# Why is the eigenwave function not periodic in the lattice? [closed]

There is such an expression in the Bloch wave: $$\psi(\mathbf{r} + \mathbf{R}_n) = e^{i\mathbf{k}\cdot \mathbf{R}_n} \psi( \mathbf{r} )$$ Where $$\mathbf{R}_n = n_1 \mathbf{a}_1 + n_2 \mathbf{a}_2 + n_3 \mathbf{a}_3$$, $$\mathbf{a}$$ is the basis vector of the unit cell.
At the same time, in the lattice, for any function$$f(\mathbf{r})$$, we have: $$f(\mathbf{r}+\mathbf{R}_n) = f(\mathbf{r})$$ Isn't this contradictory, shouldn't $$\psi$$ be an arbitrary function $$f$$?
And it is also described in the Born–von Karman boundary condition.

• @hft In the lattice, every unit cell is all the same, so this expression is true. Is this not true? Commented Aug 11, 2022 at 2:30
• It’s hard to imagine what you mean here. $f(\mathbf r)=\mathbf r$ is a function and it clearly doesn’t have the property you speak of. Why would this be true? In free space, every point is “the same” (in the same sense that in a lattice, every lattice point is the same) but that obviously doesn’t mean that every function is constant, right? Commented Aug 11, 2022 at 19:10

At the same time, in the lattice, for any function $$f(\mathbf{r})$$, we have: $$f(\mathbf{r}+\mathbf{R}_n) = f(\mathbf{r})$$

No. This is not true. Why would this be true for "any function" in a lattice? (It is not.)

Isn't this contradictory, shouldn't $$\psi$$ be an arbitrary function $$f$$?

Yes, what you wrote is contradictory, but what you wrote also is not true.

I think what you are trying to get at is that a Bloch function can be written in the form: $$\psi(\mathbf{r}) = e^{i\mathbf{k}\cdot \mathbf{r}}u(\mathbf{r})\;,$$ where the function $$u$$ is periodic: $$u(\mathbf{r}+\mathbf{R}_n) = u(\mathbf{r})\;.$$

Just because the function $$u$$ is periodic does not mean every function you ever consider is periodic. For example, the Bloch function clearly is not periodic (it changes by a phase $$e^{i\mathbf{k}\cdot\mathbf{R}_n}$$).

The quantity in a lattice that is known to be periodic is the potential: $$V_{lattice}(\vec r + \vec R_n) = V_{lattice}(\vec r)\;,$$ but this does not mean that the solutions to the Schrödinger equation for this potential have this same periodicity.

The above statement can be rephrased as: the symmetry of the potential (in this case the periodicity) is not necessarily a symmetry of the solutions to the Schrödinger equation for that potential.

By analogy, consider the hydrogen atom. The hydrogen atom potential is spherically symmetric, but the solutions are not necessarily spherically symmetric. Only the "s-wave" solutions are spherically symmetric. The "p-wave" and higher angular momentum solutions are not spherically symmetric. The symmetry is still important because the solutions transform according to irreducible representations of the group (e.g., the $$Y_{\ell m}$$ for a spherical potential).

Similarly, the solutions to the Schrödinger equation in a periodic potential are not necessarily periodic in the same way. Like the "s-wave" in hydrogen, we have one type of solution that has the same symmetry (the $$k=0$$ solutions). But the vast majority of solutions are not periodic. But in the same way we can use group theory to come up with the labels (the $$\vec k$$ is similar to the $$\ell m$$ label in hydrogen in this regard).

• Yes, there are some mistakes in my writing, but I still don't understand why$f(\mathbf{r}) =f(\mathbf{r}+\mathbf{R}_n)$ is not ture. Commented Aug 11, 2022 at 2:34
• I updated my answer. Do you get it?
– hft
Commented Aug 11, 2022 at 2:34
• Thanks for your answer, I know bloch can be written as a plane wave with periodic function grades. so why $u(\mathbf{r} + \mathbf{R}_n) = u(\mathbf{r})$, but $\psi(\mathbf{r} + \mathbf{R}_n) \ne \psi (\mathbf{r})$ , How to tell when a function is equal and when it is not equal.Thank you Commented Aug 11, 2022 at 2:39
• I've updated my answer. In general there is no way. But I've explained further what is known to be periodic and what is known to be not periodic and why. @ZhaoDazhuang
– hft
Commented Aug 11, 2022 at 15:21