Negative Partial Density of States

Abstract

We first give an introduction to Burgers circuit (BC). We then show that phase of the wavefunction in different model systems can be understood in general using BC. This includes the new discontinuous phase discussed in Chap. 2. Then the three prong potential is introduced and it is shown that the discontinuous phase has continuous counterparts that lead to the dramatic effect that PDOS can become negative. Such continuous counterparts were not known in case of classical waves. FSR-like rules can also be understood in terms of BC. General proofs are provided that make our results independent of the models used for demonstration.

Appendices

Appendix A

From Eq. (5.8), we can write

$$\begin{aligned} \rho _{pd}(3,1)= & {} -\frac{1}{2\pi }\int _{sample} d\mathbf{r} ^{3} |t_{31}|^2\frac{\delta \theta _{t_{31}}}{\delta V(\mathbf{r} )} \end{aligned}$$

(5.34)

In Eq. (5.34), using the well-known semi-classical approximation $\int _{sample} d\mathbf{r} ^{3}\frac{\delta t_{31}}{\delta V(\mathbf{r} )}\cong - \frac{dt_{31}}{dE}$, we get (see Eq. 5.13)

$$\begin{aligned} \rho _{pd}(3,1)\approx & {} \frac{1}{2\pi }|t_{31}|^2\frac{d\theta _{t_{31}}}{dE} \end{aligned}$$

(5.35)

Now, for a contour $C'$ traced when energy is varied from 0 to $E_{1}$,

(5.36)

Appendix B

We know that in a scattering problem increasing incident energy by dE is equivalent to decreasing the potential globally by a constant amount $\Delta \varepsilon$, such that $dE=-e\Delta \varepsilon$, where e is particle charge that we will set to 1 to simplify our arguments. That is, the new potential is $V'(\mathbf{r} )=V(\mathbf{r} )-\Delta \varepsilon$. Hence, if we can generate a closed subloop in the A-D by varying E, then we can also do so by globally changing the potential and for such a closed contour like ABQFA in Fig. 5.7c,

$$\begin{aligned} \oint _{ABQFA} \delta \theta _{s_{\alpha \beta }}=0 \nonumber \\ \text {i.e.,}-\oint _{ABQFA}\int _{global}\frac{\delta \theta _{s_{\alpha \beta }}}{\delta V(\mathbf{r} )}\Delta \varepsilon d\mathbf{r} ^3=0 \end{aligned}$$

(5.37)

Now we replace the global integration over $\mathbf{r}$ by an integration over the sample or the scattering region only, since we have seen that it can be done in case of closed contours or inside an integration of the type $\oint _{C}$ in Eq. (5.37). This has been discussed in Eq. (5.23).

$$\begin{aligned} \therefore -\oint _{C}\int _{sample}\frac{\vert s_{\alpha \beta }\vert ^{2}}{\vert s_{\alpha \beta }\vert ^{2}}\frac{\delta \theta _{s_{\alpha \beta }}}{\delta V(\mathbf{r} )}\Delta \varepsilon d\mathbf{r} ^3= & {} 0\nonumber \\ \text {or,}-\oint _{C}\frac{1}{\vert s_{\alpha \beta }\vert ^{2}}\Delta \varepsilon \int _{sample}\vert s_{\alpha \beta }\vert ^{2}\frac{\delta \theta _{s_{\alpha \beta }}}{\delta V(\mathbf{r} )}d\mathbf{r} ^3= & {} 0\nonumber \\ \text {or,}2\pi \oint _{C}\frac{\rho _{pd}(\alpha ,\beta )}{\vert s_{\alpha \beta }\vert ^{2}}\Delta \varepsilon= & {} 0\nonumber~\text {Using Eq. (5.8)} \end{aligned}$$

Therefore, the R.H.S of Eq. (5.25) is justified if an electronic charge is multiplied to the numerator.