Integral Operators

Note that any operator \(K\) on a function space \(U\) may be expressed as an integral,

\(K=\mathbb{I}K\mathbb{I}=\int_{a}^bdx''\int_a^b dx' w(x'')w(x')|x''\rangle\langle x''|K|x'\rangle\langle x'| = \int_a^b dx''dx' w(x'')w(x')\langle x''|K|x'\rangle |x''\rangle\langle x'| = \int_a^b dx''dx' w(x'')w(x')K(x'',x') |x''\rangle\langle x'|\)

so that \(K_xf(x) \equiv \langle x|K|f\rangle = \langle x|\int dx''dx'w(x'')w(x')K(x'',x')|x''\rangle\langle x'|f\rangle= \int dx''dx'w(x'')w(x')K(x'',x')\frac{\delta(x-x'')}{w(x)}f(x')=\int dx' w(x')K(x,x')f(x')\)

If \(K(x,x')\) is just a function (not a distribution or generalized function) we \(K\) an integral operator.

We can think of \(K(x,x')\) as a matrix element wich \(x\) the row and \(x'\) the column and \(f(x')\) the column vector. Then (as homework) \(K^\dagger(x,x') = K^*(x',x)\), just like a matrix.

\(K(x'',x')\) is called the integral kernel of \(K\).

(Example the momentum operator as seen Monday)

Inverses of Differential Operators

We may expect it to involve integration (to undo the differentiation). \(L^{-1}\) an integral operator. We get constants of integration, which can be fixed with boundary conditions. We may have to eliminate the nullspace, Differential opeators have non-empty nullspaces but sometimes the homogeneous solutions cannot satisfy the boundary conditions which would eliminate the nullspace from the domain of the operator.

So let’s procede, assume \(L=L^\dagger\) is self adjoint.

Solve \(L|u\rangle = |s\rangle\) on \(N^\perp(L)\) i.e. the subspace spanned by eigenvectors with nonzero eigenvalues. Let’s call \(L^{-1}\equiv G\). I.e. \(L^{-1}L=LL^{-1}=LG=\mathbb{I}\). When \(G\) exists, it’s called a Green’s operator. In general, there may be more than one Green’s operator depending on the boundary conditions.

Properties of Green’s Operator

\(\langle x|x'\rangle = \frac{\delta(x-'x)}{w(x')} = \langle x|LG|x'\rangle = \langle x|L|Gx'\rangle = L_x \langle x|G|x'\rangle = L_x G(x,x')\)

So, \(L_xG(x,x')=\frac{\delta(x-x')}{w(x)}\) is a differential equation for the Green’s operator. I.e. the Green’s operator satisfies the differential equation. It is the same differential equation we had before \(L|u\rangle = |s\rangle\) but with a delta function source (e.g. point charge). The specific solution depends on the bounary conditions. Lets say we found \(G(x,x')\), then \(|u\rangle = G|s\rangle\). Or, \(\langle x|u\rangle = \langle x|G|s\rangle = u(x)=\int_a^b dx' w(x')G(x,x')s(x')\). This gives the particular solution.

Finding Green Functions

Solve the differential Equation

\(L_xG(x,x')=\frac{\delta(x-x')}{w(x)}\) that satisfies the problems’ given boundary conditions. So, solves the homogeneous equation for all \(x\) not \(x'\), at \(x\) equal to \(x'\) it is discontinuous in the last order derivative in \(L\) of the Green’s function. This is called a junction condition at \(x=x'\).

D&K does this for 2nd order linear operators and does example for a string.

For Self Adjoint Operators, Expand the Green Function in the Eigenfunctions of L

Fourier Transform

Example: Simple Harmonic Oscillator
\(m\ddot{x}+m\omega^2x=F(t)\) \(\left[\frac{d^2}{dt^2}+\omega^2\right]x(t)=Tx(t)=\frac{1}{m}F(t)=s(t)\). So, \(L_tG(t,t')=\delta(t'-t)\).

The solution is \(x(t)=\int_{-\infty}^\infty dt' (G(t,t')s(t))+x_0(t)\) and fixed with boundary conditions.

Note coefficients in \(L_t=\frac{d^2}{dt^2}=\omega^2\) are independent of time. Thus, we can transform \(t\to t+\alpha\) to no effect. Hence, \(G(t,t')=G(t-t')\).

Define \(t\to t-t'\). So, \(L_tG(t)=\delta(t)\). Then, \(L_t\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dE \exp(iE t)\tilde{G}(E)=\delta(t)\). So, \(L_t\exp(iEt) = (-E^2+\omega^2)\exp(iEt)\). Then, \(L_tG(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dE (\omega^2-E^2)\exp(iEt)\tilde{G}(t)=\delta(t)\).

Then, \(\int dt\exp(-iE't)\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dE (\omega^2-E^2)\exp(iEt)\tilde{G}(t)=\int dt\exp(-iE't)\delta(t)\).

Then, \(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dE (\omega^2-E^2)\int_{-\infty}^\infty dt\exp(i(E-E')t)\tilde{G}(t)=\exp(-iE't)\).

From a definition of the Dirac Delta Function, \(\delta(t)=\frac{1}{2\pi}\int_{-\infty}^\infty\exp\pm i\omega t\).

Then, \(\sqrt{2\pi}(\omega^2-E'^2)\tilde{G}(E')=1\).

Hence, \(\tilde{G}(E)=\frac{1}{\sqrt{2\pi}}\frac{1}{\omega^2-E^2}\).

\(G(t)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}dE\exp(iEt)\tilde{G}(E) = \frac{-1}{2\pi}\int_{-\infty}^\infty\frac{\exp(iEt)}{E^2-\omega^2}\). We have an integral we will need to solve with Complex analysis.

\(\int_\mathbb{R}dE\frac{\exp(iEt)}{(E-\omega)(E+\omega)}=\int_\mathbb{R}dE g(E)\), simple poles at \(\pm\omega\).

This is a singular integral, so it is ambiguous. Thus, we must choose the contour.
- Principal Value (Non-singular bit, limit as you go straight through the singularities) [This would be a solution without the delta function]
- Small semicircles around the poles (4 possibilities) [This gives the solutions with the delta function]
5 choices! Our boundary conditions are needed to distinguish which solution to use.

Res \(g(-\omega)=\lim_{z\to-\omega}\frac{\exp(izt)}{2z}=\frac{-\exp(-i\omega t)}{2\omega}\), hence for our integral \(Res (-\omega) = \frac{\exp(-i\omega t)}{4\pi\omega}\)

Res \(g(\omega)=\lim_{z\to\omega}\frac{\exp(izt)}{2z}=\frac{\exp(i\omega t)}{2\omega}\), hence for our integral \(Res (\omega) = \frac{-\exp(i\omega t)}{4\pi\omega}\)

Since \(\exp(iEt)=\exp(itR\cos\theta)\exp(-tR\sin\theta)\) we must choose the upper contour for positive \(t\) and the lower contour for negative \(t\).

Choose the contour that goes into the negative imaginary axis. If we use \(\Gamma^+\) then the contour encloses the poles. If we use \(\Gamma^-\) then the contour does not enclose the poles.

\(t> 0\): \(G(t)=2\pi i\sum Res_{enclosed} = 2\pi i \frac{1}{\omega}\sin(\omega t)\) \(t<0\): \(G(t)=0\) Hence, \(G(t)=\frac{\sin\omega t}{\omega}\Theta(t)\).

This choice of contour gives rise to the retarded Green function. The opposite choice gives rise to the advanced Green function. The cross is the Feynman Green function.