Green's Function and Zero Modes of Differential Operators

Question

I have this question from reading QFT textbooks.

Consider harmonic oscillator as a simpler model for differential operators. The Lagrangian is given by

$$L=\frac{1}{2}\dot{x}^{2}-\frac{1}{2}\omega^{2}x^{2}.$$

After integration by parts, one can express the classical action as

$$S[x;j]=\int_{-\infty}^{\infty}dt\left[\frac{1}{2}x(t)\left(-\frac{d^{2}}{dt^{2}}-\omega^{2}\right)x(t)+x(t)j(t)\right],$$

where $j(t)$ is an external source.

One can define the following second order differential operator

$$L(t,s)\equiv\left(\frac{d^{2}}{dt^{2}}+\omega^{2}\right)\delta(t-s),$$

where $\delta(t-s)$ is the Dirac delta function. Using this notation, one can write the classical action in a compact form

$$S[x;j]=-\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dtds\frac{1}{2}x(t)L(t,s)x(s)+\int_{-\infty}^{\infty}dtx(t)j(t)\equiv-\frac{1}{2}\langle x,Lx\rangle+\langle j,x\rangle.$$

In QFT, one defines the Greens's function of the differential operator $L$ as its "right inverse", i.e.

$$\left(\frac{d^{2}}{dt^{2}}+\omega^{2}\right)G(t-s)=\delta(t-s),$$

where $\delta(t-s)$ is treated as the "continuous generalization" of the Kronecker delta symbol. In our compact notation, this is just

$$LG=.$$

Then, it is well-known that the equation of motion

$$\ddot{x}(t)+\omega^{2}x(t)=j(t) \tag{1}$$

has a unique solution

$$x(t)=x_{0}(t)+\int_{-\infty}^{\infty}dsG(t-s)j(s), \tag{$\star$}$$

where $x_{0}$ is the unique solution of the homogeneous ODE

$$\ddot{x}(t)+\omega^{2}x(t)=0. \tag{2}$$

If we use the compact notation introduced above, the solution ($\star$) is very easy to understand:

$$Lx=Lx_{0}+LGj=LGj=j=j,$$

which is precisely equation (1). However, there is a problem. In linear algebra, if an operator has an inverse, then it cannot have zero-eigenmodes. But from equation (2), it is clear that $x_{0}$ is a zero mode of the linear operator $L$. Then, we cannot define its Green's function $G(t-s)$, which is absurd.

The same problem arises when we are dealing with QFT in higher dimensions, and $L$ becomes a second order partial differential operator.

What's going on? Where did I make mistakes?

Answer 1

1. The identity $LG={\bf 1}$ implies that $L$ is surjective and $G$ is injective, but not necessarily the other way around. It might be OK for $L$ to have zero-modes.

2. Note that for OP's action functional $S$ to govern a well-posed variational problem, one should impose appropriate boundary conditions. This choice in turn uniquely specifies a Greens function $G$ for the problem.

Answer 2

You didn't make any mistake, but let's refresh some facts about linear algebra.

First, as you mention, if the spectrum of a linear operator $A$ contains the eigenvalue $0$, then the corresponding eigenspace forms its non-trivial kernel, therefore it is not an injective map and, in the end, it is not invertible.

Nevertheless, it doesn't mean that the equation $Ax=b$ doesn't have any solution. Three cases are possible : $(i)$ if $b=0$, then $x\in\mathrm{ker}A$; $(ii)$ if $b\neq0$ and $b\in\mathrm{Im}A$, then $\exists x$ such that $Ax=b$ by definition of $\mathrm{Im}A$; finally, $(iii)$ if $b\neq0$ and $b\not\in\mathrm{Im}A$, then there is no solution. This bla bla comes from the rank theorem and basically tells you that $A$ is only invertible on $\mathrm{Im}A$ $-$ pseudo-inverses are actually restricted to this subspace.

As $\delta\neq0$, the equation $LG=\delta$ falls into the case $(ii)$ or $(iii)$ and, since you succeed in finding a Green function, we are manifestly in the second case. Consequently, the general solution $G$ of this equation is made of elements $G_n$ of the kernel of $L$ (the homogeneous solutions, here: the harmonic oscillators of frequency $\omega$) and a particular solution $G_P$ which you are looking for, such that "$\,G_P=L^{-1}\delta\,$" and $LG = L(\sum_n\lambda_nG_n + G_P) = LG_P = \delta$.

Your confusion came from the fact that, too often, linear algebra teachers insist on the fact the that linear operators are not invertible but forget to recall that equations still admit solutions thanks to the rank theorem, so that students get used to think that equations with singular operators are unsolvable, while they solve them constantly without noticing anymore (when dealing with inhomogeneous ODEs in calculus courses for instance).