Part a
The physics of a potential step for the case E>U0
U(x)=[+U00 x>0x<0
u(x)=eikx+Re−ikx
k=ℏ22mE
u′(x)=Te−ik′x
k′=ℏ22mE−U0
Continuity wave function =x=0 implies
1+R=T
[ikeikx−ikRe−ikx]x=0=[ik′Teik′x]x=0
k(1−R)=k′(1+R)
(k+k′)R=(k−k′)
R=k+k′k−k′
T=1−R=k+k′2k
u(x)=[eikxk+k′k−k′e−ikx x<0
k+k′2keik′x x>0
Preflection=∣R∣2=(k+k′k−k′)2
Ptransmission=1−Preflection=(k+k′)24kk′
The transmission probability goes to 1 k=k′ (since there is no step)
The transmission probability goes for 0 k′=0 (since the kinetic energy is zero)
Part b
P(x,x+Δx)≈∣Ψ(x,t)∣2δx.
P(x,x+Δx)=∫xx+Δx∣Ψ(x,t)∣2dx
P(−∞,+∞)=∫−∞∞∣Ψ(x,t)∣2dx=1.
P(∞,+∞)=∫−∞∞∣C∣2dx=1.
P(x=0,L)=∫0L∣C∣2dx=1.
C=L1 where L=2π
P(x=0,L/2)=∫0L/2∣2π1∣2
=(0.5π1)20.5π=0.25
Comments