Green function and operator Sturm-Liouville

Question

Consider $$\mathcal{L}=\sigma_y - c\mathbb{I}$$

Where $\sigma_y$ is the Pauli matrix, and $\mathbb{I} =\left[\begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array}\right]$ is the identity matrix.

And find Green's function. Also find eigenvectors and eigenvalues.

I have no idea how to solve this problem.

The example have in the book is, whether the equation $$ \mathcal {L} y(x) = f(x) $$,

$$\mathscr{B}y(x)=0$$ on the contour and $$ \mathscr{B} = \left\lbrace \begin{array}{ll} \alpha_1 + \alpha_2 \frac{d}{dx} &, x=a\\ \beta_1 + \beta_2 \frac{d}{dx} &, x=b \end{array} \right. $$

we look for the green function $G(x,\xi)$ that satisfies

$$ \mathcal {L}y=f \rightarrow \mathcal {L}G(x, \xi)=\delta(x-\xi) \rightarrow \mathscr{B}G(x,\xi)=0 $$

Using the characteristic functions $\Phi_{\lambda}$ of the operator $\mathcal {L}$, that is, the functions that satisfy:

$$ \mathcal {L} \Phi_{\lambda} = \lambda \Phi_{\lambda} \rightarrow \mathscr{B}_{\Phi_{\lambda}} $$

If G exists and set ${\Phi_{\lambda}}$ is complete, then G can be represented as:

$$ G(x,\xi) = \sum_{\lambda}^{} \gamma_{\lambda}(\xi) \Phi_{\lambda}(x)$$

$$ \mathcal {L} G(x,\xi) = \sum_{\lambda}^{} \gamma_{\lambda}(\xi) \mathcal {L} \Phi_{\lambda}(x) = \sum_{\lambda}^{} \gamma_{\lambda}(\xi) \lambda \Phi_{\lambda}(x) = \delta(x-\xi)$$

We multiply both sides by $ \Phi_{\lambda'} (x) $ and integrate over $x$

$$ \sum_{\lambda}^{} \gamma_{\lambda}(\xi) \lambda \delta_{\lambda \lambda'}=\Phi_{\lambda'}(\xi) $$

$$ \gamma_{\lambda}(\xi) = \frac{\Phi_{\lambda'}(\xi)}{\lambda} $$

thus

$$ G(x,\xi)=\sum_{\lambda}^{} \frac{\Phi_{\lambda}(\xi)\Phi_{\lambda}(x)}{\lambda} $$

Note 1.Usually the differential equation solved is $$ \mathcal {L}y-\lambda y=f $$ where $\lambda$ is an arbitrary parameter, in this case, the eigenvalues of $\mathcal {L}$ are generally represented by $\lambda_n$, the non-functional functions by $\Phi_n(x)$ and the bilinear form is

$$ G(x,\xi)=\sum_{n}^{} \frac{\Phi_{n}(\xi)\Phi_{n}(x)}{\lambda_n - \lambda} $$ In the context of this problem we talk about the poles of the Green function, the $\lambda$ plane and the resonance phenomena.

2. in complex function spaces, the bilinear formula will be modified to $$ G(x,\xi)=\sum_{n}^{} \frac{ \overline{\Phi_{\lambda}}(\xi)\Phi_{\lambda}(x)}{\lambda} $$ and the green function will not be symmetrical, but hermitian under the exchange of x and $\xi$ positions

$$ G(x,\xi) = \overline{G}(x,\xi) $$

This is the example of the book, but applying it in the requested exercise confused me a lot and does not move.

Answer 1

The term Green's function is usually used when the linear operator in question is a differential operator. Are you sure that there is no differential operator that should appear in $\mathcal{L}$? If $\mathcal{L}$ is exactly as you have written it, continue reading.

I don't know how much you know, so I will address the basics in a not-completely-rigorous way.

In mathematical physics and applied mathematics one often encounters equations of the form \begin{equation} \mathcal{L}u = f, \end{equation} where $\mathcal{L}$ is a linear operator, $f$ is a function, and $u$ is a function to be found. The naïve most concrete way to define a Green's function is to say that it is a function $G$ with two arguments that produces a solution $u$ from the given function $f$ by the formula \begin{equation} u(x) = \int G(x,x')f(x')dx', \end{equation} so that \begin{equation} (\mathcal{L}u)(x) ~=~ \mathcal{L}\int G(x,x')f(x')dx' ~=~ f(x). \end{equation} That is, the Green's function allows one to map the right-hand side $f$ to a solution of the original inhomogeneous equation.

The case here is not the same, but it does involve a linear operator. $\mathcal{L}$ is a $2\times 2$ complex matrix, $f$ is a vector in $\mathbb{C}^2$, and $u$, if it exists, is also a vector in $\mathbb{C}^2$. We want to solve the linear algebraic equation $\mathcal{L}\mathsf{u} =\mathsf{f}$, where

\begin{eqnarray} \mathcal{L} &=& \sigma_y - cI\\ &=& \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)-\left(\begin{array}{cc} c & 0 \\ 0 & c \end{array}\right), \end{eqnarray} where $\sigma_y$ is a Pauli spin matrix. In short, we want the inverse of $\mathcal{L}$, since $\mathsf{u} = \mathcal{L}^{-1}\mathsf{f}$.

Note that if $\det\mathcal{L} = 0$, there is no inverse.

How can we call this a Green's function? It's a stretch, in my opinion, but look at the formula for the solution. Suppose that the inverse of $\mathcal{L}$ exists, and let $\mathsf{A} = \mathcal{L}^{-1}$. Denote the entries of vectors and matrices with square brackets. Then \begin{equation} \mathsf{u}[i] = \sum_{j=1}^{2}\mathsf{A}[i,j]\mathsf{f}[j]. \end{equation} Compare this to the integral involving $G(x,x')$, and I think you will see analogies. $x$ corresponds to $i$, $x'$ corresponds to $j$, the sum over column indices is the analogue of the integration over $x'$, and $\mathsf{A}[i,j]$ is the analogue of $G(x,x')$.