Question #1

1. Question #1 A gwoster travels to the left with an energy, E. It is in a potential that is 0 for x>0 and V0 > E for x<0. The probability it will be reflected and travel to the right is • 0% • between 0% and 100% • 100%

2. Question #2 A gwoster travels to the left with an energy, E. It is in a potential that is 0 for |x|>L and V0 > E for |x|<L. The probability it will be reflected and travel to the right is • 0% • between 0% and 100% • 100%

3. Classically Unbounded Motion For the case where E > V(x) as x goes to +infinity or –infinity the motion is unbounded. The eigenstate oscillates for infinite distance. Consequence: can’t be normalized in usual way. How to treat this case? Don’t worry about normalization but compare “pieces” of the wave function to obtain physical properties.

4. Case 1 V = infinity for x > 0 and V = 0 for x < 0. -(h2/2M) d2y/dx2 = E y for x<0.What is the solution? y(x) = A sin(k x) where k = (2 M E)1/2 / h How to interpret? [Hint: use eix = cos(x) + i sin(x) which gives cos(x) = (ei x + e-i x)/2 & sin(x) = (ei x – e-i x)/(2i) ]

5. Case 1 y(x) = A sin(k x) where k = (2 M E)1/2 / h sin(k x) = (ei k x – e-i k x)/(2i) y(x) = (A/2i) ei k x + (-A/2i) e-i k x Amount moving to right |A/2i|2 = A2/4Amount moving to left |-A/2i|2 = A2/4 Since amounts are the same, implies total reflection Why take | |2 ?

6. Case 2, E < V0 V = V0 for x > 0 and V = 0 for x < 0. I II -(h2/2M) d2y/dx2 = E y for x<0. Region I-(h2/2M) d2y/dx2 = (E-V0) y for x>0. Region II

7. Case 2, E < V0 -(h2/2M) d2y/dx2 = E y for x<0. Region I-(h2/2M) d2y/dx2 = (E-V0) y for x>0. Region II yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(-a x) a = (2 M [V0 – E])1/2 /h Why can choose coefficient in Region II to be 1? Why not exp(a x) in Region II? How to solve for A & B? Wave function must be continuous at x = 0 and 1st derivative of wave function must be continuous at x=0.

8. Case 2, E < V0 yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(-a x) a = (2 M [V0 – E])1/2 /h Continuity of wave function yI(0) = yII(0) a A + B = 1 Continuity of 1st derivative y’I(0) = y‘II(0) a i k1 (A – B) = -a a A – B = i a/k1 2 equations and 2 unknowns so can solve A = (k1 + i a)/(2 k1) & B = (k1 – i a)/(2 k1)

9. Case 2, E < V0 yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(-a x) a = (2 M [V0 – E])1/2 /h Amount moving to right |A|2 = (k12 + a2)/(4 k12)Amount moving to left |B|2 = (k12 + a2)/(4 k12) This implies complete reflection Note B = -A in the limit a goes to infinity.yI= 2 i A sin(k1 x) Note yII(x) is not 0 (y(x) not 0 for x>0); probability to be where E < V! But gets small exponentially fast. What would happen if h could go to 0?

10. Case 2, E > V0 V = V0 for x > 0 and V = 0 for x < 0. originally -(h2/2M) d2y/dx2 = E y for x<0. Region I-(h2/2M) d2y/dx2 = (E-V0) y for x>0. Region II

11. Case 2, E > V0 -(h2/2M) d2y/dx2 = E y for x<0. Region I-(h2/2M) d2y/dx2 = (E-V0) y for x>0. Region II yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /h Why can choose coefficient in Region II to be 1? How to solve for A & B? Wave function must be continuous at x = 0 and 1st derivative of wave function must be continuous at x=0.

12. Case 2, E > V0 yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /h Continuity of wave function yI(0) = yII(0) a A + B = 1 Continuity of 1st derivative y’I(0) = y‘II(0) a i k1 (A – B) = i k2a A – B = k2/k1 2 equations and 2 unknowns so can solve A = (k1 + k2)/(2 k1) & B = (k1 – k2)/(2 k1)

13. Case 2, E < V0 yI(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 / h yII(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /h Amount moving to right |A|2 = (k1 + k2)2/(4 k12)Amount moving to left |B|2 = (k1 – k2)2/(4 k12) Fraction reflected R = |B|2/|A|2 = (k1 – k2)2/ (k1 + k2)2 Note fraction reflected goes to 1 if either k goes to 0 For E >> V0 the reflected fraction decreases like 1/E2 The transmitted fraction, T = 1 – R, is not equal to 1/|A|2Why?