# 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.

#### 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.

Created: 2023-06-25 Sun 02:23

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