Consider a zero-temperature, one-dimensional crystal with allowed electron momenta $k_n = \frac{2\pi n}{L}$.

Question: Which is the more correct way to think about the Fermi sea?

  1. Sharp plane waves --

    $$ \prod_{\epsilon_k<\epsilon_f} c_{k\uparrow}^\dagger c_{k\downarrow}^\dagger \lvert 0\rangle$$


  2. Wave packets that are very narrow in momentum space --

    $$ \prod_{\epsilon_{k_1}<\epsilon_f, \epsilon_{k_2}<\epsilon_f} \alpha_{k_1\uparrow}^\dagger(x_{k_1}) \alpha_{k_2\downarrow}^\dagger(x_{k_2}) \lvert 0\rangle,$$


    $$\alpha_{k\uparrow}^\dagger(x_k) = \sum_q \exp\biggl[-\frac{1}{2}\biggl(\frac{q-k}{\delta}\biggr)^2 + i q x_k\biggr]\ c_{q\uparrow}^\dagger,$$

    with a small width $\delta$ and randomly distributed positions $x_{k_1},x_{k_2}$.

Also, if (2) is more correct, what determines the width $\delta$?

Discussion: I expected sharp plane waves. But wave packets seem necessary to make sense of the semiclassical equation of motion:

$$\frac{d}{dt}k = -e E\tag{3}$$

which, as I understand it, applies to the center $k$ of a given wave packet.

For a definite example where wave packets seem necessary, consider Bloch oscillations. One solves for the positions $x_k$ as a function of time using (3).



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.