Step Potentials

incomplete Finite Square Well

$$x>\frac{L}{2};e^{-\kappa x}$$ $$\kappa =\frac{\sqrt{2m(v_0-E)}}{\hslash }$$ $$x<-\frac{L}{2};e^{\kappa x}$$

Solutions for E from:

$$\tan \xi =\frac{\sqrt{{\xi }_0^2-{\xi }^2}}{\xi }$$ $$\xi =\frac{\sqrt{2mE}}{\hslash };{\xi }_0=\frac{L}{\hslash }\sqrt{\frac{m{v}_0}{2}}$$

What happens if $E>V_0$? (energy is greater then potential)

We can take the Time independent Schrodinger equation

$$-\frac{{\hslash }^2}{2m}\frac{d^2\phi (x)}{d{x}^2}+v(x)\phi =E\phi (x)$$ $$\frac{d^2\phi (x)}{d{x}^2}=-\frac{2m}{{\hslash }^2}(E-V)\phi (x)=-k_0^2\phi (x)$$

Where

$$k_0=\frac{\sqrt{2m(E-V_0)}}{\hslash },k=\frac{\sqrt{2mE}}{\hslash }$$

Now Consider Regions 1 and 2 Region 1: For $x<0$ ; $E>(V=0)$

$$k=\frac{\sqrt{2mE}}{\hslash }>0$$

Solutions:

$$A{e}^{ikx}+B{e}^{ikx}$$

Region 2: For $x>0$, $E<V_0$

$$k_0=\frac{\sqrt{2m(E-V_0)}}{\hslash }=\frac{i\sqrt{2m(v_0-E)}}{\hslash }$$ $$\xi \kappa \equiv \frac{\sqrt{2m(v_0-E)}}{\hslash }$$ $$\therefore \phi =C{e}^{i{k}_{0}x}+D{e}^{-i{k}_0}x$$ $$\phi =C{e}^{i{k}_{0}x}+0e^{-i{k}_0}x$$

What happens if E > V everywhere? Region I: $x<0;k=\frac{\sqrt{2mE}}{\hslash }$ with solution:

$$\phi \sim A{e}^{ikx}+B{e}^{-ikx}$$

Region II: $x>0;E>V_0,k_0=\frac{\sqrt{2m(E-V_0)}}{\hslash }$

with solution:

$$\phi \sim C{e}^{i{k}_{0}x}+D{e}^{-i{k}_{0}x}$$

Consider incident traveling wave from the left: In the above diagram:

Region 1

$e^{ikx}$ Represents flux coming from the left and $e^{-ikx}$ Represents the flux reflected from the potential

$\therefore$ for $x<0$ $\phi (x)=A{e}^{ikx}+b^{-ikx};k=\sqrt{\frac{2mE}{\hslash }}$ and for $x>0\phi (x)=C{e}^{i{k}_{0}x}+D{e}^{-i{k}_{0}x};k_0=\frac{\sqrt{2m(E-V_0)}}{\hslash }$

**no wave incident from $x=\infty$

There is reflection and transmission: Lets use the probability current to quantify this: Recall:

$$j_x=\frac{\hslash }{2mi}\left({\phi }^{\ast }\frac{d\phi }{dx}-\phi \frac{d{\phi }^{\ast }}{dx}\right)$$

Now, $$ R\equiv\frac{j_{ref}}{j_{inc}} ; T \equiv \frac{j_{trans}}{j_{inc}}

$$![[Pastedimage20231020155432.png]]Recallalso:$$

\phi_{(x,t)} = \int A(k)e^{i(kx-\omega t)} , dx