time dependent Schröidinger Equation

Today I decide to share how I work on a paper. As a physicist or any scientist actually, you need to read papers and see how others do things. I will read following paper: "Railway switch transport model"by Martin Horvat, Thomas Prosen Guilano Benenti and Giulio Casati. PRE 86 052102 (2012). I actually know all the authors of this paper and worked / meet with them before. First two is student- advisor from Lubiana and last two from italy,Casati being the most esteemed (and elderly) one. Lets start

I have some background on this work. In previous paper Italian group shows that assymetry play a role in thermoelectric phenemone and efficiency can be increased with it. In this paper they try a toy model to show this affect.

3 terminal for simplicity is considered. 2nd node is probe and transport is between 1 and 3. Heat and particle current leaves each node. wires are one dimentional and one momentum state can be carried at a time. Therefore transport is just permutation between nodes.

There are 6 cases. 3 Shown on left. P1 shows reflections of incoming particles back to starting terminals.

$P_1=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

P 2 shows particles coming from 1 goes back to 1 and ones coming from 2 goes to 3 and vice versa

$P_2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$

There are two currents particle and heat. Those occur due to inbalance between each node. In other words, there are Temperature and chemical potential difference between each nodes.

$P_3=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

Onsager eqn can be written between each terminal.

$L_{i,j}\equiv \; <!--\; identical\; to\; -->\left[\begin{array}{cc}{K}_{ij}& Q_{ij}\\ Q_{ij}& T_{ij}\end{array}\right]$

However, this is a small difference and so linear response theory works

Finally linear response eqn gives

$J_i=\underset{j=1}{\overset{N}{\sum \; <!--\; n-ary\; summation\; -->}}{L}_{ij}{x}_j$

where × are small bias between terminals and

$\left(T_{ij},Q_{ij},K_{ij}\right)=\int \; <!--\; integral\; -->dE{e}^{-E}{T}_{ij}\left(E\right)\times \; <!--\; multiplication\; sign\; -->\left(1,E,E^2\right)$

These eqn is very familiar to me. Instead of Botzaean weight fermi distribution is used in thermoelectric case where electrons are the mediators. In thermoelectric, the most important quantity is T, transmission. In this model T is the permutation operator. It's
a deterministic model, T is known if E and which terminal current coming from. Next, they show that for 3 terminal case 6 transmission possible, only 2 cases are time asymmetric, they are

$P_4=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$

$P_5=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$

other 4 time symmetric since each of them contains at least one terminal has reflection, on the other hand currrent transmits from one terminal to other on those assymetric ones.

Next, they look at the transport between 1 and 3, assuming 2nd terminal is a probe that means current is zero in it. Also they consider current 1 to 3 so assume × is zero for 3rd terminal. This way they connect current between 1 and 3 by a reduced matrix which is 2×2.

$L_{red}=\left[\begin{array}{cc}{l}_{11}& l_{12}\\ l_{21}& l_{22}\end{array}\right]$

$J_1=L_{red}{X}_1$

So, 3×3 Matrix is reduced to 2×2 by the condition probe current zero. It' is neces sary to do a lot of algebra after this to find the matrix element. Asymetry now arises in 2×2 matrix, off diagonal elements can be different for asymmetric cases but see me for symmetric ones. rest of the paper show their result for the effect of assymety on efficiency. First they start with 3 terminals and later for more than 3 terminal cases discussed.

Interestingly, they find figure of merit is not improved with increasing as symmetry. This is disappointing I believe, since in their early paper (PRL) they show clearly asymmetry increase efficiency. This may be due to the symplistic model they have and they argue It gets better with increasing the terminal counts.