5
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    Commented 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 $$ \frac{d \mathbf{l}}{d k_x} = \sqrt{ \left(\frac{d l_x}{d k_x}\right)^2 + \left(\frac{d l_y}{d k_x}\right)^2 } $$ $\endgroup$ Commented Sep 12, 2019 at 8:15
  • $\begingroup$ There are only 2 disconnected curves with the same energy E in 2D. $\endgroup$
    – zltn.guba
    Commented Jul 5 at 8:17

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.