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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! :)

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1 Answer 1

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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.

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