1
$\begingroup$

This question refers primarily to this paper by R. Prange. The author considers a $2D$ electron system with a single impurity modeled as a $\delta$ function. The end goal is to show that when the impurity potential is switched on and some states become localised, the current carried by the extended states increases to compensate for this loss exactly.

The Hamiltonian of the system is $H = H_0 + \lambda \delta(r)$ with $$H_0 = \frac{1}{2}\left[-\partial_x^2+(p_y+x)^2\right]-vx$$ in appropriate units. The normalised eigenfunctions of $H_0$ are given by $$\psi_{np} =\frac{1}{2^n n! L} e^{ip_y y} H_n(x+p) \phi(x+p)$$ where $L$ is the sample length in the $y$ direction, $p = p_y-v$, $H_n$ are Hermite polynomials and $\phi(x)=\frac{1}{\pi^{1/4}}e^{-x^2/2}$. The corresponding energy eigenvalues are $$E_{np}=n +vp$$ One can then consider an eigenstate $\psi^\alpha$ of the full Hamiltonian $H$. It's not hard to see that its energy $E_\alpha$ is determined by the equation $$1=2\pi\lambda \sum_{np}\frac{|\psi_{np}(0)|^2}{E_\alpha-n-pv}\tag{2}$$ At this point the author claims

To evaluate (2) and (3), the sum over p is replaced by a principal value integral plus a contribution coming from the $p$ values in the immediate neighborhood of the singular point. By introducing $k_\alpha-\delta_\alpha=LE_\alpha/2\pi v$, where $k_\alpha$ is an integer and $2|\delta_\alpha|<1$, as well as $p_\alpha =2\pi k_\alpha/L$, Eq. (2) may be rewritten as $$1=\lambda G(E_\alpha) + \lambda \phi(p_\alpha)^2 \sigma_\alpha/v\tag{2a}$$ Here $G(E)$ is the principal value integral [...]. The discrete sum is convergent and is given by $\sigma_\alpha=-\pi \cot(\pi \delta_\alpha)$

However, they do not explain what exactly they are doing and they don't give explicit expressions for the principal value integral $G(E)$. My thought was to turn the $p$ integration into an integral, $$\sum_{np} \to \sum_{n}\int \frac{dp}{2\pi}$$ The integral would then be singular whenever $E_\alpha -n-pv =0$ and one can apply the Sokhotski-Plemelj theorem to split the integral into a principal value integral and a contribution near the singular point. However, this results in a horrible infinite sum of Hermite polynomials, which appears intractable.

How does one get from $(2)$ to $(2a)$?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.