Free Electron Schrödinger Equation (Energy Method)

Question

For the simplest atom, its wave function is described by the PDE of Schrodinger equation: $$ -i h \frac{\partial u}{\partial t }=\frac{h^{2}}{2m} \Delta u + \frac{e^{2}}{r}u$$

The potential $\frac{e^{2}}{r}$ is a function of radial distance $r$.

So, as a simple warm-up problem,let’s take the free Schrodinger equation with the following Dirichlet Boundary conditions given as: $$-i \frac{\partial u}{\partial t }=\frac{1}{2} \Delta u, \quad 0<x<1,$$ $$u(0,x)=f(x),$$ $$u(t,0)=0,$$ $$u(t,1)=0,$$

In one dimension, where we’ve set $h = m = 1$ and I have dropped the potential term for the free electron case.

Thus using the energy method I will have :

\begin{align*} -i u_{t} &= \frac{1}{2} u_{xx} \\ -i uu_{t} &= \frac{1}{2} uu_{xx} \\ -i \partial_{t} \int_{0}^{1} \frac{ u^{2} }{2}dx&=\frac{1}{2} \int_{0}^{1} uu_{xx} dx\\ -i \partial_{t} \int_{0}^{1} \frac{ u^{2} }{2}dx&=\frac{1}{2} \int_{0}^{1} uu_{xx} dx\\ -i \frac{1}{2}\partial _{t}\|u\|_{2}^{2}&=\frac{1}{2} \int_{0}^{1} u u_{xx}dx \\ -i \partial_{t}\|u\|_{2}^{2}&=2\frac{1}{2} \int_{0}^{1} u u_{xx} dx\\ -i \partial _{t}\|u\|_{2}^{2}&=- \int_{0}^{1} u_{x}^2 dx \end{align*}

1. Am I right ?
2. And if I am how can I continue from here ? here it states that : "Then one would integrate in time and one would obtain the energy estimate

$$ \|u(\cdot ,t)\|_{2}\leq \|f(\cdot )\|_{2}$$" And for the lhs is simple enough to understand.But I cannot understand the result in the rhs. 3) And last how I can handle the imaginary $i$ in $t$?

Answer 1

Here $u$ is complex so the energy is $u^* u$ where $u^*$ is complex conjugate. Then you must compute $$ (u^* u)_t = u^* u_t + u^*_t u= \frac{i}{2} u^* u_{xx} - \frac{i}{2} u^*_{xx} u = \frac{i}{2}( (u^* u_x)_x - (u^*_x u)_x ) $$ Integrating over $x$ and using zero boundary conditions on $u$ $$ \frac{d}{dt}\int_0^1 u^* u dx = \frac{d}{dt}\int_0^1 |u|^2 dx = 0 $$ so that $$ \|u(\cdot,t)\|_2 = \|f(\cdot)\|_2 $$