Part a
The physics of a potential step for the case E > U 0 E>U_0 E > U 0
U ( x ) = [ + U 0 0 x > 0 x < 0 U(x)=[_{+U_0}^0 \space _{x>0}^{x<0} U ( x ) = [ + U 0 0 x > 0 x < 0
u ( x ) = e i k x + R e − i k x u(x)=e^{ikx}+Re^{-ikx} u ( x ) = e ik x + R e − ik x
k = 2 m E ℏ 2 k=\sqrt{\frac{2mE}{\hbar^2}} k = ℏ 2 2 m E
u ′ ( x ) = T e − i k ′ x u'(x)=Te^{-ik'x} u ′ ( x ) = T e − i k ′ x
k ′ = 2 m E − U 0 ℏ 2 k'=\sqrt{\frac{2mE-U_0}{\hbar^2}} k ′ = ℏ 2 2 m E − U 0
Continuity wave function = x = 0 =x=0 = x = 0 implies
1 + R = T 1+R=T 1 + R = T
[ i k e i k x − i k R e − i k x ] x = 0 = [ i k ′ T e i k ′ x ] x = 0 [ike^{ikx}-ikRe^{-ikx}]_{x=0}=[ik'Te^{ik'x}]_{x=0} [ ik e ik x − ik R e − ik x ] x = 0 = [ i k ′ T e i k ′ x ] x = 0
k ( 1 − R ) = k ′ ( 1 + R ) k(1-R)=k'(1+R) k ( 1 − R ) = k ′ ( 1 + R )
( k + k ′ ) R = ( k − k ′ ) (k+k')R=(k-k') ( k + k ′ ) R = ( k − k ′ )
R = k − k ′ k + k ′ R= \frac{k-k'}{k+k'} R = k + k ′ k − k ′
T = 1 − R = 2 k k + k ′ T=1-R= \frac{2k}{k+k'} T = 1 − R = k + k ′ 2 k
u ( x ) = [ e i k x k − k ′ k + k ′ e − i k x u(x)=[^{e^{ikx}} {\frac{k-k'}{k+k'}}e^{-ikx} u ( x ) = [ e ik x k + k ′ k − k ′ e − ik x x < 0 x<0 x < 0
2 k k + k ′ e i k ′ x \frac{2k}{k+k'}e^{ik'x} k + k ′ 2 k e i k ′ x x > 0 x>0 x > 0
P r e f l e c t i o n = ∣ R ∣ 2 = ( k − k ′ k + k ′ ) 2 P_{reflection}=|R|^2=(\frac{k-k'}{k+k'})^2 P re f l ec t i o n = ∣ R ∣ 2 = ( k + k ′ k − k ′ ) 2
P t r a n s m i s s i o n = 1 − P r e f l e c t i o n = 4 k k ′ ( k + k ′ ) 2 P_{transmission}=1-P_{reflection}=\frac{4kk'}{(k+k')^2} P t r an s mi ss i o n = 1 − P re f l ec t i o n = ( k + k ′ ) 2 4 k k ′
The transmission probability goes to 1 k = k ′ k=k' k = k ′ (since there is no step)
The transmission probability goes for 0 k ′ = 0 k'=0 k ′ = 0 (since the kinetic energy is zero)
Part b
P ( x , x + Δ x ) ≈ ∣ Ψ ( x , t ) ∣ 2 δ x . P(x,x+Δx)≈|Ψ(x,t)|^2δx. P ( x , x + Δ x ) ≈ ∣Ψ ( x , t ) ∣ 2 δ x .
P ( x , x + Δ x ) = ∫ x x + Δ x ∣ Ψ ( x , t ) ∣ 2 d x P(x,x+Δx)=∫^{x+Δx}_x|Ψ(x,t)|^2dx P ( x , x + Δ x ) = ∫ x x + Δ x ∣Ψ ( x , t ) ∣ 2 d x
P ( − ∞ , + ∞ ) = ∫ − ∞ ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 1. P(−∞,+∞)=∫^∞_{−∞}|Ψ(x,t)|^2dx=1. P ( − ∞ , + ∞ ) = ∫ − ∞ ∞ ∣Ψ ( x , t ) ∣ 2 d x = 1.
P ( ∞ , + ∞ ) = ∫ − ∞ ∞ ∣ C ∣ 2 d x = 1. P(∞,+∞)=∫^∞_{−∞}|C|2dx=1. P ( ∞ , + ∞ ) = ∫ − ∞ ∞ ∣ C ∣2 d x = 1.
P ( x = 0 , L ) = ∫ 0 L ∣ C ∣ 2 d x = 1. P(x=0,L)=∫^L_0|C|^2dx=1. P ( x = 0 , L ) = ∫ 0 L ∣ C ∣ 2 d x = 1.
C = 1 L C=\sqrt\frac{1}{L} C = L 1 where L = π 2 L= \frac{\pi}{2} L = 2 π
P ( x = 0 , L / 2 ) = ∫ 0 L / 2 ∣ 1 π 2 ∣ 2 P(x=0,L/2)=∫^{L/2}_0 |\sqrt{\frac{1}{\frac{\pi}{2}}}|^2 P ( x = 0 , L /2 ) = ∫ 0 L /2 ∣ 2 π 1 ∣ 2
= ( 1 0.5 π ) 0.5 π 2 = 0.25 =(\frac{1}{0.5 \pi}) \frac{0.5 \pi}{2} = 0.25 = ( 0.5 π 1 ) 2 0.5 π = 0.25
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