What is a spatial wavefunction in QFT?

in summary, the equation (4.68) is saying that the scattering cross-section is proportional to the square of the momentum of the particles.
  • #1
George Wu
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TL;DR Summary
In peskin P102 ,it mentions spatial wavefunction, I don't know what does it means exactly.
截屏2023-05-13 20.41.13.png

My understanding is:
$$\phi (\mathbf{k})=\int{d^3}\mathbf{x}\phi (\mathbf{x})e^{-i\mathbf{k}\cdot \mathbf{x}}$$
But what is ##\phi (\mathbf{x})## in Qft?
In quantum mechanics,
$$|\phi \rangle =\int{d^3}\mathbf{x}\phi (\mathbf{x})\left| \mathbf{x} \right> =\int{d^3}\mathbf{k}\phi (\mathbf{k})\left| \mathbf{k} \right> $$
where ##\left| \mathbf{x} \right> ## and ##\left| \mathbf{k} \right> ##are the eigenvectors of operater ##\mathbf{X}## and##\mathbf{K}##
In qft, ##\left| \mathbf{k} \right> ##is still the eigenvector of ##\mathbf{K}=-\int{d^3}x\pi (\mathbf{x})\nabla \phi (\mathbf{x})=\int{\frac{d^3k}{(2\pi )^3}}\mathbf{k}a_{\mathbf{k}}^{\dagger}a_{\mathbf{k}}##
However what about ##\left| \mathbf{x} \right> ##?
My question is:
Is there any proper definition of ##\left| \mathbf{x} \right> ##?
Can ##|\phi \rangle ## still be written as:
$$|\phi \rangle =\int{d^3}\mathbf{x}\phi (\mathbf{x})\left| \mathbf{x} \right> $$(maybe with some factors)?
If not, what does spatial wavefunction ##\phi (\mathbf{x})##mean?
 
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  • #2
This is very misleading. In QFT there are no wave functions, because QFT describes a situation where the particle number is not (necessarily) fixed. In relativistic QT of interacting particles you always can have annihilation and creation processes, which change the particle number or the kind of particles.

A normalizable single-particle state indeed has the form (4.65) with ##\phi(\vec{p})## and arbitrary square-integrable function.

Further in QFT you can define position observables for all massive particles and massless particles with spin 0 or spin 1/2. In this case you have "position eigenstates" as in non-relativsitic QT. All other massless particles do not admit a position operator (particularly for photons!).
 
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  • #3
vanhees71 said:
This is very misleading. In QFT there are no wave functions, because QFT describes a situation where the particle number is not (necessarily) fixed. In relativistic QT of interacting particles you always can have annihilation and creation processes, which change the particle number or the kind of particles.

A normalizable single-particle state indeed has the form (4.65) with ##\phi(\vec{p})## and arbitrary square-integrable function.

Further in QFT you can define position observables for all massive particles and massless particles with spin 0 or spin 1/2. In this case you have "position eigenstates" as in non-relativsitic QT. All other massless particles do not admit a position operator (particularly for photons!).
This question arise when I try to understand the equation (4.68) :
1683988964062.png

In order to understand the factor ##e^{-i\mathbf{b}\cdot \mathbf{k}_B}##:
I use the so-called "spatial wavefunction":
if$$\phi _B(\mathbf{k}_B)=\int{d^3\mathbf{x}}\phi _B(\mathbf{x})e^{-i\mathbf{k}_B\cdot \mathbf{x}}$$
then:$$\int{d^3\mathbf{x}}\phi _B(\mathbf{x}-\mathbf{b})e^{-i\mathbf{k}_B\cdot \mathbf{x}}=\phi _B(\mathbf{k}_B)e^{-i\mathbf{k}_B\cdot \mathbf{b}}$$
So,I would like to know what ##\phi (\mathbf{x})##means.
 
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  • #4
Ok! These are (asymptotic) free wave functions or scattering in and out states. The idea is to derive the scattering-matrix elements with such true states, i.e., wave packets which are normalizable to 1. The plane waves or momentum eigenstates are not "true states", because you cannot normalize them to 1 but only to a wave function. Taking normalizable wave packets, that are "narrow in momentum space" to define the S-matrix elements and then take the modulus squared, ##|S_{fi}|^2##, and only then make the incoming and outgoing wave packets plane waves, leads to the correct cross section formula in a very physically intuitive way. This is indeed very nicely treated in Peskin and Schroeder.

Mathematically you can also use a shortcut by quantizing everything first in a (large) finite volume, e.g., taking a cube of length, ##L##, and impose periodic spatial boundary conditions for the fields. Then you have a discrete set of momenta ##\vec{k}=\frac{2 \pi}{L}## and the plane wave modes are only integrated over the finite volume and are thus normalizable. Again you calculate ##|S_{fi}|^2## and then take the "infinite-volume limit" ##L \rightarrow \infty##.
 
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  • #5
vanhees71 said:
Ok! These are (asymptotic) free wave functions or scattering in and out states. The idea is to derive the scattering-matrix elements with such true states, i.e., wave packets which are normalizable to 1. The plane waves or momentum eigenstates are not "true states", because you cannot normalize them to 1 but only to a wave function. Taking normalizable wave packets, that are "narrow in momentum space" to define the S-matrix elements and then take the modulus squared, ##|S_{fi}|^2##, and only then make the incoming and outgoing wave packets plane waves, leads to the correct cross section formula in a very physically intuitive way. This is indeed very nicely treated in Peskin and Schroeder.

Mathematically you can also use a shortcut by quantizing everything first in a (large) finite volume, e.g., taking a cube of length, ##L##, and impose periodic spatial boundary conditions for the fields. Then you have a discrete set of momenta ##\vec{k}=\frac{2 \pi}{L}## and the plane wave modes are only integrated over the finite volume and are thus normalizable. Again you calculate ##|S_{fi}|^2## and then take the "infinite-volume limit" ##L \rightarrow \infty##.
Thanks for your explanation,I think I get the spirit.
 
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