# Confused about how we scale graph axis' to make the axis' dimensionless.

I am trying to understand the solution to part $$\mathrm{(iii)}$$. But, for the question I'm asking to make sense I need to include the solutions to parts $$\mathrm{(i)}$$ and $$\mathrm{(ii)}$$ also:

Consider a triangular lattice where the sides of the triangles have length $$d$$. The figure gives a choice of unit cells (dashed lines).

$$\mathrm{(i)}$$ Use the sides of the unit cells as the primitive lattice vectors, $$\boldsymbol{a}_1$$ and $$\boldsymbol{a}_2$$. Write down these vectors in Cartesian coordinates.

$$\mathrm{(ii)}$$ Write down a pair of reciprocal space vectors $$b_{1,2}$$ satisfying the condition that $$a_i\cdot b_j = 2\pi\delta_{ij}$$ . (If you want to use the explicit formula in three dimensions given in the lectures, then you should pick as $$\boldsymbol{a}_3$$ the unit vector in the direction out of the page.)

$$\mathrm{(iii)}$$ The reciprocal lattice vectors $$\boldsymbol{G}$$ are defined by $$\boldsymbol{G} = h_1b_1 + h_2b_2$$ where $$h_{1,2}$$ are integers and $$\boldsymbol{b}_1$$ and $$\boldsymbol{b}_2$$. Sketch the lattice that is formed by the reciprocal lattice vectors $$\boldsymbol{G}$$ of the triangular lattice.

Solutions:

$$\mathrm{(i)}$$ The primitive lattice vectors are $$\boldsymbol{a}_1 = (d, 0)$$ and $$\boldsymbol{a}_2 = \left(\dfrac{d}{2},\dfrac{\sqrt{3}d}{2}\right)$$.

$$\mathrm{(ii)}$$ A choice of the primitive lattice vectors (bold arrows in diagram) for the reciprocal lattice is $$\boldsymbol{b}_1=\left(\dfrac{2\pi}{d},-\dfrac{2\pi}{\sqrt{3}d}\right)$$ and $$\boldsymbol{b}_2=\left(0,\dfrac{4\pi}{\sqrt{3}d}\right)$$. Other choices are possible, such as $$−\boldsymbol{b}_1$$ and $$−\boldsymbol{b}_2$$.

$$\mathrm{(iii)}$$ $$\boldsymbol{G} = h_1\boldsymbol{b}_1 + h_2\boldsymbol{b}_2$$ with integers $$h_{1,2}$$. The diagram shows all the $$\boldsymbol{G}$$ vectors plotted as points in $$\boldsymbol{k}$$-space. All the $$\boldsymbol{G}$$ vectors form a periodic array in reciprocal space. This ‘reciprocal lattice’ for a triangular lattice in real space is itself a triangular lattice in $$\boldsymbol{k}$$-space.

When I asked my lecturer about this scaling on the $$x$$ and $$y$$ axis he just said (something like) that it is to "avoid having factors of $$\dfrac{2\pi}{d}$$ on each increment of the $$x$$ and $$y$$ axis". This makes sense since having a dimensionless $$x$$-axis looks clearer than this:

and similarly for the $$y$$ axis.

So I will first factor out $$\dfrac{2\pi}{d}$$ then the reciprocal lattice vectors are $$\boldsymbol{b}_1=\left(\dfrac{2\pi}{d},-\dfrac{2\pi}{\sqrt{3}d}\right)=\dfrac{2\pi}{d}\left(1,-\dfrac{1}{\sqrt{3}}\right)$$ and $$\boldsymbol{b}_2=\left(0,\dfrac{4\pi}{\sqrt{3}d}\right)=\dfrac{2\pi}{d}\left(0,\dfrac{2}{\sqrt{3}}\right)$$. Writing it this way, I thought the graphs $$x$$-axis should look like this:

and similarly for the $$y$$-axis.

The reason I think the graph axis label should read $$\dfrac{2\pi k_x}{d}$$ and not $$\dfrac{k_x d}{2\pi}$$ (in the solution) is simply because I have factored out the $$\dfrac{2\pi}{d}$$ above so that what is plotted does not depend on $$\dfrac{2\pi}{d}$$. Math is not my strong point and I just cannot figure out why the axis reads $$\dfrac{k_x d}{2\pi}$$ instead of $$\dfrac{2\pi k_x}{d}$$ (which is what it looks like it should be). Can anyone please explain what is going on here?

• Do you have a reference/ a book you are using for this course? Maybe I can try study it and and attempt an answer @Sirius Black May 8, 2021 at 18:45
• @Buraian Hi, can we talk in private, I am not allowed to upload the link here. It's a solid-state physics course, but I'm not sure how much use it would be to you since my question here is purely mathematical: Why axis label $x$ changes to $\dfrac{2\pi k_x}{d}$ when making the axis increments dimensionless? May 8, 2021 at 18:55
• here @Sirius Black May 8, 2021 at 19:02
• BTW, anyone can access the link... and it is not wise to spend all reputation on a bounty. May 10, 2021 at 16:48
• @TymaGaidash Why is it not wise? Reputation is just bounty credit to me. May 10, 2021 at 18:08

The formula $$\dfrac{d}{2\pi}\cdot k_x$$ (from the solution) means that the value $$k_x = \dfrac{2\pi}{d}$$, for example, maps to point $$\dfrac{d}{2\pi}\cdot \dfrac{2\pi}{d} = 1$$ on the horizontal axis, which is consistent with the math and the graph. This also means that the value of the unit step on the horizontal axis is equal to $$\dfrac{2\pi}{d}$$ i.e. the inverse of the constant in the formula $$\dfrac{d}{2\pi}\cdot k_x$$.

The convention for annotating graph axes is to indicate how the original variable maps to the axis. A formula of $$c\cdot x$$ for example, where $$c$$ is a constant and $$x$$ is the free variable, means that the value $$x=0$$ maps to the point $$c\cdot0 = 0$$ on the axis, values $$x = \pm 1$$ map to points $$c\cdot \pm1 = \pm c$$ etc.

For a simple example, consider the plot of $$y=\sin(x)$$ where the horizontal axis has been rescaled to multiples of $$\pi$$ instead of radians.

The way to read, for example, the local maximum point off this graph is: when $$\dfrac{x}{\pi} = 0.5$$ the value of $$\sin(x)$$ is $$1$$. And, indeed, $$\dfrac{x}{\pi} = 0.5 \iff x = \dfrac{\pi}{2}$$ and $$\sin\left(\dfrac{\pi}{2}\right) = 1$$.

More generally, such remapping of the axes is not limited to linear functions, and not restricted to the horizontal axis. For example, log-log plots of $$y = f(x)$$ plot $$\log(y)$$ against $$\log(x)$$ e.g. here.

The $$x$$-axis and $$y$$-axis in the solution graph have equidistant marked points $$0,1,2,\ldots$$. The blue arrows represent the vectors \begin{align*} \mathbb{\tilde{b}}_1&=\left(1,-\frac{1}{\sqrt{3}}\right)=\left(1,-0.577\ldots\right)\\ \mathbb{\tilde{b}}_2&=\left(0,\frac{2}{\sqrt{3}}\right)=\left(1,1.154\ldots\right)\\ \end{align*} The solution graph presents scaled tilde-vectors. The original vectors are \begin{align*} \mathbb{b}_1&=\frac{2\pi}{d}\left(1,-\frac{1}{\sqrt{3}}\right)\\ \mathbb{b}_2&=\frac{2\pi}{d}\left(0,\frac{2}{\sqrt{3}}\right)\\ \end{align*}

We therefore have the relation \begin{align*} \mathbb{b}_1&=\frac{2\pi}{d}\mathbb{\tilde{b}}_1 \qquad\to\qquad \mathbb{\tilde{b}}_1=\frac{d}{2\pi}\mathbb{b}_1\\ \mathbb{b}_2&=\frac{2\pi}{d}\mathbb{\tilde{b}}_2 \qquad\to\qquad \mathbb{\tilde{b}}_2=\frac{d}{2\pi}\mathbb{b}_2\ \end{align*} We see the tilde-vectors are scaled by $$\color{blue}{\frac{d}{2\pi}}$$ which is indicated by $$k_x\frac{d}{2\pi}$$ and $$k_y\frac{d}{2\pi}$$.