Lecture 16: Scattering States and the Step Potential. 1 The Step Potential 1. 4 Wavepackets in the step potential 6

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1 Lecture 16: Scattering States and the Step Potential B. Zwiebach April 19, 2016 Contents 1 The Step Potential 1 2 Step Potential with E>V Step Potential with E<V Wavepacets in the step potential 6 1 The Step Potential Figure 1: The step potential. We now begin our detailed study of scattering states. These are un-noralizable energy eigenstates. They siply cannot be noralized, just lie oentu eigenstates. These energy eigenstates are not states of particles, one ust superpose scattering states to produce noralizable states that can represent a particle undergoing scattering in soe potential. Here we exaine the step potential (Figure 1), defined by ({ 0, x < 0, V (x) = (1.1) V 0, x 0. Our solutions to the Schrödinger equation with this potential will be scattering states of definite energy E. We can consider two cases: E>V 0 and E<V 0. In both cases the wavefunction extends infinitely to the left and is non-noralizable. Let us begin with the case E>V 0. 1

2 2 Step Potential with E > V 0 Figure 2: The energy E of the stationary state is greater than the step V 0. The full x axis is classically allowed. The stationary state with energy E is of the for Ψ(x, t) = ψ(x)e iet/, (2.2) and we will focus first on the unnown ψ(x). In order to write a proper ansatz for ψ(x) we visualize a physical process in which we have a wave incident on the step barrier fro the left. Given such a wave traveling in the direction of increasing x, we would expect a reflected wave and a transitted wave. The reflected wave, oving in the direction of decreasing x, would exist for x < 0. The transitted wave, oving in the direction of increasing x, would exist for x > 0. The ansatz for the energy eigenstate ust therefore contain all three pieces: ( Ae ix + Be ix x < 0, ψ(x) = (2.3) Ce ix x > 0. Recall that e ix, with > 0, represents a wave oving in the direction of increasing x, given the universal tie dependence above. Therefore A is the coefficient of the incident wave, B is the coefficient of the reflected wave, and C is the coefficient of the transitted wave. The waves for x < 0 have wavenuber and the wave for x > 0 has wavenuber. These wavenubers are fixed by the Schrödinger equation 2 2E 2(E V 0) =, 2 = 2. (2.4) 2 There are two equations that constrain our coefficients A, B, and C: both the wavefunction and its derivative ust be continuous at x = 0. With these two conditions we can solve for B and C in ters of A. This is all we could expect to do: because of linearity the overall scale of these three coefficients ust reain undeterined. In fact, we can thin of A as the input value and B and C as output values. Let us begin: ψ(x) ust be continuous at x = 0. Thus ψ (x) ust be continuous at x = 0. Thus A + B = C. (2.5) ia ib = ic! A B = C. (2.6) 2

3 Solving for B and C in ters of A, we get B C 2 =, =. (2.7) A + A + If A is real B and C are real. For E = V 0, we have = 0 and equations (2.7) give B = A and C =2A. Therefore, for E = V 0 the energy eigenstate is {( 2A cos(x) x<0, E = V 0 : ψ(x) = (2.8) 2A x>0, and loos lie this: Figure 3: Energy eigenstate for E = V 0. We get further insight into the solution by evaluating the probability current to the left and to the right of the x = 0 step. Recall the for of the probability current for a wavefunction ψ is ψ J = I ψ (2.9) x A short calculation shows that the current J L to the left of the step is ( 2 J L = jaj 2 jbj ) = J A J B, J = jaj 2 2 A, JB = jb. (2.10) j There is no interference arising fro the incident and reflected waves. The total current to the left of the step is siply the current J A associated with the incident wave inus the current J B associated with the reflected wave. The current J R to the right of the step is j R = jcj 2 = J C. (2.11) In any stationary solution there cannot be accuulation of probability at any region of space because the probability density ρ is anifestly tie-independent. While probability is continuously flowing ρ in scattering solutions, it ust be conserved. Fro the conservation equation J x + t = 0, the tie independence of ρ iplies that the current J ust be x-independent. In particular, our solution (2.7) ust iply that J L = J R. Let us verify this: ( 2 J L = ja j2 jbj 2 ( ( ) ) ) = 1 A 2 + j j ( 4 ) 4 2 (2.12) = ( + ) 2 jaj 2 = A 2 = ( + ) 2 j j jc j2 = J R, } {{ {z } 3 } C 2

4 6 as expected. The equality of J L and J R iplies that J A J B = J C! J A = J B + J C! JB JC 1 = +. (2.13) We now define the reflection coefficient R to be the ratio of the probability flux in the reflected wave to the probability flux in the incoing wave: J B jbj 2 ( ) 2 R = = 1 JA jaj 2 +. (2.14) This ratio happens to be the nor squared of the ratio B/A, and it is anifestly less than one, as it should be. We also define the transission coefficient T to be the ratio of the probability flux in the transitted wave to the probability flux in the incoing wave: J T C j = jc = =. (2.15) J A jaj 2 ( + ) 2 ( + ) 2 The above definitions are sensible because R and T, given in ters of current ratios, add up to one: J A R + T = 1, (2.16) C 2 A 2 as follows by inspection of (2.13). Note that T = because of wavenubers to the right and to the left of the step are not the sae. Recall that for E = V 0 we found = 0. In that case we have full reflection: R = 1 and T = 0. Indeed the probability current associated with the constant wavefunction that exists for x > 0 (see (2.8)) is zero. Additionally we can give an arguent fro continuity. The coefficients R and T ust be continuous functions of the energy E. For E < V 0 we expect T = 0 since the forbidden region is all of x > 0 and an exponentially decaying wavefuntion cannot carry probability flux. If T = 0 for any E < V 0 it ust still be zero for E = V 0, by continuity. J A 3 Step Potential with E < V 0 Figure 4: The step potential barrier. When E < V 0 the region x > 0 is a classically forbidden region. Let us try to solve for the energy eigenstate without re-doing all the wor involved in solving for B and C in ters of A. For this purpose we first note that the ansatz (2.3) for x < 0 can be left unchanged. On the other hand, for x > 0 the earlier solution ψ(x) = Ce ix, 2 2(E V 0 ) =, (3.17) 4 2

5 should becoe a decaying exponential ψ(x) = Ce κx, κ 2 2(V 0 E) =. (3.18) We note that the forer becoes the latter upon the replaceent 2! i κ. (3.19) This eans that we can siply perfor this replaceent in our earlier expressions for B/A and C/A and we obtain the new expressions. In particular fro (2.7) we get Therefore with B iκ = A + iκ (3.20) B i( iκ) κ + i = = A i( + iκ) κ i = e 2iδ(E), (3.21) E δ(e) = tan 1 ( ) = tan 1r ( ). (3.22) κ E V 0 Since the agnitude of A is equal to the agnitude of B, we have J A = J B and J C = 0. Thus T = 0 and R = 1. As noted above, the ratio B/A is a pure phase. The phase of the nuerator κ + i is δ(e) and the phase of the denoinator κ i is δ(e), thus the total phase 2δ(E) for the ratio. We did not absorb the inus sign into the phase; in this way δ(e)! 0 as E! 0. Note that δ(e) is positive and does not exceed π/2. In fact a setch of δ(e) is given in Figure 5. Figure 5: The phase δ(e) as a function of energy E < V 0. The total wavefunction for x < 0 is interesting ψ(x) = Ae ix + ( Ae 2iδ(E) )e ix = Ae iδ(e) ( e iδ(e) e ix e iδ(e) e ix ) (3.23) = 2iAe iδ(e) sin(x δ(e)) This eans that the probability density is jψj 2 = 4A 2 sin 2 (x δ(e)). (3.24) 5

6 The point x 0 > 0 deterined by the condition x 0 = δ(e) is the point in the forbidden region where the extrapolation of the allowed-region solution would vanish. Of course in the forbidden region x > 0, the probability density jψj 2 is a decaying exponential. Figure 6: Nor squared for the energy eigenstate when E < V 0. For x > 0 the probability density decays exponentially with x. The point x 0 is the point where the extrapolation of the x < 0 probability density would have vanished. For future use we record the derivative of the phase δ(e) with respect to the energy s dδ(e) 1 1 δ (E) =. (3.25) de 2 E(V 0 E) Note that this derivative becoes infinite both for E! 0 and for E! V 0. Figure 7: The derivative δ (E) as a function of energy E < V 0. 4 Wavepacets in the step potential Now we exaine the ore physical scenario. As we ve seen with the free particle, the stationary states are not noralizable, and physical particles are actually represented by wavepacets built with an infinite superposition of oentu eigenstates. We can do siilarly thing with our energy eigenstates. We will consider energy eigenstates with E > V or equivalently with 2 > ˆ 0 2, where 2 2E 2V 0 = > ˆ 2, (4.26) 2 2 and we will superpose the. To begin we write the energy eigenstates in a slightly different for, including the tie dependence. Setting A = 1 and using the values for the ratios B/A and C/A we find the solution 8 ( ) < e ix + e ix e ie()t/, x < 0, + Ψ(x, t) = (4.27) : 2 e ix e ie()t/, x >

7 We can for a superposition of these solutions by ultiplying by a function f() and integrating over 8 R ( ) < d f() e ix + e ix ie ˆ e ()t/, x < 0, Ψ(x, t) = : R + (4.28) ˆ d f() 2 eix t/ e ie(), x > 0. + Here f() is a real function of that is essentially zero except for a narrow pea at = 0. Note that we have only included oentu coponents with energy greater than V 0 by having the integral s lower liit set equal to. ˆ The integral only runs over positive because only in that case the e ix waves are oving towards positive x, and are therefore genuine incident waves. The above is guaranteed to be a solution of the Schrödinger equation. We can split the solution into incident, reflected and transitted waves, as follows. 8 < Ψ inc (x, t) + Ψ ref (x, t), x < 0, Ψ(x, t) = (4.29) : Ψ trans (x, t), x > 0. Naturally both Ψ inc (x, t) and Ψ ref (x, t) exist for x < 0 and Ψ trans (x, t) exists for x > 0. We then have, explicitly Z Ψ inc (x < 0, t) = d f()e ix e ie()t/, ˆ Z ( ) Ψ ref (x < 0, t) = d f() e ix e ie()t/, (4.30) ˆ + Z ( ) 2 Ψ trans (x > 0, ) = t d f() e ix e ie()t/. ˆ + How does the pea of Ψ inc (x, t) ove? For this we loo for the ain contribution to the associated integral which occurs when the total phase in the integrand is stationary for 0. We therefore require ( d 2 2 ) t 0 0 x d 2 = 0! x t = 0 =) x = t. (4.31) 0 This is the relation between t and x satisfied by the pea of Ψ inc. It describes a pea oving with constant velocity 0 / > 0. Since Ψ inc (x, t) requires that x < 0, the above condition shows that we get the pea only for t < 0. The pea of the pacet gets to x = 0 at t = 0. For t > 0, Ψ inc (x, t) is not zero, but it ust be rather sall, since the stationary phase condition cannot be satisfied for any x in the doain x < 0. Consider now Ψ ref (x, t). This tie the stationary phase condition is d ( 2 2 ) t x = 0! x + t = 0 = ) x = d t. (4.32) The relation represents a pea oving with constant negative velocity 0 /. Since Ψ ref (x, t) requires that x < 0, the above condition shows that we get the pea only for t > 0, as it befits a reflected wave. For t > 0, Ψ ref (x, t) is not zero, but it ust be rather sall, since the stationary phase condition cannot be satisfied for any x in the doain x < 0. 7

8 Finally, let us consider Ψ trans. The stationary phase condition reads: d ( x 2 2 t = 0 d 2 ) d x 0 t = 0 (4.33) 0 d 0 Using 2 = 2 2V0 2 d d =, (4.34) bac to the earlier equation we quicly find that Transitted wave pea: x = t, (4.35) with evaluated for = 0. Since x > 0 is the doain of Ψ trans this describes a pea oving to the right with velocity / for t > 0. For t < 0, Ψ trans (x, t) is not zero, but it ust be rather sall, since the stationary phase condition cannot be satisfied for any x in the doain x > 0. In suary, for large negative tie Ψ inc doinates and both Ψ ref and Ψ trans are very sall. For large positive tie, both Ψ ref and Ψ trans doinate and Ψ inc becoes very sall. These situations are setched in figures 8 and 9. Of course for sall ties, positive or negative, all three waves exist and together they describe the coplex process of collision with the step in which a reflected and a transitted wave are generated. Figure 8: At large negative ties an incoing wavepacet is traveling in the +x direction. Figure 9: At large positive ties we have a reflected wavepacet traveling in the xˆ direction and the transitted wavepacet traveling in the +xˆ direction. Let us now exaine a wavepacet built with energies E < V 0. Recall that in this situation B/A = e 2iδ(E). Therefore for an incident wave, all of whose oentu coponents have energy less 8

9 than V 0, Ψ inc (x < 0, t) = the associated reflected wavefunction is Ψ ref (x < 0, t) = Z 0 Z ˆ ˆ 0 d f() e ix e iet/, (4.36) d f() e 2iδ(E) e ix e iet/. (4.37) Using the ethod of stationary phase again to find the evolution of the pea, d ( Et ) 2 0 0t 2δ(E) x = 0! 2δ (E) x d = 0. (4.38) 0 Fro this we quicly find 0 ( x = t 2 δ ) (E), (4.39) where the derivative is evaluated at E( 0 ). The reflected wave pacet is oving towards ore negative x as tie grows positive. This is as it should. But there is a tie delay associated with the reflected pacet, evident when we copare the above equation with x = 0 t. The tie delay is given by tie delay = 2 δ (E). (4.40) The derivative δ (E) was evaluated in (3.25) and it is positive. We see that the delay is particularly large for wave pacets of little energy or those with energies just below V 0. We conclude the analysis of the step potential by discussing what it eans to observe the particle in the forbidden region. It would be contradictory if the observer could ae the following two stateents: 1. The particle is located in the forbidden region. 2. The particle has energy less than V 0. Both stateents taen to hold siultaneously would iply that the particle has negative inetic energy, soething that is inconsistent. In particular with E < V 0 we would have a negative inetic of agnitude V 0 E. Figure 10: The step potential with potential energy V 0. If we could observe a particle in the forbidden region with energy E then the inetic energy would be negative. 9

10 First note that in the solution the particle penetrates into the forbidden region a distance of about 1/κ, where, you will recall that 2 2(V 0 E) κ =. (4.41) 2 To be sure the particle is in the forbidden region its position uncertainty x ust be saller than the penetration depth: 1 x. (4.42) κ The particle acquires soe oentu p due to the position easureent: p κ. (4.43) x Due to this oentu induced by the position easureent there is soe additional contribution E to the inetic energy p 2 2 κ 2 E = = V 0 E, (4.44) 2 2 where we used (4.41). Fro this inequality we find that the total energy will exceed V 0 E tot = E + E E + (V 0 E) = V 0. (4.45) While the arguent is heuristic, it gives soe evidence that no negative inetic energy will be detected for a particle that is found in the forbidden region. Sarah Geller transcribed Zwiebach s handwritten notes to create the first LaTeX version of this docuent. 10

11 MIT OpenCourseWare Quantu Physics I Spring 2016 For inforation about citing these aterials or our Ters of Use, visit:

Chapter 17: Resonant transmission and Ramsauer Townsend. 1 Resonant transmission in a square well 1. 2 The Ramsauer Townsend Effect 3

Contents Chapter 17: Resonant transmission and Ramsauer Townsend B. Zwiebach April 6, 016 1 Resonant transmission in a square well 1 The Ramsauer Townsend Effect 3 1 Resonant transmission in a square well