上个学期期末做的有限温场论大作业。学习并讲了一下一直有兴趣的Keldysh场论。
主要就是学习了这篇文章的一部分：Sieberer, L. M., Buchhold, M., & Diehl, S. (2016). Keldysh field theory for driven open quantum systems. Reports on Progress in Physics, 79(9), 096001.
还有之前入门学的一本书：Kamenev, A. (2011). Field Theory of Non-Equilibrium Systems. Cambridge: Cambridge University Press.

有驱动开放系统的Keldysh场论

Basics of Keldysh Field Theory

Motivation: Why do we need Keldysh field theory?

In some open quantum systems, besides coherence, driven–dissipative dynamics also occur on equal footing.
Dynamical non-equilibrium quantum many-body phases emerge in such systems.
Studying such systems calls for a merger of many-body field theory in equilibrium and quantum mater equation.

Closed Time Contour

Equilibrium field theory use only one-way time path, while for non-equilibrium case we need to consider a closed time contour.
Let be an evolution from to , then the density at time is .
The expected value of an observable at time is:

$\langle\hat{\mathcal{O}}\rangle(t) \equiv \frac{\operatorname{Tr}[\hat{\mathcal{O}} \hat{\rho}(t)]}{\operatorname{Tr}[\hat{\rho}(t)]}=\frac{1}{\operatorname{Tr}[\hat{\rho}(t)]} \operatorname{Tr} [\hat{\mathcal{U}}_{-\infty, t} \hat{\mathcal{O}} \hat{\mathcal{U}}_{t,-\infty} \hat{\rho}(-\infty) ]\tag{1}$

In equilibrium field theory, only the forward evolution is considered, because the backward path can be eliminated by

$\hat{\mathcal{U}}_{+\infty,-\infty}|0\rangle = e^{iL}|0\rangle\tag{2}$

which means the adiabatic evolution takes the GS at back, up to a phase .

With the condition in , the value of can be written in only forward evolution:

$\langle\mathrm{GS}|\hat{\mathcal{O}}| \mathrm{GS}\rangle =\frac{ \langle 0 |\hat{\mathcal{U}}_{+\infty, t} \hat{\mathcal{O}}_{t,-\infty} | 0 \rangle}{ \langle 0 |\hat{\mathcal{U}}_{+\infty,-\infty} | 0 \rangle}\tag{3}$

The normalization denominator offsets , and it is amounting to subtracting all disconnected vacuum diagrams.
In non-equilibrium cases, is not satisfied, so a closed time contour is needed.
The expectation value of in closed time contour:

$\langle\hat{\mathcal{O}}\rangle(t)=\frac{1}{\operatorname{Tr}[\hat{\rho}(-\infty)]} \operatorname{Tr} [\hat{\mathcal{U}}_{-\infty,+\infty} \hat{\mathcal{U}}_{+\infty, t} \hat{\mathcal{O}} \hat{\mathcal{U}}_{t,-\infty} \hat{\rho}(-\infty) ]$

\begin{figure}
    \centering
    \includegraphics[width=0.8\textwidth]{pic/ClosedContour.png}
    \label{fig:contour}
\end{figure}

The closed time contour is the central idea of non-equilibrium field theory.

Keldysh Partition Function

With a closed time contour , the partition function is defined as:

$Z = \frac{\operatorname{Tr}[\hat{\mathcal{U}}_{\mathcal{C}}\hat\rho(-\infty)]}{\operatorname{Tr}[\hat\rho(-\infty)]}$

With , the partition function is normalized to .
Adding a source , the partition function becomes:

$Z[J] = \frac{\operatorname{Tr}[\hat{\mathcal{U}}_{\mathcal{C}}[J]\hat\rho(-\infty)]}{\operatorname{Tr}[\hat\rho(-\infty)]}$

And the expected value of is the derivative of :

$\langle \hat{\mathcal{O}}\rangle (t)=\left. \frac{i}{2} \frac{\delta Z[J]}{\delta J(t)} \right|_{J=0}$

In contrast, in finite temperature field theory, the exp value is the derivative of , which subtracts all disconnected terms.

Markovian Keldysh Action from the Master Equation

Now we proceed from the master equation to Keyldysh action.
The open quantum system is described by quantum master equation:

$\partial_{t} \rho=\mathcal{L} \rho=-\mathrm{i}[H, \rho]+\sum_{\alpha} \gamma_{\alpha} (L_{\alpha} \rho L_{\alpha}^{\dagger}-\frac{1}{2} \{L_{\alpha}^{\dagger} L_{\alpha}, \rho \} )$

Using the coherent states as basis, the Keldysh (closed contour) version of is:

$\begin{aligned} &\mathcal{L} (\psi_{+}^{*}, \psi_{+}, \psi_{-}^{*}, \psi_{-} )=-\mathrm{i} (H_{+}-H_{-} ) \\ &\quad+\sum_{\alpha} \gamma_{\alpha} [L_{\alpha,+} L_{\alpha,-}^{*}-\frac{1}{2} (L_{\alpha,+}^{*} L_{\alpha,+}+L_{\alpha,-}^{*} L_{\alpha,-} ) ] \end{aligned}$

The subscript represents the forward/backward part of the contour.

\begin{frame}{Keldysh Action}
\small
\begin{itemize}
\item The partition function written in functional integral is then:
\begin{equation}
\begin{aligned}
Z&=\int \mathcal{D}[\psi_+,\psi_+^,\psi_-,\psi_-^] e^{iS}=1,\
S&=\int_{-\infty}^{+\infty}dt (\psi_+^i\partial_t \psi_+-\psi_-^i\partial_t \psi_- -i\mathcal{L})
\end{aligned}
\end{equation}
Note that all the cross terms of $\psi_\sigma^ \psi_{\sigma’}, \sigma \neq \sigma’\mathcal{L}J_\pm = (j_\pm,j^_\pm)\Psi_\pm=(\psi_\pm,\psi^*_\pm)$, the partition function is:

\end{itemize}
\end{frame}

$Unknown environment 'frame'\begin{frame}{Keldysh Rotation and Green’s Functions} \small \begin{itemize} \item The action using basis $\Psi_\pm$ has redundancy. \item The redundancy can be avoided by performing Keldysh rotation and using new basis $\phi_c,\phi_q$: \begin{equation} \begin{aligned} \phi_c &= \frac{1}{\sqrt{2}}(\psi_++\psi_-),\\ \phi_q &= \frac{1}{\sqrt{2}}(\psi_+-\psi_-). \end{aligned} \end{equation} \item The subscript $c$ means classical, $q$ means quantum. The partition function can then be written in the new basis: \begin{equation} Z\left[J_{c}, J_{q}\right]=\int \mathcal{D}\left[\Phi_{c}, \Phi_{q}\right] \operatorname{exp}[\mathrm{i} S+\mathrm{i} \int_{t, \mathbf{x}}\left(J_{q}^{\dagger} \Phi_{c}+J_{c}^{\dagger} \Phi_{q}\right)] \end{equation} \end{itemize} \end{frame}$

\begin{frame}{Keldysh Rotation and Green’s Functions}
\small
\begin{itemize}
\item After the Keldysh rotation, we can calculate the correlation, or Green’s function, between the classical and quantum field, and we’ll get a matrix like:
\begin{equation}
\begin{aligned}
&\left(\begin{array}{cc}
\langle\phi_{c}(t, \mathbf{x}) \phi_{c}^{} (t^{\prime}, \mathbf{x}^{\prime} ) \rangle_{d} & \langle\phi_{c}(t, \mathbf{x}) \phi_{q}^{} (t^{\prime}, \mathbf{x}^{\prime} ) \rangle_{d} \
\langle\phi_{q}(t, \mathbf{x}) \phi_{c}^{} (t^{\prime}, \mathbf{x}^{\prime} ) \rangle_{d} & \langle\phi_{q}(t, \mathbf{x}) \phi_{q}^{} (t^{\prime}, \mathbf{x}^{\prime} ) \rangle_{d}
\end{array} \right)\
&=-\left(\begin{array}{cc}
\frac{\delta^2Z}{\delta j_q^(t,\mathbf{x})\delta j_q(t’,\mathbf{x’})} & \frac{\delta^2Z}{\delta j_q^(t,\mathbf{x})\delta j_c(t’,\mathbf{x’})} \
\frac{\delta^2Z}{\delta j_c^(t,\mathbf{x})\delta j_q(t’,\mathbf{x’})} & \frac{\delta^2Z}{\delta j_c^(t,\mathbf{x})\delta j_c(t’,\mathbf{x’})}
\end{array}\right)
=\mathrm{i} \left( \right)
\end{aligned}
\end{equation}
\item Here the block of 0 shows directly the redundancy in basis.
\item The superscript means ‘Keldysh’, means ‘retarded’, and means ‘advanced’.
\item And the subscript means the disconnected diagrams are included, since it is calculated by the derivations of .
\end{itemize}
\end{frame}

\section{Keldysh Field Theory by Example}
\begin{frame}{Model of a Single Mode Cavity}
\small
\begin{itemize}
\item Now we show the Keldysh field theory on a simple model.
\item The Hamiltonian of single mode photons in the cavity is

with and the creation and annihilation operators of photons, and the frequency of the cavity mode.
\item For the master equation () that describe this model, the operator has a single term , describing the decay with rate .
\item The Keldysh action is:
\begin{equation}
\begin{aligned}
S=\int_t&\{a_+^(i\partial_t-\omega_0)a_+-a_-^(i\partial_t-\omega_0)a_-\
&-i\kappa[2a_+a_-^-(a_+^a_++a_-^*a_-)]\}
\end{aligned}
\end{equation}
\end{itemize}
\end{frame}

\begin{frame}{Model of a Single Mode Cavity}
\small
\begin{itemize}
\item Performing Keldysh rotation and Fourier trans into the frequency space, the action is:
\begin{equation}
S=\int_\omega\left(a_c^(\omega),a_q^(\omega)\right)\left(\right)
\left(\right)
\end{equation}
\item Here is the inverse propagator. And the three parts in can be calculated:

\item From the here of the simple model, key properties of the Keldysh Green’s functions will be discussed. And they are also generally true for other systems.
\end{itemize}
\end{frame}

\begin{frame}{Properties of Keldysh Green’s Function}
\small
\begin{itemize}
\item \textbf{Conservation of probability.} The partition func is a manifestation of the conservation of probability. ()
\item \textbf{Analytic structure.} The poles of are , which are below the rational axis. (For poles are above the rational axis).
\item The inverse Fourier transformation of the retarded Green’s function is .
\item With the Heaviside func , if perturbed at the is only non-zero for . This is why the $\langle \phi_c \phi_q^\rangleG^A=\langle \phi_q \phi_c^\ranglet<0$ and is called ‘advanced’.
\end{itemize}
\end{frame}

\begin{frame}{External Source}
\small
\begin{itemize}
\item The partition function with source is:
\begin{equation}
\begin{aligned}
Z[j_c,j_q]&=\int \mathcal{D}[a_c^,a_q^,a_c,a_q] \exp(iS-iS_j),\
S_j&=\int_t j_c^a_q+j_q^a_c+c.c.
\end{aligned}
\end{equation}
\item And from the derivation of and , can be calculated.
\item Note that is proportional to . \textbf{Due to causality, the ‘quantum’ source has to be zero in any physical setup.} The is only a calculation tool here.
\item The ‘classical’ source , however, can have physical meanings. For example, the coupling to the photon out of the cavity in this simple cavity model.
\end{itemize}
\end{frame}

\begin{frame}{Response and Correlation Functions}
\small
\begin{itemize}
\item \textbf{The retarded Green’s function represents the linear response of the system.}
\item For the case of single mode cavity, if the external photons interact with the photons inside,

\item Then the and can be seen as and $j_c^$. And the Keldysh action is:
\begin{equation}
S-j = S-\int_t (j^(t)a_q(t)+j(t)a_q^(t))
\end{equation}
\item Assume the coupling is weak. To the first order of the current, the photon field in the cavity is then
\begin{equation}
\begin{aligned}
\langle a_c(t)\rangle_j &= \langle a_c(t)\rangle_{j=0}-i\sqrt \kappa\int_{t’}\langle a_c(t)a_q^(t’)\rangle_{j=0}j(t’)\
&=\langle a_c(t)\rangle_{j=0}+\sqrt \kappa\int_{t’}G^R(t-t’)j(t’)
\end{aligned}
\end{equation}
\end{itemize}
\end{frame}

$Unknown environment 'frame'\begin{frame}{Response and Correlation Functions} \small \begin{itemize} \item \textbf{The Keldysh Green’s function $G^K$ represents statistical information on excitations: how are the excitations occupied?} \item In this cavity example, the mode occupation number is: \begin{equation} iG^K(t,t) = 2\langle a^\dagger a \rangle +1 \end{equation} \item And using the previous result for $G^K(\omega)$, we can easily get: \begin{equation} \langle a^\dagger a \rangle=\frac{1}{2}(i\int_\omega G^K(\omega)-1)=0, \end{equation} which means the cavity is empty without pumping. \end{itemize} \end{frame}$

$Unknown environment 'frame'\begin{frame}{Driven-dissipative Condensate} \small \begin{itemize} \item Now consider a bosonic many-body system with interaction, non-linear loss, and linear dissipation. \item A basic experimental setup of photons in a cavity interact with the Wannier-Mott excitons. The Hamiltonian is: \begin{equation} H_0=H_C+H_X+H_{C-X} \end{equation} with C-cavity photons, X-excitons. \item $H_C$ and $H_X$ takes the same quadratic form ($\alpha=C/X$): \begin{equation} H_{\alpha}=\int \frac{\mathrm{d} \mathbf{q}}{(2 \pi)^{2}} \sum_{\sigma} \omega_{\alpha}(q) a_{\alpha, \sigma}^{\dagger}(\mathbf{q}) a_{\alpha, \sigma}(\mathbf{q}) \end{equation} \item And the interaction part: \begin{equation} H_{C-X}=\Omega_{R} \int \frac{\mathrm{d} \mathbf{q}}{(2 \pi)^{2}} \sum_{\sigma}\left(a_{X, \sigma}^{\dagger}(\mathbf{q}) a_{C, \sigma}(\mathbf{q})+\text { H.c. }\right) \end{equation} \end{itemize} \end{frame}$

\begin{frame}{Driven-dissipative Condensate}
\small
$Unknown environment 'columns'\begin{columns} \column{0.6\textwidth} \begin{itemize} \item The quadratic H can be diagonalized by two new modes: lower and upper exciton-polaritons. \item Experimentally, it is sufficient to consider only lower polaritons. The Keldysh action is: \end{itemize} \column{0.4\textwidth} \begin{figure}[htpb] \centering \includegraphics[width = \linewidth]{pic/polaritons.png} \end{figure} \end{columns}$

    \begin{equation}\label{eq:realAction}
        \begin{aligned}
            S=&\int_{t,x}\{\phi_q^*(i\partial_t+K_c\nabla^2-r_c+ir_d)\phi_c+c.c\\
            &-[(u_c-iu_d)(\phi_q^*\phi_c^*\phi_c^2+\phi_q^*\phi_c^*\phi_q^2)+c.c]\\
            &+i2(\gamma+2u_d\phi_c^*\phi_c)\phi_q^*\phi_q\}
        \end{aligned}
    \end{equation}
    \begin{itemize}
    \item Here $K_c=1/(2m_{LP})$, $r_c=\omega_{LP}^0$. And the noise parameter is defined as $\gamma=(\gamma_l+\gamma_p)/2$ and $r_d=(\gamma_l-\gamma_p)/2$. l-loss, p-pumping.
\end{itemize}

\end{frame}

\begin{frame}{Driven-dissipative Condensate}
\small
\begin{itemize}
\item Solving for this action is hard. Now we consider a simple mean-field analysis that can lead us to the condensation. Mean-field means fluctuations of the field is neglected:
\begin{equation}\label{eq:meanField}
\frac{\delta S}{\delta \phi_c^}=0,
\frac{\delta S}{\delta \phi_q^}=0
\end{equation}
\item Inserting mean-field condition into the action () gives:

\item For the Im part of (): if , the solution is in symmetric phase, ; for , there is a finite condensation .
\item This result shows a driven-dissipative condensate.
\end{itemize}
\end{frame}

\section{Semi-classical Limit}
\begin{frame}{Canonical Scaling Analysis}
\small
\begin{itemize}
\item To study the long-wavelength phenomenon, we need to consider the \textbf{semiclassical limit}. And this could be done by RG that ignores small- contribution.
\item For the Keldysh action talked earlier in (), we can do the RG using canonical scaling analysis near the critical point, where functions scaling in .
\item After the scaling analysis, any quartic term that includes more than one quantum field is irrelevant. The action becomes:
\begin{equation}
\begin{aligned}
S=&\int_{t,x}\{\phi_q^(i\partial_t+(K_c-iK_d)\nabla^2-r_c+ir_d)\phi_c+c.c\
&-[(u_c-iu_d)\phi_q^\phi_c^\phi_c^2+c.c]
+i2\gamma\phi_q^\phi_q\}
\end{aligned}
\end{equation}
\end{itemize}
\end{frame}

\begin{frame}{Canonical Scaling Analysis}
\small
\begin{itemize}
\item Using the Hubbard–Stratonovich transformation (detailed in class), the Gauss integral is changed to basis $[\phi_c^,\phi_c,\xi^,\xi]$.
\begin{equation}
\begin{aligned}
Z &=\int \mathscr{D}\left[\phi_{c}, \phi_{c}^{}, \xi, \xi^{}\right] \mathrm{e}^{-\frac{1}{2 \gamma} \int_{t, \mathrm{x}} \xi \xi} \
& \times \delta\left(\left[\mathrm{i} \partial_{t}+\left(K_{c}-\mathrm{i} K_{d}\right) \nabla^{2}-r_{c}+\mathrm{i} r_{d}-\left(u_{c}-\mathrm{i} u_{d}\right)\left|\phi_{c}\right|^{2}\right] \phi_{c}-\xi\right) \
& \times \delta\left(\left[-\mathrm{i} \partial_{t}+\left(K_{c}+\mathrm{i} K_{d}\right) \nabla^{2}-r_{c}-\mathrm{i} r_{d}-\left(u_{c}+\mathrm{i} u_{d}\right)\left|\phi_{c}\right|^{2}\right] \phi_{c}^{}-\xi^{}\right) .
\end{aligned}
\end{equation}
Here has the physical meaning of a Markovian Gaussian noise source with zero mean, and second momentum .
\item The functions restrict to be the solution of a macro-scale equation, the \textbf{Langevin equation}:

\end{itemize}
\end{frame}

\section{Symmetries of the Keldysh Action}
$Unknown environment 'frame'\begin{frame}{Thermodynamic Equilibrium} \small Thermodynamic equilibrium condition for a system with $H$ at temperature $T=1/\beta$: \begin{itemize} \item (i) density matrix is in $\rho = \exp{-\beta H}/Z$ and $Z=\operatorname{tr}[\text{e}^{-\beta H}]$. \item (ii) the unitary time evolution of the system is $U=\text{e}^{-iHt}$. \item Static properties are not sufficient for identifying equilibrium. So a relation of dynamical response is needed: a fluctuation-dissipation relationship (FDR). \item The FDR can appear in the form of two-point correlation functions. For the single-component Bose system (\ref{eq:GreenFunc}): \begin{equation} G^K(\omega,\mathbf{q})=\coth(\frac{\omega}{2T})(G^R(\omega,\mathbf{q})-G^A(\omega,\mathbf{q})) \end{equation} \end{itemize} \end{frame}$

\begin{frame}{Thermodynamic Equilibrium}
\small
\begin{itemize}
\item The FDRs holds for correlation functions of arbitrary order, not just the 2-point Green’s functions in .
\item And the -order of FDR is related to a \textit{symmetry} of the Keldysh action: thermal symmetry .

\item Careful analysis gives a precise form of the symmetry transformation:
\begin{equation}\label{eq:thermalSym}
\begin{aligned}
&\mathcal{T}_{\beta} \psi_{\sigma}(t, \mathbf{x})=\psi_{\sigma}^{}(-t+\mathrm{i} \sigma \beta / 2, \mathbf{x}) \
&\mathcal{T}_{\beta} \psi_{\sigma}^{}(t, \mathbf{x})=\psi_{\sigma}(-t+\mathrm{i} \sigma \beta / 2, \mathbf{x})
\end{aligned}
\end{equation}
is the composition of complex conjugation and inversion of the sign of time . Here .
\item So thermal equilibrium can be diagnosed by a simple symmetry test on the Keldysh action.
\end{itemize}
\end{frame}

\begin{frame}{Semi-classical Limit of the Thermal Symmetry}
\small
\begin{itemize}
\item Semi-classical limit: typical frequency scale is much smaller than , .
\item Firstly, the symmetry operation in () can be expanded by and save to the leading order. This leads to a FDR of Raleigh-Jeans form:

\item Secondly, the equivalent Langevin equation in () can be written as two parts: reversible (coherent) and irreversible (dissipative).
\begin{equation}
i\partial_t \psi = \frac{\delta H_c}{\delta \psi^}-i\frac{\delta H_d}{\delta \psi^} +\xi
\end{equation}
\item The effective is:

\end{itemize}
\end{frame}

$Unknown environment 'frame'\begin{frame}{Semi-classical Limit of the Thermal Symmetry} \small \begin{itemize} \item The thermal symmetry as well as the detailed balance requires: \begin{equation} H_c=rH_d, r=\frac{K_c}{K_d}=\frac{u_c}{u_d} \end{equation} \item This means the equilibrium evolution of reversible and irreversible parts are locked to a single value. \item In complex plane, the vectors $K=K_c+iK_d$ and $u=u_c+iu_d$ point at the same direction at equilibrium: \begin{figure} \centering \includegraphics[width=0.8\textwidth]{pic/complexplane.png} \label{fig:CPlane} \end{figure} \end{itemize} \end{frame}$

\begin{frame}{Noether’s Theorem in Keldysh Formalism}
\small
\begin{itemize}
\item Assume a transformation of field $\Psi=(\psi_+,\psi_+^,\psi_-,\psi_-^)\alphat\mathbf{x}T_0=\mathbb{I}T_\alphaj^\mu\langle \partial_\mu j^\mu \rangle=0\mathscr{D}=\int_{\mathbf{x}}j^0\mathscr{D}\langle \operatorname{d}\mathscr{D}/\operatorname{d}t\rangle =0$.
\end{itemize}
\end{frame}

$Unknown environment 'frame'\begin{frame}{Noether’s Theorem in Keldysh Formalism} \small \begin{itemize} \item Proof: for $\alpha \ll 1$, $T_\alpha$ can be expanded: \begin{equation} T_\alpha=\Psi +\alpha \mathfrak{G} \Psi +O(\alpha^2) \end{equation} Here $\mathfrak{G}$ is called the generator of the transformation. \item LHS of (\ref{eq:symmetry}): after the expansion and the coefficient of the first order has to vanish, then we get: \begin{equation} \int_{t, \mathbf{x}}\left(\frac{\partial \mathcal{L}}{\partial \Psi^{T}} \mathfrak{G} \Psi+\frac{\partial \mathcal{L}}{\partial \partial_{\mu} \Psi^{T}} \partial_{\mu}\mathfrak{G} \Psi \right)\equiv \int_{t, \mathbf{x}}\partial_\mu f^\mu =0. \end{equation} with $f^\mu$ a vector field. \item Expanding the RHS of (\ref{eq:symmetry}) gives the Noether current $j^\mu$: \begin{equation}\label{eq:NoetherCurrent} \int_{t, \mathbf{x}} \alpha \partial_{\mu}\left\langle f^{\mu}-\frac{\partial \mathscr{L}}{\partial \partial_{\mu} \Psi^{T}} \mathfrak{G} \Psi\right\rangle=\int_{t, \mathbf{x}} \alpha \partial_{\mu} \langle j^\mu \rangle=0 \end{equation} in which the mean comes from the integral in partition integral, $Z=\int \mathcal{D}[\Psi]\exp(iS[\Psi])$. \end{itemize} \end{frame}$

\begin{frame}{Extended Continuity Equation in Open Systems}
\small
\begin{itemize}
\item In open system, the particle number is not conserved.
\item Keldysh action is symmetry under classical phase rotation with terms like $\phi^_\sigma \phi_{\sigma’}\sigma \neq \sigma’U_qU_q$, if the partition function is invariant, then

\item Now consider the action (). After expansion and integrate by parts, we get:
\begin{equation}
\left\langle\partial_{\mu} j_{c}^{\mu}\right\rangle-\gamma_{p}\left\langle\psi_{+}^{} \psi_{-}\right\rangle+\gamma_{l}\left\langle\psi_{-}^{} \psi_{+}\right\rangle+4 u_{d}\left\langle\left(\psi_{-}^{} \psi_{+}\right)^{2}\right\rangle=0
\end{equation}
\item Here we have terms proportional to , and from the pumping and loss and are incompatible with .
\item In open systems, however, we can still add a dissipative current and measure the local mass flow by .
\end{itemize}
\end{frame}

$Unknown environment 'frame'\begin{frame}{Spontaneous Symmetry Breaking and the Goldstone Theorem} \small \begin{itemize} \item A finite average value $\left\langle\phi_{c}\right\rangle \neq 0$ breaks the $U_c$ symmetry and results in a massless mode, the Goldstone boson. \item The massless mode is a pole at $\omega = \mathbf{q} =0$ for $G^{R/A}$, as discussed before. \item Poles of the Green’s functions correspond to roots of $\det[G^{-1}(\omega, \mathbf{q})]$. So the existence of a zero mode means $\det M = 0$, with $M=-G^{-1}$. \item By detailed calculation, it is shown that $\det M=0$ with only $U_c$ invariance. This infers that $U_c$ invariance is sufficient for the existence of a massless mode. \item This tells us the Goldstone mode can occur in open systems with dissipation and loss as well. \end{itemize} \end{frame}$

Cameologist

有限温场论-Keldysh场论

有驱动开放系统的Keldysh场论

Basics of Keldysh Field Theory

Closed Time Contour

Keldysh Partition Function

Markovian Keldysh Action from the Master Equation