8.04 Quantum Physics Lecture XVOne-dimensional potentials: potential stepFigure I: Potential step of height V0. The particle is incident from the left with energy E. We analyze a time independent situation where a current of particles with a well-defined energy is incident on the barrier. The time-independent SE is ˆHu(x)= Eu(x) (15-1) h¯2 d2u −2m dx2 (x)+ V (x)u(x)= Eu(x) (15-2) d2u 2m dx2 = − h¯2 [E − V (x)]u(x) (15-3) Qualitative features of solutions for regions of constant V1: If E − V1 > 0, the solutions are of the form e±ik1x with ¯h2k2 = E − V1, k1 real.2m ¯Interpretation. h2k2 is the KE of the particle with total energy E in a region of 2m potential V1, the e±ikx wavefunctions correspond to particles traveling left /right. Figure II: In a region where the particle energy is greater than the (constant) potential, the solutions of the SE are plane waves e±ikx, where E − V1 =¯h2k2/2m is the kinetic energy of the particle in that region. If E − V1 < 0, the solutions are of the form e±κ1x with ¯h22κm 12 = V1 − E, κ1 real. These are damped exponentials with a decay length constant κ1 (decay length κ1−1), where ¯h2κ12 = V1 − E represents the “missing” kinetic energy of the particle 2m As EV , the decay length κ1−1 becomes longer and longer. → Massachusetts Institute of Technology XV-1 8.04 Quantum Physics Lecture XVFigure III: In a region where the particle energy is less than the (constant) potential, the solutions of the SE are exponentially growing or decaying functions, e±κx, where V1 − E =¯h2κ2/2m is the ”missing kinetic energy” of the particle in that region. Figure IV: When a light wave experiences total internal reflection on a glass-vacuum interface, an evanescent (non-traveling, exponentially decaying wave) builds up inside the vacuum. The closer we are to the critical angle for total internal reflection, the longer the decay length of the evanescent wave. This phenomenon is analogous to a particle entering a classically forbidden region with V1 >E. The less forbidden the region, the longer the decay length. Note. There is a non-zero probability to find the particle with energy E in a “classiccall forbidden region” with EV0: define Massachusetts Institute of Technology XV-2 8.04 Quantum Physics Lecture XVFigure V: The light field ”tunneling” through the forbidden region can be detected as it emerges on the other side in a second prism. Figure VI: As a particle tunnels through a barrier and emerges from the other side, the energy E and the Broglie wavelength 2π/k remain the same. The amplitude of the emerging wave is smaller than that of the incident wave. Figure VII: Potential step ¯h2k2 2m ¯h2 q2 2m = E = E − V0 (KE in region x < 0) (KE in region x > 0) (15-4) (15-5) Massachusetts Institute of Technology XV-3 � � � � � � � ¯ � � 8.04 Quantum Physics Lecture XVThe most general solution is Aeikx + Be−ikx in the region x< 0 (15-6)Ceiqx + De−iqx in the region x> 0 (15-7)If we choose as the initial condition a particle incident from the left (A = 0), then the particle can be transmitted to the RHS (C = 0), or, as we shall see, partially reflected by the barrier in spite of E>V0 (B = 0). However, if no particle is incident from the right then D = 0. Calculate the particle current (or flux) In region x< 0: h¯du du∗ j< =2im u∗ dx − dx u (15-8) h¯��� �� =2im A∗e−ikx + B∗e ikx ikAeikx − ikBe−ikx − c.c. (15-9) =2hkm |A|2 + AB∗e 2ikx − A∗Be−2ikx −|B|2 − c.c. (15-10) hk¯� 22 � = m |A|−|B|→ net current for x< 0 (15-11) We define the reflection amplitude r = B , and the reflection coefficient as R = r2 = ��A ||�B �2 .A For x> 0: ¯hqj> = m |C|2 (15-12) Continuity of wavefunction at x = 0: ψ(x 0) = A + B = ψ(x 0) = C (15-13) →← In spite of the potential step, the derivative of the wavefunction must also be continuoous � du � � du � � � d � du � dx x=� − dx x=−� = −� dx dx 2m � � dx (15-14) = − ¯h2 −� dx[E − V (x)]u(x) = 0 (15-15) For future applications, we note that if the potential contains a delta function term λδ(x − a), with some magnitude of the delta function λ, then the same calculation Massachusetts Institute of Technology XV-4 � � � � 8.04 Quantum Physics Lecture XVgives���� � a+�du du 2m dx x=a+� − dx x==a−� = h¯2 a−� dxλδ(x − a)u(λ) (15-16) 2m = h2 λu(a) (15-17) ¯To summarize, we have the following rules: Rule 1. The wavefunction u(x) is always continuous duRule 2. The first spatial derivative of the wavefunction dx is continuous if the potenntia does not contain δ-function like terms. (It may contain potential steps). Rule 2.1. if the potential contains a term λδ(x − a), the the first derivative du isdx discontinuous at x = a amnd satisfies the relation du du 2m dx x=a+� − dx x=a−� = h¯2 λu(a) (15-18) Figure VIII: A discontinuity in the slope of the wavefunction occurs at a delta function potential. The difference in wavefunction slopes is proportional to the strength of the δ potential, and to the value of the wavefunction at the cusp. Continuity of ψ: A + B = C (15-19)Continuity of ψ�: ik(A − B)= iqC (15-20)Massachusetts Institute of Technology XV-5 ���������������������(A − B) 8.04 Quantum Physics Lecture XV Solve for B, C in terms of A k C = A + B = (15-21) qk k A 1 − q = −B 1+ q (15-22) Aq − k = −Bq + k (15-23) qq B = A (15-24) k + q 2 � k − qk − qCABAAA (15-25) ++=== k + qk + q Reflection amplitude r = B = k − q (15-26) Ak + q C 2k Transmission amplitude t = = (15-27) Ak + q�22Bk − qReflection coefficient r2 (15-28)||==Ak + q2 4k2CATransmission coefficient t2 (15-29)||==(k + q)2 �2hk Reflection current j= hk¯B2 =¯k − qA2 (15-30) ← m ||mk + q ||¯2 ¯4kq 22hq hk Transmission current j→,x>0 = m |C|= m (k + q)2 |A|(15-31) hk hk Net current for x< 0 j< =¯m (|A|2 −|B|2)= ¯m |A|2 (k 4+ kq q)2 (15-32) hq hk Net current for x> 0 j> =¯m |C|2 =¯m (k 4+ kq q)2 |A|2 (15-33) The current obeys the continuity equation (see problem set) ∂j ∂ ∂x + ∂t|ψ|2 = 0 (15-34) Here we are considering stationary states, ∂ ψ2 = 0 (no change of probability density ∂t ||in time), = j = const, current is continuous across the potential step, ⇒ j< = j>, (15-35) Massachusetts Institute of Technology XV-6 8.04 Quantum Physics Lecture XVor ¯hk 2jinc = j→,x<0 = m |A|= jrefl + jtrans (15-36) = j,x<0 + j,x>0 (15-37) ←→hk¯2 ¯2hq = m |B|+ m |C|. (15-38) Note. |r|2 + |t|2 = 1 because the particle velocity is different for �x> 0 from that for x< 0. Discussion of results In contrast to classical mechanics, there is some reflection at the potential step even though the energy of the particle is sufficient to surpass it. This is familiar from optics, where a step-like change in the index of refraction (e.g., air-glass interface) leads to partial reflection. The particle reflection is a consequence of the matching of the wavefunction and its derivative at the boundary. Again, this is similar to optics where the matching of th electromagnetic fields at the boundary results in a reflected field. Note. For a very smooth change of potential (or refractive index in optics) there is not reflection. What is smooth? A change over many wavelengths. Changes of the potential over a distance l short compared to a wavelength λ = 2kπ result in reflection. Slow changes of potential over many λ do not result in reflection if particle energy exceeds barrier height. Figure IX: A potential that varies smoothly over many de Broglie wavelengths does not produce partial reflection if the particle energy is sufficient to surpass it. Intermediate region lλ: we expect resonance phenomena (non-monotonic ∼changes of reflection probability with particle energy). For the potential step, the Massachusetts Institute of Technology XV-7 ���������������������������������� � � 8.04 Quantum Physics Lecture XVreflection probability |r|22 → 0 for k → q (E � V1), and (15-39) |r|→ 1 for q → 0(E � V1), as expected. (15-40) (15-41) Interestingly, the reflection probability can be written as |r|2 =√E −√E − V1 �2 √E + √E − V1 (15-42)i.e. it does not depend explicitly on ¯h. However, the reflection is still inherenetly nonclasssica in that the potential needs to change abruptly compared to the particle’s de Broglie wavelength, that depends on ¯h. Solution for E 0 (15-46) The e+κx term is not normalizable, D =0 We can go through the same procedure as before using the continuity of ψ1ψ� at x = 0, or use the previous calculation if we set q iκ (Ceiqx Ce−κx then).→→Consequently, 2 2 k2 + κ2BAk − iκ2 = 1 (15-47)|r|===k2 + κ2k + iq2 2 4k2 + κ2CA2kk + iκ2 = 0 (15-48) |t|===k2 + κ2 (15-49) A part of the wave penetrates the barrier, which is why the ’transmission’ amplitude does not vanish. Note, however, that there is no associated particle current: Since Ce−kx does not have a spatially varying phase, the particle current j = ¯h 2im ψ∗ ∂ψ ∂x − c.c. (15-50) Massachusetts Institute of Technology XV-8 8.04 Quantum Physics Lecture XVvanishes for x> 0, ¯hk 2j< = m (|A|−|B|2) = 0 (15-51) j> = 0 (15-52) The net current is zer0 in steady-state because all particles are reflected. Note. The reflected wave has an energy-dependent phase shift r = B = k − iκ (15-53) Ak + iκ (k − iκ)2 = (15-54) k2 + κ2 k2 − κ2 − 2ikκ = (15-55) k2 + κ2 = e iφ (15-56) 2kκ with tan φ = − −κ2k2The phase shift of the wave is important in 3D scattering problems. Can we localize the particle in the forbidden region? Figure X: The wavefunction for E 0. Can the particle be observed there? To be sure that we have measured the particle inside the barrier, and not outside, we must measure its position at least with accuracy Δx ≈ κ−1 . Then according to ¯Heisenberg uncertainty, a momentum kick exceeding Δp ≥ h hκ will be transΔΔ ∼ ¯ferred onto the particle. Massachusetts Institute of Technology XV-9 8.04 Quantum Physics Lecture XVHow much energy do we transfer? ΔE = E(p +Δp) − E(p) (15-57) (p +Δp)2 p2 = (15-58) 2m − 2m pΔp (Δp)2 = + (15-59) m 2m p =¯hk (15-60) pΔp can be positive or negative, (Δp)2 is always positive. the transferred energy is on average (Δp)2 h¯2 h¯2κ2 �ΔE� =2m =2m(Δx)2 =2m = V0 − E (15-61) According to Heisenberg uncertainty, the measurement that localizes the particle inside the barrier transfers enough energy to allow the particle to be legitimately there. Rule. A positive KE E − V1 > 0 corresponds to a spatially oscillating wavefunction e±ikx with rate constant k (oscillation period λ = 2kπ ). A negative (“missing”) KE E − V1 < 0 corresponds to a spatially decaying or growing wavefunction e±κx with decay rate constant κ (decay length κ−1). The “missing” KE is associated with the size of the region (κ−1) that the particle occupies in the classically forbidden space. Massachusetts Institute of Technology XV-10