waves are no normalizable states but generalized states in the sense of distributions!

Ah, thanks a lot for making this a bit more clear. So it's a box. I'm reading Mandl and Shaw, but the authors have a tendency to hide a lot of details. A couple of questions to your response;

What kind of troubles can arise when dealing with continuous momentum spectra?

the plane wave mode functions, are they solutions to the Klein-Gordon equation, or any differential equation? I know they are complete, but how are they used in field theory?

When going to the continuum case for which I am more familiar, would it be correct to say that 1/\sqrt{V} \to 1/(2\pi)^{1/2} and \delta_{\vec p, \vec p'} \to \delta^{(3)} (\vec p - \vec p')?

Thanks again! :)

 

[vanhees71]

If you study QFT (I think Mandl and Shaw isn't so bad, but I've not looked into it for quite some time; my recommendation for beginners is usually Ryder and for advanced students Weinberg's three-volume book, which is for sure the best), the first time where you really encounter trouble with the full-space treatment is when it comes to the calculation of cross sections. This gives you S-matrix elements with an "energy-momentum conserving \delta distribution":
S_{fi} =\mathrm{i} T_{fi} (2 \pi)^4 \delta^{(4)}(p-p').
Of course, the \delta distribution is rather a feature than a bug, but it's making trouble when you want to calculate the cross section, because you must formally take the S-matrix element's modulus squared, and the square of a \delta distribution doesn't make sense whatsoever.

There are (at least) two ways out:

(a) the physics solution (a bit tedious) is to start not with plane-wave asymptotically free in states for the colliding particles but with the realistic description of asymptotically free wave packets. Then everything becomes unproblematic and at the very end you find that you simply do not have to square the (2\pi)^4 \delta but only the regular T-matrix element and put only one facgtor (2 \pi)^4 \delta^{(4)}(p-p') into your cross section formula (or transition matrix element for that matter). For a very careful treatment of this solution see Peskin/Schroeder, Introduction to Quantum Field Theory (which book must be read with great care, because it's pretty sloppy at other places, but for this (and not only this) issue it's just great).

(b) the mathematics solution (elegant but not very intuitive) is to regularize the problem by taking a large finite four-dimensional volume, where large means it should be large compared all considered wavelengths of the particle's wave function (in both the initial and the final states under consideration). Then you calculate the transition-matrix elements, convert them to rates per volume by deviding through the four volume and then (!) take the infinite-volume limit.

Taking this limit among other things means that all sums over three-momenta have to be substituted by integrals. If you take your three volume very large, of course the possible momenta in the single-particle plane-wave solution become continuous and any momentum-volume element \mathrm{\mathrm{d}}^3 \vec{p} contains [L/(2 \pi)]^3 \mathrm{d}^3 \vec{p} momentum states. Thus in the limit L \rightarrow \infty you formally set
\sum_{\vec{p}} \rightarrow V \int_{\mathbb{R}^3} \frac{\mathrm{\mathrm{d}}^3 \vec{p}}{(2 \pi)^3},
where V means a fictitious spatial volume. In the infinite-volume limit, however such sums make only sense if you go over to a "density" rather than a absolute number, and thus you divide this "substitution rule" by L^3, and then everything makes sense for densities, i.e., the correct limit it
\frac{1}{V} \sum_{\vec{p}} \rightarrow \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \quad \text{for} \quad V \rightarrow \infty.