Green’s Functions
Introduction
An inhomogeneous differential equation can be written as
The Green’s function solves
So that
The operator D does not operate on r’ so that
(1.4)
When D contains a potential term, physics texts frequently usually assume that the complete set of solutions to (1.1) with ρ(r) = 0 is known. This complete set includes discrete bound states in addition to continuum states and a sum notation has evolved to sum over the discrete states and integrate over the continuum states. The complete set of states in this document is a Fourier series. ..\Fourier\Welcome.doc .htm This series can become a Fourier integral when certain mathematical conditions are satisfied and for simplicity the integral form will usually be used below, but in the case of doubt, the Fourier series is meant so that the underlying summation is always over a set of box normalized plane wave states.
Green functions for various D’s and boundary conditions are treated in classical physics references such as the 1953 edition of Morse and Feshbach. [Morse, P. M. and Feshbach, H., Methods of Theoretical Physics, McGraw-Hill, New York, (1953) – Chapter 7]
In a 1951 book David Bohm [ Bohm, D, Quantum Theory, Prentice-Hall (1951) p.543] in a section on the “Theory of Scattering” gives a rather complicated derivation of the Green’s function solving;
(1.5)
Both references find the standard solution ready to be used in making The_Born_Approximation.doc for scattering
(1.6)
Neither of the two references remark on the fact that this form makes (1.3) into a convolution integral. ..\Fourier\DiscreteConvolution.doc .htm. This integral is most easily evaluated by Fourier transforming g and ρ, multiplying the product and then back transforming. Though the FFT can be traced way back in history, it was not in common usage at the time that these authors worked.
Fourier Transforms
Equation (1.3) is a convolution integral. ( ..\Fourier\convolution.doc .htm) The value of ψ(r) can be found by multiplying the Fourier transform of G times the Fourier transform of ρ(r) and then back transforming. The introduction of the Fast Fourier Transform, FFT, made a qualitative difference in the use of Green’s functions. [http://www.boulder.nist.gov/div853/greenfn/tutorial.html ]
The present work differs from that given by the
The Fourier transform is defined in Press to be
(1.7)
One Dimension, First Derivative
The original equation is
The Greens function equation is
The Fourier transform of the delta function is
Or
In terms of the unknown G(f), the Green’s function is
And
Substitute (1.11), (1.12), and (1.13) into (1.9)
(1.14)
This solves as
(1.15)
Analytic method
The authors of our basic references found g In the time domain using the following method
(1.16)
The poles of the integrand are at the locations where . This is in the negative f plane. For t > 0 , f needs to go to -jR to converge as a contour integral, while for t < 0, it goes to +jR. Since the value of the integral is -2pj times the sum of the residues at the enclosed poles (for a contour in the negative plane), g(t) = 0 for t < 0. For t > 0
(1.17)
The integral in (1.3) must be evaluated, but for out test case ρ is a delta function making this trivial.
FFT method
The integral in (1.3) is a convolution integral ..\Fourier\DiscreteConvolution.doc . It is fastest to evaluate it by multiplying the Fourier transforms of the two functions and then back transforming.
The left side of (1.8) needs to be Fourier transformed. – Though for our test case ρ is a delta function making P[m] always 1.
Then there is a simple multiplication to give
A back transform gives the final solution. This requires two transforms of N log2N operations and N multiplications. The upper limit of N is some small number times 106 which is determined by the compiler.
When G is back transformed
Figure 1 g(t) - red with N=64- Blue is the theoretical solution. Code is in DrivenDecay.zip
Second Derivative
One Dimension – Scattering
A wave wanders back and forth as . In regions where the potential is zero
(2.1)
The value of k > 0 determines the oscillation frequency and hence the energy of the wave. If there is a driving force ρ(x)
The Green’s function equation similar to (1.9) is
Substituting the Fourier expansions for each term (2.3) becomes
or
There is a pole at f = k/2π. The value is proportional to -1/e for f = k/2π - e and to 1/e for f = k/2π + e.
This can be back transformed analytically
(2.6)
Let m=2p|x|, and a = k/2p so that a tabulated integral can be used [W. Grobner & N. Hofreiter, INTEGRALTAFEL –ZWEITER TEIL BESTIMMTE INTEGRALE, Springer Verlag (1961) 333-69a] - This is for the principle value, the integral to k-e + the integral from k+e.
One way to take a principle value numerically is to let G(f) become
(2.8)
This is transformed using residues in Residue Evaluation of g.docx .htm. The G in (2.8) does not solve (2.4) at f = k/2π. This is only one point, but so is a delta
function. The g given by (2.7) satisfies (2.3) for x > 0 and for x <
0.
For x = e and -e
(2.9)
The second partial at exactly 0 is
(2.10)
Thus
(2.11)
The integral of g. which is continuous over this same region, is e. This demonstrates that G(f) as given by (2.8) does transform to a g(x) which satisfies (2.3).
One Dimension – Bound
. This has a maximum of 1 for x =0 and then starts down. In regions where the potential is zero
(3.1)
(2.2)
The Green’s function equation similar to (1.9) is
(2.3)
Substituting the Fourier expansions for each term (2.3) becomes
(2.4)
or
(2.5)
This can be back transformed analytically
(2.6)
Let m=2p|x|, and to put the integrand in the form found in Dwight’s table or definite integrals 859.001 so that
It can also be back transformed numerically – in this case using N=2048
Figure 2 Symmetric back transform of G (gtr) and Theory as given in (2.7) Code in OneDBoundG-g.zip
This shows that (1.3) will normally produce a wave function that extends only a short distance from the potential in (2.2). The driven part of the wave function in the first Born approximation is
Figure 3 Sample bound state potential
If v(x) is a negative delta function at x0, this becomes
(3.3)
This says that in this case the driven part of the wave function looks very much like figure 2 and does satisfy the boundary conditions.