# Delta Potential (Homework 10-325)

Consider a particle moving in a potential

$V(x)=left{begin{matrix}infty & text{ for } x < 0 \V_0delta(x-a) & text { for }x>aend{matrix}right.$

a) For $V_0<0$ find the energies of the bound state(s) and determine how their number depends on $|V_0|$

b) For $V_0>0$ find the phase of the reflection coefficient assuming that the wave is incident from the right propagating toward zero.

We make the translation $y=x-a$ so that

$V=$

$=∞$ ,if  $y<-a$

$=V_0 δ(y-0)$,  if  $y>a$

Which means that the potential is

Index 1 is the transmitted wave, index 2 is the reflected wave. A is wave from left side, B is wave from right side.

$ψ(y)=A_1*e^{iky}+A_2*e^{-iky}$ for $y<0$

$psi(y)=B_1*e^{iky}+B_2*e^{-iky}$, for $y>0$

Continuity condition for wave at $y=0$:

$A_1 e^{ik0}+A_2 e^{-ik0}=B_1 e^ik0+B_2 e^{-ik0}$ or $A_1+A_2=B_1+B_2$

Write Schroedinge equation:

$-ℏ^2/2m*ψ’ (y)+V_0 δ(y-0)*ψ(y)=Eψ(y)$

Integrate Sch equation on a small y interval around y=0

$-ℏ^2/2m*∫_{-ε}^ε ψ” (y)dy+∫_{-ε}^ε V_0 δ(y-0)ψ(y)dy=E*∫_{-ε}^ε ψ(y)dy$

When $ε→0$ we have $E*∫_{-ε}^ε ψ(y)dy→0$ so that from above it remains

$-ℏ/2m*[ψ’ (ε)-ψ’ (-ε) ]+V_0 ψ(0)=0$

From the definition of $ψ(y)$ we have when $ε→0$

$-ℏ/2m ik[B_1-B_2-(A_1-A_2 ) ]+V_0 (A_1+A_2 )=0$

So that the second condition the coefficients NEED to fulfill is

$B_1-B_2-(A_1-A_2 )=(2mV_0)/(ikℏ^2 )(A_1+A_2)$

a)

Bound state $(V_0<0)$. The wavevector k above is complex

$k=√2mE/ℏ=i*√(2m|E|)/ℏ$ so that $ψ(y)$ increases and decreases exponentially

To not be divergent when $y→±∞$ one needs

$ψ(y)=A_1*e^iky$ ,for $y<0$

$ψ(y)=B_2*e^(-iky)$ ,for $y>0$)

with $A_2=B_1=0$

From red conditions above we have

$A_1=B_2$ and $-B_2-A_1=(2mV_0)/(ikℏ^2 )*A_1$

or$-ik=(mV_0)/ℏ^2$

or $sqrt{(2m|E|)}/ℏ=(mV_0)/ℏ^2$

or  $|E|=(mV_0^2)/2ℏ$

b)

Scattering on delta function ($V_0>0$):

We take wave incident from the right towards zero

$A_1=t$ (transmitted) and $A_2=0$ (no incoming particle from left);

$B_1=r$ (reflected) and $B_2=1$ (incoming wave from right ride towards zero)

From red conditions above we obtain

$(1+r=t)$

$(r-1)-t=(2mV_0)/(ikℏ^2 ) t)$

Solving we get

$(1+r=t)$

$-1+r=t((2mV_0)/(ikℏ^2 )+1)$

$(2r=t((2mV_0)/(ikℏ^2 )+2)@2r=2t-2$
$2t-2= (2mV_0)/(ikℏ^2 )*t+2t$
so that $t=(ikℏ^2)/(mV_0)$ and $r=t-1= (ikℏ^2)/(mV_0 )-1$
$tan ⁡α=-Im(r)/Re(r) =ℏk/(mV_0 )$