# Step Potential

1. Nov 18, 2004

### UpQuark

Ok, I am need of some serious help here. ,I don't want an answer just some guidance.
In the attachment you'll see a diagram of a step potential. The partical is travelling from the left. There is nothing incident from the right. We need to find the probability current density in region 2 in terms of V1, V2, a, E, m, and the incident probability current density.

What I have (hopefully correctly so far) is that we need seven equations to find the seven variables.

Simplified I've come up with,
K1=sqrt(2mE)/(hbar)
K2=sqrt(2m(E+V1))/(hbar)
K3=sqrt(2m(E-V2))/(hbar)

psi(x)=Aexp(i(K1)x) + Bexp(-i(K1)x) from negative infinity to zero
psi(x)=Cexp(i(K2)x) + Dexp(-i(K2)x) from zero to a
psi(x)=Eexp(i(K3)x) + Fexp(-i(K3)x) from a to infinity
also note that the scattering matrix is
S(Energy)=Aexp(i(K1)x) - Dexp(-i(K2)x) - Fexp(-i(K3)x) from a to infinity

Setting the derivatives from either side equal and setting up to solve the constants we find,

K1(A-B)=K2(C-D)
C(K2)exp(i(K2)a)=E(K3)exp(i(K3)a) - F(K3)exp(-1(K3)a)
A-D-F=AS(Energy)
E(K3)exp(i(K3)a) - F(K3)exp(-i(K3)a)=AS(Energy)(K1)exp(i(K1)a)
S(Energy)=Aexp(i(K1)a) - Dexp(-i(K2)a) - Fexp(-i(K3)a)

those I'm pretty sure of, except maybe in third and fifth equations adding in the B term..... not sure though..... and lastly, the ones that I keep changing my mind over,
A-B=C+D
C-D=E+F

should that read A+B=C and C+D=E? perhaps, but it seems that the transmission coefficient minus the reflection coefficient of one region should equal the transmission coefficient for the next.

If this much is right, I'll go on and use a computer to solve the system for each variable and then, go ahead solve for the probability current density in region 2 using
J2= abs(S(Energy)^2))J1
where J1 is the initial probability current density and J2 is the final probability current density.

If anyone can throw me a bone that'd be great.

#### Attached Files:

• ###### step.doc
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2. Nov 18, 2004

### xepma

Well, my first impression is that you have too much constants A,B,C etc.

I'm guessing that for the region where V = 0, you should get psi[x] = A exp[kx] + B exp[-kx]. So in that case A must be 0, because that term blows up when x -> minus infinity. This also applies for the region V2. So then you will get 4 constants, which you can solve by stating that psi[x] and it's derative must be continious (which you already did ofcourse).