0
$\begingroup$

Consider a parabolic dispersion $\varepsilon_{\boldsymbol q} = \frac{\boldsymbol |q|^2}{2m}$.

The two-particle density of states $\rho_2(\boldsymbol k, \varepsilon)$ is zero for $\varepsilon < \varepsilon_{\textrm{min},\boldsymbol k} := \frac{|\boldsymbol k|^2}{4m}$.

Above the threshold, the DOS has the following behaviour in 1D, 2D and 3D: \begin{align} \rho_2(\boldsymbol k,\varepsilon) &\sim \frac{1}{\sqrt{\varepsilon-\varepsilon_{\textrm{min},\boldsymbol k}}} \qquad \textrm{(in 1D)}, \\ \rho_2(\boldsymbol k,\varepsilon) &\sim \textrm{const} \qquad \qquad\quad \textrm{(in 2D)}, \\ \rho_2(\boldsymbol k,\varepsilon) &\sim \sqrt{\varepsilon-\varepsilon_{\textrm{min},\boldsymbol k}} \qquad \textrm{(in 3D)} . \end{align}

One might hence expect that, more generally, $\rho_2(\boldsymbol k,\varepsilon) \sim \left( \varepsilon-\varepsilon_{\textrm{min},\boldsymbol k} \right)^{\frac{D}{2}-1}$.

However, this seems to be wrong. A calculation shows that for $D>3$, $$\rho_2(\boldsymbol k,\varepsilon) \sim \left\{ \begin{array}{cll} \varepsilon^{\frac{D}{2}-1} & & \boldsymbol k = 0, \\ k^{D-3} \sqrt{\varepsilon-\varepsilon_{\textrm{min},\boldsymbol k}} & & \boldsymbol k \neq 0. \end{array} \right.$$

Is there an intuitive way to understand that the behaviour should qualitatively change when going from 3D to 4D? (Note that I am not asking to calculate the above brute-force, as I have already done so.)

NB: In case one is wondering how the above $\rho_2(\boldsymbol k, \varepsilon)$ interpolates between $\boldsymbol k =0$ and $\boldsymbol k \neq 0$, the resolution lies in the fact that the $\boldsymbol k \neq 0$ solution is only valid for $\varepsilon_{\textrm{min},\boldsymbol k} < \varepsilon < \varepsilon^*_{\boldsymbol k}$ where $\varepsilon^*_{\boldsymbol k} \to \varepsilon_{\textrm{min},\boldsymbol k}$ as $\boldsymbol k \to 0$.




ADDENDUM/EDIT: I was asked to include my calculation for $D>3$. Firstly (taking the spherical coordinates such that $\boldsymbol k$ is along one of the axes; for this it is essential that $\boldsymbol k \neq 0$),

$$ \rho_2(\boldsymbol k, \varepsilon) \propto \int \delta(\varepsilon_{\boldsymbol q} + \varepsilon_{ \boldsymbol k - \boldsymbol q} - \varepsilon ) \; \mathrm d^D \boldsymbol q \propto \int \left. \left( \frac{1}{\left|\partial_\theta \varepsilon_{\boldsymbol k - \boldsymbol q} \right|} \;\sin \theta \right) \right|_{\varepsilon_{\boldsymbol q} + \varepsilon_{ \boldsymbol k - \boldsymbol q} = \varepsilon} \; q^{D-1} \mathrm d q .$$

Now note that $|\boldsymbol k-\boldsymbol q|^2 = k^2 + q^2 - 2kq \cos \theta$. Hence, $\partial_\theta \varepsilon_{\boldsymbol k - \boldsymbol q} = kq/m \times \sin \theta $. Plugging this in, we obtain $$ \rho_2(\boldsymbol k, \varepsilon) \propto \frac{m}{k} \int q^{D-2} \mathrm dq.$$ One has to be careful with the domain of integration. At the threshold energy $\varepsilon_{\textrm{min},\boldsymbol k}$, we have $q = k/2$. Close to the threshold energy, $\varepsilon = \varepsilon_{\textrm{min},\boldsymbol k} + \delta$, we have the condition that $k/2 - \sqrt{m \delta } \leq q \leq k/2 + \sqrt{ m \delta}$. This condition follows from demanding that $\cos \theta \leq 1$ (using the fact that $\varepsilon = \varepsilon_{\boldsymbol q} + \varepsilon_{ \boldsymbol k - \boldsymbol q}$). Hence, we have that $$ \begin{align} \rho_2(\boldsymbol k, \varepsilon) &\propto \frac{m}{k} \left. q^{D-1} \right|^{k/2 + \sqrt{m \delta } }_{k/2 - \sqrt{m \delta } } \\ &= \frac{m}{k} \; k^{D-1} \; \left[ \left( 1 +\frac{\sqrt{m \delta } }{k} \right)^{D-1} - \left( 1 -\frac{\sqrt{m \delta } }{k} \right)^{D-1} \right] \\ &\approx \frac{m}{k} \; k^{D-1} \; \left[ \left( 1 +(D-1)\frac{\sqrt{m \delta } }{k} \right) - \left( 1 -(D-1)\frac{\sqrt{m \delta } }{k} \right) \right] \\ & \propto k^{D-3} \sqrt{\delta} = k^{D-3} \sqrt{\varepsilon-\varepsilon_{\textrm{min},\boldsymbol k}}. \end{align}$$

$\endgroup$
  • $\begingroup$ Where does your formula for $D>3$ come from? Note that the dimensions are inconsistent, so it is most likely wrong. $\endgroup$ – AccidentalFourierTransform Sep 6 '18 at 3:00
  • $\begingroup$ @AccidentalFourierTransform I think you might have inadvertently given a partial answer to my question: if $\boldsymbol k = 0$, there is only one unitful quantity in the game (i.e. $\varepsilon$) and hence $\rho_2$ must scale as $\varepsilon^{D/2-1}$ (note that $\rho_2(\boldsymbol k, \omega)$ has dimensions due to the measure), whereas for $\boldsymbol k \neq 0$ we can use $\boldsymbol k$ to have different powers. Note that I have left out prefactors depending on $\boldsymbol k$, as I was only interested in the asymptotic dependence on $\varepsilon$. $\endgroup$ – Ruben Verresen Sep 6 '18 at 3:03
  • $\begingroup$ I see. Still, I am somewhat skeptical of that formula. Do you think you can include a reference? $\endgroup$ – AccidentalFourierTransform Sep 6 '18 at 3:05
  • $\begingroup$ @AccidentalFourierTransform I have not found a reference to it, so I added my calculation to the original post. Note that the factor of $k^{D-3}$ nicely pops up to assuage your dimensional concerns. $\endgroup$ – Ruben Verresen Sep 6 '18 at 3:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.