question #1
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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 xTRANSCRIPT
Question #1A 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
(a) 0%
(b)between 0% and 100%
(c) 100%
Question #2A 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
(a) 0%
(b)between 0% and 100%
(c) 100%
Classically Unbounded MotionFor 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 1V = infinity for x > 0 and V = 0 for x < 0.
-(2/2M) d2dx2 = 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 eix = cos(x) + i sin(x) which gives cos(x) = (ei x + e-i x)/2 & sin(x) = (ei x – e-i x)/(2i) ]
Case 1
(x) = A sin(k x) where k = (2 M E)1/2 /
sin(k x) = (ei k x – e-i k x)/(2i)
(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 ?
Case 2, E < V0
V = V0 for x > 0 and V = 0 for x < 0.
-(2/2M) d2dx2 = E for x<0. Region I-(2/2M) d2dx2 = (E-V0) for x>0. Region II
I II
Case 2, E < V0
-(2/2M) d2dx2 = E for x<0. Region I-(2/2M) d2dx2 = (E-V0) for x>0. Region II(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(- x) = (2 M [V0 – E])1/2 /
Why can choose coefficient in Region II to be 1?
Wave function must be continuous at x = 0 and 1st 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 < V0
(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(- x) = (2 M [V0 – E])1/2 /
I(0) = II(0) A + B = 1
’I(0) = ‘II(0) i k1 (A – B) = - A – B = i k1
A = (k1 + i /(2 k1) & B = (k1 – i /(2 k1)
Continuity of wave function
Continuity of 1st derivative
2 equations and 2 unknowns so can solve
Case 2, E < V0
(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(- x) = (2 M [V0 – E])1/2 /
Amount moving to right |A|2 = (k12 + 2)/(4 k1
2)Amount moving to left |B|2 = (k1
2 + 2)/(4 k12)
This implies complete reflection
Note B = -A in the limit goes to infinity.= 2 i A sin(k1 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 > V0
V = V0 for x > 0 and V = 0 for x < 0.
-(2/2M) d2dx2 = E for x<0. Region I-(2/2M) d2dx2 = (E-V0) for x>0. Region II
originally
Case 2, E > V0
-(2/2M) d2dx2 = E for x<0. Region I-(2/2M) d2dx2 = (E-V0) for x>0. Region II(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /
Why can choose coefficient in Region II to be 1?
Wave function must be continuous at x = 0 and 1st derivative of wave function must be continuous at x=0.
How to solve for A & B?
Case 2, E > V0
(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /
I(0) = II(0) A + B = 1
’I(0) = ‘II(0) i k1 (A – B) = i k2 A – B = k2k1
A = (k1 + k2/(2 k1) & B = (k1 – k2 /(2 k1)
Continuity of wave function
Continuity of 1st derivative
2 equations and 2 unknowns so can solve
Case 2, E < V0
(x) = A exp(i k1 x) + B exp(-i k1 x) k1 = (2 M E)1/2 /
II(x) = exp(i k2 x) k2 = (2 M [E – V0])1/2 /
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|2
Why?