### Lesson from Condensed Matter Physics (CMP)

In condensed matter systems, it is useful to study the*collective response*to external fields. For example, when adding an electric field $\vec E$ to a metal, a current $\vec J$ would be generated. We call the response the

*conductivity*, which is unique for each material, thus reflects internal properties of the system. Quantitatively, the conductivity is defined in terms of the external field and the respond current \[

J(t, \vec x) = \int \mathrm d^3x' \int_{-\infty}^t \mathrm dt' \sigma(t-t', \vec x - \vec x') E(t', \vec x').

\] The upper limit of the temporal integral is due to causality. In general, the conductivity may depends on the external field $\vec E$ (non-linear effect). In the weak field limit $E \to 0$, however, it is independent to $E$ (linear response) and provides an important probe for the property of the material. There are also other linear responses of the condensed matter systems. The Green-Kubo formula (Melville Green 1954, Ryogo Kubo 1957) provides a neat recipe to compute the linear response of the condensed matter system to an external field.

Suppose a weak external field $F(t, \vec x)$ couples to a condensed matter system through $H_\mathrm{int} = \int \mathrm d^3x F(t, \vec x) A(t, \vec x)$. The system was originally described by the density operator $\varrho =Z^{-1} \exp \big[ -\beta (H - \mu N) \big]$, where $Z = \mathrm{tr} \exp \big[ -\beta (H - \mu N) \big]$. The linear response theory concerns the expectation value of an observable $B(t, \vec x)$: \[

\langle B(t, \vec x) \rangle_F - \langle B(t, \vec x) \rangle= \mathrm{tr} \left[ \big(\varrho_F(t)-\varrho(t)\big) B(t, \vec x) \right] \equiv \int \mathrm dt' \int \mathrm d^3x' \chi_{AB}(t-t',\vec x-\vec x') F(t', \vec x')

\] where $\varrho_F(t)$ is the density matrix in the perturbed system. The Green-Kubo formula asserts that the linear transportation coefficient \[

\chi_{AB}(t-t', \vec x - \vec x') = -i \Theta(t-t') \langle [B(t, \vec x), A(t', \vec x')] \rangle.

\] For example, the conductivity is related to the current-current correlation function*, \[

\sigma_{AB}(t-t', \vec x - \vec x') = -i \Theta(t-t') \langle [\partial_t^{-1}J(t, \vec x), \partial_{t'}^{-1}J(t', \vec x')] \rangle.

\]

*

*In general, the current is a four-vector hence the response is a tensor. One should consider the contribution from all components, which gives rise to several transport coefficients: the longitudinal conductivity, the density responses as well as the Hall conductivity etc*.

### Quantum Field Theory (QFT)

The vacuum state of QFT $|\Omega\rangle$ is intrinsically many-body (even the free field theory!). Let's disturb the QFT vacuum and measure the linear response.*classical source*. We couple it to the quantum field $\varphi(x)$ through $\mathscr H_\mathrm{int}(x) = A_r(x) J_r(x)$, where $A$ is a local operator constructed from $\varphi$. Then we measure the vacuum expectation value (VEV) of an observable $B_s(x)$ (To measure this observable, we may, for example, couple the field to a classical field. That's why we often consider the case $A = B$, which we sill talk about later.): $\langle B_s(x) \rangle_J \equiv Z_J^{-1} \langle \Omega | B_s(x) | \Omega \rangle_J$, where $Z_J = \langle \Omega | \Omega\rangle_J$ is the perturbed partition function. In the weak source limit ($J\to 0$), the response should be linear, similar to the condensed matter system: \[

\langle B_s(x) \rangle = \int \mathrm d^4 x' \chi_{sr}^{BA}(x,x') J_r(x') + \mathcal O(J^2).

\] Let $Z$ be the unperturbed partition function. In the weak external field limit $J \to 0$, \[

Z_J = Z \Big(1 - i \int \mathrm d^4x \langle A_r(x) \Omega \rangle J_r(x) - \frac{1}{2} \int \mathrm d^4x\mathrm d^4y \langle \mathcal T\big\{ A_r(x) A_s(y) \big\} \rangle J_r(x) J_s(y) \Big).

\] and $\langle \Omega | B_s(x) | \Omega \rangle_J$ is, \[

\langle \Omega | B_s(x) | \Omega \rangle_J = \langle \Omega | B_s(x) | \Omega \rangle

-i \int \mathrm d^4 x' \langle \Omega | \mathcal T \big\{B_s(x)A_r(x')\big\} | \Omega\rangle J_r(x').

\] Then the vacuum expectation value (VEV) \[

\langle B_s(x) \rangle_J

= \langle B_s(x) \rangle + i \int \mathrm d^4x' \langle A_r(x') \rangle \langle B_s(x) \rangle J_r(x') - i \int \mathrm d^4 x' \langle \mathcal T \big\{B_s(x) A_r(x')\big\} \rangle J_r(x').

\] It is convenient to work with "renormalized" operators that have vanishing VEV without the presence of the external source. For example, we may define: $B^R_s(x) = B_s(x) - \langle B_s(x)\rangle$. It is easy to see,\[

\langle B^R_s(x) \rangle_J

=- i \int \mathrm d^4 x' \langle \mathcal T \big\{B^R_s(x) A^R_r(x')\big\} \rangle J_r(x').

\] Therefore, the transport coefficient $\chi(x,x') \propto \langle \mathcal T \big\{B^R_s(x) A^R_r(x')\big\} \rangle$. Note that causality is preserved by time-ordering operator $\mathcal T$.

From now on, we'll assume all the operators have been properly renormalized, unless elsewhere stated.

#### Example 1: Field Propagation

$A = B = \varphi$, where $\varphi(x)$ is the renormalized field such that $\langle \Omega |\varphi(x)|\Omega\rangle = 0$. $\langle\varphi(x)\rangle_J$ represents the amplitude for finding a physical particle in the disturbed vacuum. \[\langle\varphi_a(x)\rangle_J =- i \int \mathrm d^4 x' \langle \mathcal T \big\{\varphi_a(x) \varphi_b(x')\big\} \rangle J_b(x'),

\] Here $D_{ab}(x-x') \equiv i\langle \mathcal T \big\{\varphi_a(x) \varphi_b(x')\big\} \rangle$ is nothing but the Feynman propagator. This means sense physically: the classical source $J$ creates a physical particle at $x'$, and then the particle propagate to $x$ to be detected.

Note that we are not doing perturbation theory. The graphical representations are not necessarily Feynman diagrams.

#### Example 2: Vacuum Polarization

Let $A = B = J^\mu(x)$, where $J^\mu$ is the electromagnetic current. We couple a classical electromagnetic field $\mathcal A_\mu(x) e^{-\epsilon |x^0|}, (\epsilon\to 0^+)$ to the vacuum and measure the current: \[\langle J^\mu(x) \rangle_{\mathcal A} = -i \int \mathrm d^4x' \left< \mathcal T \big\{J^\mu(x)J^\nu(x') \big\} \right> \mathcal A_\nu(x') e^{-\epsilon |x'^0|}.

\] The linear transport coefficient is called the polarization tensor: \[

\Pi^{\mu\nu}(x-x') \equiv \left< \mathcal T \big\{J^\mu(x)J^\nu(x') \big\} \right>.

\] Consider a

*free Dirac field*, \[

\psi(x) = \sum_{s=\pm} \int\frac{\mathrm d^3 k}{(2\pi)^32\omega_p} \Big[ u_s(k) b_s(k) e^{ik\cdot x} + v_s(k) d^\dagger_s(k) e^{-ik\cdot x} \Big].

\] The electromagnetic current is $J^\mu(x) = \bar\psi(x)\gamma^\mu\psi(x)$. Applying Wick theorem, the polarization tensor is \[

\Pi^{\mu\nu}(x-x') = \mathrm{tr} \Big[ \gamma^\mu S(x-x') \gamma^\nu S(x-x') \Big] - \mathrm{tr} \Big[\gamma^\mu S(x-x)\Big] \mathrm{tr} \Big[\gamma^\nu S(x'-x')\Big].

\]

Now consider a

*perturbative spinor electrodynamics*. Let's denote the free Dirac action as $S_0$, the

interaction as $S_\mathrm{int} = \int\mathrm d^4x \bar\psi(x)\gamma^\mu\psi(x) A_\mu(x)$.

Here we are only concerning the linear effect. One may well ask the question of the induced current by a strong classical source field. The problem is called the

*Schwinger effect*. It turns out, in the semi-classical approximation, the partition function is \[

Z_\mathcal{A} \equiv \langle \Omega | \Omega \rangle_\mathcal{A} \approx e^{iS_\mathrm{eff}} \] Therefore, the pair production probability (or rather the vacuum decay probability) \[

P = 1 - e^{-2\mathrm{Im}S_\mathrm{eff}}. \] Considering only the one-loop effect in a constant E-field, Schwinger calculated the vacuum decay rate (Phys. Rev.

**82**, 664 (1951)), \[

\frac{\mathrm dN}{\mathrm dV \mathrm dt} = \frac{(eE)^2}{4\pi^3}\sum_{n=1}^\infty \frac{1}{n^2} e^{-n\pi E_c/E} \] where $E_c = \frac{m_e^2c^3}{e\hbar} \sim 10^{18} \text{V/m}$ is a super-duperly strong field! However, it may be found in: a) heavy ion collision; b) magnetar; c) condensed matter emulated QED (e.g., graphene); d) high energy lasers (still a long way to go).

#### Example 3: Hadron Structure

Let $A = B = J^\mu(x)$, where $J^\mu$ is the electromagnetic current. The linear response can also be used to study a bound state. The key is to "create" and "annihilate" a bound state from the vacuum with the field operator, \[|\psi(z)\rangle \equiv \lim_{z^0\to-\infty}\psi(z) |\Omega\rangle, \\

\langle\psi(y) | \equiv \lim_{y^0\to+\infty} \langle \Omega | \bar\psi(y).

\] We couple a classical electromagnetic field $\mathcal A_\mu(x) e^{-\epsilon |x^0|}, (\epsilon\to 0^+)$ to a physical particle through the minimal coupling $\mathcal A_\mu J^\mu$. Then we measure the charge distribution: \[

\langle \psi(y) | J^\mu(x) |\psi(z)\rangle_{\mathcal A} = -i \int \mathrm d^4x' \left< \psi(y)\right| \mathcal T \big\{J^\mu(x)J^\nu(x') \big\} \left| \psi(z)\right> \mathcal A_\nu(x') e^{-\epsilon |x'^0|}

\] The particle propagation before and after the experiment is no interest to us. Let's do Fourier transform:\[

\psi(z) |\Omega\rangle = \int \frac{\mathrm d^3 p}{(2\pi)^32\omega_p} \tilde\psi(p) e^{ip\cdot x} |p,\sigma\rangle, \quad (\omega_p = \sqrt{m^2+\mathbf p^2})

\] Therefore, let's study the current distribution of plane wave modes: \[

\langle p',\sigma' | J^\mu(x) | p,\sigma \rangle_{\mathcal A} = -i \int \mathrm d^4x' \langle p',\sigma' | \mathcal T \big\{J^\mu(x)J^\nu(x') \big\} | p,\sigma \rangle \mathcal A_\nu(x') e^{-\epsilon |x'^0|}

\]

#### Beyond Linear Response

The linear response can be use to formulate the perturbation theory. Let's disturb a scalar field $\varphi(x)$ with a classical source $j$. The new partition function is, \[\begin{split}

Z_j &= \int \mathcal D_\varphi \, \exp \Big( iS + i\int\mathrm d^4x\, j(x) \varphi(x) \Big)

=\langle\Omega\mid \mathcal T \exp\Big( i\int \mathrm d^4x\, j(x)\varphi(x)\Big) \mid \Omega\rangle\\

&= \sum_{n=0}^\infty \frac{i^n}{n!}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) \langle\Omega\mid\mathcal T\varphi(x_1)\varphi(x_2)\cdots \varphi(x_n)\mid\Omega\rangle \\

&= \sum_{n=0}^\infty \frac{i^n}{n!}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) G(x_1, x_2, \cdots, x_n)

\end{split}

\] where $G(x_1,x_2,\cdots,x_n) = \langle\Omega\mid\mathcal T\varphi(x_1)\varphi(x_2)\cdots \varphi(x_n)\mid\Omega\rangle$ is the $n$-point correlation function (aka. $n$-point causal correlator). The translational symmetry implies $G(x_1,\cdots,x_n) = G(x_1-a,\cdots,x_n-a)$. In the momentum space, \[

G(x_1, \cdots, x_n) = \int\frac{\mathrm d^4p_1}{(2\pi)^4} \cdots \frac{\mathrm d^4p_n}{(2\pi)^4} \exp\Big( ip_1\cdot x_1+\cdots + ip_n\cdot x_n\Big) G(p_1,\cdots,p_n)

\] The translational symmetry implies $G(p_1,\cdots,p_n) = (2\pi)^4\delta^4(p_1+\cdots+p_n)\tilde G(p_1,\cdots,p_n)$.

The correlators $G(x_1,x_2,\cdots,x_n)$ can be represented by the sum of graphs subject to $n$ external legs, $G(x_1,\cdots,x_n) = \sum_{g} D_g(x_1,\cdots, x_n)$, known as the Feynman diagrams.

A diagrammatic representation of a n-point causal correlator |

Next, it is convenient to work with the irreducible diagrams (represented by connected graphs), $C$, and the corresponding correlator $G_c(x_1,x_2,\cdots,x_n)$. Apparently, any diagram $D_g$ with topology $g$ can be written as a product of its connected components, \[

\frac{1}{n_g!} \frac{i^{n_g}}{|\mathrm{Aut}\,g|}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_{n_g}\, j(x_1) \cdots j(x_{n_g}) D_g(x_1, \cdots, x_{n_g}) \\

= \prod_{s\in S_g} \frac{1}{m_s!} \bigg( \frac{1}{n_s!}\frac{i^{n_s}}{|\mathrm{Aut}\, s|}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_{n_g}\, j(x_1) \cdots j(x_{n_g}) D_s \bigg)^{m_s}

\] where $S_g$ be the set of connected subgraphs of graph $g$, $m_s$ is the multiplicity of $s\in S_g$ in $g$, that is, $g$ consists of $m_s$ copies of $s$; $n_s$ evaluates the number of external legs in graph $s$ and $n_g = \sum_{s\in S_g} m_s n_s$. The factor $\frac{1}{n_s!}$ came from the multinomial coefficients ${n_g \choose n_{s_1}, n_{s_2}, \cdots}}$, which represents the number of ways to split the $n_g$ external sources.

Then the disturbed partition function $Z_j$, \[

\begin{split}

Z_j &= \sum_{n=0}^\infty \frac{i^n}{n!}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) G(x_1, \cdots, x_n) \\

&= \sum_{n=0}^\infty \frac{i^n}{n!} \int\mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) \sum_{g, n_g=n} \frac{1}{|\mathrm {Aut}\,g| } D_g(x_1, \cdots, x_n) \\

&= \sum_{g} \prod_{s\in S_g} \frac{1}{m_s!} \bigg(\frac{1}{n_s!}\frac{i^{n_s}}{|\mathrm{Aut}\, s|} \int\mathrm d^4x_1\,\cdots\mathrm d^4x_{n_s}\, j(x_1) \cdots j(x_{n_s}) D_s \bigg)^{m_s} \\

\end{split}

\] In the last line, $s \sim \frac{1}{n_s!}\frac{i^{n_s}}{|\mathrm{Aut}\, s|} \int\mathrm d^4x_1\,\cdots\mathrm d^4x_{n_s}\, j(x_1) \cdots j(x_{n_s}) D_s$ is the expression for connect graph $s$. Compare the last line with multinomial expansion $(x_1+x_2+\cdots + x_k)^n = \sum_{m_1,m_2,\cdots,m_k} {n \choose m_1,m_2,\cdots,m_k} x_1^{m_1}x_2^{m_2}\cdots x_k^{m_k}$. Therefore, we can change the order of summation and multiplication and \[

Z_j = \exp \sum_{n=0}^\infty \frac{i^n}{n!} \int \mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) \sum_{g\in C, n_c=n} D_c(x_1,x_2,\cdots,x_n).

\] Here $C$ is the collection of all connected Feynman diagrams. This result is the

**linked cluster theorem**, which relates the free energy with the connect diagrams, \[

iW_j = \ln Z_j = \sum_{n=0}^\infty \frac{i^n}{n!}\int\mathrm d^4x_1\,\cdots\mathrm d^4x_n\, j(x_1) \cdots j(x_n) G_c(x_1, x_2, \cdots, x_n).

\] This theorem can be proven more rigorously from induction, by defining the connected correlator $G_c(x_1, \cdots, x_n) = G(x_1, \cdots, x_n) - \sum_{P}\prod_{p\in P} G_c(\{x\}_p)$.

Consider the vacuum expectation value (VEV) of the field $\varphi(x)$, \[

\langle \varphi(x) \rangle_j = Z_j^{-1} \frac{1}{i}\frac{\delta}{\delta j(x)} Z_j

= \frac{\delta}{\delta j(x)} W_j

\] Let's do a Legendre transformation to introduce a new quantity (the minus sign is the convention): \[

-\Gamma_j = \int \mathrm d^4x\, \delta/\delta j(x) W_j \cdot j(x) - W_j

\]