## Oct 29, 2014

### 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.

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. 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).

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

## Oct 23, 2014

### Cosmology and the First Law of Thermal Dynamics

According to the standard cosmology, our universe is expanding. During the process, the content of the Universe is also changing. For example, about 13.7 billion years ago, the Universe was dominated by dark matter; however, today it is the dark energy that rules the sky. According to the standard cosmological model, Lambda-CMD, the energy density of the dark energy $\rho_\Lambda = \frac{\Lambda}{8\pi G}$ ($\Lambda$ is Einstein's cosmological constant) does not change during the expansion of the Universe. That is why the dark energy is increasing as the Universe becomes larger. There is no known  interaction between the dark energy and the ordinary matter (including dark matter) except for gravity - which according to the general relativity is merely the curvature of the Universe. There is no exchange of energy. So where does this large amount of energy come from? Does the first law of thermal dynamics (the law of energy conservation) still hold? Given the "dark" nature of the dark energy, one may turn to consider physical matters, such as photons - the cosmic microwave background radiation (CMB). As the Universe expands, the photons are redshifted, i.e., their frequencies $\nu$ become smaller $\nu/(1+z), z>0$. Since the Universe is extremely empty, the loss of photons due to reaction is negligible. Therefore, the total energy of the radiation $E = N_\gamma h \nu$ would decrease as the Universe expands. So, again, where does the energy go?

 Fig. 1, Left: an artist's impression of the chronology of the Universe. Right: the contents of the Universe according to WMAP project. Credit: NASA / WMAP Science Team
The answer is in the Friedmann's equations themselves. Recall the Robertson-Walker metric $\mathrm ds^2 = \mathrm dt^2 - a^2(t) \left( \frac{\mathrm dr^2}{1-k r^2} + r^2 \mathrm d\theta^2 + r^2 \sin^2\theta\mathrm d\phi^2 \right)$ where, $k$ represent the curvature of the space, and $a(t)$ is the scale factor that satisfies the Friedmann's equations state, $\begin{split} \left(\frac{\dot a}{a} \right)^2 &= \frac{8\pi G}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2} \\ \frac{\ddot a}{a} &= -\frac{4\pi G}{3} (\rho + 3p) + \frac{\Lambda}{3} \end{split}$ Take the derivative on the first equation and substitute the second one into it and then we get, $\dot \rho = -3\frac{\dot a}{a} (\rho + p ) \Rightarrow \mathrm d\rho = -3\frac{\mathrm da}{a}(\rho + p).$ The volume of the Universe expands as $V = V_0 a^3$ or $\frac{\mathrm dV}{V} = 3 \frac{\mathrm da}{a}$. Here $\rho$ is the energy density of the matters (baryon matter, radiation and dark matter etc). Then the internal energy is $U = \rho V$ or $\mathrm dU = \mathrm d\rho V + \rho \mathrm dV$. Plug in the expression for $\mathrm dV$ and $\mathrm d\rho$. We arrive, $\mathrm dU + p \mathrm dV = 0!$ This is precisely the first law of thermal dynamics (for matters) for an adiabatic expansion. Therefore, the loss of radiation energy is due to the work of the CMB ratiation done to the Universe, which makes the Universe expands.

Now, what about the dark energy? Where does the dark energy come from? Simply consider the internal energy of the dark energy: $\mathrm d U_\Lambda = \rho_\Lambda \mathrm d V$. This is because the energy density of the dark energy is constant. Add the work term, $\mathrm dU_\Lambda + p_\Lambda \mathrm dV= (\rho_\Lambda + p_\Lambda) \mathrm dV$. Now we need a relation that relates the energy density and the pressure, which is called the equation of state (EoS). To obtain the EoS for dark energy, we can look at the second Friedmann's equation. If the cosmological constant is treated as some energy, the right-hand side of the equation should be written as $\frac{\ddot a}{a} = -\frac{4\pi G}{3} (\rho + 3p) -\frac{4\pi G}{3} (\rho_\Lambda + 3p_\Lambda)$ Namely, $4\pi G (\rho_\Lambda + 3p_\Lambda) = -\Lambda$. But $\rho_\Lambda = \frac{\Lambda}{8\pi G}$. Therefore, $p_\Lambda = - \rho_\Lambda$. This is the EoS for the dark energy, which says the dark energy has negative pressure! Turn back to the thermal dynamical equation, clearly $\mathrm dU_\Lambda + p_\Lambda \mathrm dV = 0$. Again, the dark energy also obeys the first law of thermal dynamics. It turns out the dark energy was created from the negative pressure and the work done by the Universe.