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We know that two parallel planes have the same Miller indices so we can have an infinite number of parallel planes close to one another all of which have the same Miller indices. But it's claimed that

the interplanar distance between two adjacent planes is given as $$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$.

I can't understand what adjacent means here? to me the next parallel plane to any plane is infinitely close to it, hence the $$d_{h k l }=0$$

Can someone please explain what exactly is 'adjacent'? Thank you.

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2 Answers 2

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A reciprocal lattice plane is a plane that passes through the reciprocal lattice points and whose successive stacking generates the entire reciprocal lattice.

Let the real space lattice translation vector be $\mathbf{T}$ which can be written as a linear combination of the primitive lattice vectors $\mathbf{a_1}$, $\mathbf{a_2}$, $\mathbf{a_3}$. Note that this linear combination is restricted in the sense that the coefficients can only be integers. Thus considering an arbitrary origin, the location of a point (similar to the position vector in classical mechanics) is given by $\mathbf{T}=p\mathbf{a_1}+q\mathbf{a_2}+r\mathbf{a_3}$ with $p,q,r \in Z$. You might think of reaching any point in the real space lattice by moving in multiples of only $\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}$. Note that you cannot reach any arbitrary point in space but can only reach discrete points. This is the essence of a lattice.

Now, the reciprocal lattice is constructed by the same mechanism, but with lattice vectors $\mathbf{b_1}$, $\mathbf{b_2}$, $\mathbf{b_3}$ with the condition that $\mathbf{a_i}.\mathbf{b_j}=2\pi \delta_{ij}$. Hence, the reciprocal lattice translation vector $\mathbf{G}=h\mathbf{b_1}+k\mathbf{b_2}+l\mathbf{b_3}$. By the same reasoning, the reciprocal lattice is again discrete. It is important to realise that the reciprocal lattice doesn't live in the same space as the real lattice.

Now, a real/reciprocal lattice plane is one which contains the real/reciprocal lattice points. When viewed from "above", that is, the direction of the normal, all the lattice planes corresponding to the same Miller indices look alike. Since the lattice is discrete and not continuous, there is a separation between the lattice planes. This can be calculated to be $$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ Note that the above formula is valid only for cubic systems.

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  • $\begingroup$ Thank you, but we did not study reciprocal lattices, my text went from defining Miller indices for planes to interplanar spacing without using reciprocal lattices. And it is there that my confusion lies. Is it possible for you to explain it without alluding to reciprocal lattices? If not, then I'll study it and come back? $\endgroup$
    – Kashmiri
    Commented Nov 23, 2020 at 14:57
  • $\begingroup$ Thats okay. You just have to realise that the lattice is not continuous but a discrete set of points. Thus any two successive lattice planes will be separated by a distance. $\endgroup$
    – AlphaBaal
    Commented Nov 23, 2020 at 15:00
  • $\begingroup$ For a visual representation, try having a look at this physics.stackexchange.com/questions/269087/… $\endgroup$
    – AlphaBaal
    Commented Nov 23, 2020 at 15:02
  • $\begingroup$ Thank you once again, I do know that the lattice is a discrete set of points but what confuses me is the definition of Miller planes $hkl$. Is it that the Miller indices of a plane are only defined only for planes passing via the actual lattice points? $\endgroup$
    – Kashmiri
    Commented Nov 23, 2020 at 15:04
  • $\begingroup$ Umm... how would you define a plane that doesn't pass through any points? Remember, the space is discrete now. $\endgroup$
    – AlphaBaal
    Commented Nov 23, 2020 at 15:05
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A plane with Miller indices ($h$ $k$ $l$) is the nearest lattice plane from the origin. To prove this, I have to show there is no lattice plane before the plane with Miller indices ($h$ $k$ $l$)

Let the primitive axes be $a$, $b$, $c$. So the plane with Miller indices ($h$ $k$ $l$) cuts the axes at $1/h$, $1/k$, $1/l$. We denote another set of axes $a^{\prime} = a/h$, $b^{\prime} = b/k$ and $c^{\prime} = c/l$. Suppose there exists a parallel lattice plane $x$ which locates before ($h$ $k$ $l$) and intersects the axes at $1/h^{\prime}$, $1/k^{\prime}$ ,$1/l^{\prime}$. As it locates before, it must intersect $a^{\prime}$, $b^{\prime}$ and $c^{\prime}$ in same ratio. Let it cut the axes at $1/n$, $1/n$ and $1/n$ where $n$ is an integer.

As the plane $x$ is a lattice plane, it must touch a lattice point at $T = u a^{\prime} + v b^{\prime} + w c^{\prime}$, where $u$, $v$, $w$ are integers.

Now $\left( a^{\prime} - b^{\prime} \right)$ and $\left( a^{\prime} - c^{\prime} \right)$ are two vectors which are parallel to the plane $x$. For suitable $u$, $v$, and $w$, we can get a vector on axis $a^{\prime}$. So, $T + v \left( a^{\prime} - b^{\prime} \right) + w \left( a^{\prime} - c^{\prime} \right) = \left( u + v + w \right) a^{\prime}$.

This vector touches the plane $x$ but we see the plane intersects $a^{\prime}$ at $1/n$. So, $\left( u + v + w \right) a^{\prime} = a^{\prime}/n$, here $u$, $v$, $w$ and $n$ are integers... This is not possible. So the pane $x$ can not exist and ($h$ $k$ $l$) is the nearest plane.

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