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! :)


1 Answer 1


Let me simplify notation and the case without losing too much generality.

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.


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.