What is the energy of a particle?

Question

I do not mean to ask a naive question though it seems to be.

I have been impressed for a long time that the higher energy $E$ a particle has, the smaller structure we can detect with it. This impression is extremely vivid when we think about renormalization. We often think the momentum cut-off $\Lambda$ as in some hand-waved sense the smallest distance 1/$\Lambda$. And all these come from the uncertainty principle: $\Delta x\cdot\Delta p\sim\hbar$, or $\Delta E\cdot \Delta t\sim\hbar$.

The key problem is that there is an essential difference between the energy uncertainty $\Delta E$ and the energy $E$ itself, and between the momentum uncertainty $\Delta p$ and the momentum $p$ itself. Actually, there is no any relation between $p$ and $x$ or $E$ and $t$, except that they are Fourier-transformed variables to each other.

Now when we talk about the energy of a particle, we actually mean the energy $E$ rather than the energy uncertainty $\Delta E$. When we talk about the momentum cut-off $\Lambda$, we also mean the largest momentum $p$ rather than some momentum uncertainty.

Then how to explain the impression I mentioned in the second paragraph? Do they keep to be ture?

My own answer: It is incorrect that the fact that higher energy particles detect smaller distance comes from the uncertainty principle. The higher energy one particle has, the smaller de Broglie's wave length is has. And the de Broglie's length limit the resolution of one detector. Similarly, the cut-off $\Lambda$ in renormalization is also related to the de Broglie's wave length and hence the resolution of the particles

Answer 1

There is actually a second mystery here and we can solve both of them at the same time.

The second problem is that the energy of a particle depends on the relative motion of the observer (that is it is frame dependent). So whose energy do you use in deciding what you can probe?

We can solve both problems together. First we note that probing some structure involves an interaction between the probe and the subject, and that interaction means an exchange of energy and momentum. And second we note that the squares of four vectors are Lorentz invariants. So we use the square of a energy-momentum four-vector characterizing the interaction between probe and subject (well, actually the magnitude of that four-vector to get the units right).

That leave the matter of choosing what "energy-momentum four-vector characterizing the interaction" we should use, and for two-body interactions Stanley Mandelstam already figured out our choices.

A result of all this is that I can use the same 6 GeV electron beam to probe large(ish) scale structure in low-scattering angle experiments and small(ish) scale structure in large scattering angle experiments (which, in fact, I did for my dissertation work: $Q^2 = -t$ in the range between $3.3$ and $8.1\,\mathrm{GeV}^2$).

Answer 2

One has to realize that the Heisenberg Uncertainty Principle (HUP) is an observationally validated principle, it has been incorporated in the mathematical theory of quantum mechanics in the algebra of the commutators of the corresponding operators.

The deltas arise because there is an inherent wavelength in the probability densities , and the Δ(χ) and Δ(p) of the HUP are a reflection of these wavelengths in the structure of quantum mechanical dimension systems.

It is easier to understand with photons: The higher the momentum of the photon(same as energy in units of c=1) the higher the frequency and the smaller the wavelength. Starting with x rays one can see , because of the small wavelength, collective details of crystals,as long as the uncertainty is within the HUP bounds: small distances in crystals require large photon frequencies. On hadrons, collective details appear with higher energy gamma scatterings.

The probability density wave structure is inherent in all quantum mechanical systems, so your "there is no any relation between p and x or E and t, except that they are Fourier-transformed variables to each other" is at the core of the HUP , rather than being an exception