# Van Hove singularity (calculating the logarithm)

I am trying for hours to understand this calculation, I hope someone can help me with it. In the paper of Van Hove himself (https://doi.org/10.1103/PhysRev.89.1189) he derived the logarithmic divergence of the density of states, but I am having trouble to reproduce his result. I am focussing on the 2 dimensional case, d=2.

He starts with the density of states as

$$g(\nu) \mathrm{d}\nu = c \int \mathrm{d}^dq$$

$$\Rightarrow g(\nu)= c \int_{S(\nu)}\mathrm{d}^{d-1}q \Big[\sum_{i=1}^d \big(\frac{\partial\nu(\mathbf q)}{\partial q_i}\big)^2\Big]^{-\frac{1}{2}}$$

with c being some constant, $$\nu(\mathbf q)$$ being the dispersion relation and $$S(\nu)$$ being the path in the first Brillouin zone (here 2-torus) on which $$\nu(\mathbf q) = \nu$$ which is a constant w.r.t. the integral, since it appears as the argument on the l.h.s. of the equation. From Morse theorem we know, that $$\nu(\mathbf q)$$ has 4 extrema, of which 2 are saddle points. I expanded $$\nu(\mathbf{q})$$ around the critical arguments as follows, with $$\nu_c \equiv \nu(\mathbf q_c)$$ and $$\boldsymbol \xi$$ having a small norm.

$$\nu(\mathbf q_c + \boldsymbol \xi) = \nu_c + \xi_x^2 ~\partial_x^2 \nu + \xi_y^2 ~\partial_y^2 \nu + \mathcal{O}(\delta) \\ \qquad \quad \,\, = \nu_c \pm a_x ~ \xi_x^2 \mp a_y ~ \xi_y^2 + \mathcal{O}(\delta)$$

where the different signs correspond to the two possible saddle points I want to expand the density of states around (only for the saddle points the logarithmic divergence seems to appear).

Treating only one saddle point for now, with $$\boldsymbol \xi = \mathbf q - \mathbf q_c$$ and thus $$\partial_{q_i} \xi_i = \partial_{\xi_i} \xi_i$$, I plugged one of them into the density of states from above:

$$g(\nu) = c \int_{S(\nu)}\mathrm{d}\mathbf s \Big[\sum_{i=1}^d \big(\frac{\partial}{\partial \xi_i}(\nu_c + a_x ~ \xi_x^2 - a_y ~ \xi_y^2)\big)^2\Big]^{-\frac{1}{2}} = \frac{c}{2} \int_{S(\nu)}\mathrm{d}\mathbf s \Big[a_x^2 ~ \xi_x^2 + a_y^2 ~ \xi_y^2\Big]^{-\frac{1}{2}}\\ = \frac{c}{2} \int_{S(\nu)}\mathrm{d}\mathbf \xi_y \Big[a_x^2 ~ \xi_x^2(\xi_y) + a_y^2 ~ \xi_y^2\Big]^{-\frac{1}{2}} ~\Big|\frac{\mathrm d \mathbf s}{\mathrm d \xi_y}\Big|$$

where in the last line I parametrized $$\mathbf s = \left(\xi_x(\xi_y),~\xi_y\right)^T$$ and changed the integration variable to $$\xi_y$$ for convenience. Now this is equal to (adding a zero):

$$\frac{c}{2} \int_{S(\nu)}\mathrm{d}\mathbf \xi_y \Big[a_x^2 ~ \xi_x^2(\xi_y) + a_y^2 ~ \xi_y^2 + a_x ~ a_y ~ \xi_y^2 - a_x ~ a_y ~ \xi_y^2\Big]^{-\frac{1}{2}} ~\Big|\frac{\mathrm d \mathbf s}{\mathrm d \xi_y}\Big| \\ = \frac{c}{2} \int_{S(\nu)}\mathrm{d}\mathbf \xi_y \Big[a_x (\nu - \nu_c) + (a_x+a_y)a_y \xi_y^2\Big]^{-\frac{1}{2}} ~\Big|\frac{\mathrm d \mathbf s}{\mathrm d \xi_y}\Big|$$

where I made use of the expansion of $$\nu$$ above. From there we also get $$\xi_x(\xi_y) = \frac{1}{\sqrt{a_x}}\sqrt{\nu-\nu_c + a_y ~ \xi_y^2} \approx \frac{1}{\sqrt{a_x}}\Big(\sqrt{\nu-\nu_c} + \frac{a_y ~ \xi_y^2}{2 \sqrt{\nu-\nu_c}}\Big)$$

where I used smallness of $$||\xi||$$ to hopefully get a nicer integral. Plugging in the parametrization and supressing the "y" in $$\xi_y$$ for a leaner notation gives

$$g(\nu) = \frac{c}{2}\int_{S(\nu)} \mathrm d \xi \frac{\sqrt{1- \frac{a_y^2 ~ \xi^2}{a_x(\nu-\nu_c)}}}{\sqrt{a_x(\nu-\nu_c)+(a_x+a_y)a_y ~ \xi^2}}$$

So I hope there are no major mistakes in there. Is this the right way until here? I am a bit stuck at this point, because I don't know how to incorporate the constraint, that $$S(\nu)$$ is the path along the $$\mathbf q$$ values, for which $$\nu(\mathbf q) \equiv \nu = const.$$ holds, other than treating $$\nu$$ as a constant. Also I cannot see, how the result of Van Hove should come out in the end (C is the constant from integration)

$$g(\nu) = C - c ~ \mathrm{log}\left|1-\frac{\nu}{\nu_c}\right|$$

because I am not integrating over $$\nu$$, but over the momenta.

I would be very happy, if someone could help me! :)

Let's suppose that our dispersion relation is given by $$E(r) = E_0+\frac{1}{2} a(x^2 -y^2), \quad a>0$$ so that we have a saddle point situation at $$r\equiv (x,y)=0$$. Then the LDOS (in the thermodynamic limit in $$d=2$$ dimensions) would be given by $$\text{LDOS}= \frac{1}{(2\pi)^2}\int_{E(r)=E} \frac{ds}{|\nabla E|}$$ Let's assume that we only care about $$E$$ near $$E_0$$, so that we write $$E=E_0 +\epsilon$$ where $$\epsilon >0$$ (same proof holds for $$<0$$). Then $$E(r)=E$$ would give us $$x^2-y^2=R^2\equiv \frac{2\epsilon}{a}$$ Let us parametrize $$x,y$$ so that $$x=R \cosh{\theta}$$ and $$y =R \sinh{\theta}$$. Then we see that $$|\nabla E| = aR\sqrt{\cosh^2\theta+\sinh^2{\theta}}$$ And that $$ds = Rd\theta\sqrt{\cosh^2\theta+\sinh^2{\theta}}$$ Therefore, we have $$\text{LDOS}= \frac{2}{(2\pi)^2 a} \int d\theta$$ The extra factor $$2$$ is because we have the "symmetric" parametrization $$x=-R \cosh{\theta}$$. A first glance at the integral and you would think it is $$=\infty$$. However, notice that we are integrating over the first Brillouin zone, i.e., $$[-\pi,\pi]^2$$. Hence, we see that $$|x| \le \pi$$ and thus $$\theta$$ must be integrated from $$-\theta_0 \to \theta_0$$ where $$R \cosh{\theta_0} = \pi \Rightarrow \cosh{\theta_0}=\sqrt{\frac{\pi^2 a}{2\epsilon}}$$ Notice that we only care about small $$\epsilon>0$$. Hence, the right-hand-side is very large and thus we can approximate $$\cosh{\theta_0}\approx e^{\theta_0}/2$$. Hence, $$\theta_0=\frac{1}{2}\log{ \left(\frac{2\pi^2 a}{\epsilon}\right)}$$ Hence, $$\text{LDOS}= \frac{1}{2\pi^2 a} \left( \log {2\pi^2a} -\log\epsilon\right)$$ Hence, we have logarithmic divergence.