This is another one of those examples where textbooks always just gloss over it with the remark that it "can be done" and then just state the result.

I want to compute the general form of a van Hove singularity if the dispersion relation expanded to second order has a saddle point. As a simple example, consider

$$E = E_0 + a_x k_x^2 - a_y k_y^2$$ where the coefficients $a_x$ and $a_y$ are both positive.

I understand how the derivation works when all coefficients are negative or all coefficients are positive, because then the surface of constant energy is actually finite: It's an ellipsoid and we can estimate its volume.

But how to proceed for saddle point?

In the 2D-example above, I have $$D(E) \propto \int_{E(k_x,k_y) = E} \frac{1}{\sqrt{a_x^2 k_x^2 + a_y^2 k_y^2}}$$

and I'm not sure how to go from there. Do I actually find a parametrization for that path and compute that integral or is there a better way to arrive at the approximate form?


1 Answer 1


The complete expression for the integral is $$ D(E) = \frac{1}{4\pi^2}\int_{E(k_x,k_y)=E} \frac{d\mathbf{l}}{\left|\nabla_k E(k_x,k_y)\right|} $$ This is the integration over a curve which in your case is described by the following equation: $$ E = E_0 + a_x k_x^2 - a_y k_y^2 $$

Depending on what energy value you substitute, you should select either $k_x$ or $k_y$ as the integration variable. The variable selected should change continuously from $-\infty$ to $+\infty$ on the curve.

This is just for convenience. One variable is convenient on the one side of the singularity, another is on the other side.

If you select say $k_x$ the integral will look as follows: $$ D(E) = 4\cdot\frac{1}{4\pi^2}\int_0^\infty \frac{1}{\left|\nabla_k E\bigl(k_x,k_y(k_x)\bigr)\right|} \left|\frac{d\mathbf{l}(k_x)}{dk_x}\right| dk_x $$ This is a usual one-dimensional integral.

The factor $4$ is here because there are four identical parts of the curve while we are integrating over only one.

  • $\begingroup$ Could you explain $\frac{d{l(k_{x})}}{d{k_{x}}}$ is in this. I have been looking in to this for some time and am very confused. If you know of some references for me in this topic that would be great too. $\endgroup$
    – Hari
    Aug 14, 2019 at 12:34
  • $\begingroup$ @Hari, this factor appears when we change the integral over a curve to integral over the straight axis. Essentially, it is a relation between the infinitesimal segments of the curve and the axis that is calculated as follows $$ \frac{d \mathbf{l}}{d k_x} = \sqrt{ \left(\frac{d l_x}{d k_x}\right)^2 + \left(\frac{d l_y}{d k_x}\right)^2 } $$ $\endgroup$ Sep 12, 2019 at 8:15

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