Relativity Problem Set 9 - Solutions

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1 Relativity Problem Set 9 - Solutions Prof. J. Gerton October 3, 011 Problem 1 (10 pts.) The quantum harmonic oscillator (a) The Schroedinger equation for the ground state of the 1D QHO is ) ( m x + mω x ψ(x) = E 0 ψ(x). (1) Evaluating the kinetic operator gives m ψ(x) = x m B (1 B x ) ψ(x), () If we want the Schroedinger equation to be satisfied it must be ( B m + m ) ( ) ω x B + m E 0 = 0. (3) This is a polynomial equation, in which each coefficient of the polynomial in the variable x must equate zero. The coefficient of x gives The other equation then gives which is the correct form for the ground state energy. (b) We impose the normalization condition B = mω. (4) E 0 = 1 ω, (5) + dx ψ(x) = 1, (6) which leads to A + dx e B x = 1. (7) 1

2 To evaluate the integral we use the result for Gaussian integrals + dy e y = π, (8) together with the substitution y = x B that leads to or A + A = dy B e y = ( ) 1/4 B = π A B π = 1, (9) ( m ω ) 1/4. (10) π Problem (10 pts.) Expectation values for a harmonic oscillator (a) Since ψ 1 (x) is an odd function, ˆpψ 1 (x) = i x ψ 1 (x) is even. Then the integr (ψ 1 (x)) (ˆpψ 1 (x)) is a product of an even function 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 integr (ψ 1 (x)) (ˆxψ 1 (x)) is odd in x thus its integral over the even domain specified is zero. (c) We need to compute ˆK = + ψ 1 (x) ( 3 β x ) ω ψ 1 (x) = 3 ω. (11) 4 This result is expected, because of the following fact: we know that the total energy must be E 1 = 3/ ω, that for the virial theorem applied to the harmonic field we have K = V. (1) So, E 1 = K + V = K, (13) K = 1 E 1 = 3 ω. 4 (14)

3 3 Problem 3 (10 pts.) The Evanescent Wave (a) The momentum in the region x > 0 is k = m(e V 0 )/ = i m E V 0 /, the last equality holding because V 0 > E. We introduce κ = m E V 0 /, so that k = iκ the wave function is with C a constant. ψ(x) = C e κx, (15) (b) The number of particles dn within the positions x x + dx is given by dn = ψ(x) 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) = e κx. (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, R = p k p + k = p iκ p + iκ. (18) Now, computing this quantity gives R = p iκ p + iκ = p iκ p + iκ = 1. (19) p + iκ p iκ So, for any value of κ p the reflection coefficient is one. (d) The answers to (b) (c) are in contrast, because in (b) one finds that there is a transmitted current in x > 0 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, T = J transmitted J incident, (0) where J is a current. So, the expression for T is T = T 0 e κx, (1) with T 0 a constant. This expression holds for large x 1/κ is particularly good for E V 0, where T 1. So, in these cases the transmission coefficient is much smaller than R, energy conservation is not violated. However, in the region x 1/κ or for E V 0 this approximation is no longer true.

4 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] [a, b], the potential is V (x) = 0. Energy conservation then gives p m = E, or p = me. () In the region [ b, b] we have V (x) = V 0, with the condition V 0 > E. Indicating the momentum in this region with p, the energy conservation relation is thus p m + V 0 = E, or p m = E V 0 < 0. (3) So, in the last equation, the right h side is always negative, by assumption. This means that the momentum has to be imaginary, as suggested in the text. We write p = iκ, with κ a real quantity with the dimension of a momentum. If we do so, the kinetic energy is p m = (iκ) m = κ m. (4) The term E V 0, on the other h, can be written as E V 0 = E V 0, because we know that E V 0 < 0. Here, x indicates the absolute value of x. Finally, putting together these expressions, we find the expression for κ ss κ m = E V 0 or κ = m E V 0. (5) 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 KE 0 we obtain E V (x) 0. (6) 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] [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, 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.

5 5 Problem 5 (10 pts.) The double well - Part (a) We begin with the region [ a, b], where the momentum is p. The wave function is then ψ 1 (x) = C 1 e ipx + C e ipx, (7) with C 1 C some normalization constants to be found. In the region [ b, b], the momentum is iκ the wave function is then ψ (x) = D 1 e i(iκ)x + D e i(iκ)x = D 1 e κx + D e κx (8) with D 1 D constants to be fixed. Finally, in the region [b, a] the momentum is p the wave function is ψ 3 (x) = C 1 e ipx + C e ipx, (9) with C 1 C some normalization constants which are different from C 1 C. Summing up, the wave function is oscillating in the classically allowed region 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 ψ (x) = D 1 e κx + D e κx, (30) ψ ( x) = D 1 e κx + D e κx. (31) Imposing ψ ( x) = ψ (x) in this region gives D 1 = D, so that the wave function is ψ (x) = D 1 [ e κx + e κx]. (3) We can use the function cosh(x) = (e x + e x )/, so that finally ψ (x) = D 1 cosh(κx). (33) To impose the symmetrization in the other regions, we impose that the wave function ψ 1 (x) in [ a, b] being equal to the wave function ψ 3 ( x) in [b, a], that is, the wave function is symmetric in the symmetric domain [ a, b] + [b, a]. We have ψ 1 (x) = C 1 e ipx + C e ipx, in [ a, b], (34) ψ 3 ( x) = C 1 e ipx + C e ipx, in [b, a]. (35) Imposing the condition ψ 3 ( x) = ψ 1 (x) gives C 1 = C C 1 = C 1. Thus, if we decide to get rid of the constants C 1 C, the symmetrization gives the expression for the wave function in [b, a] as ψ 3 (x) = C e ipx + C 1 e ipx, (36)

6 6 while the wave function ψ 1 (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 This gives the relation 0 = C 1 e ip( a) + C e ip( a) = C 1 e ipa + C e ipa. (37) C = C 1 e ipa. (38) We impose the continuity at x = b. This means that the expressionψ(x) = C 1 e ipx + C e ipx, valid in the region [ a, b], must be equal to the expression ψ(x) = D 1 cosh(κx), valid in the region [ b, b], when x = b. Imposing this, we have C 1 e ip( b) + C e ip( b) = D 1 cosh(κ( b)). (39) This gives the relation C 1 e ipb + C e ipb = D 1 cosh(κ( b)) = D 1 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 ψ 1 (x) x = ψ (x) x at x = b. (41) Since ψ 1 (x) = ip C 1 e ipx + ip C e ipx, x (4) ψ (x) = D 1 κ sinh(κx), x (43) the continuity of the derivative at x = b yields ip C 1 e ip( b) + ip C e ip( b) = D 1 κ sinh(κ( b)). (44) Using sinh( x) = sinh(x), we finally obtain the third relation (c) We rewrite the three conditions below: ip C 1 e ipb + ip C e ipb = D 1 sinh(κb). (45) C = C 1 e ipa continuity at x = a, (46) C 1 e ipb + C e ipb = D 1 cosh(κb) continuity at x = b, (47) ip C 1 e ipb +ip C e ipb = D 1 κ sinh(κb) continuity of the derivative at x = b. (48)

7 7 Plugging the first relation in the second gives C 1 e ipa [ e ipb ipa e ipb+ipa] = D 1 cosh(κb), (49) while plugging the first relation in the third gives Now, since the two relations above are ip C 1 e ipa [ e ipb ipa + e ipb+ipa] = D 1 κ sinh(κb). (50) Dividing these two relations gives or e ipb ipa + e ipb+ipa = cos p(a b), (51) e ipb ipa e ipb+ipa = i sin p(a b), (5) ic 1 e ipa sin p(a b) = D 1 cosh(κb), (53) ip C 1 e ipa cos p(a b) = D 1 κ sinh(κb). (54) C 1 e ipa sin p(a b) p C 1 e ipa cos p(a b) = D 1 cosh(κb) D 1 κ sinh(κb), (55) κ tan p(a b) = p tanh κb. (56) (d) In terms of the given quantities, this is [ ] [ ] E tan me(a b) tanh m E V0 b = E V 0. (57) This condition gives the value of the energy E, which then results quantized. The above relation is thus a quantization requirement.

Lecture 5. Potentials

Lecture 5 Potentials 51 52 LECTURE 5. POTENTIALS 5.1 Potentials In this lecture we will solve Schrödinger s equation for some simple one-dimensional potentials, and discuss the physical interpretation