Delta Potential (Homework 10-325)

Consider a particle moving in a potential
$V(x)=\left\{\begin{matrix}\infty & \text{ for } x < 0 \\V_0\delta(x-a) & \text { for }x>a\end{matrix}\right.$
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  $y<-a$

Scattering on delta Potential

$=V_0 δ(y-0)$,  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
$(1+r=t)$
$-1+r=t((2mV_0)/(ikℏ^2 )+1)$ 
we add:
$(2r=t((2mV_0)/(ikℏ^2 )+2)@2r=2t-2$
So that
$2t-2= (2mV_0)/(ikℏ^2 )*t+2t$
so that $t=(ikℏ^2)/(mV_0)$ and $r=t-1= (ikℏ^2)/(mV_0 )-1$

The phase of the reflection coefficient is
$\tan ⁡α=-Im(r)/Re(r) =ℏk/(mV_0 )$


valentin68

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