Add page 7.

src/differential_equations_2.md

@@ -399,13 +399,98 @@ PDE was supplied with an initial condition.
 Putting the solutions to the two ODEs together and redefining 
 $\Tilde{A}=A \cdot c_1$, we arrive at the solutions for theb PDE,
 
-$\psi_n(x,t) = \Tilde{A} e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
+$\psi_n(x,t) = \Tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
 
 Notice that there is one solution $\psi_{n}(x,t)$ for each natural number $n$. 
 These are still very special solutions. We will begin discussing next how to 
 obtain the general solution in our example. 
 
+## Self-adjoint differential equations: Connection to Hilbert spaces! ##
 
+As we hinted was possible earlier, let us re-write the previous equation by 
+defining a linear operator, $L$, acting on the space of functions which satisfy
+$\phi(0)=\phi(L)=0$:
+
+$$L[\cdot]:= \frac{- \hbar^2}{2m} \frac{d^2}{dx^2}[\cdot]. $$
+
+Then, the ODE can be writted as 
+
+$$L[\phi]=\lambda \phi.$$
+
+This equation looks exactly like, and turns out to be, an eigenvalue equation!
+
+!!! info "Connecting function spaces to Hilbert spaces"
+    
+    Recall that a space of functions can be transformed into a Hilbert space by 
+    equipping it with a inner product,
+    
+    $$\langle f, g \rangle = \int^{L}_{0} dx f*(x) g(x) $$
+    
+    Use of this inner product also has utility in demonstrating that particular 
+    operators are *Hermitian*. The term hermitian is precisely defined below.
+    Of considerable interest is that hermition operators have a set of nice 
+    properties including all real eigenvalues and orthonormal eigenfunctions. 
+    
+The nicest type of operators for many practical purposes are hermitian 
+operators. In quantum physics for example, any physical operator must be 
+hermitian. Denote a hilbert space $\mathcal{H}$. An opertor 
+$H: \mathcal{H} \mapsto \mathcal{H}$ is said to be hermitian if it satisfies
+
+$$\langle f, H g \rangle = \langle H f, g \rangle \ \forall \ f, \ g \ \epsilon \ \mathcal{H}.$$
+
+Now, we would like to investigate whether the operator we have been working with,
+$L$ satisfies the criterion of being hermitian over the function space 
+$\phi(0)=\phi(L)=0$ equipped with the above defined inner product (i.e. it is a
+Hilbert space). Denote this Hilbert space $\mathcal{H}_{0}$ and consider let 
+$f, \ g \ \epsilon \ \mathcal{H}_0$ denote two functions from the Hilbert space.
+Then, we can investigate
+
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} \int^{L}_{0} dx f*(x) \frac{d^2}{dx^2}g(x).$$
+
+As a first step, it is possible to do integration by parts in the integral,
+
+$$\langle f, L g \rangle = \frac{+ \hbar^2}{2m} ( \int^{L}_{0} dx \frac{d f*}{dx} \frac{d g}{dx} - [f*(x)\frac{d g}{dx}] \big{|}^{L}_{0} )$$
+
+The boundary term vansishes, due to the boundary conditions $f(0)=f(L)=0$, 
+which directly imply $f*(0)=f*(L)=0$. Now, intergrate by parts a second time
+
+$$\langle f, L g \rangle = \frac{- \hbar^2}{2m} (\int^{L}_{0} dx \frac{d^2 f*}{dx^2} g(x) - [\frac{d f*}{dx} g(x)] \big{|}^{L}_{0} ).$$
+
+As before, the boundary term vanishes, due to the boundary conditions 
+$g(0)=g(L)=0$. Upon cancelling the boundary term however, the expression on 
+the right hand side, contained in the integral is simply 
+$\langle L f, g \rangle$. Therefore,
+
+$$\langle f, L g \rangle=\langle L f, g \rangle. $$
+
+We have demonstrated that $L$ is a hermitian operator on the space 
+$\mathcal{H}_0$. As a hermitian operator, $L$ has the property that it's 
+eigenfunctions form an orthonormal basis for the space $\mathcal{H}_0$. Hence it
+is possible to expand any function $f \ \epsilon \ \mathcal{H}_0$ in terms of
+the eigenfunctions of $L$. 
+
+!!! info "Connection to quantum states"
+    
+    Recall that q quantum state $\ket{\phi}$ can be written in an orthonormal 
+    basis $\{ \ket{u_n} \}$ as 
+    $$\ket{\phi} = \underset{n}{\Sigma} \bra{u_n} \ket{\phi} \ket{u_n}.$$ 
+    
+    In terms of hermitian operators and their eigenfunctions, the eigenfunctions
+    play the role of the orthonormal basis. In reference to our running example,
+    the 1D Schr\"{o}dinger equation of a free particle, the eigenfunctions 
+    $sin(\frac{n \pi x}{L})$ play the role of the basis functions $\ket{u_n}$.
+    
+To close our running example, consider the initial condition 
+$\psi(x,o) = \psi_{0}(x)$. Since the eigenfunctions $sin(\frac{n \pi x}{L})$ 
+form a basis, we can now write the general solution to the problem as 
+
+$$\psi(x,t)  = \overset{\infinity}{\underset{n}{\Sigma}} c_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}),$$
+
+where in the above we have defined the coefficients using the Fourier 
+coefficient,
+
+$$c_n:= \int^{L}_{0} dx sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
+    
 
 
     
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