- #1
moso
- 14
- 0
Dear Community,
I am trying to figure out what is happening in this article (https://journals.aps.org/prb/abstract/10.1103/PhysRevB.29.130) when they are calculating the Fredholm Determinant (Section IV). The basic idea is that you want to solve
$$
k = |\frac{det(1+h_0)}{det'(1+h_0+v)}|
$$
where ´means that the zero eigenfunction is not taking into account. And with
$$
h_0 \psi(u) = -\frac{d^2 \psi}{d u^2} + \frac{2\alpha}{\pi} \int_{-\infty}^{\infty} du' \frac{\psi(u)-\psi(u´)}{(u-u´)^2}
$$
and with a potential on the form
$$
v= -3z_c(u)
$$
which for $$\alpha=0$$ is $$Sech^2(u/2)$$. They then go on to define a phase shift and find the eigenvalues which are depicted in figure 6. My problem is understanding how they achieve equation 4.28 and what is $$< E_1 |V|E>$$ in the equation. I have attached the relevant pages. And also what are they doing in order to get the ratio.
I am trying to figure out what is happening in this article (https://journals.aps.org/prb/abstract/10.1103/PhysRevB.29.130) when they are calculating the Fredholm Determinant (Section IV). The basic idea is that you want to solve
$$
k = |\frac{det(1+h_0)}{det'(1+h_0+v)}|
$$
where ´means that the zero eigenfunction is not taking into account. And with
$$
h_0 \psi(u) = -\frac{d^2 \psi}{d u^2} + \frac{2\alpha}{\pi} \int_{-\infty}^{\infty} du' \frac{\psi(u)-\psi(u´)}{(u-u´)^2}
$$
and with a potential on the form
$$
v= -3z_c(u)
$$
which for $$\alpha=0$$ is $$Sech^2(u/2)$$. They then go on to define a phase shift and find the eigenvalues which are depicted in figure 6. My problem is understanding how they achieve equation 4.28 and what is $$< E_1 |V|E>$$ in the equation. I have attached the relevant pages. And also what are they doing in order to get the ratio.