Quantum Physics- Potential Well and Resonance

8.04 Quantum Physics Lecture XVIThe potential barrier: tunnelingFigure I: Tunneling through a potential barrier. Assume E a, (16-2)(16-3)where we have omitted the term De−ikx that corresponds to an incident waveform the right. Inside the barrier the SE is d2u 2m (x)=+ (V0 − E)u(x)= κ2 u(x) (16-4) dx2 h¯2mwith κ2 = h2 (Vo − E). As before, κ is the decay constant in the classically forbidden ¯region (κ−1 is the decay length) that is associated with the “missing” KE necessary ¯to surpass the barrier classically, h2κ2 = V0 − E. Consequently inside the barrier 2m u(x)= Ee−κx + Feκx , for |x|≤ a (16-5) As before, we need to match the solution u(x) and its derivative u�(x) at the boundariies • At x = −a: Ae−ika + Beika = Ee+κa + Fe−κa for u (16-6) +ikAe−ika − ikBeika =+κEe+κa + κFe−κa for u� (16-7) At x = a:• Ceika = Ee−κa + Feκa for u (16-8) ikAeika = −κEe−κa + κFeκa for u� (16-9) Massachusetts Institute of Technology XVI-1 ������ 2 8.04 Quantum Physics Lecture XVIB (or the reflection probabilityWe are interested in the reflection amplitude r = A 2��C A 2 B C) and the transmission amplitude t=(or transmission probability|r|2 =A ) from the barrier. Remember that A2 determines the incident current, |t|and is a free parameter. It is useful to divide the equation for u� by the equation for ||=A 1 du d u (or alternatively, match u(x) dx = dx (ln u(x)) directly. Then we write • At x = −a: +ikAe−ika − ikBe+ika = −κEeκa + κFe−κa (16-10) Ae−ika + Beika Eeκa + Fe−κa At x = a:• +ikCeika −κEe−κa + κFeκa ik = Ceika = Ee−κa + Feκa (16-11) (matching of d (ln u(x)) = 1 du at boundaries). dx u(x) dx Now we proceed to eliminate E,F (Eq. 16-11): ikEe−κa + ikFeκa = −κEe−κa + κFeκa (16-12) (κ + ik)Ee−κa =(κ − ik)Feκa (16-13) E = κ − ikFe2κa (16-14) κ + ik Substitute into Eq. 16-10: κ+ikRHS = −κκ−ik Fe3κa + κFe−κa (16-15) κ−ik Fe3κa + Fe−κa κ+ik −κ(κ − ik)e+2κa + κ(κ + ik)e−2κa = (16-16) (κ − ik)e2κa +(κ + ik)e−2κa −κ2(e2κa − e−2κa)+ ikκ(e2κa + e−2κa) = (16-17) κ(e2κa + e−2κa) − ik(e2κa − e−2κa) κ2 sinh(2πa)+ ikκ cosh(2κa) = − κ cosh(2κa) − ik sinh(2κa) (16-18) Massachusetts Institute of Technology XVI-2 � � � � � � 8.04 Quantum Physics Lecture XVIConsequently, Eq. 16-10+ikAe−ika − ikBeika [κ cosh(2κa) − ik sinh(2κa)] (16-19) ��� � = Ae−ika + Beika −κ2 sinh(2κa)+ ikκ cosh(2κa) (16-20) = Ae−ika(+ikκ cosh(2κa)+ k2 sinh(2κa)+ κ2 sinh(2κa) − ikκ cosh(2κa)) (16-21) = Beika(+ikκ cosh(2κa)+ k2 sinh(2κa) − κ2 sinh(2κa)+ ikκ cosh(2κa)) (16-22) Ae−ika (k2 + κ2)sinh(2κa)= Beika 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) (16-23) B r = (16-24) A (k2 + κ2)sinh(2κa) = e−2ika (16-25) 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) reflection amplitude from barrier. To calculate the transmission amplitude C , we use the continuity of u at x = a:A Ceika = Ee−κa + Fe+κa (16-26) = κ − ikFeκa + Feκa (16-27) κ + ik 2κ = Feκa (16-28) κ + ik Massachusetts Institute of Technology XVI-3 � � � � 8.04 Quantum Physics Lecture XVIWe find F from the continuity of u at x = −a: RHS = Eeκa + Fe−κa (16-29) = κ − ikFe3κa + Fe−κa (16-30) κ + ik = Feκa κ − ike 2κa + κ + ike−2κa (16-31) κ + ik κ + ik = Feκa 2κ cosh(2κa) − 2ik sinh(2κa) (16-32) κ + ik RHS = Ae−ika + Beika (16-33) = Ae−ika + Ae−ika (k2 + κ2)sinh(2κa) (16-34) 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) = Ae−ika 1+ (k2 + κ2)sinh(2κa) (16-35) 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) 2ikκ cosh(2κa)+2k2 sinh(2κa) = Ae−ika 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) . (16-36) Then, C 2κF = e κa−ika (16-37) A Aκ + ik2κ Ae−2ika (2ikκ cosh(2κa)+2k2 sinh(2κa)= (16-38) A 2κ cosh(2κa) − 2ik sinh(2κa)2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa) 1 =2κe−2ikaik (16-39) 2ikκ cosh(2κa)+(k2 − κ2)sinh(2κa)C= (16-40) A 2kκ = e−2ika (16-41) 2kκ cosh(2κa) − i(k2 − κ2)sinh(2κa) Figure II: Tunneling through the potential barrier. Consequently, we have the results for the barrier Massachusetts Institute of Technology XVI-4 8.04 Quantum Physics Lecture XVIh2k2¯= E• 2m¯h2κ2 • 2m = V0 − EB = e−2ika −i(k2+κ2)sinh(2κa)r = • A 2kκ cosh(2κa)−i(k2−κ2)sinh(2κa)t = C = e−2ika 2kκ• A 2kκ cosh(2κa)−i(k2−κ2)sinh(2κa) Since the energy and particle velocity are the same on both sides of the barrier, here we have |r|2 + |t|2 = 1. Figure III: The sinh function. Let us look at |t|2 |t|2 = (2kκ)2 +(k2 (2+ kκκ)22 )2 sinh2(2κa) (16-42) where we have used cosh2(x) = 1+sinh2(x). Since, sinh is a monotonically increasing 2mfunction, and κ = h2 √V0 − E, the transmission is monotonically decreasing with ¯barrier height V0. In the limit of small transmission, κa � 1 (barrier width large compared to decay ��2 ��2 length κ−1), we have sinh(2κa) ≈ 1 e2κa = 1 e4κa and |t|2 → 4kκ e−4κa . In this 24 k2+κ2 limit the tunneling probability falls off exponentially with barrier thickness (in units of decay length κ−1). → This exponential dependence explains the extremely wide variation in, e.g., lifetimes of unstable nuclei (µs to 109 years, corresponding to a variation by a factor of 1022). Massachusetts Institute of Technology XVI-5 8.04 Quantum Physics Lecture XVIFigure IV: The transmission through the barrier as a function of decay wavevector κ.Figure V: In the limit of large barrier height or width, the transmission falls off exponentially because the wavefunction inside the barrier is dominated by the exponenttiall decaying term. Potential well: resonance phenomena We first consider scattering (E> 0) x ≤−a : Aeikx + B−ikx (16-43) −a ≤ x ≤ a : Ee+iqx + Fe−iqx (16-44) x ≥ a : Ceikx (16-45) Figure VI: The potential well.Massachusetts Institute of Technology XVI-6 8.04 Quantum Physics Lecture XVIh2k2¯= E• 2mh22¯q= V0 + E• 2m Instead of going through the calculation again, we note that these equations are equivalent to those of the potential barrier (for E 0. We then obtain the →∞, → · attractive delta potential V (x)= −λδ(x). We are interested in bound states: E< 0 Define• h2κ2¯2m =0 − E = −E = |E|, κ> 0 (16-51) Massachusetts Institute of Technology XVI-7 � � � 8.04 Quantum Physics Lecture XVIFigure VII: If the potential well is sufficiently deep or wide, it can support bound states with discrete energies −V0 0:• + D−κx�Ce�κx� (16-53) • Continuity of wavefunction at x = 0: A = D (16-54) • Derivative obeys (Lecture XV) 2m u�(�) − u�(−�)= − h2 λu(0) (16-55) ¯2m κD − κA = − h2 λA (16-56) ¯2m −2κ = − h2 λ (16-57) ¯ Massachusetts Institute of Technology XVI-8 8.04 Quantum Physics Lecture XVIm κ1 = λ (16-58) h¯2 h¯2κ2 h¯2 m2 m E1 = − = − h4 λ2 = − h2 λ2 (16-59) 2m 2m¯2¯→ Binding energy for attractive δ-function. The δ potential supports Figure IX: Comparison of bound states as the potential evolves from a very deep to a very shallow potential. In the very deep potential, like in the infinite well, the wave function oscillates sinusoidally inside the well, and decays exponentially in the forbidden region. In the very shallow potential, the wavefunction is is mostly located in the ”forbidden” region outside the well. exactly one bound state of energy E mλ2 . For a finite-size well, this result = −2¯h2 corresponds to the limiting case of a weak potential that supports only one 2¯h2 ma˜bound state (V0 � ma˜) with energy E = −2¯h22 V 2 2 0 . Massachusetts Institute of Technology XVI-9 � � � � 8.04 Quantum Physics Lecture XVIFigure X: Solutions in different regions. Two attractive δ-potentials We could proceed as before, or simplify slightly by making use of the fact that the potential is symmetric x →−x, and therefore we expect solutions of definite parity. The even solution in the middle region is 2B cosh(κx), and A = D, which eliminates two parameters. • Continuity of u: 2B cosh(κa)= Ae−κa (16-60) Derivative: • 2m −κAe−κa − κ2B sinh(κa)= − h2 λAe−κa (16-61) ¯2m h2 λ − κ Ae−κa =2κB sinh(κa) (16-62) ¯2m h2 λ − κ 2B cosh(κa)=2κB sinh(κa) (16-63) ¯2ma h¯2κaλ − 1 = tanh(κa) (16-64) There is always exactly one solution of the eigenvalue equation (16-64) for even parity. From the figure we see that for the bound state κa < 2maλ , which is where h2 2maλ 1 ¯the function κa − 1 intersects zero. On the other hand, since tanh(x) ≤ 1, we h21 m¯need 2maλ κa − 1 < 1, or κ> h2 λ. Larger κ means larger magnitude of binding energy h2¯h2κ2 m 2m¯ ¯E = − 2m . We have h¯2 m λ<κ< h¯2 λ If we compare this to the binding-energy in single δ-potential, κ1 = h2 λ we see that the particle is more strongly bound in the ¯double-well potential. Massachusetts Institute of Technology XVI-10 8.04 Quantum Physics Lecture XVIFigure XI: Graphic solution of the eigenvalue equation 16-64. Reason. Given the discontinuity in slope due to the potential, it is possible to choose a steeper wavefunction (larger κ larger binding energy) when the two δ-functions →are close. Variation of binding energy with well separation a: As we decrease a, the Figure XII: Comparison of the wavefunction for two different well spacings. If the wells are close, for the same wavefunction discontinuity at each δ function the wavefunction outside the two wells can decay faster (larger κ), resulting in larger binding energy |E| =¯h2κ2/2m. Figure XIII: Graphic comparison of the binding energies for large and small separation 2a between the binding sites. mλbinding energy increases from the value given by κ = h2 (binding energy of a single ¯well attained at a →∞) towards the value κ = 2¯mλ , attained as a → 0. Thus h2 the binding energy quadruples. the possibility of the wavefunction in a double-well system to change so as to decrease the kinetic (and possibly potential) energy is at the origin of chemical bonds in molecules. Massachusetts Institute of Technology XVI-11