## Oct 29, 2014

### Lesson from Condensed Matter Physics (CMP)

In condensed matter systems, it is useful to study the response to external fields. For example, when a metal is imposed to an electric field $\vec E$, a current $\vec J$ is generated. The response reflects a unique property of the material, known as the conductivity, which is defined in terms of the external field and the response $J(t, \vec x) = \int \mathrm d^3x' \int_{-\infty}^t \mathrm dt' \sigma(t-t', \vec x - \vec x') E(t', \vec x').$ 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, $\sigma_{AB}(t-t', \vec x - \vec x') = -i \Theta(t-t') \langle [J(t, \vec x), J(t', \vec x')] \rangle.$

### 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.

Let $J_r(x)$ be a 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. 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, $\langle B^R_s(x) \rangle_J = - i \int \mathrm d^4 x' \left< \mathcal T \big\{B^R_s(x) A^R_r(x')\big\} \right> J_r(x').$ 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).

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