Solving Delta Potential Barrier Problem with Schrodinger Equation

[Petar Mali]

Homework Statement

In delta potential barrier problem Schrödinger equation we get

$$\psi(x)=Ae^{kx}, x<0$$

$$\psi(x)=Ae^{-kx}, x>0$$

We must get solution of

$$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx$$

Homework Equations

The Attempt at a Solution

$$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \frac{d\psi}{dx}|^{\epsilon}_{-\epsilon}$$ and get the solution

I can say that the whole function is

$$\psi(x)=Ae^{-k|x|}$$

I don't have first derivative in 0.

$$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=0$$Why I don't get same solution different then zero like in case

$$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \frac{d\psi}{dx}|^{\epsilon}_{-\epsilon}$$

?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[gabbagabbahey]

First, I'd like to point out how disappointed I was when I clicked on this thread and found that it wasn't a Chuck Norris movie

I can say that the whole function is

$$\psi(x)=Ae^{-k|x|}$$

I don't have first derivative in 0.

No you cant. The wavefunction [tex]\psi(x)=\left\{\begin{array}{lr}Ae^{kx}, & x<0 \\ Ae^{-kx}, & x>0\end{array}\right.[/itex] is undefined at $x=0$ (as it should be for a delta function potential). The wavefunction $\psi(x)=Ae^{-k|x|}$ <b>is</b> defined at $x=0$; the two wavefunctions are not equivalent.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=0$$Why I don't get same solution different then zero like in case </div> </div> </blockquote><br /> I don't see how you are getting zero for that limit. Show the rest of your steps.[/tex]

 

[Petar Mali]

$$lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{-\epsilon}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=0$$

$$lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=lim_{\epsilon \rightarrow 0}\frac{d\psi}{dx}|^{0}_{-\epsilon}+lim_{\epsilon \rightarrow 0}\frac{d\psi}{dx}|^{\epsilon}_{0}$$

$$\frac{d\psi}{dx}=kAe^{kx}$$ for $$x<0$$

$$\frac{d\psi}{dx}=-kAe^{-kx}$$ for $$x>0$$

$$lim_{\epsilon \rightarrow 0} \int^{0}_{-\epsilon}\frac{d^2\psi}{dx^2}dx+lim_{\epsilon \rightarrow 0} \int^{\epsilon}_{0}\frac{d^2\psi}{dx^2}dx=<br /> kA-lim_{\epsilon \rightarrow 0}kAe^{k\epsilon}-lim_{\epsilon \rightarrow 0}kAe^{-k\epsilon}+kA=2kA-2kA=0$$

 

[BerryBoy]

Double check your exponentials (hint hint) and maybe expand them out ignoring terms higher than $\epsilon^1$