2

Let's consider the following:

We have a Green function $G$, and we want to know what linear differential equation is solved by $G$.

How to do this? The question is: If I know $G$, then is there a method that allow to solve equation $LG=\delta$ with respect to $L$? In other words, normally we have the differential equation, and we try to get Green function $G$ in order to solve it. I know the Green function of the equation, but try to obtain the right equation that is solved by $G$.

Qmechanic
  • 201,751
Tomek
  • 63
  • 2
    Might [math.se] be better suited for this maths question? – Kyle Kanos Aug 11 '19 at 20:58
  • 2
    Suppose $\sum_{n}|n\rangle \langle n|=1$, and $L=\sum_{n} \lambda_{n} |n\rangle \langle n|$. Then $G=\sum_{n} \frac{1}{\lambda_{n}} |n\rangle \langle n|$ and it follows $GL=1$. Since $G$ exists it follows $G=L^{-1}$. But since you're doing it backwards, assume $L$ exists and try reversing the flow. – Cinaed Simson Aug 11 '19 at 21:45
  • As far as I understand, you describe the operator L as matrix. Is it a proper thing to do since L need to be differential operator, i.e. constructed from derivatives? – Tomek Aug 11 '19 at 23:34
  • 2
    This would be akin to solving for the matrix $A$ given one of its eigenvectors -- you do not have nearly enough degrees of freedom. Impossible, uniquely. – AccidentalFourierTransform Aug 11 '19 at 23:49
  • Given an arbitrary (say, symmetric) function $G$, the sought-for operator $L$ will rarely be a differential operator of finite order. – Qmechanic Aug 12 '19 at 08:01

1 Answers1

1

The following is a bit of an inductive approach and it would probably not work for all Green functions. The basic equation that you want to solve is $$ \hat{D} G(\mathbf{x}) = \delta(\mathbf{x}) , $$ where $\hat{D}$ is the differential operator that you want to find and $G$ is the Green function, which is known. Say for instance your Green function is given by $$ G(\mathbf{x})=\int \frac{1}{m^2+|\mathbf{k}|^2}\exp(i \mathbf{k}\cdot\mathbf{x}) d^3k . $$ if one can somehow get rid of the denominator inside the integral, one can see that the result would produce the Dirac delta function. So, the differential operator must produce $m^2+|\mathbf{k}|^2$ when it operates on the exponential function inside the integral. For each $\mathbf{k}$, we need a gradient operator, which would bring down a $i\mathbf{k}$. So, it then follows that the required differential operator is $$ \hat{D}=m^2-\nabla^2 . $$

This inductive approach is perhaps not very useful for a general case, but it should cover most of the typical cases that one finds. If there is another case that you are interested in that cannot be treated in this way, please include it in the question and then we can think how to deal with it.

flippiefanus
  • 14,614