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How the Green's functions and the Quantum Mechanics are related? Do they can be used to solve the Schrödinger equation of an particle subjet to some potential that is not a Dirac's delta? And the proprieties of some Green's functions that are symmetrical, i.e. $ G(x|\xi) = G(\xi|x)^{\ast} $, has some relation with the propriety of the inner product $ \langle \alpha \vert \beta \rangle = \langle \beta \vert \alpha \rangle^{\ast} $?

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Schrödinger equation is a linear partial differential equation, so sure, you can use the usual formalism of Green's functions to solve it.

First let's recall how the stuff works. Suppose $L$ is the linear operator and $D$ are the boundary conditions and we want to solve equations $Lu = f$ and $Du = 0$ for $u$. Using the identity property of the convolution $g*\delta = g$ one is motivated to solve the simpler equation $LG = \delta$ and then one finds $u = G*f$ because $$L(G*f) = (LG)*f = \delta*f = f$$.

Now, for the time-independent Schrödinger equation the following should be useful. If the operator (understood also with the given boundary conditions) also has a complete basis of eigenvectors $\left\{\left|\phi_n\right>\right\}$ corresponding to eigenvalues $\left\{\lambda_n\right\}$ then the Green's function can easily be seen to be $$G(x, x') = \sum_n {\phi_n(x)^* \phi_n(x') \over \lambda_n}$$ (just apply the operator $L$ to it and use that $L \left|\phi_n\right> = \lambda_n \left|\phi_n\right>$. So again we can see that $G$ is in a sense an inverse of $L$ (and indeed it is often written simply as $L^{-1}$).

Now, it turns out there is a deeper connection between Green's functions and quantum mechanics via Feynman's path integral if we pass to the time dependent Schrödinger equation. I am not going to derive all the stuff here but suffice it to say that Green's function takes on the meaning of a propagator of the particle. Namely, the probability amplitude that the particle gets from the event (t, x) to the event (t', x') is a Green's function of the time-dependent Schrödinger equation $G(x,t;x',t') = \left<x\right| U(t,t') \left|x'\right>$. So yes, the fact that the Green's function is symmetric is precisely because it can be interpreted as an inner product.

This stuff generalizes further to quantum field theory and Green's functions are among the basic objects of study there.

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Nice job buddy +1 –  user346 Feb 5 '11 at 11:10
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An interesting system to study to understand this is the Poincare half disk or half plane. ${\cal H}^2$. It also illustrates the role of Laplacian operators, Green’s functions and resolvents. The Laplace-Beltrami operator on a Riemannian manifold with a Gaussian metric is $$ \Delta~=~\sum_{ij}{1\over\sqrt{g}}{\partial\over{\partial x^i}}\Big(\sqrt{g}g^{ij}{\partial\over{\partial x^j}}\Big), $$ for $g~=~|det(g_{ij}|$. The Laplacians for the Poincare half plane and disk are $$ \Delta_{{\cal H}^2}~=~y^2\Big({{\partial^2}\over{\partial x^2}}~+~{{\partial^2}\over{\partial y^2}}\Big),~\Delta_{{\cal D}}~=~(\alpha^2~-~x^2~-~y^2)^2\Big({{\partial^2}\over{\partial x^2}}~+~{{\partial^2}\over{\partial y^2}}\Big). $$ The Laplacian commutes with all group elements $g~\in~Iso({\cal H}^2)$ $\Delta T_g~=~T_g\Delta$, so the metric is invariant under these isometries. The Laplacian satisfies the the differential equation $(\Delta~+~\lambda)f(z)~=~0$ defined as the kernel of the resolvent $(\Delta~+~\lambda)^{-1}$ by the equation $$ (\Delta~+~\lambda)^{-1}f(z)~=~\int (z,~z^\prime,~\lambda)f(z^\prime)d\mu(z^\prime), $$ with the harmonic condition $(\Delta~+~\lambda)G(z,~z^\prime,~\lambda)~=~\delta(z,~z^\prime)$.

This space is such that the Laplaican $\Delta$ has eigenvalue $-2$, or a negative Gaussian curvature. This space is a model for the $AdS_2$ spacetime. I hope that with this example you can see answers to you questions, such as the symmetry under interchange.

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Perhaps, in the interests of clarity, you could make explicit the definition of the Green's function. I can see why someone might find this answer unhelpful. @Rodrigo asks about Green's functions in QM and the Schrodinger equation and you start off with the Poincare half-plane. Technically, of course, your answer is helpful for someone who actually bothers to read it. +1 –  user346 Feb 5 '11 at 8:10
    
Hm, not that this isn't interesting stuff but I don't see any connection with the question whatsoever. -1 –  Marek Feb 5 '11 at 8:55
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@Rodrigo asks How the Green's functions and the Quantum Mechanics are related?. @Lawrence provides a concrete example of how to calculate a Green's function, albeit for a free field. The Laplace-Beltrami operator is nothing more than the kinetic term of the Schrodinger equation. How is that not related to the question? The question has other parts, but every answer does not have to answer every single subpart of a question. @Lawrence's answers tend to have more analytical content and are superior to many "hand-wavy" answers. –  user346 Feb 5 '11 at 9:30
    
@space_cadet: be so kind and point to me the place where there is any quantum mechanics in this example. Also show me the place where there is an explicit connection between QM and GF. –  Marek Feb 5 '11 at 9:34
    
@Marek did you not read my previous comment? –  user346 Feb 5 '11 at 9:37
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In more 'down-to-earth' QM, you use Green's functions to find the density of states. I'm deprived of my books so at a loss for giving a good reference, but the idea is to calculate $$G(x,x';E) = \langle x, (E - H)^{-1} x' \rangle,$$ where $H$ is the system's Hamiltonian. You can then define a spectral function $F(x,x',E) = -\frac{1}{\pi} lim_{\epsilon \rightarrow 0} \text{Im } G(x,x',E+i\epsilon),$ whose trace is the density of states: $$\mathcal{N}(E) = \int F(x,x,E) dx.$$ Finally, you can also use this formalism to calculate other expectation values, with formulas like (modulo an incorrect prefactor) $\langle A \rangle = -\frac{1}{\pi} \text{Im Tr}(AG).$ So yes, they are symmetrical, but they can not really be used to 'solve' a Schrödinger's equation, only on a formal level. That's why they're useful though: they're used all the time in many-body QM/solid state physics, where you'll never 'solve' the problem but can learn lots of interesting stuff by indirect approaches, as the one used above.

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+1 all correct except for your statement that they can not really be used to 'solve' a Schrödinger's equation, only on a formal level. For scattering from a potential the Green's function is exactly calculable for many important cases and encodes the full physical content of the solutions. –  user346 Feb 5 '11 at 9:36
    
@space_cadet: forgot about that entirely, good job mentioning it. –  Gerben Feb 5 '11 at 17:25
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