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

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

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.

Case 1 V = infinity for x > 0 and V = 0 for x < 0. -(  2 /2M) d 2  dx 2 = E  for x<0. What is the solution?  (x) = A sin(k x) where k = (2 M E) 1/2 /  How to interpret? [Hint: use e ix = cos(x) + i sin(x) which gives cos(x) = (e i x + e -i x )/2 & sin(x) = (e i x – e -i x )/(2i) ]

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

Case 2, E < V 0 V = V 0 for x > 0 and V = 0 for x < 0. -(  2 /2M) d 2  dx 2 = E  for x 0. Region II III

Case 2, E < V 0 -(  2 /2M) d 2  dx 2 = E  for x 0. Region II    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(-  x)  = (2 M [V 0 – E]) 1/2 /  Why can choose coefficient in Region II to be 1? Wave function must be continuous at x = 0 and 1 st derivative of wave function must be continuous at x=0. Why not exp(  x) in Region II? How to solve for A & B?

Case 2, E < V 0    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(-  x)  = (2 M [V 0 – E]) 1/2 /   I (0) =  II (0)  A + B = 1  ’ I (0) =  ‘ II (0)  i k 1 (A – B) = -   A – B = i  k 1 A = (k 1 + i  /(2 k 1 ) & B = (k 1 – i  /(2 k 1 ) Continuity of wave function Continuity of 1 st derivative 2 equations and 2 unknowns so can solve

Case 2, E < V 0    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(-  x)  = (2 M [V 0 – E]) 1/2 /  Amount moving to right |A| 2 = (k  2 )/(4 k 1 2 ) Amount moving to left |B| 2 = (k  2 )/(4 k 1 2 ) This implies complete reflection Note B = -A in the limit  goes to infinity.    = 2 i A sin(k 1 x) Note  II (x) is not 0 (  (x) not 0 for x>0); probability to be where E < V! But gets small exponentially fast. What would happen if  could go to 0?

Case 2, E > V 0 V = V 0 for x > 0 and V = 0 for x < 0. -(  2 /2M) d 2  dx 2 = E  for x 0. Region II originally

Case 2, E > V 0 -(  2 /2M) d 2  dx 2 = E  for x 0. Region II    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(i k 2 x) k 2 = (2 M [E – V 0 ]) 1/2 /  Why can choose coefficient in Region II to be 1? Wave function must be continuous at x = 0 and 1 st derivative of wave function must be continuous at x=0. How to solve for A & B?

Case 2, E > V 0    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(i k 2 x) k 2 = (2 M [E – V 0 ]) 1/2 /   I (0) =  II (0)  A + B = 1  ’ I (0) =  ‘ II (0)  i k 1 (A – B) = i k 2   A – B = k 2  k 1 A = (k 1 + k 2  /(2 k 1 ) & B = (k 1 – k 2  /(2 k 1 ) Continuity of wave function Continuity of 1 st derivative 2 equations and 2 unknowns so can solve

Case 2, E < V 0    (x) = A exp(i k 1 x) + B exp(-i k 1 x) k 1 = (2 M E) 1/2 /   II (x) = exp(i k 2 x) k 2 = (2 M [E – V 0 ]) 1/2 /  Amount moving to right |A| 2 = (k 1 + k 2 ) 2 /(4 k 1 2 ) Amount moving to left |B| 2 = (k 1 – k 2 ) 2 /(4 k 1 2 ) Fraction reflected R = |B| 2 /|A| 2 = (k 1 – k 2 ) 2 / (k 1 + k 2 ) 2 Note fraction reflected goes to 1 if either k goes to 0 For E >> V 0 the reflected fraction decreases like 1/E 2 The transmitted fraction, T = 1 – R, is not equal to 1/|A| 2 Why?