Non-uniqueness of the k-vector in Bloch state

How to understand that Bloch wave solutions can be completely characterized by their behaviour in a single Brillouin zone? Given Bloch wave: \begin{equation*} \psi_{\mathbf{k}}(\mathbf{r}) = u_{\mathbf{k}}(\mathbf{r}) \exp (i\mathbf{k}\mathbf{r}) \end{equation*} I can write wavefunction for momentum $\mathbf{k}' = \mathbf{k} + \mathbf{G}$. \begin{equation*} \psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) = u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r})\exp (i\mathbf{k}\mathbf{r}) \end{equation*} As I understand $u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r}) \neq u_{\mathbf{k}}(\mathbf{r})$, so: \begin{equation*} \psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) \neq \psi_{\mathbf{k}}(\mathbf{r}). \end{equation*} Or I am wrong?

P.S. If Bloch functions are periodic in $\mathbf{k}$ space, then why free particle solution is not a Bloch function ($V=0=V(\mathbf{r}) = V(\mathbf{r} + \mathbf{G})$).

• Why do you think free particle wavefunctions are not Bloch functions? They are, and in the first Brillouin zone their periodic part (which you denote by $u_k$) is constant. – Ruslan Oct 14 '15 at 18:11
• However $\exp(ikr) \neq \exp(i[k + G]r)$ and energy is parabolic $\neq$ periodic. – Baranas Oct 14 '15 at 19:23
• I made a wikipedia image on this topic: commons.wikimedia.org/wiki/File:BlochWaves1D.svg See the caption too. Does that help? If not, what's confusing about it? – Steve Byrnes Oct 16 '15 at 17:54

2 Answers

Let's consider the simplest problem: electron in free space. As free space is completely periodic, we can take an arbitrary value $$a$$ as a period of the potential. That is $$V(x)=\textrm{const.}=V(x+a),$$ which allows the application of Blochs theorem. Now Schrödinger's equation for such electron looks (in dimensionless units) like

$$-\psi''(x)=K^2\psi(x),\tag1$$

where $$K$$ is (vacuum) wavenumber of the electron, $$K^2=E$$ is total energy. Using Bloch's theorem, we can substitute $$\psi(x)=u(x)e^{ikx}$$:

$$-u''(x)-2iku'(x)+k^2u(x)=K^2u(x).\tag2$$

Here $$k$$ is quasiwavenumber. General solution of $$(2)$$ is

$$u(x)=A\exp(-i(k+K)x)+B\exp(-i(k-K)x).\tag3$$

By Bloch's theorem we know that $$u(x)=u(x+a)$$, the same goes for its derivative. Thus, we can impose periodic boundary conditions on $$(2)$$ and find the particular solution. From $$u(0)=u(a)$$ we can get using $$(3)$$:

$$A=B\left(\frac{e^{2iaK}-1}{e^{ia(k+K)}-1}-1\right),\tag4$$ thus

$$\frac{u(x)}B=e^{-i(k-K)x}+e^{-i(k+K)x}\left(\frac{e^{2iaK}-1}{e^{ia(k+K)}-1}-1\right).\tag5$$

$$B$$ is normalization constant, for our purposes we can ignore it, taking $$B=1$$.

Now, from $$u'(0)=u'(a)$$ we have restriction on $$K$$:

$$K=k-\frac{2\pi n}a,\,n\in\mathbb{Z}.\tag6$$

Substituting $$(6)$$ into $$(5)$$, we get:

$$u(x)=\exp\left(-i\frac{2\pi n}ax\right).$$

Finally, we get our Bloch solution for free electron:

$$\psi(x)=\exp\left(-i\frac{2\pi n}ax\right)\exp(ikx)=\exp\left(i\left(k-\frac{2\pi n}a\right)x\right).$$

You may see that this function is parametrized by two values: $$k\in\mathbb{R}$$ and $$n\in\mathbb{Z}$$. $$k$$ is quasiwavenumber, while $$n$$ indexes branches of $$E(k)$$ dependence.

For each $$k$$ we have a set of possible energies, each corresponding to some band. These bands are uniquely (for given $$k$$) determined by $$n$$. Now what happens if you replace $$k\to k-G=k-\frac{2\pi}a$$? Simple:

$$K_n=k-\frac{2\pi n}a\;\to\; \left(k-\frac{2\pi}a\right)-\frac{2\pi n}a,$$

and right hand side of the above is nothing but

$$RHS=\left(k-\frac{2\pi}a\right)-\left(\frac{2\pi (n+1)}a-\frac{2\pi}a\right)=k-\frac{2\pi(n+1)}a=K_{n+1}.$$

I.e. if you increase $$k$$ so far as to go out of first Brillouin zone, you just end up in another band. I.e. you don't get new states this way.

This just means that the band structure can be seen in extended zones scheme:

If you look at the picture above, you'll see that the fact that you appeared in a different band is just an artifact of our band indexing and the form of the solution. You could as easily continuously move along the lines in the picture periodically. And it appears that this motion is actually more natural for crystals, due to the places where the bands split. Here's how it looks for non-free electron in crystal:

• Thank you, your explanation is very helpfull. My problem was that I never considered to solve Schrodinger equation with Bloch wave form. I thougt that $Ae^{ikx} + Be^{-ikx}$ is the solution for periodic = constant potencial. However if I undestand correctly the difference between one and many parabolas is Born-von Karman boundary conditions. I still need to contemplate on this difference, but again thank you, for me it was big jam. Now it is clear what means to define $u_{k+G,n}(x)\exp(iGx)=u_{k,n}(x)$ so that $\psi_k(x)$ are periodic in $k$ space. – Baranas Oct 16 '15 at 21:16

$u_k(r)$ has the periodicity of the crystal. Therefore, its Fourier expansion only includes reciprocal lattice vectors:

$$u_k(r)= \sum_GC_{k-G}e^{iG\cdot r}.$$ Therefore, $$u_{k'}(r)\exp(iGr)=u_k(r).$$

• Sorry, I still don't see why $\sum_{\mathbf{G}_i} C_{\mathbf{k} + \mathbf{G}, \mathbf{G}_i} \exp(i\mathbf{G}_i\mathbf{r}) \exp(i\mathbf{G}\mathbf{r}) = \sum_{\mathbf{G}_i} C_{\mathbf{k}, \mathbf{G}_i} \exp(i\mathbf{G}_i\mathbf{r})$ – Baranas Oct 7 '15 at 16:09
• By change of variables, you can define $G->G+G'$ Since we are summing over all $G$s, both summation becomes equivalent. Reference: Ashcroft and Mermin – Goobs Oct 7 '15 at 16:49
• Actually I think that you are wrong. In wiki it is written that for different $k$'s, $u$'s can be different (look at picture)! – Baranas Oct 8 '15 at 9:02
• Yes, $u's$ can be different. I haven't said that $u's$ are same. I have multiplied it by a term. – Goobs Oct 8 '15 at 11:38