# Nearly free electron model and perturbation theory: How to determine matrix elements of perturbation?

In Steven Simons The oxford Sold state basics, the Nearly free electron model is tackled by treated the weak periodic potential as a perturbation to the free electron model. That is, the full hamiltonian is of the form $$H=H_0+V(\vec{r})$$ where $$H_0=\frac{\hat{p}^2}{2m}$$ and $$V(\vec{r})=V(\vec{r}+\vec{R})$$ where $$\vec{R}$$ is any direct lattice vector. The eigenstates of the unperturbed free electron hamiltonian are simply plane waves of the form $$|\vec{k}\rangle=\frac{1}{\sqrt{L^3}}e^{i\vec{k}\cdot \vec{r}}$$ where $$\vec{k}$$ satisfies the BVK boundary conditions and $$L^3$$ is the volume of the entire crystal. Now to use perturbation theory, we must determine the matrix elements of $$V(\vec{r})$$ in the unperturbed basis. The elements are of the form (I am omitting the vector over-bars in my notation for brevity) : $$\langle k'|V|k\rangle =\frac{1}{L^3}\int e^{-i(k'-k)\cdot r}\,\,V(r)\,d^3r \tag{1}$$ Now my first question is whether this integral is over a single unit cell within the crystal or if its over the entire crystal volume? Steven Simons then goes on to say that this integral is zero unless $$\vec{k'}-\vec{k}$$ is a reciprocal lattice vector. Why is this the case though? I realize that because of the periodicity of $$V$$, its fourier coefficients are of the form $$V_G=\frac{1}{V_{cell}}\int_{cell}e^{-i\vec{G} \cdot \vec{r}}V(\vec{r})d^3r \tag{2}$$ But as far as I'm aware, this does not imply that any integral of the form in eq 2 must equal to zero unless $$k'-k=G$$? My other issue is that he claims that the integrals in equation 1 are the fourier coefficients of $$V(\vec{r})$$ but they can't be because eq 1 goes over the entire crystal (?) while eq 2 integrates only over a unit cell in the lattice. So what is going on here? Why is equation to zero unless $$\vec{k'}-\vec{k}$$ is a reciprocal lattice vector?

Any help on this would be most appreciated!

Starting from $$\langle k|k^\prime\rangle = \delta_{kk^\prime}$$ it is easy to deduce that $$\langle k|V|k^\prime\rangle = \sum\limits_G V_G \, \delta_{k^\prime-k,G} \quad ,$$

by inserting $$\displaystyle V(x)=\sum\limits_G e^{iGx}\, V_G$$. We see that indeed the non-zero contributions of these matrix elements, namely the ones of the form

$$V_G = \langle k+G|V|k\rangle = \frac{1}{\Omega} \, \int_{\Omega} \mathrm{d}x\,e^{-iGx}\, V(x) \tag{*}$$

are the Fourier coefficients of the potential. Here, $$\Omega = L^3$$ is the total volume.

Regarding your second doubt, why both expressions for $$V_G$$ coincide. This is basically the property of a lattice periodic function:

We can replace the integral over the whole domain by the sum over the integrals of all unit cells, i.e. $$\int_{\Omega} \mathrm{d}x \,f(x) \rightarrow \sum\limits_R \, \int_{\omega}\mathrm{d}x \,f(x+R)$$ and $$\Omega = N \, \omega$$, where $$N$$ is the number of unit cells and $$\omega$$ is the volume of such a unit cell. You can find a more detailed treatment of these issues in these lecture notes, equations $$(4)$$-$$(14)$$ or in these lecture notes, equations $$(2.26)$$ - $$(2.29)$$, but take care of possibly different notations and conventions.

To this end, note that here the integrand is periodic with respect to $$R$$ and we thus find:

$$V_G = \underbrace{\frac{1}{N}\sum_R 1}_{=1}\,\frac{1}{\omega} \,\int_\omega \mathrm{d}x \, e^{-iGx} \, V(x) \quad ,$$

which shows that your equation $$(2)$$ coincides with the result $$(*)$$:

$$V_G = \frac{1}{\Omega} \,\int_{\Omega} \mathrm d x \,e^{-iGx}\, V(x) = \frac{1}{\omega}\, \int_{\omega} \mathrm d x \, e^{-iGx}\, V(x)$$

Since some time has passed and no specialists turned up, I am going to attempt a hand-wavy explanation.

Lets stick with integral over the whole crystal and assume that its of some volume $$\mathcal{V}$$, with the bulk containing many unit cells.

Now:

\begin{align} \langle \mathbf{k'} | V| \mathbf{k} \rangle&=\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right)\,V\left(\mathbf{r}\right)\\ &=\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right)\,V\left(\mathbf{r}+\mathbf{R}\right) \\ &=\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right)\,\exp\left(\mathbf{R}.\boldsymbol{\nabla}\right)V\left(\mathbf{r}\right) \\ &=\frac{1}{\mathcal{V}}\oint_\mathcal{\partial V} d^3r\,\dots V\left(\mathbf{r}\right)\\&\quad-\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,V\left(\mathbf{r}\right)\,\exp\left(\mathbf{R}.\boldsymbol{\nabla}\right)\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right) \\ &\approx +\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,V\left(\mathbf{r}\right)\,\exp\left(-\mathbf{R}.\boldsymbol{\nabla}\right)\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right) \end{align}

The way I would make the first term dissapear would be something along the lines of it being a surface term, since there is always at least one $$\boldsymbol{\nabla}$$ involved, and therefore either $$V$$ goes to zero on the surface, or this term becomes un-important for large enough volumes. You can do it with more rigour by actually evaluating it, at least for some low-order cases.

Assuming you are happy with it,

\begin{align} \langle \mathbf{k'} | V| \mathbf{k} \rangle&\approx \frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,V\left(\mathbf{r}\right)\,\exp\left(-\mathbf{R}.\boldsymbol{\nabla}\right)\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right) \\ &= \exp\left(i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{R}\right)\cdot\frac{1}{\mathcal{V}}\int_\mathcal{V} d^3r\,V\left(\mathbf{r}\right)\,\exp\left(-i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{r}\right) \\ &= \exp\left(i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{R}\right)\cdot\langle \mathbf{k'} | V| \mathbf{k} \rangle \end{align}

So you have $$\exp\left(i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{R}\right)\approx 1$$ as a condition. This gives you the selection rules for lattice vectors, I think,

More explicitly, if $$\langle \mathbf{k'} | V| \mathbf{k} \rangle\neq 0$$:

$$1-\exp\left(i\left(\mathbf{k}'-\mathbf{k}\right).\mathbf{R}\right)=\frac{\oint_\mathcal{\partial V} d^3r\,\dots V\left(\mathbf{r}\right)}{\langle \mathbf{k'} | V| \mathbf{k} \rangle}$$

Clearly $$\frac{1}{\mathcal{V}}\oint_\mathcal{\partial V} d^3r\,\dots V\left(\mathbf{r}\right)\to 0$$ may not work for small enough volumes. This is, for example, why quantum dot behaviour is not trivial. Also region $$\langle \mathbf{k'} | V| \mathbf{k} \rangle\approx 0$$, i.e. close to the reciprocal vector is of great interest, and will be affected by surface terms (according to above expression). More detail on good approximations is needed to make progress here, but, I would think, this is where Physics begins