--- jupytext: formats: ipynb,md:myst,py:light text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.13.8 kernelspec: display_name: Python 3 language: python name: python3 --- ```{code-cell} :tags: [hide-cell] import mmf_setup; mmf_setup.nbinit() import os from pathlib import Path FIG_DIR = Path(mmf_setup.ROOT) / '../Docs/_build/figures/' os.makedirs(FIG_DIR, exist_ok=True) import logging; logging.getLogger("matplotlib").setLevel(logging.CRITICAL) %matplotlib inline import numpy as np, matplotlib.pyplot as plt try: from myst_nb import glue except: glue = None ``` (sec:GreensFunctions)= # 10. Green's Functions The coverage of Green's functions in {cite}`Arfken:2013` is quite reasonable. Here are some highlights presented from a different perspective, analogous to that in {cite}`Hassani:2013`. I highly recommend comparing these two approaches -- learning to translate from vectors and matrices to differential equations is an extremely valuable skill. The general idea is to find particular solutions to a linear differential equation. Recall that this is a functional equivalent of solving a linear system: \begin{gather*} \mathcal{L} y(x) = f(x), \qquad \mat{L}\ket{y} = \ket{f}. \end{gather*} Green's functions are the functional equivalent of inverting $\mat{L}$: \begin{gather*} \mathcal{L}G(x, y) = \delta(x-y), \qquad \mat{L}\mat{G} = \mat{1}\\ y(x) = \int G(x, y)f(y)\d{y}, \qquad \ket{f} = \mat{G}\ket{f}. \end{gather*} :::::{admonition} $G(x,y) = G(x-y)$ as a [Toeplitz matrix][]. :class: dropdown Green's functions are often presented in the form \begin{gather*} \mathcal{L}G(x) = \delta(x) \end{gather*} which obscures the matrix nature of $G$. This can be done when $\mathcal{L}$ is independent of $x$ -- i.e., translationally invariant. In this case \begin{gather*} G(x, y) = G(x-y). \end{gather*} This corresponds to a [Toeplitz matrix][] or diagonal-constant matrix $\mat{G}$ where the entries depend only on their distance from the diagonal. I.e. the diagonal has one constant value, the next diagonal has another constant value etc. \begin{gather*} [\mat{G}]_{ab} = g(a - b). \end{gather*} [Toeplitz matrices][Toeplitz matrix] have many nice properties, including the fact that they are diagonalized by plane waves \begin{gather*} \braket{m|f_n} = e^{2\pi I m n/N}. \end{gather*} This is why Fourier techniques are so useful for finding Green's functions. :::{doit} Show that $\mat{G}\ket{f_m} = \ket{f_m}\lambda_m$ if $\mat{G}$ is a [Toeplitz matrix][]. :class: dropdown *(Incomplete)* ::: While this often simplifies things, it is best to keep in mind the nature of Green's functions as matrices, not as vectors. ::::: ```{code-cell} :tags: [margin, hide-input] beta = 0.4 w0 = 0.5 ws = wp, wm = 1j*beta + np.sqrt(w0**2 - beta**2)*np.array([1, -1]) eps = 0.01 fig, ax = plt.subplots() ax.spines[:].set_visible(False) r = 0.05 th = 0.2 z0 = -1+0.01j z1 = wm.real+0.01j + r*np.exp(1j*(-th - np.pi/2)).real z2 = wm + r*np.exp(1j*(-th - np.pi/2)) z23 = wm + r*np.exp(1j*(-np.linspace(th, 2*np.pi - th)-np.pi/2)) z3 = wm + r*np.exp(1j*(+th - np.pi/2)) z4 = wm.real+0.01j + r*np.exp(1j*(+th - np.pi/2)).real z5 = wp.real+0.01j + r*np.exp(1j*(-th - np.pi/2)).real z6 = wp + r*np.exp(1j*(-th - np.pi/2)) z67 = wp + r*np.exp(1j*(-np.linspace(th, 2*np.pi - th)-np.pi/2)) z7 = wp + r*np.exp(1j*(+th - np.pi/2)) z8 = wp.real+0.01j + r*np.exp(1j*(+th - np.pi/2)).real z9 = 1.0+0.01j zs = np.concatenate([[z0, z1, z2], z23, [z3, z4, z5, z6], z67, [z7, z8, z9]]) ax.plot(ws.real, ws.imag, 'xk') ax.plot(zs.real, zs.imag,'-C0') ax.axhline([0], ls='-', lw=0.2, c='k') ax.axvline([0], ls='-', lw=0.2, c='k') ax.set(xticks=[], yticks=[], xlim=(-1,1), ylim=(-0.1, 0.6), aspect=1); ``` :::{margin} Contour used to obtain the acausal (advanced) Green's function for a damped harmonic oscillator. ::: :::::{doit} Damped Harmonic Oscillator $\mathcal{L}=\partial_{tt}+2\beta\partial_{t}+\omega_0^2$. Find the general Green's function for the damped harmonic oscillator. Recall that $\mathcal{L}$ is independent of time (invariant under shifts of time), so we can write $G(t, \tau) = G(t-\tau)$, thus find the general homogeneous solutions $h_1(t)$ and $h_2(t)$ such that \begin{gather*} G(t) = \begin{cases} h_{1}(t), & t < 0\\ h_{2}(t), & t > 0 \end{cases}, \qquad \mathcal{L}G(t) = \delta(t). \end{gather*} Note that there are many such solutions since $G(t) + h(t)$ is also a valid Green's function. Find the causal Green's function where $G_{+}(t < 0) = 0$. :::{solution} Here we solve the slightly simpler case without damping where $\beta = 0$. The homogeneous solutions are simply $e^{\pm \I \omega_0 t}$ so we have \begin{gather*} G(t) = \begin{cases} a_{+}e^{\I \omega_0 t} + a_{-}e^{-\I \omega_0 t}, & t < 0,\\ b_{+}e^{\I \omega_0 t} + b_{-}e^{-\I \omega_0 t}, & t > 0. \end{cases} \end{gather*} $G(t)$ must be continuous at $t=0$, so $a_+ + a_- = b_+ + b_-$. Likewise, the derivatives must be discontinuous with a unit step so that we obtain $\delta(t)$: \begin{gather*} \I\omega_0(b_{+} - b_{-}) - \I\omega_0(a_{+} - a_{-}) = 1. \end{gather*} We can express these conditions as $\braket{D_0|A} = 0$ and $\braket{D_1|A} = 1$ in terms of the following vectors \begin{gather*} \ket{D_0} = \begin{pmatrix} +1 \\ +1 \\ -1 \\ -1\\ \end{pmatrix}, \qquad \ket{D_1} = -\I\omega_0 \begin{pmatrix} -1 \\ +1 \\ +1 \\ -1\\ \end{pmatrix}, \qquad \ket{A} = \begin{pmatrix} a_+\\ a_-\\ b_+\\ b_-\\ \end{pmatrix}. \end{gather*} Noting that \begin{gather*} \braket{D_0|D_0} = 4, \qquad \braket{D_1|D_1} = 4\omega^2, \qquad \braket{D_0|D_1} = 0, \end{gather*} we can write the solution as \begin{gather*} \ket{A} = \frac{\ket{D_1}}{\braket{D_1|D_1}} + \ket{B} \end{gather*} where $\ket{B}$ is any vector orthogonal to the subspace $\{\ket{D}_0, \ket{D}_1\}$. This subspace is easily constructed by inspection, but could be found using the Gram-Schmidt procedure (QR decomposition): \begin{gather*} \ket{A} = \frac{1}{4\I\omega_0} \begin{pmatrix} -1 \\ +1 \\ +1 \\ -1 \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} +1 \\ -1 \\ +1 \\ -1 \end{pmatrix}. \end{gather*} To find the causal Green's function, we want $a_+ = a_- = 0$. This requires $b_+ = -b_- = b$ and then $2\I \omega_0 b = 1$ \begin{gather*} G_{+}(t) = \Theta(t)\frac{e^{\I\omega_0 t} - e^{-\I\omega_0 t}}{2\I\omega_0} = \Theta(t)\frac{\sin(\omega_0 t)}{\omega_0}. \end{gather*} The full solution can be found most simply using Fourier techniques and contour integrals as follows. Express \begin{gather*} G(t) = \mathcal{F}^{-1}(\tilde{G}_\omega) = \int \frac{\d{\omega}}{2\pi} e^{\I \omega t} \tilde{G}_{\omega}, \qquad \delta(t) = \mathcal{F}^{-1}(1) = \int \frac{\d{\omega}}{2\pi} e^{\I \omega t}. \end{gather*} Then, taking the Fourier transform of $\mathcal{L}G = \delta$ gives \begin{gather*} \mathcal{F}(\mathcal{L}G) = (-\omega^2 + 2\I \beta \omega + \omega_0^2)\tilde{G}_\omega = 1, \\ \tilde{G}_{\omega} = \frac{-1}{(\omega - \omega_{-})(\omega - \omega_{+})}, \qquad \omega_{\pm} = \I \beta \pm \sqrt{\omega_0^2 - \beta^2}. \end{gather*} The Green's function can be computed by a contour integral \begin{gather*} G(t) = \mathcal{F}^{-1}(\tilde{G}_\omega) = \int_{-\infty}^{\infty}\frac{\d{\omega}}{2\pi} \frac{-e^{\I \omega t}}{(\omega - \omega_{-})(\omega - \omega_{+})} = \oint_{C}\frac{\d{\omega}}{2\pi \I} \frac{-\I e^{\I \omega t}}{(\omega - \omega_{-})(\omega - \omega_{+})} \end{gather*} where the contour is closed along a circular arc going to infinity. To ensure that the additional contributions vanish (justifying the last equality), we must close the contour up if $t > 0$ and down if $t<0$ to ensure that the exponential factor decays (see Jordan's lemma). If $0 < \beta$, then both poles like in the upper half plane, so the integral is zero if $t<0$ and the sum of both residues if $t>0$: \begin{gather*} G(t) = \Theta(t)\left( \frac{-\I e^{\I \omega_- t}}{\omega_- - \omega_+} + \frac{-\I e^{\I \omega_+ t}}{\omega_+ - \omega_-}\right) = \Theta(t) \frac{e^{\I \omega_+ t} - e^{\I \omega_- t}}{\I(\omega_+ - \omega_-)}\\ = \Theta(t) \frac{e^{-\beta t}\sin(\bar{\omega}t)}{\bar\omega}, \qquad \bar{\omega} = \sqrt{\omega_0^2 - \beta^2}, \end{gather*} which approaches our previous solution as $\beta \rightarrow 0$. If we strictly have $\beta = 0$, then the two poles like on the real axis. In this case, we must choose how we go around them. There are four choices here: going below both gives the causal (retarded) Green's function as above. Going over both gives the acausal (advanced) Green's function that is zero for $t>0$. Going over one and under the gives a Green's function that is non-zero for all times. These mixed forms are often better suited for perturbation theory and play a key role in computing Feynman diagrams. Does the previous calculation mean that there is only one Green's function if $\beta > 0$? No: there are always others we can find by suitably deforming the contour as shown on the right. ::: ::::: Although formally $\mat{G} = \mat{L}^{-1}$, the formalism presented this way works even if $\mat{L}$ is singular. In particular, we are generally faced with a problem where $\mathcal{L} h(x) = 0$ has homogeneous solutions. ## Self-Adjoint Operators We now explain {cite}`Arfken:2013` Eq. (10.18) (recast slightly): \begin{gather*} G(x, y) = A\begin{cases} h_{1}(x)h_{2}(y), & x < y,\\ h_{2}(x)h_{1}(y), & x > y,\\ \end{cases} \tag{10.18} \end{gather*} for a second-order self-adjoint operator $\mathcal{L}$ where $h_1(x)$ is the homogeneous solution satisfying the left boundary condition and $h_2(x)$ is a homogeneous solution satisfying the right boundary condition. The argument is simply that for $x \neq y$, the functional dependence on $x$ must be a homogeneous solution satisfying the boundary conditions, and $\mat{G} = \mat{G}^\dagger$ hence $G(x, y) = G(y, x)^*$. *(The form here assumes that $h_{i}$ are real.)* The constant $A$ is found by ensuring that the discontinuity at $x=y$ gives the delta-function. Note that in your solution for the causal Green's function $G_+(t)$ fo the Harmonic oscillator above, one cannot express $G(t, \tau) = G_{+}(t-\tau) = \Theta(t-\tau)\sin\bigl(\omega_0(t-\tau)\bigr)/\omega_0$ in this form: this lack of symmetry is because the operator $\mathcal{L}$ is not self-adjoint with the specified boundary conditions. :::::{admonition} Details :class: dropdown This follows from the following arguments, made in terms of linear algebra but tagged with the corresponding equation in {cite}`Arfken:2013`. We start with the definition of the Green's function: \begin{gather*} \mat{L}\mat{G} = \mat{1}. \tag{10.6} \end{gather*} Since $\mat{L}$ is Hermitian, it has a complete set of orthonormal eigenstates \begin{gather*} \mat{L}\ket{n} = \ket{n}\lambda_n, \qquad \braket{m|n} = \delta_{mn}. \tag{10.10} \end{gather*} This means we can express both $\mat{G}$ and $\mat{1}$ in the basis \begin{gather*} \mat{G} = \sum_{nm}\ket{n}g_{nm}\bra{m}. \tag{10.11}\\ \mat{1} = \sum_{m}\ket{m}\!\bra{m}. \tag{10.12} \end{gather*} Inserting these into (10.6) we have \begin{gather*} \sum_{nm}\mat{L}\ket{n}g_{nm}\bra{m} = \sum_{nm}\ket{n}\lambda_n g_{nm}\bra{m} = \sum_{m}\ket{m}\!\bra{m}. \tag{10.13} \end{gather*} Compute the matrix elements of this expression $\braket{n|\cdots|m}$ (rename the inner dummy indices while you do this) to find *(no summation convention)* \begin{gather*} \lambda_n g_{nm} = \delta_{nm}, \qquad g_{nm} = \frac{\delta_{nm}}{\lambda_n},\\ \mat{G} = \sum_{n}\ket{n}\frac{1}{\lambda_n}\bra{n}. \tag{10.14}\\ \mat{G} = \mat{G}^\dagger. \tag{10.15} \end{gather*} Thus, we see that $\mat{G}$ is also Hermitian. ::::: [Toeplitz matrix]: [Fuch's theorem]: [method of undetermined coefficients]: [delta function]: [Riemann-Stieltjes Integral]: [Heaviside step function]: [distribution]: [variation of parameters]: [Wronskian]: [osculating]: [integrating factor]: [envelope]: