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