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  Expectation values in Doi-Peliti formalism

+ 3 like - 0 dislike

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

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