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Reflection and Transmission of Gwosters in Different Potentials

This text explains the reflection and transmission probabilities of gwosters in two different potentials. It also discusses unbounded motion, the treatment of eigenstates, and the solutions for wave functions in different potential regions.

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Reflection and Transmission of Gwosters in Different Potentials

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  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?

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