I have a 2D square crystal which has a continuous potential $U(\vec r)$. It stretches to infinity. I want to find the diffraction pattern for this crystal. The potential I use is $$U(\vec r )=U(x \hat x+ y \hat y)=2 U_0(\cos(q \; x)+\cos(q \; y)) = \\2 U_0 \cos(\vec q \cdot \vec r ).$$ Where $\vec q = q\hat x + q\hat y$.

This potential can be written in a more useful form,

$$U(\vec r )= \sum_{\vec G} U_{\vec G} e^{i \vec G \cdot \vec r }\;\;\;\;\;\;\;\;\;\;\;\;\;\tag{1}$$ The sum can be taken over $\vec G = \vec q_1, \vec q_2, \vec q_1 + \vec q_2$ and so on, i.e. $\vec G = \sum_{i=1,2,3,...} n_i \vec q_i$. In this case, $U_{\vec G} = U_{0} $ if $\vec G \in \{ \pm q \hat x,\pm q\hat y\} $ and $U_{\vec G} =0$ otherwise.

Using this, it is easy and well known how to find the diffraction pattern.

The amplitude that a photon with wavevector $\vec k$ scatters to $\vec k'$ due to the crystal is $$F=F_{\Delta \vec k=\vec k' - \vec k}=\langle \vec k'|\hat U| \vec k \rangle \propto \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k)\cdot \vec r} U(\vec r)\\$$ Substituting in the $U(\vec r)$ from Eq.(1), we see that the integral is equal to $$\int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k)\cdot \vec r} U(\vec r)\\=\sum_{\vec G}U_{\vec G} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G)\cdot \vec r}\\ =\sum_{\vec G \in \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G)\cdot \vec r} \;\;\;\;\;\;\;\;\; \tag{2}$$ In the last equality I used the fact that all but 4 of the $U_{\vec G}$ are $0$, so we only have to sum over the 4 relevant $\vec G$.

From Eq.(2) scattering only happens when F is nonzero, i.e. when the arguments in the exponential are 0. Thus,$$\Delta \vec k = \{ \pm q \hat x,\pm q\hat y\}$$

This result seems wrong. Considering a square lattice potential of just delta functions instead of cosines, we would have expected scattering equally at every reciprocal lattice site to infinity, $$\Delta \vec k = \vec G$$

Clearly, the potential I am using should have a similar result to the delta function potential. What is the correct diffraction pattern here, and what am I doing wrong?

  • $\begingroup$ Why would you expect your cosine potential to produce the same result as delta functions? Isn't the delta function an infinite sum of cosines? $\endgroup$
    – nasu
    Mar 4, 2022 at 4:58
  • $\begingroup$ Are you suggesting an infinitely periodic crystal produces a diffraction pattern with only 4 peaks? $\endgroup$
    – Mondo Duke
    Mar 4, 2022 at 7:22
  • $\begingroup$ Your formulation is somewhat slappy. First set up the two basis vectors $\vec a$, $\vec b$, each lattice point can be wriiten as $\vec R(n, m) = n \vec a + m \vec b$, Then reduce the scattering integral into a single unit cell. $\endgroup$
    – ytlu
    Mar 4, 2022 at 9:31
  • $\begingroup$ @MOndo Duke I did not suggest anything. Just asked a question. A real crystal potential is not described by a single cosine so it has nothing to do with it. You are discussing the difference between a cosine and a delta function. $\endgroup$
    – nasu
    Mar 4, 2022 at 16:35
  • $\begingroup$ @ytlu I understand that I could have done that. But I am trying to do it a more general way that doesn't rely on unit cells. $\endgroup$
    – Mondo Duke
    Mar 5, 2022 at 0:49

1 Answer 1


My following standard procedure of crystallography analysis verify your concerns. The form factor vanishes except for $\Delta \vec k$ be one of the $\vec G_1$'s.

Let's continue analyse your integral.

\begin{align} F&=F_{\Delta \vec k=\vec k' - \vec k}\\ &=\langle \vec k'|\hat U| \vec k \rangle\\ & \propto \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k)\cdot \vec r} U(\vec r)\\ &=\sum_{\vec G}U_{\vec G} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G)\cdot \vec r}\\ &=\sum_{\vec G_1 = \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G_1} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G_1)\cdot \vec r} \tag{1} \end{align}

I build the unit cell by defining two basis vectors: \begin{align} \vec a_1 &= \frac{2\pi}{q} \hat x;\\ \vec a_2 &=\frac{2\pi}{q} \hat y.\\ \vec R(n_1,n_2) &= n_1 \, \vec a_1 + n_2 \,\vec a_2. \end{align} and the reciproal lattice vector: \begin{align} \vec b_1 &= q \hat x;\\ \vec b_2 &= q \hat y.\\ \vec G(h,k) &= h \, \vec b_1 + k \,\vec b_2. \end{align} Note that $$ \vec R \cdot \vec G = \text{integer} \,\times\, 2\pi.$$

With these arrangements, we replace the position $\vec r$ with a coordination within the unit cell: $ \vec r = \vec R(n_1, n_2) + \vec\rho$, an re-write Eq.(1) as \begin{align} F &=\sum_{\vec G_1 = \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G_1} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G_1)\cdot \vec r}\\ &=\sum_{\vec G_1 = \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G_1}\sum_{\vec R(n_1, n_2)}\int_{\text{unit cell}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G_1)\cdot \left( \vec R + \vec \rho\right)} \\ &= \sum_{\vec R(n_1, n_2)} e^{-i(\Delta \vec k\cdot \vec R)}\,\,\sum_{\vec G_1 = \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G_1}\int_{\text{unit cell}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G_1)\cdot \vec \rho} \\ &= \sum_{\vec R(n_1, n_2)} e^{-i(\Delta \vec k\cdot \vec R)}\, f_1(\Delta\vec k). \end{align}

The first exponential phase determines the Bragg's condition $$\Delta k = \vec G(h, k)\,\, \text{ for arbitrary integrers } \, h, k.$$

Finally examine the form factor for $\Delta k = \vec G(h, k) = h q \hat x+ k q \hat y$ \begin{align} f_1(\Delta \vec k)&= \sum_{\vec G_1}\int_{unit cell} d\vec\rho\,U_{\vec G_1} e^{-i(\Delta \vec k -\vec G_1)\cdot \vec \rho}\\ & = U_o \int_0^{2\pi/q} dx \int_0^{2\pi/q} dy e^{iq(h+1)x + i(k+1)y)}\\ & + U_o \int_0^{2\pi/q} dx \int_0^{2\pi/q} dy e^{iq(h-1)x + i(k-1)y)}\\ & + U_o \int_0^{2\pi/q} dx \int_0^{2\pi/q} dy e^{iq(h+1)x + i(k-1)y)}\\ & + U_o \int_0^{2\pi/q} dx \int_0^{2\pi/q} dy e^{iq(h-1)x + i(k+1)y)}\\ \end{align}

The form factor vanishes, except for $\Delta \vec k$ being one of $\vec G_1$'s.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.