Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Expectation values in Doi-Peliti formalism

+ 3 like - 0 dislike
627 views

I'm trying to understand how the Doi-Peliti (DP) action is constructed, and specifically how they compute expectation values. To this end, I've been using the book by Taüber as a reference (Critical Dynamics: A Field Theory Approach to Equilibrium and Non-Equilibrium Scaling Behaviour).

The point I seem to be missing is during the discretized $\rightarrow$ path integral procedure of the expectation value of an observable. For a single chemical species with lattice occupation numbers $\{n_i\}$ this is defined as $$ \langle A(t)\rangle =\sum_{\{n_i\}}A(\{n_i\})P(\{n_i\}, t) $$ where $P(\{n_i\},t)$ denotes the probability of observing a configuration $\{n_i\}$ and follows a master-type equation. In the DP formalism, the chemical reactant is assigned a site-specific bossonic ladder algebra $a_i$, $a_i^{\dagger}$ and one finds that the expectation value above may be expressed in this language by \begin{equation} \label{eq:expect} \langle A(t)\rangle=\langle\mathcal{P}|A\left(\{a_i^{\dagger}a_i\}\right)|\Phi(t)\rangle \end{equation} Here the projection operator $\langle\mathcal{P}|=\langle 0|\prod_ie^{a_i}$, and the state vector $|\Phi(t)\rangle=\sum_{\{n_i\}}P(\{n_i\},t)|\{n_i\}\rangle$ satisfies the imaginary time Schrödinger equation $$ \partial_t|\Phi(t)\rangle=-H(\{a_i^{\dagger}\},\{a_i\})|\Phi(0)\rangle $$ ($H(\{a_i^{\dagger}\},\{a_i\})$ is meant to indicate that $H$ is normal-ordered).

By shifting the operator $\prod_ie^{a_i}$ in the above expression for $\langle A(t)\rangle$ over to the right, one obtains $$ \langle A(t)\rangle=\langle0|\tilde{A}\left(\{a_i^{\dagger}\rightarrow1\},\{a_i\}\right)e^{-H(\{a_i^{\dagger}\rightarrow 1+a_i^{\dagger}\},\{a_i\})t}|\tilde{\Phi}(0)\rangle $$ in which $\tilde{A}(\{1\},\{a_i\})$ is obtained from $A$ by normal ordering and replacing $a_i$ by $1$ (e.g. $a_i^{\dagger}a_ia_j^{\dagger}a_j\rightarrow a_i\delta_{ij}+a_ia_j$), and $$ |\tilde{\Phi}(0)\rangle=\prod_ie^{a_i}|\Phi(0)\rangle $$

Here comes the part I seem to fail to understand. If we denote by $$ U(t_2,t_1)=e^{-H(\{a_i^{\dagger}\rightarrow 1+a_i^{\dagger}\},\{a_i\})(t_2-t_1)} $$ then clearly $U(t_2,t_1)=U(t_2,t')U(t',t_1)$. We may thus split the time-evolution operator $U$ in $\langle A(t)\rangle$ up into many pieces and insert the completeness relation $$ 1=\int\prod_i\frac{d\phi_i^*d\phi_i}{2\pi i}e^{\sum_i\phi_i^*\phi_i}|\phi\rangle\langle\phi| $$ ($i$ in the denominator is the imaginary unit and $|\phi\rangle$ is a coherent state) inbetween each time-step to obtain $$ \langle A(t)\rangle=\int\left(\prod_{i,k}\frac{d\phi_i^*(t_k)d\phi_i(t_k)}{2\pi i}\right)\langle0|\tilde{A}\left(\{1\},\{a_i\}\right)|\phi(t_f)\rangle\left(\prod_j\langle\phi(t_j)|U(t_j,t_{j-1})|\phi(t_{j-1})\rangle\right)\times\langle\phi(t_0)|\tilde{\Phi}(0)\rangle $$ The matrix elements $$ \langle\phi(t_j)|U(t_j,t_{j-1})|\phi(t_{j-1})\rangle $$ are easily calculated. However, to me it seems that $$ \tilde{A}\left(\{1\},\{a_i\}\right)|\phi(t_f)\rangle=\tilde{A}\left(\{1\},\{\phi_i(t_f)\}\right)|\phi(t_f)\rangle $$ since $a_i|\phi\rangle=\phi_i|\phi\rangle$. In particular, I don't find it obvious how the above tends to the path integral $$ \langle A(t)\rangle=\int\prod_i\mathcal{D}[\phi_i^*,\phi_i]\tilde{A}(\{1\},\{\phi_i(t)\})e^{-\mathcal{A}[\phi_i^*,\phi_i]} $$ (for some action $\mathcal{A}$ I leave unspecified), as it seems as though the observable $\tilde{A}$ should only be evaluated at the final point $\phi(t_f)$.

Sorry for the very long message. Any help would be greatly appreciated!


This post imported from StackExchange Physics at 2017-08-11 12:47 (UTC), posted by SE-user john

asked Apr 21, 2017 in Theoretical Physics by john_utf8 (15 points) [ revision history ]
edited Aug 11, 2017 by Dilaton

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...