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Download Relativity Problem Set 9 - Solutions Prof. J. Gerton October 23, 2011
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Relativity Problem Set 9 - Solutions Prof. J. Gerton October 23, 2011 Problem 1 (10 pts.) The quantum harmonic oscillator (a) The Schroedinger equation for the ground state of the 1D QHO is ~2 ∂ 2 mω 2 x2 − + ψ(x) = E0 ψ(x). 2m ∂x2 2 (1) Evaluating the kinetic operator gives − ~2 ~2 ∂ 2 ψ(x) = B (1 − 2B x2 ) ψ(x), 2m ∂x2 m If we want the Schroedinger equation to be satisfied it must be 2 2~2 B 2 m ω 2 ~ B 2 − + − E0 = 0. x + m 2 m (2) (3) This is a polynomial equation, in which each coefficient of the polynomial in the variable x must equate zero. The coefficient of x2 gives B= mω . 2~ (4) E0 = 1 ~ω, 2 (5) The other equation then gives which is the correct form for the ground state energy. (b) We impose the normalization condition Z +∞ dx |ψ(x)|2 = 1, (6) −∞ which leads to A 2 Z +∞ 2 dx e−2B x = 1. −∞ 1 (7) 2 To evaluate the integral we use the result for Gaussian integrals Z +∞ √ 2 dy e−y = π, (8) −∞ √ together with the substitution y = x 2B that leads to Z +∞ dy −y A2 √ 2 √ A e =√ π = 1, 2B 2B −∞ or A= Problem 2 2B π 1/4 = m ω 1/4 π~ . (9) (10) (10 pts.) Expectation values for a harmonic oscillator (a) Since ψ1 (x) is an odd function, p̂ψ1 (x) = −i~∂x ψ1 (x) is even. Then the integrand (ψ1 (x)) (p̂ψ1 (x)) is a product of an even function and an odd function, so it is odd. The integral is thus zero because it is the integral of an odd function over an even domain. (b) With the same reasoning, the integrand (ψ1 (x)) (x̂ψ1 (x)) is odd in x thus its integral over the even domain specified is zero. (c) We need to compute Z +∞ hK̂i = −∞ 3 3 ψ1 (x) ( − β x2 ) ~ω ψ1 (x) = ~ ω. 2 4 (11) This result is expected, because of the following fact: we know that the total energy must be E1 = 3/2~ω, and that for the virial theorem applied to the harmonic field we have hKi = hV i. (12) So, E1 = hKi + hV i = 2hKi, and hKi = course name 1 3 E1 = ~ω. 2 4 (13) (14) PS # 3 Problem 3 (10 pts.) The Evanescent Wave p p (a) The momentum in the region x > 0 is k = 2m(E − V0 )/~ = i p 2m|E − V0 |/~, the last equality holding because V0 > E. We introduce κ = 2m|E − V0 |/~, so that k = iκ and the wave function is ψ(x) = C e−κx , (15) with C a constant. (b) The number of particles dN within the positions x and x + dx is given by dN = |ψ(x)|2 dx. (16) So, in the case discussed in part (a), the probability density to find the particle at the point x > 0 is |ψ(x)|2 = e−2κx . (17) This probability, though small, is non-zero, so that there is a certain small probability that the particle be at some x > 0. This probability decays exponentially with x, so further away from the origin the probability drops to zero. (c) This probability is given by the reflection coefficient, p − k 2 p − iκ 2 = R = p + iκ . p + k Now, computing this quantity gives p − iκ 2 p − iκ p + iκ = R = = 1. p + iκ p + iκ p − iκ (18) (19) So, for any value of κ and p the reflection coefficient is one. (d) The answers to (b) and (c) are in contrast, because in (b) one finds that there is a transmitted current in x > 0 and in (c) one finds that all of the incident particles are reflected. This ambiguity is solved by thinking of the definition for the transmitted wave, Jtransmitted 2 , (20) T = Jincident where J is a current. So, the expression for T is T = T0 e−2κx , (21) with T0 a constant. This expression holds for large x 1/κ and is particularly good for E V0 , where T 1. So, in these cases the transmission coefficient is much smaller than R, and energy conservation is not violated. However, in the region x . 1/κ or for E ∼ V0 this approximation is no longer true. course name PS # 4 Problem 4 (10 pts.) The double well - Part 1 (a) Since the potential is infinite in these regions, the wave function is zero. (b) In the regions [−a, −b] and [a, b], the potential is V (x) = 0. Energy conservation then gives √ p2 = E, or p = 2mE. (22) 2m In the region [−b, b] we have V (x) = V0 , with the condition V0 > E. Indicating the momentum in this region with p0 , the energy conservation relation is thus p02 + V0 = E, 2m or p02 = E − V0 < 0. 2m (23) So, in the last equation, the right hand side is always negative, by assumption. This means that the momentum has to be imaginary, as suggested in the text. We write p0 = iκ, with κ a real quantity with the dimension of a momentum. If we do so, the kinetic energy is (iκ)2 κ2 p02 = =− . 2m 2m 2m (24) The term E − V0 , on the other hand, can be written as E − V0 = −|E − V0 |, because we know that E − V0 < 0. Here, |x| indicates the absolute value of x. Finally, putting together these expressions, we find the expression for κ ss − p κ2 = −|E − V0 | or κ = 2m|E − V0 |. 2m (25) Compare the expression for κ with the expression for p above. (c) In classical mechanics, the particle cannot stay in the region [−b, b] because this would violate the conservation of energy constraint. In fact, since the kinetic energy has to be always positive in classical mechanics, from KE + V (x) = E and KE ≥ 0 we obtain E − V (x) ≥ 0. (26) The expression above yields the domain x where the particle can move in the classical picture. In our case, this domain is the union of the two regions [−a, −b] and [a, b], but not the region [−b, b]. In quantum mechanics instead, the wave function extends in the classically forbidden region as well, so there is a certain probability to find the particle in this region. This phenomenon is peculiar of quantum mechanics, and gives rise to cute effects like the quantum tunneling, which explains the α-decay in nuclei or how the tunnel diode works, among many other things. course name PS # 5 Problem 5 (10 pts.) The double well - Part 2 (a) We begin with the region [−a, −b], where the momentum is p. The wave function is then ψ1 (x) = C1 e−ipx + C2 eipx , (27) with C1 and C2 some normalization constants to be found. In the region [−b, b], the momentum is iκ and the wave function is then ψ2 (x) = D1 e−i(iκ)x + D2 ei(iκ)x = D1 eκx + D2 e−κx (28) with D1 and D2 constants to be fixed. Finally, in the region [b, a] the momentum is p and the wave function is ψ3 (x) = C10 e−ipx + C20 eipx , (29) with C10 and C20 some normalization constants which are different from C1 and C2 . Summing up, the wave function is oscillating in the classically allowed region and exponentially suppressed or enhanced in the classically forbidden region. (b) Imposing the symmetry of the wave function ψ(−x) = ψ(x) allows us to get rid of three of the six constants above. In the region [−b, b], which is already symmetric, we have ψ2 (x) = D1 eκx + D2 e−κx , (30) and ψ2 (−x) = D1 e−κx + D2 eκx . (31) Imposing ψ2 (−x) = ψ2 (x) in this region gives D1 = D2 , so that the wave function is ψ2 (x) = D1 e−κx + eκx . (32) We can use the function cosh(x) = (e−x + ex )/2, so that finally ψ2 (x) = 2D1 cosh(κx). (33) To impose the symmetrization in the other regions, we impose that the wave function ψ1 (x) in [−a, −b] being equal to the wave function ψ3 (−x) in [b, a], that is, the wave function is symmetric in the symmetric domain [−a, −b] + [b, a]. We have ψ1 (x) = C1 e−ipx + C2 eipx , in [−a, −b], (34) and ψ3 (−x) = C10 eipx + C20 e−ipx , in [b, a]. (35) Imposing the condition ψ3 (−x) = ψ1 (x) gives C1 = C20 and C1 = C10 . Thus, if we decide to get rid of the constants C10 and C20 , the symmetrization gives the expression for the wave function in [b, a] as ψ3 (x) = C2 e−ipx + C1 eipx , course name (36) PS # 6 while the wave function ψ1 (x) in [−a, −b] is unchanged. Imposing boundary conditions, we first impose that the wave function in x = −a is zero. Thanks to the symmetrization, this already includes the condition that the wave function being zero at x = a. At x = −a we have 0 = C1 e−ip(−a) + C2 eip(−a) = C1 eipa + C2 e−ipa . (37) This gives the relation C2 = −C1 e2ipa . (38) We impose the continuity at x = −b. This means that the expressionψ(x) = C1 e−ipx + C2 eipx , valid in the region [−a, −b], must be equal to the expression ψ(x) = 2D1 cosh(κx), valid in the region [−b, b], when x = −b. Imposing this, we have C1 e−ip(−b) + C2 eip(−b) = 2D1 cosh(κ(−b)). (39) This gives the relation C1 eipb + C2 e−ipb = 2D1 cosh(κ(−b)) = 2D1 cosh(κb), (40) where in the last step we use the fact that the function cosh(x) = cosh(−x), as required by the symmetry. Imposing now the continuity of the derivative, we have Since ∂ψ1 (x) = ∂ψ2 (x)∂x at x = −b. ∂x (41) ∂ψ1 (x) = −ip C1 e−ipx + ip C2 eipx , ∂x (42) and ∂ψ2 (x) = 2D1 κ sinh(κx), ∂x the continuity of the derivative at x = −b yields −ip C1 e−ip(−b) + ip C2 eip(−b) = 2D1 κ sinh(κ(−b)). (43) (44) Using sinh(−x) = − sinh(x), we finally obtain the third relation −ip C1 eipb + ip C2 e−ipb = −2D1 sinh(κb). (45) (c) We rewrite the three conditions below: C2 = −C1 e2ipa continuity at x = −a, C1 eipb + C2 e−ipb = 2D1 cosh(κb) continuity at x = −b, (46) (47) −ip C1 eipb +ip C2 e−ipb = −2D1 κ sinh(κb) continuity of the derivative at x = −b. (48) course name PS # 7 Plugging the first relation in the second gives C1 eipa eipb−ipa − e−ipb+ipa = 2D1 cosh(κb), (49) while plugging the first relation in the third gives −ip C1 eipa eipb−ipa + e−ipb+ipa = −2D1 κ sinh(κb). (50) Now, since eipb−ipa + e−ipb+ipa = 2 cos p(a − b), (51) eipb−ipa − e−ipb+ipa = −2i sin p(a − b), (52) and the two relations above are −2iC1 eipa sin p(a − b) = 2D1 cosh(κb), (53) −2ip C1 eipa cos p(a − b) = −2D1 κ sinh(κb). (54) and Dividing these two relations gives 2C1 eipa sin p(a − b) 2D1 cosh(κb) = , 2p C1 eipa cos p(a − b) −2D1 κ sinh(κb) or κ tan p(a − b) = − p . tanh κb (55) (56) (d) In terms of the given quantities, this is tan h√ s i hp i 2mE(a − b) tanh 2m|E − V0 |b = − E . |E − V0 | (57) This condition gives the value of the energy E, which then results quantized. The above relation is thus a quantization requirement. course name PS #