Quantum field theory 2, lecture 05

Inverse propagators as derivative operators

In the following we need an understanding of the physical meaning of two-point functions and their inverses, so it is a good point to recall this here.

Consider a quantum effective action of the form \[\Gamma[\Phi] = \int d^dz \sqrt{g}(z) \left\{ \frac{1}{2}g^{\mu\nu}(z) \partial_\mu \Phi_n(z) \partial_\nu \Phi_n(z) + U(\Phi(z)) \right\}\] A summation over field components with \(n=1,\ldots, N\) is implied here. Let us determine the inverse propagator first in position space, through the definition \[P_{jk}(x,y) = \left[ \frac{1}{\sqrt{g}(x)} \frac{\delta}{\delta \Phi_j(x)} \right] \left[ \frac{1}{\sqrt{g}(y)} \frac{\delta}{\delta \Phi_k(y)} \right] \Gamma[\Phi].\] We find from \(\delta\Phi_n(z)/\delta\Phi_j(x) = \delta^{(d)}(x-z)\delta_{nj}\) \[\begin{split} & P_{jk}(x,y) = \frac{1}{\sqrt{g}(x)}\frac{1}{\sqrt{g}(y)}\int d^dz \sqrt{g}(z)\\ & \times \left\{g^{\mu\nu}(z) \frac{\partial}{\partial z^\mu} \delta^{(d)}(x-z)\delta_{nj} \frac{\partial}{\partial z^\nu} \delta^{(d)}(y-z)\delta_{nk} +\frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k} \delta^{(d)}(x-z) \delta^{(d)}(y-z) \right\}. \end{split}\] Derivatives of Dirac distributions are again distributions and one can work with them well, albeit with some care. The derivatives with respect to \(z\) can be replaced by derivatives with respect to \(x\) and \(y\), and the integral over \(z\) and sum over \(n\) can then be done, leading to \[\begin{split} P_{jk}(x,y) = & \frac{1}{\sqrt{g}(x)}\frac{1}{\sqrt{g}(y)}\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial y^\nu}\int d^dz \sqrt{g}(z) \left\{g^{\mu\nu}(z) \delta^{(d)}(x-z)\delta_{nj} \delta^{(d)}(y-z)\delta_{nk} \right\} \\ & +\frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k} \frac{1}{\sqrt{g}(x)} \delta^{(d)}(x-y) \\ = & \delta_{jk}\frac{1}{\sqrt{g}(x)}\frac{1}{\sqrt{g}(y)}\frac{\partial}{\partial x^\mu}\frac{\partial}{\partial y^\nu} \sqrt{g}(x) g^{\mu\nu}(x) \delta^{(d)}(x-y) +\frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k} \frac{1}{\sqrt{g}(x)} \delta^{(d)}(x-y) \\ = & \frac{1}{\sqrt{g}(y)}\left\{\delta_{jk}\left[\frac{1}{\sqrt{g}(x)}\frac{\partial}{\partial x^\mu} \sqrt{g}(x) g^{\mu\nu}(x) \frac{\partial}{\partial x^\nu} \right] +\frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k} \right\} \delta^{(d)}(x-y). \end{split}\] The result is written as a distribution, with the expression in square brackets in the last line being the covariant Laplace-Beltrami operator, \[\Box_x = \frac{1}{\sqrt{g}(x)}\frac{\partial}{\partial x^\mu} \sqrt{g}(x) g^{\mu\nu}(x) \frac{\partial}{\partial x^\nu}.\]

Propagators as Greens functions

The propagator is now a Greens function to this derivative operator such that \[\begin{split} \int d^d z \sqrt{g}(z) P_{jn}(x,z) \Delta_{nk}(z,y) = & \left\{ \delta_{jn} \Box_x +\frac{\partial^2 U}{\partial\Phi_j\partial\Phi_n} \right\} \Delta_{nk}(x,y) \\ = & \frac{1}{\sqrt{g}(x)} \delta^{(d)}(x-y) \delta_{jk}. \end{split}\] As usual, the precise form of a Greens function is also determined by boundary conditions. We will discuss this in more detail later.

Dispersion relations

Speciallizing now to cartesian coordinates in Minkoski space, we can write both inverse propagator and propagator in a Fourier expansion \[\begin{split} P_{jk}(x,y) = & \int \frac{d^{d}p}{(2\pi)^d} e^{ip(x-y)} P_{jk}(p),\\ \Delta_{jk}(x,y) = & \int \frac{d^{d}p}{(2\pi)^d} e^{ip(x-y)} \Delta_{jk}(p),\\ \end{split}\] and find the algebraic relation \[P_{jn}(p) \Delta_{nk}(p) = \delta_{jk}.\] Propagating particles correspond to poles of the Greens function in momentum space \(\Delta_{jk}(p)\). These in turn correspond to vanishing eigenvalues of the inverse propagator \(P_{jn}(p)\). They can be detected through the characteristic equation \[\det(P_{jk}(p))=0,\] which encodes the dispersion relations.

Mass squared matrix

Concretely we find here \[P_{jk}(p) = \delta_{jk} p^2 + \frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k},\] and the vanishing eigenvalues, or poles of the relativistic propagator, occur when equations of the form \[p^2 + m^2 = -(p^0)^2 + \mathbf{p} + m^2 = 0,\] are fulfilled, where \(m^2\) is an eigenvalue of the mass square matrix \[M^2_{jk} = \frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k}.\] We will investigate this matrix in more detail below.

Correlation lengths in classical statistical field theory

For static classical statistical field theories one has in a similar way an inverse propagator or correlation function of the form \[P_{jk}(p) = \delta_{jk} \mathbf{p}^2 + M^2_{jk},\] and an eigenvalue \(m^2\) of the matrix \(M^2_{jk}\) encodes now the information of a correlation length for a specific field, \(\xi=1/m\). A Eucliden correlation function or propagator of the form \((\mathbf{p}^2 +m^2)^{-1}\) becomes large for \(\mathbf{p}^2\to 0\) if \(m^2=0\) and describes long range correlations. By analogy this is also called a massless field then. In contrast, for \(m>0\) the correlation length \(\xi=1/m\) is finite and correlations decay exponentially for distances much larger than the correlation length.

Energy gap in non-relativistic quantum field theories

For complex non-relativistic quantum fields, the inverse propagator is typically of the form \[\Delta_{jk}(p) = \delta_{jk} \left[-p^0 + \frac{\mathbf{p}^2}{2m}\right] + M^2_{jk}.\] Now an eigenvalue \(\nu\) of \(M^2_{jk}\) has the significance of an offset in the energy of particle – or quasi-particle – dispersion relation, \[p^0 = \frac{\mathbf{p}^2}{2m} + \nu.\] One should keep in mind that the overall energy scale in non-relativistic quantum mechanics has no physical significance, but relative energies do, and oftentimes \(\nu\) has the significance of an energy gap that needs to be overcome to create a specific excitation.

Goldstone bosons

Massless or almost massless fields play an important role in high-energy physics because they can be created with little energy and cannot decay for kinematic reasons. In classical statistical field theory they describe long range correlations. Finally, in condensed matter physics or non-relativistic quantum field theory the “massless modes” correspond to gapless excitations, i. e. those that can be excited without much energy cost. Accordingly they dominate thermodynamics and many material properties at small temperatures.

Spontaneous breaking of a continuous global symmetry always introduces massless particles or gapless modes. If the ground state leads to spontaneous breaking of a continuous global symmetry, massless scalar excitations have to be present. They are called “Goldstone bosons".

More specifically, this is a property of the matrix \[M^2_{jk} = \frac{\partial^2 U}{\partial\Phi_j\partial\Phi_k},\] and follows from symmetry arguments. Intuitively the reason is clear: A flat direction in the potential (valley in the Mexican hat potential) is dictated by invariance of \(U\) with respect to a continuous symmetry. This direction corresponds to a vanishing eigenvalue \(m^2\) of the matrix \(M^2_{jk}\).

For an example with \(U(1)\) symmetry we have \[U=\frac{1}{2}\lambda \left( \rho-\rho_0 \right)^2,\] with \[\rho=\Phi^*\Phi=\frac{1}{2}\left(\Phi_1^2 +\Phi_2^2 \right) ,\] where the real fields \(\Phi_1\), \(\Phi_2\), are related to the complex field \(\Phi=(\Phi_1 + i\Phi_2)/\sqrt{2}\). We can choose the expansion point \(\Phi_{1}= \sqrt{2}\Phi_0\), \(\Phi_{2}= 0\), \(\rho=\Phi^2_0\) since the direction of the expectation value is arbitrary. The mass matrix is now easily computed. With the first derivatives \[\begin{split} \frac{\partial U}{\partial\Phi_1}&= \frac{\partial U}{\partial \rho} \frac{\partial \rho}{\partial\Phi_1}=\lambda (\rho-\rho_0)\Phi_1,\\ \frac{\partial U}{\partial\Phi_2}&= \frac{\partial U}{\partial \rho} \frac{\partial \rho}{\partial\Phi_2}=\lambda (\rho-\rho_0)\Phi_2, \end{split}\] one obtains \[\begin{split} M_{11}^2&=\frac{\partial^2 U}{\partial\Phi_1^2}=\lambda (\rho-\rho_0)+ \lambda\Phi_1^2,\\ M_{22}^2&=\frac{\partial^2 U}{\partial\Phi_2^2}=\lambda (\rho-\rho_0)+ \lambda\Phi_2^2,\\ M_{12}^2&=M_{21}^2=\frac{\partial^2 U}{\partial\Phi_1 \partial\Phi_2}= \lambda\Phi_1\Phi_2. \end{split}\] The mass matrix for \(\rho=\rho_0\), \[M^2=\lambda\begin{pmatrix} \Phi_1^2& \Phi_1 \Phi_2\\ \Phi_1 \Phi_2& \Phi_2^2 \end{pmatrix},\] always has one vanishing eigenvalue. In particular, the evaluation at \(\Phi_1=\sqrt{2}\Phi_0\), \(\Phi_2=0\) yields a diagonal matrix \[M_{ab}^2= \begin{pmatrix} 2 \lambda\rho_0 & 0 \\ 0 & 0 \end{pmatrix}.\] The radial mode \(\Phi_1\) has mass squared \(m^2=2 \lambda \rho_{0}\), while the Goldstone mode \(\Phi_2\) is massless, \(m^2=0\).

The massless field is called “Goldstone boson”. This Goldstone boson emerges for example in connection with superfluid Helium-4 or ultracold Bose gases with repulsive interactions. For a non-relativistic spin zero complex field \(\varphi(x)\) the \(U(1)\) symmetry \(\varphi(x) \to e^{i \alpha}\varphi(x)\) is related to particle number conservation. The field equation for \(\varphi(x)\) is the Gross-Pitaevskii equation. For the relativistic case it is a Klein-Gordon equation with interaction, \[\Box_x\Phi(x) - \lambda \left[\Phi^*(x)\Phi(x) - \rho_{0}\right]\Phi(x) = 0.\]

More generally, the number of massless Goldstone modes that emerges in connection with spontaneous symmetry breaking is given by the number of Lie algebra generators of the symmetry group before spontaneous symmetry breaking minus the number of generators afterwards. Intuitively this is the number of flat directions of the effective potential, and the number of vanishing eigenvalues of the matrix \(M^2_{jk}\).

Gauged scalar field

Let us now generalize our considerations to a field with electromagnetic charge such that the action \(\Gamma[\Phi,A]\) is invariant under local \(\text{U}(1)\) transformations, \[\begin{split} \Phi(x) & \to e^{i\alpha(x)}\Phi(x),\\ A_\mu(x) & \to A_\mu(x) + \frac{1}{e} \partial_\mu \alpha(x). \end{split}\] This means that derivatives of the complex fields appear in terms of covariant derivatives \[D_\mu\Phi(x) = [\partial -ieA_\mu(x)]\Phi(x).\] What happens now with spontaneous symmetry breaking?

Let us consider the following effective action for scalar QED, playing here the role of a Landau theory, \[\Gamma[\Phi,A] =\int_x \left\{ D^{\mu}\Phi^*D_{\mu}\Phi+\frac{\lambda}{2}(\Phi^*\Phi-\rho_{0})^2+\frac{1}{4} F^{\mu\nu}F_{\mu\nu} \right\}.\]

Now assume first a constant background scalar field \(\Phi(x)=\Phi_{0}\), and we can take \(\Phi_{0}\) to be real without loss of generality. For this configuration of the scalar field we find \(D_\mu \Phi = -ieA_\mu\Phi_0\), and \(D^\mu\Phi^* D_\mu\Phi = e^2 \Phi_0^2 A^\mu A_\mu\). But this looks like a mass term for the photons! Indeed, we can evaluate \(F^{\mu\nu}F_{\mu\nu} = 2 \partial^\mu A^\nu \partial_\mu A_\nu - 2 \partial^\nu A^\mu \partial_\mu A_\nu\). With two partial integrations the second term becomes \((\partial_\mu A^\mu)^2\) and when we work in Landau gauge where \(\partial_\mu A^\mu=0\) we can drop it. The part of the action depeding on the gauge field becomes then with another partial integration \[\Gamma[A] = \int_x \left\{ \frac{1}{2} A^\nu(x) \left[-\partial^\mu \partial_\mu + 2 e^2 \Phi_0^2 \right] A_\nu(x) \right\}.\] This is the action of a massive vector field!

Higgs mechanism: photon mass

The field equation for the photon field replaces Maxwells equations in vacuum. In momentum space it reads \[\left[p^2+2e^2 \Phi^2_0\right] A_\nu(p)=0.\] The solutions are plane waves with \(p^2\) obeying \[p^2+m^2= -(p^0)^2 + \mathbf{p}^2 + m^2 = 0,\] where \(m^2=2e^2 \Phi_0^2\) is an effective photon mass squared.

This is an example for the Higgs mechanism: the photon acquires a mass term through the spontaneous breaking of a gauge symmetry!