Physics 580 Homework 5: Quantum Mechanics of a Double Square Well Potential, Assignments of Quantum Mechanics

Download Physics 580 Homework 5: Quantum Mechanics of a Double Square Well Potential and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 580: Homework 5 Seungmin Hong December 3, 2007 1 LQM 4-6 (a) Since there is an infinite potential beyond x = b, the wavefunction vanishes. So the transmission coefficient S(E) must be zero. In consequence, the modulus |N(E)| is a unit, which means δ(E) must be real. (b) In the region 0 < x < a, the wavefunction takes the form ψ(x) = Aeipx/~ +Be−ipx/~ where p = √ 2mE. At the boundary x = 0, the wavefunction vanishes so we have a condition A + B = 0, while at x = a we know N(E) = −e2iδ(E) = Be−ipa/~/Aeipa/~. From these conditions, we find pa ~ = nπ − δ(E) = p0a ~ − δ(E) , n ∈ Z+ (1.1) For δ(E) ≪ 1, the correction to the energy δE = E − E0 becomes small compared to E0 = p 2 0 2m . So to the first order, we can approximate δ(E) ≃ δ(E0). E = p2 2m = E0 − p0~ ma δ(E) +O(δ(E))2 ≃ E0 − p0~ ma δ(E0) (1.2) (c) Since the potential V (x) is finite in the region a < x < b, the uncertainty in the position space ∆x increases, which means longer period. In consequence, the energy level should be lowered, so that δ(E0) must be positive. (Or, think of uncertainly principle ∆x∆p ∼ ~.) 2 LQM 4-7 (a) The symmetry in the shape of potential tells us the energy eigenstate should have a definite parity. Usually the odd parity states tend to have higher energy for the presence of the nodes. The presence of the infinite potential in between, however, leads degeneracy, since the the nodes, which would be existing in the middle region, are suppressed. Solving the Schrödinger equation for each parity states under boundary conditions ψ(a) = ψ(b) = 0, we find ψeven(x) =      1√ b−a sin ( nπ(x−a) b−a ) a < x < b ψeven(−x) −b < x < −a 0 elsewhere (2.1) ψodd(x) =      1√ b−a sin ( nπ(x−a) b−a ) a < x < b −ψodd(−x) −b < x < −a 0 elsewhere (2.2) and each of these is a solution of energy En = π2~2 2m(b−a)2n 2 (n ∈ Z+). So the energy levels of the double square well is doubly degenerate. 1 (b) In the case of infinite barrier, N(E) must be −1 since an incident wave would be perfectly reflected. In an extension to this, we can expand N(E) as a series, to an order of the inverse of the barrier. N(E) = −1 +N (1)(E/V ) +N (2)((E/V )2) + · · · ≡ −e2iδ(E) (2.3) Since ∣ ∣N(E)) ∣ ∣ 2 ≤ 1, we can set iN (1)(E/V ) ≡ 2δ(E) ≪ 1. On the other hand, similar method can be repeated for S(E). S(E) = 0 + S(1)(E/V ) + S(2)((E/V )2) + · · · (2.4) Since ∣ ∣N(E) ∣ ∣ 2 + ∣ ∣S(E) ∣ ∣ 2 = 1, S(1)(E/V ) ≃ √ 1 − ∣ ∣ − 1 +N (1)(E/V ) ∣ ∣ 2 = √ −(2δ(E))2 ∼ iJ(E) , |J(E)| ≪ 1 (2.5) In the last expression J(E), we take into the higher order corrections into account, so that it is not necessarily equal to 2δ(E). Now, let us consider the energy corrections. The wave function in each region should take the following form. ψ(x) = { Aeikx +Be−ikx −b < x < −a Ceikx +De−ikx a < x < b where, ~k = √ 2mE (2.6) Since potential is symmetric, the above wavefunction should be either parity even (A = D, B = C) or parity odd (A = −D, B = −C). Imposing the boundary conditions ψ(−b) = 0, we have B = −Ae−2ikb (2.7) The part B e−ikx is a linear superposition of the reflected wave from Aeikx and the transmitted part from the wave D e−ikx. Similarly, C eikx can be thought of a linear sum of a reflected and a transmitted wave. Hence, we obtain x = −a : B eika = N(E)Ae−ika + S(E)D e−ika (2.8) Solving the equations, we find e−2ik(b−a) = 1 + 2iδ(E) ∓ iJ(E) ≃ e−2i ( k0(b−a)−(δ∓J/2) ) (2.9) ⇒ k = k0 − δ(E) ∓ J(E)/2 b− a , k0 ≡ p0 ~ = nπ b− a (2.10) Note we have exponentiated δ(E) and J(E) since both are small enough. Therefore, to the level of approxi- mation, we find E± = E0 − p0~ m(b− a) { δ(E0) ∓ J(E0) 2 } (2.11) Here the minus sign corresponds to the parity even state and the plus is for the odd states. Note the odd parity state must be energetically unfavorable, since its wavefunction has higher curvature. (c) At an equilibrium, the protons in the H+2 molecule ion are 0.106 nm apart and the binding energy is known to be 2.65 eV (K. Krane, Modern Physics 2nd ed., Wiley, 1996). Roughly estimating, b− a ≃ 0.90 Å (a ≃ 0.16 Å), p0~ m(b− a) = π~2 2ma0(b− a) = (hc)2 8πmc2a0(b− a) ≃ (1240 eV/nm) 2 8π × 0.511MeV × 0.53Å× 0.90Å ≃ 25.1 eV (2.12) Using the WKB approximation, we can estimate J(E) = √ T (E) = e− 1 ~ ∫ a −a dx √ 2m(V−E) ≃ e− 2a~ √ 2m|E| ≃ e−0.82 ≃ 0.44 (2.13) Since 1 − (δ(E))2 ≃ |N(E)| = √ 1 − (J(E))2, δ(E) ≃ 0.32. Hence, the binding energy becomes E+ − E0 = 25.1 eV × { 0.32 − 0.44 2 } ≃ 2.51 eV (2.14) This result is consistent with the known value, at least to an order of magnitude. Note the above estimation lacks accuracy since the assumptions δ(E), J(E) ≪ 1 are not successfully satisfied. 2 On the other hand, at the top of the lowest band, ka = qa = π, hence we find ( dq dk )∣ ∣ ∣ ∣ k= π a = 0 , ( d2q dk2 )∣ ∣ ∣ ∣ k= π a = −aπ A =⇒ m ∗ m ∣ ∣ ∣ ∣ k= π a = − A π2 = −mv0a π2~2 < 0 (4.12) The effective mass tells us how the electron response to an external field. As can be seen, near the Bragg plane, the behavior of electron is rather counterintuitive. It moves to the opposite direction to the field. In some sense, this can be interpreted as the movement of the particle with opposite charge, often called ‘hole’ in the context of condensed matter physics. In the bottom of the band, one can observe the effective mass significantly depends on the potential exerted by the lattices. If the potential is high enough, then the particles mostly trapped between the two lattices so that it looks as if electrons acquire infinite mass. In this limit, the eigenfunction has zeros at the lattice points, so that energy levels are quantized. The other extreme, v0 → 0 has already been already discussed. 5 LQM 4-16 (a) Using the conservation of the probability, we obtain the following. |ψ1|2 + |ψ2|2 = 1 ⇒ ψ1 ∂ψ∗1 ∂t + ∂ψ1 ∂t ψ∗1 + ψ2 ∂ψ∗2 ∂t + ∂ψ2 ∂t ψ∗2 = 0 (5.1) Substituting i~∂ψ2∂t = Kψ1 and its complex conjugate into the above equation, we get ψ1 ( i~ ∂ψ∗1 ∂t +Kψ∗2 ) + ψ∗1 ( i~ ∂ψ1 ∂t −Kψ2 ) = 0 (5.2) Note the two terms are complex conjugate to each other, hence the real part of both terms must be zero. For an arbitrary ψ1, this condition is satisfied if i~ ∂ψ1 ∂t = Kψ2 (5.3) (b) Let us define a two-component vector, Ψ = ( ψ1 ψ2 ) . Then, the coupled equation reads i~ ∂tΨ = Kσ1Ψ (5.4) In order to diagonalize the matrix σ1, we can implement the similarity transformation. Once transformed, one can easily read off the eigenstates. Ψ+(t) = 1√ 2 ( 1 1 ) e−iKt/~ (5.5) Ψ−(t) = 1√ 2 ( 1 −1 ) eiKt/~ (5.6) As is obvious, the state Ψ+ has an even parity with respect to the exchange between ψ1 and ψ2, while Ψ− has an odd parity. (c) This problem is similar to the two level problem. Here the driving ‘force’ of oscillation is nothing but a phase difference between the two region. The probability changes in the both box can be connected with the number of particles, in which sense, the phase difference is associated with the number of particles via the uncertainty principle. Note there is an conjugacy relation between the phase and the particle number in 5 Quantum Mechanics. Ψ(0) = 1√ 2 ( e−iδ/2 eiδ/2 ) = e−iδ/2 + eiδ/2 2 · 1√ 2 ( 1 1 ) + e−iδ/2 − eiδ/2 2 · 1√ 2 ( 1 −1 ) (5.7) ⇒ Ψ(t) = cos ( δ 2 ) √ 2 ( 1 1 ) e−iKt/~ − i sin ( δ 2 ) √ 2 ( 1 −1 ) eiKt/~ (5.8) ⇒ ψ1(t) = 1√ 2 ( cos ( δ 2 ) e−iKt/~ − i sin ( δ 2 ) eiKt/~ ) (5.9) ∣ ∣ψ1(t) ∣ ∣ 2 = 1 2 [ 1 + sin δ sin (2Kt ~ ) ] ⇒ ∂t ∣ ∣ψ1(t) ∣ ∣ 2 = K ~ sin δ cos (2Kt ~ ) (5.10) 6 (a) 0 0.5 1 1.5 2 qaΠ -1 -0.5 0 0.5 1 C o s @ q a D + v S i n @ q a D  q a -1 -0.5 0 0.5 1 kaΠ 0 5 10 15 20 25 E H k L As is obvious, the bottom band does have a cutoff on k. In the lowest band, the free electron should meet the following conditions. Below the critical momentum, electrons are trapped by the lattices. kc ≡ arccos ( 1 − ma|v0| ~2 ) ≤ ka ≤ π (6.1) (b) By an analytic continuation, q → iκ, we obtain cos ka = coshκa− ma|v0| ~2 sinhκa κa , E = −~ 2κ2 2m < 0 (6.2) Drawing the above, on top of the free scattering case, we have 0 1 2 3 4 ΚaΠ -1 -0.5 0 0.5 1 C o s h @ Κ a D + v S i n h @ Κ a D  Κ a -1-0.5 0 0.5 1 kaΠ -25 -20 -15 -10 -5 0 E H k L One can easily see the purple curve fills the “blank” below the critical momentum. Between ±kc, it forms a negative energy band. (c) As |v0| increases more and more portion of the lowest band will be absorbed into the negative part, bound states. Once ma|v0| ~2 > 2, the previously lowest free band entirely disappears and the 2nd band replaces it. When ma|v0| ~2 ≫ 1, the purple curve would pass through ±1 so sharply that the band become almost flat. So, we would see the energies of particles in this band approaches that of a particle in the bound state of an 6 attractive potential of strength v0. Let us examine this in a qualitative way. From eq (6.2), (ξ ≡ qa , A ≡ ma|v0| ~2 ≫ 1), 0 ≃ cos ka cosh ξ = 1 −A tanh ξ ξ =⇒ tanh ξ ≃ ξ A =⇒ ξ ≃ A (6.3) E = − ~ 2ξ2 2ma2 ≃ −mv 2 0 2~2 (6.4) we get the desired result. Because of strong attraction, particles are tightly bound to the δ-function potential well, so that it cannot reach the nearest lattice points. Thus, each particles behaves as if independently bound to respective well. Consequently, the energy becomes highly degenerate. 7