Graduate Correlation functions of quantum Ising model

Click For Summary
The discussion centers on the insertion of the operator ##\hat{\sigma}_{j}^{z}## into the correlation function ##\mathcal{G}## of the quantum Ising model as described in the paper "Dynamical quantum correlations of Ising models on an arbitrary lattice and their resilience to decoherence." It explains that when ##\alpha_j = 1##, the operator can be incorporated by recognizing the coupling of the variable ##\phi_j## to ##\hat{\alpha}_{j}^{z}## as a source term. This leads to the conclusion that inserting ##\hat{\alpha}_{j}^{z}(t)## into ##\mathcal{G}## is equivalent to applying the operator ##i \frac{\partial}{\partial \phi_j}##. The reasoning is based on the fact that this operator serves as the generator of the transformation associated with the source term ##\hat{\alpha}_{j}^{z}##. Understanding this relationship is crucial for analyzing the correlation functions in the context of the quantum Ising model.
Danny Boy
Messages
48
Reaction score
3
TL;DR
In the referenced paper, an explanation as to how to insert $\hat{\sigma}_{j}^{z}$ into a correlation function is given. I am seeking clarity on this technical point mentioned in the paper.
I have a single technical question regarding a statement on page 7 of the paper "Dynamical quantum correlations of Ising models on an arbitrary lattice and their resilience to decoherence". The paper up until page 7 defines a general correlation function ##\mathcal{G}## of a basic quantum Ising model (with only an interaction term in the Hamiltonian). The correlation function ##\mathcal{G}## up to page 7 deals only with raising and lowering operators of the form ##\sigma^{\pm}_{j}## on sites ##j## (##\alpha_j = 0## if ##\hat{\sigma}_{j}^{\pm}## appears in the correlation function ##\mathcal{G}## and ##\alpha_j = 0## otherwise). To insert operators of the form ##\hat{\sigma}_{j}^{z}## into the correlation function ##\mathcal{G}##, the following is stated:
The insertion of an operator ##\hat{\sigma}_{j}^{z}## inside a correlation function ##\mathcal{G}##, which we denote by writing ##\mathcal{G} \mapsto \mathcal{G}^{z}_{j}##, is relatively straightforward. If ##\alpha_j = 0##, then clearly the substitution ##\hat{\alpha}_{j}^{z} \mapsto \alpha_{j}^{z}(t)## does the trick. If ##\alpha_j = 1##, ##\hat{\alpha}_{j}^{z}## can be inserted by recognizing that the variable ##\phi_j## couples to ##\hat{\alpha}_{j}^{z}## as a source term, and thus the insertion of ##\hat{\alpha}_{j}^{z}(t)## is equivalent to applying ##i \frac{\partial}{\partial \phi_j}## to ##\mathcal{G}##.
Can anyone see the reasoning behind the last sentence:
If ##\alpha_j = 1##, ##\hat{\alpha}_{j}^{z}## can be inserted by recognizing that the variable ##\phi_j## couples to ##\hat{\alpha}_{j}^{z}## as a source term, and thus the insertion of ##\hat{\alpha}_{j}^{z}(t)## is equivalent to applying ##i \frac{\partial}{\partial \phi_j}## to ##\mathcal{G}##.

Thanks for any assistance.
 
Physics news on Phys.org
The reasoning behind the last sentence is that the operator ##\hat{\alpha}_{j}^{z}## can be inserted in the correlation function ##\mathcal{G}## by recognizing that the variable ##\phi_j## couples to ##\hat{\alpha}_{j}^{z}## as a source term. This means that the insertion of ##\hat{\alpha}_{j}^{z}(t)## into the correlation function is equivalent to applying the operator ##i \frac{\partial}{\partial \phi_j}## to ##\mathcal{G}##. This is because ##i \frac{\partial}{\partial \phi_j}## is the generator of the transformation for which ##\hat{\alpha}_{j}^{z}## is the source term.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 87 ·
3
Replies
87
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
3K