In the process of defining crystal momentum $\hbar k$, I found these formulas below.
By the definition of group velocity, $$v_g=\frac{d\omega_{nk}}{dk}=\frac{1}{\hbar}\frac{dE_{nk}}{dk}$$ Also if an electron is in the electric field, it'll get the electric force $F$.
(Idk why but) The group velocity $v_g$ is actually same with the real velocity of classical particle so, $$\delta E=Fv_g\delta t$$
Comparing two formulas, below holds. $$\delta E=\frac{dE_{nk}}{dk}\delta k=\hbar v_g\delta k$$ Then we can find that $$\frac{d(\hbar k)}{dt}=F$$
Here, I'm confusing why we should use the notation $\delta$ here. (for $t$, $k$ and $E$)
I thought $\delta$ could be differed from $d$ for two ways,
- Inexact differential
- Calculus of variations
But I think here $\delta$ is not used for both purposes. Actually, $\delta E$ might be the calculus of variations, but in case of $t$, it's not even a funtional.
Also in case of inexact differential, they should be path dependent - I think $t$ couldn't be.
Actually, if it was $\delta E = F\delta s$, then I could just ignore this(it might be some notation for one of two above) - But I can't just ignore $\delta s = v_g\delta t$. Why it's not $v_gdt$ instead?
At the end, $\frac{\delta k}{\delta t}$ becomes $\frac{dk}{dt}$. This confuses me more.
Why they use $\delta$ notation here instead of $d$?
I slightly heard that in physics, we use two notations for differential forms, differs by possibility of precise measurement, but I have no idea for that. I couldn't find any information.
