sol4_PHYS3316

# sol4_PHYS3316 - PHYS 3316 Solutions Problem Set 4 1...

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Unformatted text preview: PHYS 3316 Solutions: Problem Set 4 1 Propagation of wavepackets in an external potential (a) The motion of a particle in a potential U (x) is described by the Hamiltonian H= p2 + U (x) . 2m (1) The force acting on this particle due to the potential is F = −∂U/∂x. If we suppose that there is a second, external force F ext (t) also acting on the particle, then by the Newtoninan deﬁnition of a force dp = dt F = F ext (t) − ∂U [x(t)] . ∂x (2) This relation then deﬁnes the particle momentum p(t). (For those of you who have seen the Lagrangian/ Hamiltonian formulation of classical mechanics, the momentum deﬁned in this way is not the same as the canonical momentum deﬁned by H: the idea here is to ‘derive’ the Hamilton equations of motion from a Newtonian starting point.) (b) The external force F ext does work on the particle, changing its total energy H . (Since H is timeindependent, it turns out that it is conserved if there is nothing else aﬀecting the particle motion.) The diﬀerential work done dH = F ext dx, so dH = F ext x . ˙ (3) dt However, explicitly H = H (p, x), so ∂H dp ∂H dx dH = + dt ∂p dt ∂x dt ∂H ∂U = F ext (t) − ∂p ∂x From eqs. (3) and (4) it follows that ∂U ∂H F ext (t) − ∂p ∂x = x F ext (t) − ˙ ∂U ∂x =⇒ x = ˙ ∂H . ∂p (5) + ∂U x. ˙ ∂x (4) This derivation clearly holds when F ext = 0. (c) Now suppose the external force F ext = 0. From eq. (2), since ∂H/∂x = ∂U/∂x, then p=− ˙ ∂H . ∂x (6) (Please note that one can properly derive the Hamilton canonical equations of motion from more fundamental principles than the ideas we have used here to reproduce them.) (d) Now let’s consider a semiclassical wavepacket description of this particle. Since we have a spatially dependent potential, the most general dispersion relation is ω = ω (k, x) , and then the wavepacket can be written generally as ψ (x, t) = dkA[ω (k, x)]eikx−iω(k,x)t . (8) (7) What we ultimately want to do is impose the classical Hamilton equations of motion in order to determine the dispersion relation. To do this, we ﬁrst must ﬁnd the group velocity and ‘force’ x and ˙ p. We assume the following: ˙ 1 • The wavelength of the wavepacket mode is much smaller that the typical length scale over which U changes, so nearby to some point x0 , the wavenumber k = k0 is a constant. • The wavepacket envelope is sharply peaked at k = k0 , x = x0 , so that A[ω (k, x)] ≃ A[ω (k0 , x0 )] nearby to x0 . Applying these assumptions means that we can write the wavepacket as ψ (x, t) = dkA[ω0 ] exp[iΦ(k ; x, t)] , (9) where ω0 is a constant, and the phase function Φ(k ; x, t) = kx − ω (k, x)t ≃ kx − ω (k, x0 )t − (x − x0 ) ∂ω (k0 , x0 )t . ∂x (10) In the second line we have expanded the phase function in a small neighborhood of x0 , and assumed k = k0 is constant in the subleading order (i.e O(x − x0 )) term of the expansion: this is the ﬁrst assumption above. The stationary phase condition is ∂ Φ/∂k = 0, that is x− ∂ω ∂ω ˙ (k0 , x0 )t = 0 =⇒ x = (k0 , x0 ) . ∂k ∂k (11) (e) Just as the stationary phase condition in momentum space ∂ Φ/∂k = 0 deﬁnes the group velocity x, ˙ ˙ the stationary phase condition in real space ∂ Φ/∂x = 0 deﬁnes k. Hence, applying the condition ∂ Φ/∂x = 0, we have ∂ω ∂ω ˙ (k0 , x0 )t =⇒ k = (k0 , x0 ) . 0=k− (12) ∂x ∂x Using the de Broglie relations, these expressions can be rewritten as x= ˙ ∂ω , ∂p ∂ω p= ˙ . ∂x (13) There is however, some ambiguity as to the sign of p: that is, whether it is the ‘force’ acting on the ˙ particle or produced by it. It turns out to be the latter, so p = − ∂ ω/∂x. Applying the Hamilton ˙ equations of motion, for the Hamiltonian above, these relations become p ∂ω = ∂p m ∂U ∂ω =− , − ∂x ∂x from which we conclude ω= as expected. p2 + U (x) = H . 2m (14) (15) 2 Normalization and Probability Currents Consider a wavefunction ψ satisfying the Schr¨dinger equation o 2 i ∂t ψ (x, t) = − 2m 2 2 ∂x + U (x) ψ (x, t) . (16) Let Pab be the probability of the particle being in the rangle a < x < b. Then, since , dPab ∂ = dt ∂t b b dxψ ∗ (x, t)ψ (x, t) a (Pab has no x dependence) =i a dx − i(∂t ψ ∗ )ψ − ψ ∗ (i∂t ψ ) b a = i dx (Hψ )∗ ψ − ψ ∗ (Hψ ) b a 2 2 dx (∂x ψ ∗ )ψ − ψ ∗ (∂x ψ ) b (Schr¨dinger equation) o (U terms cancel) (17) b =− i 2m i 2m =− (∂x ψ ∗ )ψ − ψ ∗ (∂x ψ ) a − a dx ∂x ψ ∗ ∂x ψ − ∂x ψ ∗ ∂x ψ (integrating by parts) (18) = J (a, t) − J (b, t) . where J (x, t) = i [∂x ψ ∗ (x, t)]ψ (x, t) − ψ ∗ (x, t)∂x ψ (x, t) . (19) 2m J is called the probability current. In words, this relation is the statement that the rate of change of the probability of ﬁnding a particle between a and b is equal to the diﬀerence of the probability currents ‘ﬂowing’ in and out at a and b respectively. The physical meaning of this relation is that the total probability of ﬁnding the particle somewhere on the real line is conserved (and equal to 1, of course.) Now consider U (x) = U0 (x) + iΓ(x). In this case, the imaginary parts of U no longer cancel in step (17). Instead, i dPab =− dt 2m b a 2 2 dx (∂x ψ ∗ )ψ − ψ ∗ (∂x ψ ) + b b a dx − iΓψ ∗ ψ − ψ ∗ iΓψ (20) = J (a, t) − J (b, t) + m dx Γ(x)ψ ∗ (x, t)ψ (x, t) . a (Here total probability is no longer conserved. The imaginary part of the complex potential corresponds to particle decay into some other state(s).) 3 Expectation Values and Ehrenfest’s Theorem (a) As shown in class, it follows from the the Schr¨dinger equation that the expectation value of any o operator O satisﬁes i dO (21) H, O , = dt where H is the Hamiltonian operator. Hence we must ﬁnd the expectation value of this commutator with O = x or p. ˆ ˆ (Explicit Way) Calculating the expectation explicitly, we have 2 H, x ˆ = =− =− dx ψ ∗ − 2 2m 2 ∂x + U (x), x ψ ˆ 2m 2 2 2 dx ψ ∗ ∂x (xψ ) − ψ ∗ x∂x ψ (U (x) commutes with x) ˆ dx ψ ∗ 2∂x ψ 2m i =− dx ψ ∗ pψ ˆ m 3 (22) Hence dx ˆ = dt p ˆ m . (23) (Extracurricular Way) This can also be done just by calculating commutators. That is H, x = ˆ 12 p ,x ˆˆ 2m 1 = p p, x + p, x p ˆˆˆ ˆˆˆ 2m 1 = pi + i p ˆ ˆ 2m p ˆ . =i m (a commutator identity) (since [ˆ, x] = i ) pˆ (24) (25) The result follows immediately. (b) For O = p we have ˆ H, p ˆ = = −i = −i Hence dx ˆ = dt − ∂x U (x) . (27) dx ψ ∗ p2 ˆ + U (x), p ψ ˆ 2m (p commutes with powers of itself) ˆ (26) dx ψ ∗ U (x)∂x ψ − ψ ∗ ∂x [U (x)ψ ] dx − ψ ∗ [∂x U (x)]ψ . 4 Momentum Space Representation of Wavefunctions (a) Consider the wavepacket 2 ˜ ψ (k ) = N e−ikx0 e−D(k−k0 ) . (28) Applying the normalization condition, we have 1= ˜ dk ψ (k ) 2 (29) 2 = |N |2 = |N |2 dke−2D(k−k0 ) eikx0 e−ikx0 dle−2Dl , 2 changing integration variable such that l = k − k0 . But, since dxe−ax = it follows that 1 = |N |2 π =⇒ |N | = N = 2D 2D π 1/4 2 π , a (30) , (31) assuming N is real. We require the normalization condition (29) since we interpret |ψ |2 as a probability distribution function, and the sum of probabilities must be unity. 4 (b) The Fourier transform ψ (x) = dk ˜ √ eikx ψ (k ) 2π N dk exp[−Φ(k )] , =√ 2π where the phase Φ(k ) = D(k − k0 )2 + ik (x0 − x) = D k − k0 + i (x0 − x) 2D 2 + ik0 (x0 − x) + (x − x0 )2 , 4D (32) completing the square. Changing variables such that l = (k − k0 ) + i(x − x0 )/2D and applying eq. (30), the position space representation of the wavepacket is N ψ (x) = √ 2π = 1 2Dπ (x − x0 )2 π exp − ik0 (x0 − x) − D 4D 1/4 exp − ik0 (x0 − x) − (x − x0 )2 4D . (33) It remains to check the normalization of ψ (x). We have, again from eq. (30) dx ψ (x) 2 = = =1, 1 2Dπ dx exp − (x − x0 )2 2D 1√ 2Dπ 2Dπ (34) so the position space representation of the wavepacket is properly normalized. An underlying axiom in quantum mechanics is that there is a well-deﬁned probability distribution ˜ on the phase space of a particle. The real space |ψ (x)|2 and momentum space |ψ (k )|2 are then the marginal probability distributions with respective random variables being position and momentum. If normalization of one representation were not equivalent to the normalization of the other, we could not adopt such a probabilistic interpretation. (c) The wavenumber of oscillations in position space is the location of the envelope in momentum space, and vice versa. (d) Consider now a general wavefunction in momentum space, deﬁned such that ˜ ψ (k ) = The expectation value of position ∞ ˜ ψ (k ) , 0, k≥0 k<0. (35) x= ˆ dx ψ ∗ (x)xψ (x) = 0 ˜ dk ψ ∗ (k )ˆψ (k ) . x˜ (36) which is clearly real in the position space representation. As shown in question 5 below, the momentum space representation of the position operator is simply x = i∂k , ˆ 5 (37) 1.0 1.0 0.5 0.5 2 4 6 4 6 8 10 0.5 0.5 1.0 1.0 Figure 1: Real parts of: position space representation (left); momentum space representation (right) so the imaginary part of x in the momentum space representation is explicitly ˆ Im x = ˆ x − x∗ ˆ ˆ 2i 1∞ ˜ ˜ ˜ ˜ dk ψ ∗ (k )i∂k ψ (k ) − − i∂k ψ ∗ (k )ψ (k ) = 2i 0 1∞ ˜ ˜ = dk ∂k ψ ∗ (k )ψ (k ) 20 ∞ 1˜ ˜ . = ψ ∗ (k )ψ (k ) 2 0 2 2 k→∞ (38) Since x is real, this imaginary part must be zero, so we then have ˆ ˜ ψ (0) ˜ = ψ (k ) . (39) ˜ But by deﬁnition, ψ (0) = 0, so it follows that the modulus of the momentum space representation goes to zero as k → ∞, which is equivalent to the momentum space wavefunction itself going to zero. That is, in order for x to be real ˆ ˜ lim ψ (k ) = 0 . (40) k→∞ 5 Momentum Space Representation of Operators ˜ (a) First, in the momentum space representation, by deﬁnition the Fourier transform ψ (k ) of ψ (x) is an eigenfunction of p. That is ˆ ˜ pψ (k ) = k ψ(k ) =⇒ p = k , ˆ˜ ˆ (41) in momentum space. Now consider the position operator expectation value x= ˆ dx ψ ∗ (x)xψ (x) = ˜ dk ψ ∗ (k )ˆψ (k ) , x˜ (42) ∀ψ which implies (by a mathematical property called completeness) that xψ (k ) is the Fourier transform ˆ˜ of xψ (x). Hence xψ (k ) = ˆ˜ dx √ e−ikx xψ (x) 2π dx √ e−ikx ψ (x) = i∂k 2π ˜ = i∂k ψ (k ) =⇒ x = i∂k , ˆ 6 (43) in the momentum space representation. (b) Suppose we deﬁne the following operators via their expectation values xn = pn = dxψ ∗ (x)xn ψ (x) , ˜ ˜ dpψ ∗ (p)pn ψ (p) . (44) Note in the latter deﬁnition we have written the Fourier transform in terms of the momentum p = k . This just amounts to a rescaling of the Fourier transform: it becomes ψ (x) = Clearly in position space xn = xn . ˆ In momentum space, following the same reasoning as in part (a), we must have xn ψ (p) = ˆ˜ dx −ipx/ n √ e x ψ (x) , 2π dx −ipx/ n √ ˆ e ψ (x) =⇒ xn = i ∂p = i ∂p 2π pn = pn , ˆ and in position space we must have pn ψ (x) = ˆ dp ipx/ n ˜ √ e p ψ (p) , 2π dp ipx/ ˜ n √ ψ (p) =⇒ pn = − i ∂x e = − i ∂x ˆ 2π (46) dp ipx/ ˜ √ ψ (p) . e 2π (45) n = i∂k n . (47) Similarly, in momentum space (48) n . (49) (c) Finally, consider the analytic function f (x) = n an xn , (50) and deﬁne the operators f (x) and f (p) respectively by f (x) = f (p) = Clearly f (x) = n dxψ ∗ (x)f (x)ψ (x) , ˜ ˜ dpψ ∗ (p)f (p)ψ (p) . (51) an ψ |xn |ψ , an xn , ˆ = n = n an xn ˆ by linearity (52) f (x) = f (ˆ) . x (53) = f (ˆ) , x so then in position space the operator By identical reasoning, in momentum space f (p) = f (ˆ). p 7 ...
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## This note was uploaded on 03/29/2010 for the course PHYS 3318 taught by Professor Flanagan during the Spring '08 term at Cornell.

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