1
$\begingroup$

The formal definition of a Green's function is: \begin{equation} L(\mathbf{r})G(\mathbf r,\mathbf r^\prime) = \delta(\mathbf r-\mathbf r^\prime), \tag 1 \end{equation} where L is a time linear differential operator.

Now I am reading the book by E. L. Economou's book titled "Green's Functions in Quantum Physics" which gives the definition as, $$ [z - L(\mathbf r)]G(\mathbf r,\mathbf r^\prime;z) = \delta (\mathbf r-\mathbf r^\prime),\tag 2 $$ where $z = \lambda + is$ and L is a time independent, linear, hermitian differential operator that has eigenfunctions $\phi_n (r)$ $$ L(\mathbf r) \phi_n (\mathbf r) = \lambda_n \phi_n (\mathbf r),\tag 3$$ where $\lambda_n$ are the eigenvalues of L.

Surely, Eq. (2) can be considered as the Green function equation for the differential equation, $$ [z - L(\mathbf r)]u(\mathbf r) = f(\mathbf r),\tag 4 $$ such that we can get, $$ u(\mathbf r) = \int f(\mathbf r^\prime) G(\mathbf r,\mathbf r^\prime;z) dr^\prime.\tag 5 $$ Now I am confused!

Why he selected such a form instead of the simpler form in Eq. (1) ? What are the advantages ?

What is really z here ? It is given as a complex variable?

What are intermediate steps that are missing in going from Eq. (1) to (2) ?

$\endgroup$

1 Answer 1

2
$\begingroup$

For the differential equation $$[z−L(r)]u(r)=f(r), $$ given that the eigenvalues are $ \lambda $ as per $$ L(r)ϕ_{n}(r)=λ_{n}ϕ_{n}(r), $$ the equation for the Green's function would be $$ [λ_{n}−L(r)]G(r,r′;z)=δ(r−r′). $$ The given expression matches this except for the extra complex factor of $ \iota s $. The mathematical justification for this is to shift the poles of the Green's Function to the complex plane. Without this complex value the Green's function would be of the form $ \frac{1}{(λ-L)^{-1}}$ which has poles at the eigenvalues. The inclusion of this complex factor avoids this inconvenience by shifting the poles to the complex plane.

This is quite a common trick in physics and can be seen in a lot of places.

$\endgroup$
4
  • $\begingroup$ Thanks! However, I have some more doubts being a novice in this field. How and why can you write equation 3 of your answer ? Could you explain more elaborately. $\endgroup$ Jul 11, 2018 at 11:07
  • $\begingroup$ From equation (2) of your answer one gets: $\left[λ_{n} - L(r)\right]\phi_{n}(r)=0$ so the factor $λ_{n} - L(r)$ comes out. But, how to link it to the actual equation (equation 1 of your answer ) for which Green's function is to be defined ? $\endgroup$ Jul 11, 2018 at 11:18
  • $\begingroup$ I am also a novice in the field, just happened to know this minute detail. We are seeking to find how the response of the system (encoded in u(r)) is to any function given by f(r). The Green's function gives the response of the system to a delta function and using this, we can calculate the response to any arbitrary f(r) using equation (5) in the question. $\endgroup$
    – Hari
    Jul 11, 2018 at 13:33
  • $\begingroup$ For instance if you have the Schrodinger equation, and you need to find how an energy eigenfunction responds to a potential. Here the potential would be f(r), the eigenvalue would be $\lambda$ and the eigenfunction would be u(r). $\endgroup$
    – Hari
    Jul 11, 2018 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.