I'm a little unclear as to how QFT differs from Elementary particle physics. They both use pictorials of Feynman graphs, is it that Elementary particle physics assumes the point particle perspective, while QFT treats them as fields?

# [Physics] the difference between QFT and elementary particle physics

elementary-particlesfeynman-diagramsparticle-physicsquantum-field-theory

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Somewhat surprisingly, the "generic" particle of QFT is in fact totally delocalized.

More precisely, *particles* are thought to come from the mode expansion of *free fields*. Since every free relativistic field $\phi$ fulfills the Klein-Gordon equation $(\partial^\mu\partial_\mu - m^2)\phi = 0$, a Fourier transform shows that it can be expanded as

$$ \phi(x) = \int \frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2p^0}}(a(\vec p)\mathrm{e}^{\mathrm{i}px} + a^\dagger(\vec p)\mathrm{e}^{-\mathrm{i}px})$$

where Lorentz invariance is not manifest, but can nevertheless be shown. A quantum field is *operator-valued*, and the operator valued objects $a(\vec p),a^\dagger(\vec p)$ fulfill exactly the correct commutation relations to be interpreted as creation and annihilation operators. The $n$-particle state of particles that are associated with the field $\phi$ is now *defined* as

$$ \lvert n;p_1,\dots,p_n \rangle := a^\dagger(p_1)\dots a^\dagger(p_n)\lvert \Omega \rangle$$

where $\lvert \Omega \rangle$ is the (mostly) unique vacuum state. In this way, you first create all particle states that are sharply localized in *momentum space* (and hence completely delocalized in position space) and you can build localized particle states by the usual building of "wavepackets" with fuzzy momentum out of the sharp momentum states:

A QM wavepacket of width $\sigma_x$ localized at $x_0$ is constructed out of the pure momentum states $\lvert \vec p \rangle$ as something like $$\lvert x_0,\sigma_x\rangle = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\mathrm{e}^{\mathrm{2i\sigma_x^2(x - x_0)^2}}\lvert p \rangle$$It works exactly the same for localized QFT particles, except that one should multiply the measure with $\frac{1}{\sqrt{2p^0}}$ to have a Lorentz invariant integration, and, of course, $\lvert p \rangle = a^\dagger(p)\lvert \Omega \rangle$.

The idea that "particles are local excitations of the fields" comes from the observation that this mode expansion is almost completely analogous to a classical field fulfilling a wave equation like the Klein-Gordon equation, where the $a(\vec p),a^\dagger(\vec p)$ would directly represent an excitation of the field of wavenumber $\vec p$. It *cannot* be made precise in the context of QFT because the quantum field is operator-valued and has no definite values, so it is wholly unclear what rigorous sense could be given to it being "excited". It is a nice picture, but nothing you should take too literally.

Also, take note that this is for the *free field*. The true interacting field of a QFT cannot be mode expanded in this way, and particle states are (through the LSZ formalism) only obtained in the asymptotic past and future (when they were far enough apart for interactions to be effectively non-existent) of the theory - the Hilbert space (and hence any states you could or could not identify as particles) of interacting QFTs is essentially *unkown*.

Furthermore, more mathematical methods of constructing QFTs often first construct the $a,a^\dagger$ and the Fock space of particle states, and then define the field out of it - then, the roles of particle and field as "fundamental" and "derived" are somewhat reversed.

Composite particles in QFT have size in the sense that cross sections are not scale-independant (because they have a radius that breaks that invariance).

The radius of the proton was first measured by Robert Hofstadter. He studied the scattering of electrons and atomic nuclei. The Fourier transform of the cross-section is just (proportional to) the charge density. He found that, after a plateau, the charge density droped exponentially to zero. And that width of the transition zone was almost the same for all the nuclei. Including the proton. This means, of course, that the cross-sections weren't scale independent.

Increasing the energy of the incoming electrons, we are dealing with deep inelastic scattering. Now the electrons don't see protons, but their constituent quarks. And now, the cross sections are scale-invariant! This phenomenon is known as Bjorken scattering (in fact, this scaling is somewhat broken by quantum corrections).

In more mathematical terms, the cross section for this scattering is given by the Rosenbluth formula $$\sigma =\sigma _{0}\left[ W_{2}+2W_{1}\tan ^{2}(\frac{ \theta }{2})\right]$$ where $\sigma_0$ is the classical cross section (Rutherford for spinless particles, Mott for spin-1/2 particles) and $W_1$ and $W_2$ are the form factors. A particle is called point-like if the form factors don't depend on the momentum transfer $Q^2$. Otherwise, the size of the particle is related to the Fourier transform of the form factors.

Note that, although the calculations needed to compute it are QFT, the concept of size of particles come from scattering theory and aren't inherently quantum. Quantum mechanics doesn't change the picture. Quantum mechanics adds other length scales, like the Compton length or the Bohr radius. But the size that I've discussed is much closer to the classic concept of size for macroscopic objects.

## Best Answer

Quantum Field Theory is the framework that we normally use to study particle physics. It is the idea that the world is described by fields and each field can move into excited states which correspond to particles.

Since we are working with fields, you not constricted to a fixed particle number as in Quantum Mechanics. This is very useful since due to $E=mc^2$, particles constantly appear and disappear in the real world.

Particle physics on the other hand, is the study of the particles that make up our world. Since QFT's are terrific at describing Nature, we use them to describe particle physics; they are tool we need to study particle physics. However, some people study QFT for the sake of QFT since its a very deep and subtle subject.