A Numerical implementation of creation and annhilation operators in the SSH model

  • Thread starter Thread starter JangMilad
  • Start date Start date
Click For Summary
The discussion focuses on implementing annihilation and creation operators in the SSH model using the density matrix formalism. The user seeks guidance on defining these operators for a lattice model and understanding the physical implications of expressions involving the density matrix, specifically a_1 ρ a_1† and a_1† ρ a_1. Additionally, they request tips or references for constructing these operators in matrix form for numerical simulations. The conversation emphasizes the need for clarity on both the mathematical construction and physical interpretation of these operators. Insights shared will aid in effectively simulating the SSH model.
JangMilad
Messages
1
Reaction score
0
Hi all,

I'm working on a numerical simulation involving the SSH model and the density matrix formalism. I'm using annihilation and creation operators at the first site, denoted by a_1 and a_1^\dagger, and I'm trying to understand how to construct and compute expressions like:

a_1 \rho a_1^\dagger
a_1^\dagger \rho a_1

where \rho is the density matrix of the system.

My goal is to implement this numerically. I would appreciate any insights on:

How to define the annihilation/creation operators for a lattice model like SSH.

The physical meaning of the above expressions (e.g., in the context of Lindblad dissipators).

Any tips or references for constructing these operators explicitly in matrix form.

Thanks in advance for your help!
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

Similar threads

  • · Replies 0 ·
Replies
0
Views
683
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
23
Views
6K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
2
Views
2K
  • · Replies 87 ·
3
Replies
87
Views
8K