Integral Operators

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

$K = I K I = \int_{a}^{b} d x^{″} \int_{a}^{b} d x^{'} w (x^{″}) w (x^{'}) | x^{″} ⟩ ⟨ x^{″} | K | x^{'} ⟩ ⟨ x^{'} | = \int_{a}^{b} d x^{″} d x^{'} w (x^{″}) w (x^{'}) ⟨ x^{″} | K | x^{'} ⟩ | x^{″} ⟩ ⟨ x^{'} | = \int_{a}^{b} d x^{″} d x^{'} w (x^{″}) w (x^{'}) K (x^{″}, x^{'}) | x^{″} ⟩ ⟨ x^{'} |$

so that $K_{x} f (x) \equiv ⟨ x | K | f ⟩ = ⟨ x | \int d x^{″} d x^{'} w (x^{″}) w (x^{'}) K (x^{″}, x^{'}) | x^{″} ⟩ ⟨ x^{'} | f ⟩ = \int d x^{″} d x^{'} w (x^{″}) w (x^{'}) K (x^{″}, x^{'}) \frac{δ (x - x^{″})}{w (x)} f (x^{'}) = \int d x^{'} w (x^{'}) K (x, x^{'}) f (x^{'})$

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) $K^{†} (x, x^{'}) = K^{*} (x^{'}, x)$ , 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 $L = L^{†}$ is self adjoint.

Solve $L | u ⟩ = | s ⟩$ on $N^{⊥} (L)$ i.e. the subspace spanned by eigenvectors with nonzero eigenvalues. Let’s call $L^{- 1} \equiv G$ . I.e. $L^{- 1} L = L L^{- 1} = L G = I$ . 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

$⟨ x | x^{'} ⟩ = \frac{δ (x -^{'} x)}{w (x^{'})} = ⟨ x | L G | x^{'} ⟩ = ⟨ x | L | G x^{'} ⟩ = L_{x} ⟨ x | G | x^{'} ⟩ = L_{x} G (x, x^{'})$

So, $L_{x} G (x, x^{'}) = \frac{δ (x - x^{'})}{w (x)}$ 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 $L | u ⟩ = | s ⟩$ 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 $| u ⟩ = G | s ⟩$ . Or, $⟨ x | u ⟩ = ⟨ x | G | s ⟩ = u (x) = \int_{a}^{b} d x^{'} w (x^{'}) G (x, x^{'}) s (x^{'})$ . This gives the particular solution.

Finding Green Functions

Solve the differential Equation

$L_{x} G (x, x^{'}) = \frac{δ (x - x^{'})}{w (x)}$ 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
$m \ddot{x} + m ω^{2} x = F (t)$ $[\frac{d^{2}}{d t^{2}} + ω^{2}] x (t) = T x (t) = \frac{1}{m} F (t) = s (t)$ . So, $L_{t} G (t, t^{'}) = δ (t^{'} - t)$ .

The solution is $x (t) = \int_{- \infty}^{\infty} d t^{'} (G (t, t^{'}) s (t)) + x_{0} (t)$ and fixed with boundary conditions.

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

Define $t \to t - t^{'}$ . So, $L_{t} G (t) = δ (t)$ . Then, $L_{t} \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} d E \exp (i E t) \tilde{G} (E) = δ (t)$ . So, $L_{t} \exp (i E t) = (- E^{2} + ω^{2}) \exp (i E t)$ . Then, $L_{t} G (t) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} d E (ω^{2} - E^{2}) \exp (i E t) \tilde{G} (t) = δ (t)$ .

Then, $\int d t \exp (- i E^{'} t) \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} d E (ω^{2} - E^{2}) \exp (i E t) \tilde{G} (t) = \int d t \exp (- i E^{'} t) δ (t)$ .

Then, $\frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} d E (ω^{2} - E^{2}) \int_{- \infty}^{\infty} d t \exp (i (E - E^{'}) t) \tilde{G} (t) = \exp (- i E^{'} t)$ .

From a definition of the Dirac Delta Function, $δ (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} \exp \pm i ω t$ .

Then, $\sqrt{2 π} (ω^{2} - E^{' 2}) \tilde{G} (E^{'}) = 1$ .

Hence, $\tilde{G} (E) = \frac{1}{\sqrt{2 π}} \frac{1}{ω^{2} - E^{2}}$ .

$G (t) = \frac{1}{\sqrt{2 π}} \int_{R} d E \exp (i E t) \tilde{G} (E) = \frac{- 1}{2 π} \int_{- \infty}^{\infty} \frac{\exp (i E t)}{E^{2} - ω^{2}}$ . We have an integral we will need to solve with Complex analysis.

$\int_{R} d E \frac{\exp (i E t)}{(E - ω) (E + ω)} = \int_{R} d E g (E)$ , simple poles at $\pm ω$ .

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 $g (- ω) = lim_{z \to - ω} \frac{\exp (i z t)}{2 z} = \frac{- \exp (- i ω t)}{2 ω}$ , hence for our integral $R e s (- ω) = \frac{\exp (- i ω t)}{4 π ω}$

Res $g (ω) = lim_{z \to ω} \frac{\exp (i z t)}{2 z} = \frac{\exp (i ω t)}{2 ω}$ , hence for our integral $R e s (ω) = \frac{- \exp (i ω t)}{4 π ω}$

Since $\exp (i E t) = \exp (i t R \cos θ) \exp (- t R \sin θ)$ 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 $Γ^{+}$ then the contour encloses the poles. If we use $Γ^{-}$ then the contour does not enclose the poles.

$t > 0$ : $G (t) = 2 π i \sum R e s_{e n c l o s e d} = 2 π i \frac{1}{ω} \sin (ω t)$ $t < 0$ : $G (t) = 0$ Hence, $G (t) = \frac{\sin ω t}{ω} Θ (t)$ .

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.