# Electron dynamics in a periodic potential: Why wavepackets? Why $\vec{v}=(1/\hbar)\nabla_{\vec{k}}E_n$ is its group velocity?

Electrons in a periodic potential, $$V(\vec{ r})$$, satisfying $$V(\vec{r}+\vec{R}) = V(\vec{ r})$$, are described by Bloch states $$\psi_{n,\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}u_{n,\vec{k}}(\vec{r}), \quad u_{n,\vec{k}}(\vec{r}+\vec{R}) = u_{n,\vec{k}}(\vec{r}),\quad \forall\vec{R}$$ which are eigenstates of the periodic potential. If $$E_n(\vec{k})$$ is the energy of an electron in the $$n$$th band with crystal momentum $$\hbar k$$, the quantity $$\vec{v}(\vec{k})=\frac{1}{\hbar}\nabla_{\vec{k}}E_n(\vec{k}),$$ strictly speaking, gives the expectation value of the velocity operator, ($$-i\hbar\nabla/m$$), in the Bloch state $$\psi_{n,\vec{k}}$$ i.e. $$\vec{v}(\vec{k})\equiv \frac{1}{m}\langle\psi_{n,\vec{k}}|-i\hbar\nabla|\psi_{n,\vec{k}}\rangle.$$ This is rigorous. So far there are no wavepackets and electrons are described by delocalized Bloch waves.

However, when one talks about the semiclassical dynamics of electrons, one describes the electrons not in terms of delocalized Bloch waves but by a somewhat localized wavepacket. Why is this so? The justification of this is not clear to me. In addition, the wavepacket is so constructed that its group velocity is equal to the $$\vec{v}(\vec{k})$$ given above. The justification of this part is also not clear.

In short, the question is: Why wavepackets? Then, why $$\vec{v}=(1/\hbar)\nabla_{\vec{k}}E_n(\vec{k})$$ is its group velocity of that packet?

• The semiclassical approximation is when we can take $x$ and $p$ as both simultaneously well-defined. This is never truly possible in a quantum system because of the uncertainty relation $\Delta x \Delta p \geq \hbar/2$. It stands to reason that semiclassics will be most well-defined when the uncertainty relation is saturated, i.e., $\Delta x \Delta p = \hbar/2$. This is exactly true only for Gaussian wavepackets $$\psi(x) \propto \exp(-(x-X)^2/a + i P x),$$ where the factor $a$ quantifies the squeezedness of the wavepacket. So Gaussian Commented Dec 27, 2023 at 16:09
• It can be demonstrated by applying the Hellmann-Feynman theorem: en.wikipedia.org/wiki/Hellmann%E2%80%93Feynman_theorem, with $H_{u}(\vec{r},\vec{p},\vec{k})=\frac{1}{2m}(\vec{p}+\hbar \vec{k})^{2}+V(\vec{r})$ Commented Dec 27, 2023 at 18:32
• Closely related physics.stackexchange.com/q/370669/389318 Useful background physics.stackexchange.com/q/663974/389318 Commented Dec 27, 2023 at 19:01
• @TobiasFünke Let me know if this is clearer. Commented Dec 30, 2023 at 16:13
• @hft I think if you build wave packets from Bloch states --as superpositions of states with $k$-vectors near some $k_0$-- you will see that the group velocity of this wave packet is indeed just $\partial_k E$. IIRC Ashcroft & Mermin discusses this. The answer to "why wave packets?" however is more complicated, depending on what one wants to take as granted. Commented Jan 1 at 14:22

...the quantity $$\vec{v}(\vec{k})=\frac{1}{\hbar}\nabla_{\vec{k}}E_n(\vec{k}),$$ ... ... $$= \frac{1}{m}\langle\psi_{n,\vec{k}}|-i\hbar\nabla|\psi_{n,\vec{k}}\rangle.$$

Yes, this is derived in Ashcroft and Mermin's Appendix E.

So far there are no wavepackets and electrons are described by delocalized Bloch waves.

Yes, and remember that you can roughly think of Bloch waves in a periodic potential as analogous to plane waves for free particles (no potential).

However, when one talks about the semiclassical dynamics of electrons, one describes the electrons not in terms of delocalized Bloch waves but by a somewhat localized wavepacket. Why is this so?

Both Bloch waves and plane waves are completely delocalized; they have non-zero amplitudes everywhere in space, so they can not describe a localized particle.

In order to describe a localized particle in a periodic potential one can use a wave packet of Bloch waves, just like one can use a wave packet of plane waves to describe a localized free particle. Both Bloch waves and plane waves are complete sets of functions, so the choice is a matter of convenience. Bloch waves can be more convenient to use when the problem statement includes a periodic (non-constant) potential.

In short, the question is: Why wavepackets?

Because we have to use wave packets to describe localized particles. Plane waves and Bloch waves are delocalized and unphysical, in the sense that they are unnormalizable (unless you put them in a box). For example, with plane waves, the probability amplitude to find the particle anywhere in space is: $$P_{pl}(\vec r) = |e^{i\vec k \cdot \vec r}|^2 = 1\;.$$

Or, if we use "box normalization" where the box has a size $$V$$, the probability amplitude is still constant and the same everywhere in space: $$P_{pl,V}(\vec r) = \frac{1}{V}\;.$$

Similarly, with Bloch waves, the probability amplitude is: $$P_{bl}(\vec r) = |e^{i\vec k \cdot \vec r}u(\vec r)|^2 = |u(\vec r)|^2\;,$$ which has the same non-zero values in any unit cell in space, since $$u(\vec r +\vec R)=u(\vec r)$$.

We have grown used to thinking of plane waves as describing free particles, but really they do not; they happen to solve the free-particle time-independent Schrödinger equation, but they are not normalizable (unless forced into a box). But wave packets made up of plane waves can be normalized. Similarly, wave packets made up of Bloch waves can be normalized.

Then, why $$\vec{v}=(1/\hbar)\nabla_{\vec{k}}E_n(\vec{k})$$ is its group velocity of that packet?

Suppose you have some physical/normalizable wave packet $$\Psi(\vec r, 0)$$ where $$\Psi(\vec r, 0 ) = \sum_{n\vec k}c_{n\vec k}e^{i\vec k \cdot \vec r}u_{n\vec k }(\vec r)\;.$$

The time evolution is straight forward to figure out since we expanded in terms of Bloch waves: $$\Psi(\vec r, t) = \sum_{n\vec k}c_{n\vec k}e^{-iE_{n}(\vec k) t}e^{i\vec k \cdot \vec r}u_{n,\vec k }(\vec r)\;.$$

Now suppose that the packet is peaked about some specific $$\vec k = \vec k_0$$ and $$n=n_0$$ and write: $$\Psi(\vec r, t) \approx e^{-i(E_{n_0}(\vec k_0)+\left.\frac{\partial E_{n_0}}{\partial \vec k}\right|_{\vec k_0}\cdot (\vec k - \vec k_0)+\ldots) t}e^{i\vec k_0 \cdot \vec r}\sum_{\vec k} c_{n_0,\vec k}e^{i(\vec k - \vec k_0)\cdot r}u_{n_0,\vec k }(\vec r)\;,$$ where we can think of $$c_{n_0\vec k}$$ as something like $$e^{-\alpha(\vec k - \vec k_0)^2}$$ so that only $$\vec k$$ values near $$\vec k_0$$ are important for large $$\alpha$$. Continuing, we can thus write $$\Psi(\vec r, t)\approx e^{-i\vec k_0\cdot (\vec r + \vec v_0 t)}e^{-i E_{n_0}(\vec k_0)t}A(\vec r)\;,$$ where $$\vec v_0 \equiv \left.\frac{\partial E_{n_0}(\vec k)}{\partial \vec k}\right|_{\vec k_0}$$ and where $$A(\vec r)$$ is some function that is localized and we can think of it as looking something like $$A(\vec r) \approx e^{-\frac{r^2}{2\alpha}}\;.$$

So, our wave packet "looks like" a plane wave, but with $$\vec r$$ shifted to $$\vec r + \vec v_0 t$$ and some overall envelope function $$A(\vec r)$$ keeping it localized.

• I’d also like to add that in many cases that I’m familiar with (like an electron’s cyclotron motion in a magnetic field) the wavepacket interpretation reproduces the classical solution of a point particle sized electron. Commented Jan 3 at 22:43