10. Green’s Functions

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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

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10. Green’s Functions#

The coverage of Green’s functions in [Arfken et al., 2013] is quite reasonable. Here are some highlights presented from a different perspective, analogous to that in [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 \(\mat{L}\):

\[\begin{gather*} \mathcal{L}G(x, y) = \delta(x-y), \qquad \mat{L}\mat{G} = \mat{1}\\ y(x) = \int G(x, y)f(y)\d{y}, \qquad \ket{f} = \mat{G}\ket{f}. \end{gather*}\]

\(G(x,y) = G(x-y)\) as a Toeplitz matrix.

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 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.

Do It! Show that \(\mat{G}\ket{f_m} = \ket{f_m}\lambda_m\) if \(\mat{G}\) is a Toeplitz matrix.

(Incomplete)

While this often simplifies things, it is best to keep in mind the nature of Green’s functions as matrices, not as vectors.

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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);
_images/b06343167ef25e99cf8b0f9894b1495be33caa1f06c14597095a5d7c61b4dcd1.png

Do It! 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 [Arfken et al., 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.

Details

This follows from the following arguments, made in terms of linear algebra but tagged with the corresponding equation in [Arfken et al., 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.

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