Linear Differential Operator

The Formal Linear Differential Operator¶

We define the differentian opeartor as:

Now we can define the formal linear operator by some opeartor algebra:

We call it formal since we want to contrast with concrete linear differential opeartor which is a formal opeartor endorsed with boundary conditions(This convention is used in Mathematical Physics by Michael Stone and Paul Goldbart). We can think of the difference between them as:

The nullspace of formal opeartor is the space of general solution of homogenous diffferential equation
The nullspace of concrete operator is the space of particular solution to homogenous differential equation with boundary conditions

We’ll explore the nullspace of formal operators with a bit more detail now.

Nullspace of formal operator¶

We first observe something very nice about solutions to homogenous differential equations: they belong to $C^{\infty}$ (infinitely differentiable).

The Bootstrapping Lemma¶

Proof 1

Suppose $y\in \mathrm{Null}(L)$ , note trivially $y\in C^n$ . Then:

\begin{align*} a_0y+a_1y^{(1)}+a_2y^{(2)}...+a_n y^{(n)} &=0 \\ y^{(n)} &= -(b_0y+b_1y^{(1)}+b_2y^{(2)}...\underbrace{b_{n-1}{y}^{n-1}}_{\in C^{\min(k, 1)}}) \end{align*}

(4)

Where $b_{\nu}=a_{\nu}/a_{n}$ . Now $b_{\nu}\in C^k$ by closedness under polynomial division. Also $y^{(\nu)} \in C^{n-\nu}$ hence the R.H.S is at least one time continously differentiable(the bottle neck being $a_{n-1}y^{(n-1)} \in C^{\min(k, 1)}$ ). Therefore $y^{(n)}\in C^1$ . But this implies $y\in C^{n+1}$ hence we repeat the above argument to get $y^{(\nu)}\in C^{n+1-\nu}$ . The R.H.S has the bottleneck $a_{n-1}y^{(n-1)}\in C^{\min(k, n+1-(n-1))}=C^{\min(k, 2)}$ . We can repeat this procedure $m$ times and yield that $y^{(n)}\in C^{\min(k, m)}$ . This means the best we can get is $y^{(n)}\in C^k$ or $y\in C^{n+k}$ (after which the coeficient becomes the bottleneck).

Nullspace of $L$ is an $n$ -dimensional subspace of $C^{\infty}$ ¶

Linear Independence: Wronskian¶

Solving $Ly=0$ with constant coefficients¶

$Ly=0$ with general $a_\nu(x)$ ¶

Exercise 1 (1-dimensional Scattering Problem)

Consider the 1d Schrodinger equation:

-\frac{d^2\psi}{dx^2}+V\psi = E\psi

(5)

This is an eigenvalue problem of the operator $H=-\frac{d^2}{dx^2}+V$ :

H\psi = E\psi

(6)

Where $V$ has finite support $[-a, a], a\in \mathbb{R}$ . Denote:

$L:=\{x\in \mathbb{R}:x<-a\}$
$R:=\{x\in \mathbb{R}:x>a\}$

Problems:

Show that in $L,R$ , the general solution to the formal operator $H-E$ where $E=k^2$ is:
$\begin{align*} \psi_k &=A\exp(ikx)+B\exp(-ikx) \quad x\in L \\ \psi_k &=C\exp(ikx)+D\exp(-ikx) \quad x\in R \end{align*}$
(7)
Now consider two asymptotic boundary condition:
$\begin{cases} \Pi_L: D=0, x\to \infty \\ \Pi_R: C=0, x\to -\infty \end{cases}$
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where $\Pi_L$ depicts a left-incident wave with no incoming wave from the right and $\Pi_R$ is the opposite. Show that the solution(in $L,R$ ) for the BC $\Pi_1$ and $k>0$ is:
$\psi_k^L(x) = \begin{cases} \exp(ikx)+ r_L(k)\exp(-ikx) & x\in L \\ t_L(k)\exp(ikx) & x\in R \end{cases}$
(9)
Similarily, for $\Pi_2$ , $k<0$ :
$\psi_k^R(x)=\begin{cases} t_R(k)\exp(ikx)& x\in L \\ \exp(ikx) + r_R(k)\exp(-ikx) & x\in R \end{cases}$
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where $t, r$ denotes the transmission and reflection coefficient.
Note that $\psi_k^*$ is also a solution. Use properties of Wronskian to shwo that:
$|t_{L/R}|^2 + |r_{L/R}|^2=1$
(11)

Solution to Exercise 1

Fix a $k$ . Consider the basis $\psi_k^L, {\psi_k^L}^*$ . Form the Wronskian in $L$ :
$W_L=\begin{vmatrix} \exp(ikx)+r_L\exp(-ikx) & \exp(-ikx)+r_L^*\exp(ikx) \\ ik\exp(ikx)-ikr_L\exp(-ikx) & -ik\exp(-ikx)+ikr_L^*\exp(ikx) \end{vmatrix}$
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Evaluate the determinant:
$\begin{align*} W_L &= (-ik)\cancel{-ikr_L\exp(-2ikx) + ikr_L\exp(2ikx)}+ik|r_L|^2 \\ &-[ik\cancel{-ikr_K\exp(-2ikx)+ikr_L\exp(2ikx)}-ik|r_L|^2] \\ &= 2ik(|r_L|^2-1) \end{align*}$
(13)
Next form Wronskian in $R$ :
$W_R = \begin{vmatrix} t_L\exp(ikx) & t_L\exp(-ikx) \\ ikt_L\exp(ikx) & -ikt_L\exp(-ikx) \end{vmatrix} = -2ik|t_L|^2$
(14)
Since we would expect $W_R=W_L=0$ , henece:
$\begin{align*} W_R-W_L&=0 \\ 2ik(|r_L|^2+|t_L|^2-1) &=0 \end{align*}$
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Since $k>0$ we get as desired.

4.Fix $k$ . We repeat the same procedure with $\{\psi_k^L,\psi_{(-k)}^R\}$ . We evaluate $W_L$ :

\begin{align*} W_L &= \begin{vmatrix} \exp(ikx)+r_L(k)\exp(-ikx) & t_R(-k)\exp(-ikx) \\ ik\exp(ikx)-ikr_L(k)\exp(-ikx) & -ikt_R(-k)\exp(-ikx) \end{vmatrix} \\ &= -2ikt_R(-k) \end{align*}

(16)

Similarily,

W_R = -2ikt_L(k)

(17)

Again, $W_L=W_R=0$ so we get as desired.