In solid state physics, it is helpful for some analytical results to assume that $\boldsymbol{k}$-space is a continuum and perform the replacement

$$ \sum_{\boldsymbol{k}} f\left(\boldsymbol{k}\right) \rightarrow \frac{V}{\left(2\pi\right)^3} \int_\Omega \mathrm{d}^3k\ f\left(\boldsymbol{k}\right) \tag{1}$$

for $\boldsymbol{k}$-sums over the first Brillouin zone $\Omega$ to integrals. Here, $V$ is the volume of the crystal.
Next, we have the following othogonality relation for Bloch functions:

$$ \left\langle n\boldsymbol{k} \middle| n'\boldsymbol{k}' \right\rangle = \delta_{nn'}\delta_{\boldsymbol{k}\boldsymbol{k}'} \tag{2}$$

if $\boldsymbol{k}$-space is discrete.
My question is now, how does this relation change, if $\boldsymbol{k}$-space is assumed to be continuous? My first guess was, that

$$ \left\langle n\boldsymbol{k} \middle| n'\boldsymbol{k}' \right\rangle = \delta_{nn'}\delta\left(\boldsymbol{k}-\boldsymbol{k}'\right) \tag{3}$$

where the second delta is now the Dirac delta distribution, instead of the Kronecker delta. However, this does not seem to work, because the right hand side has now the dimension of "Volume", because the delta function has the inverse dimension of its argument, while the dimension of the left hand side should have remained unchanged. Written it in terms of wavefunctions, the left hand side is

$$ \left\langle n\boldsymbol{k} \middle| n'\boldsymbol{k}' \right\rangle = \frac{1}{V}\int \mathrm{d}^3r\ \mathrm{e}^{\mathrm{i}\left(\boldsymbol{k}-\boldsymbol{k}'\right)\cdot\boldsymbol{r}} u^*_{n\boldsymbol{k}}\left(\boldsymbol{r}\right) u_{n'\boldsymbol{k}'}\left(\boldsymbol{r}\right)\tag{4}$$

From this step on, I do not really see a proper way of how to calculate the integral, if $\boldsymbol{k}$ is continuous. One way would be to use the periodicity of the Bloch factors and rewrite this as a sum over the elementary cells times an integral over the elementary cells. This leads to the expression

$$ \frac{1}{N}\sum_\boldsymbol{R} \mathrm{e}^{\mathrm{i}\left(\boldsymbol{k}-\boldsymbol{k}'\right)\cdot\boldsymbol{R}} =\delta_{\boldsymbol{k}\boldsymbol{k}'} \tag{5}$$

and I do not known if this holds in the continuous case. Obviously, from simply writing it out, this holds true, even if $\boldsymbol{k}$ is a continuous variable and the right hand side would remain the Kronecker delta. But would this not lead to wrong results for integrals?


1 Answer 1


The correct way to take the limit is the following $$\sum_{n} = \left(\frac{L}{2\pi}\right)^d\sum_{n}\left(\frac{2\pi}{L}\right)^d = \frac{V}{(2\pi)^d}\sum_{n}{\rm d}^dk\to V\int\frac{{\rm d}^d k}{(2\pi)^d}\,.$$ Thus, when you perform the integral over your states $$V\int\frac{{\rm d}^d k}{(2\pi)^d}\langle k| k'\rangle = V\int\frac{{\rm d}^d k}{(2\pi)^d}\frac{(2\pi)^d}{V}\delta(k - k') = 1\,.$$ Thus the correct normalization is $$\langle k| k'\rangle = \frac{(2\pi)^d}{V}\delta(k - k')\,,$$ which has the correct dimensions. Note, you can arrive at this result by taking the continuum limit using the procedure outlined above: the normalization does not need to be imposed by hand.

To see this, let's consider a simple example in one dimension. Our states will be $$\langle x|n\rangle =\frac{1}{\sqrt{L}}e^{{\rm i} \frac{2\pi n}{L} x}\,.$$ Let's check that the normalization is correct in the finite limit $$\langle n|n'\rangle = \langle n | x\rangle\langle x| n'\rangle = \frac{1}{L}\int_0^L{\rm d}x\, e^{{\rm i} \frac{2\pi}{L} x (n - n')} = \delta_{n,n'}\,.$$ This is your desired normalization. Now let's take the continuum limit $$\langle x|x'\rangle = \langle x|n\rangle\langle n|x'\rangle = \sum_{n = -\infty}^\infty\frac{1}{L}e^{{\rm i}\frac{2\pi n}{L}(x - x')}\to L \int\frac{{\rm d} k}{(2\pi)}\frac{1}{L}e^{{\rm i}k(x - x')} = \delta(x - x')\,.$$ Indeed this is the correct normalization

  • $\begingroup$ Can you show how to reach the result using the continuum limit or point me to some resource that shows this? $\endgroup$
    – schade96
    Aug 14, 2020 at 23:33
  • 1
    $\begingroup$ just updated the answer $\endgroup$
    – Guy
    Aug 15, 2020 at 0:05

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