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In Quantum Field Theory for the Gifted Amateur by Tom Lancaster and Stephen J. Blundell in Sec. 23.2, while deriving an expression for the path integral, it derives it by demoting functions to vectors and operators to matrices. At the end, the expression contains a term which is the "inverse" of an operator $A(x, y)$ satisfies: $$ \int dz\, A(x, z) A^{-1} (z, y) = \delta (x, y).\tag{23.31} $$ The book stated that this is just another way to say that $A^{-1}$ is the Green's function of $A$. However, I couldn't see the equivalence between this with the usual definition of the Green's function: $$ A G(x, s) = \delta (s-x). $$ Therefore, I would like to ask if someone could explain why they are equivalent. It would be even better if one can provide an example with $A=\frac{d^{2}}{dx^{2}}.$

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  • $\begingroup$ Delta function is identity matrix and $A$ is operator matrix. Variable is row or column, then $A^{-1}$ is inverse of operator. In terms of functions it is green function that is inverse of differential operator. $\endgroup$ Commented Dec 25, 2022 at 8:35

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Note: The following is not mathematical rigorous. It rather sketches some of the relevant concepts.

Consider a linear operator $L$ with kernel $K$, i.e. $L$ acts on some function $f$ via

$$(Lf)(x) = \int\mathrm dy\, K(x,y)\, f(y) \tag{1}\quad . $$

Now consider a Green's function $G_z$ which satisfies $$(LG_z)(x) = \delta(x-z) \quad ,\tag{2}$$ and therefore $$(LG_z)(x) \overset{(1)}{=}\int \mathrm dy\, K(x,y)\, G_z(y) \overset{(2)}{=}\delta(x-z) \quad . \tag{3}$$

By writing $G_z(y) \equiv G(y,z)$, we thus arrive at

$$ \int \mathrm d y \,K(x,y)\, G(y,z) = \delta(x-z)\tag{4} \quad .$$

In this sense we have $G=K^{-1}$. The relation between $L$ and $K$ is similar to the relation of an operator $A$ and its matrix representation $M_A$, where we would replace the integral by a sum.

Regarding the example: If $L$ denotes the second derivative, i.e. we have that $(Lf)(x) = f^{\prime\prime}(x)$, then a Kernel can be defined via the derivative of the Dirac delta distribution. Indeed, we have that

$$f^{\prime\prime}(x) = \int \mathrm d y \, \delta(x-y) f^{\prime\prime}(y) = \int \mathrm dy\, \delta^{\prime\prime}(x-y) f(y) \tag{5} \quad ,$$

where we have used a property of the delta distribution.

By comparing with equation $(1)$ we hence find $K(x,y)=\delta^{\prime\prime}(x-y)$. A corresponding Green's function can be given in terms of the Heavisde step function.

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